# Is there any easy algorithm/table for computing homotopy groups?

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### Jason

Jul 21, 2003, 2:36:08 PM7/21/03
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I'm working a bit with solitons like strings, monopoles and instantons
in arbitrary dimensions. So, one thing which would be very helpful is to
be able to compute/look up the 2nd, 3rd, 4th and 5th homotopy groups of
various manifolds (often, simply Lie groups or the factor space of two
Lie groups). Unfortunately, I know next to nothing about algebraic
topology and I'm not sure if I really want to spend tons of time
studying it just to be able to compute a couple of homotopy groups. So,
do you guys know any simple algorithm(s)/table(s) to help me? (Proofs
aren't necessary unless they're simple... :) )

### John Baez

Jul 22, 2003, 6:44:19 AM7/22/03
to

There's no simple algorithm to compute homotopy groups, at least not
one that is also fast enough to be worth doing by hand. This is why
physicists like you need to have friends who are algebraic topologists.
But if you can't find any friendly algebraic topologists in your neighborhood,
you can always ask questions on sci.physics.research - some of us *love*
need to know, you can look these up here:

Mamoru Mimura, Homotopy theory of Lie groups,
Handbook of Algebraic Topology, ed. I. M. James,
North-Holland, Amsterdam, 1995, pp. 951-991.

Helpful hints: pi_2 of every Lie group is trivial. pi_3 of
every simple Lie group is Z. pi_n of any Lie group is the
same as that of any maximal compact subgroup, so you only
need to know homotopy groups of compact Lie groups.

### Marc Nardmann

Jul 26, 2003, 6:30:05 PM7/26/03
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Jason wrote:

It would be useful to know the exact homotopy sequence of a fibration;
cf. e.g. [Bredon: Topology and geometry], Theorem VII.6.7. You can use
this to get information about the homotopy groups of (Lie group modulo
subgroup), for example. The following article contains some tables:

@incollection {MR97c:57038,
AUTHOR = {Mimura, Mamoru},
TITLE = {Homotopy theory of {L}ie groups},
BOOKTITLE = {Handbook of algebraic topology},
PAGES = {951--991},
PUBLISHER = {North-Holland},
YEAR = {1995},
}

The nth homotopy group pi_n(X x Y; (x,y)) of a product X x Y with
respect to the base point (x,y) is just pi_n(X; x) x pi_n(Y; y) . Maybe
these facts and references are already all you need. It is in general
not easy to calculate all homotopy groups of a given topological space,
even for experts in algebraic topology! In particular there is no
algorithm to decide whether an arbitrary triangulated manifold is simply
connected. (The problem is that there is --- provably --- no algorithm
for deciding whether an arbitrary finitely presented group is trivial.)

-- Marc Nardmann (posted July 22)

To reply, remove every occurrence of a certain letter from my e-mail

### Jason

Jul 28, 2003, 1:07:30 AM7/28/03
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John Baez wrote:
> And if it's homotopy groups of *Lie groups* you
> need to know, you can look these up here:
>
> Mamoru Mimura, Homotopy theory of Lie groups,
> Handbook of Algebraic Topology, ed. I. M. James,
> North-Holland, Amsterdam, 1995, pp. 951-991.
Unfortunately, this extremely expensive book isn't in the library. Do
you know of any other references?

> Helpful hints: pi_2 of every Lie group is trivial. pi_3 of
> every simple Lie group is Z. pi_n of any Lie group is the
> same as that of any maximal compact subgroup, so you only
> need to know homotopy groups of compact Lie groups.

How about Lie groups which aren't their own universal cover? Are these
statements also true for them? And how about the factor spaces of two
Lie groups, G/H (where G is a compact Lie group and H is a Lie subgroup)?

The kinds of manifolds I'm interested in are like SU(3)XSU(2)XU(1),
[SU(3)XSU(2)XU(1)]/Z_6, SU(5), SO(10), [SU(3)^3]/Z_3, E_6, E_7, E_8,
[SU(4)XSU(2)XSU(2)]/Z_2, [SU(5)XU(1)]/Z_5 and the factor spaces of these
groups. You know, the usual GUT groups and factor spaces.

### Sean Carroll

Jul 28, 2003, 1:07:50 AM7/28/03
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As John says, there is no easy algorithm. Different tricks are
appropriate to different dimensionalities. For the first, second,
and third homotopy groups of different quotient spaces, we calculated
them in this paper:

We were interested in the spaces you actually get in (3+1) dimensions
from spontaneously breaking symmetries. For the higher homotopy

Sean

### Jason

Jul 28, 2003, 1:09:49 AM7/28/03
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I just browsed casually through an online INTRODUCTORY(!!!) book on
ELEMENTARY homotopy theory and frankly, it just looks way too
complicated and scary... I'm surprised it's even stated as being
don't know any friendly algebraic topologists, unfortunately... :(

### John Baez

Jul 30, 2003, 12:58:10 AM7/30/03
to
In article <bfjjvt$f6k$1...@news.udel.edu>, Jason <pri...@excite.com> wrote:

>John Baez wrote:

>> Mamoru Mimura, Homotopy theory of Lie groups,
>> Handbook of Algebraic Topology, ed. I. M. James,
>> North-Holland, Amsterdam, 1995, pp. 951-991.

>Unfortunately, this extremely expensive book isn't in the library.

If you're talking about a university library, try to
get them to buy it. University libraries are often
quite willing to buy books! They often don't know
what math and physics books to buy, and enjoy getting
suggestions. Journals are a different matter, because
they're much more expensive, and go on costing more and
taking up more room forever.

>Do you know of any other references?

I seem to remember some tables of the homotopy groups of
Lie groups in the appendices of that delightful Encylopedic
Dictionary of Mathematics, put out by the Mathematical
Society of Japan. If your library doesn't have *that*,
either shoot the librarian or buy your own copy - or both.
It's incredibly useful.

>> Helpful hints: pi_2 of every Lie group is trivial. pi_3 of
>> every simple Lie group is Z. pi_n of any Lie group is the
>> same as that of any maximal compact subgroup, so you only
>> need to know homotopy groups of compact Lie groups.

>How about Lie groups which aren't their own universal cover? Are these
>statements also true for them?

Being a mathematician by training, when I said "every Lie group"
I actually meant "every Lie group" - unlike a physicist, for whom
it would mean "every Lie group except the one you're interested in".

Anyway, taking the universal cover of a space doesn't affect any
of its homotopy groups except the first - which becomes the trivial
group.

>And how about the factor spaces of two
>Lie groups, G/H (where G is a compact Lie group and H is a Lie subgroup)?

For these you may want to use the long exact sequence of homotopy
groups

... -> pi_n(H) -> pi_n(G) -> pi_n(G/H) ->

pi_{n-1}(H) -> pi_{n-1}(G) -> pi_{n-1}(G/H) -> ...

to reduce the question to one involving the homotopy groups of G
and H and also certain maps between them. Sometimes this is quite
a pain, but if you're lucky a bunch of homotopy groups in this
sequence will happen to be trivial near the pi_n(G/H) you're
trying to compute, and things simplify. This happens surprisingly
often in real life.

If you don't know what a long exact sequence is or how to use
one, well, this is where your friendly local algebraic topologist
comes in handy. It takes a few lessons, but it's fun and worth
knowing.

>The kinds of manifolds I'm interested in are like SU(3)XSU(2)XU(1),
>[SU(3)XSU(2)XU(1)]/Z_6, SU(5), SO(10), [SU(3)^3]/Z_3, E_6, E_7, E_8,
>[SU(4)XSU(2)XSU(2)]/Z_2, [SU(5)XU(1)]/Z_5 and the factor spaces of these
>groups. You know, the usual GUT groups and factor spaces.

Maybe some physicist somewhere has compiled a table specially
for this purpose. I wouldn't be surprised.

Another good thing to know is that pi_n(G x H) = pi_n(G) x pi_n(H).
Also, modding out by discrete subgroups like Z_n only affects pi_1
of a Lie group, thanks to that long exact sequence I wrote down.

### Jason

Aug 1, 2003, 5:04:02 PM8/1/03
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John Baez wrote:
> For these you may want to use the long exact sequence of homotopy
> groups
>
> ... -> pi_n(H) -> pi_n(G) -> pi_n(G/H) ->
>
> pi_{n-1}(H) -> pi_{n-1}(G) -> pi_{n-1}(G/H) -> ...
>
> to reduce the question to one involving the homotopy groups of G
> and H and also certain maps between them. Sometimes this is quite
> a pain, but if you're lucky a bunch of homotopy groups in this
> sequence will happen to be trivial near the pi_n(G/H) you're
> trying to compute, and things simplify. This happens surprisingly
> often in real life.
>
> If you don't know what a long exact sequence is or how to use
> one, well, this is where your friendly local algebraic topologist
> comes in handy. It takes a few lessons, but it's fun and worth
> knowing.
I just decided to look over the online algebraic topology textbook once
again, skipping hundreds of pages (True, they spend hundreds of pages on
things like the first homotopy groups, simplexes, and tons of theorems
about them before moving on to higher homotopy groups). Earlier, I had
no idea what to look for in this extremely long textbook, short of
reading everything, which I certainly don't have the time,
determination, or need to do, but now I decided go straight to the exact
sequence, after reading the previous pages for prelimanaries. The way
it's presented in that textbook is so different that if I hadn't seen
the long exact sequence in the other physics paper mentioned on this
thread (without proof), I wouldn't have guessed it had anything to do
with factor spaces, especially since it talks about relative homotopy
groups, instead of homotopy groups proper!!!!

In the version mentioned in the book, they defined the nth relative
homotopy group, pi_n(X,A,x_0) for a subset A of X and an element x_0 of
A as the homotopy classes of continuous maps from [0,1]^n to X such that
the boundary of [0,1]^n maps to A and the set of values with the last
coordinate equal to 0 maps to {x_0}. After some thought, I realized
their pi_n(G,H,x_0) is the same thing as pi_n(G/H) if G/H is connected,
even though this was stated nowhere. And instead of your sequence, they
gave, for B as a subset of A as a subset of X,

...->pi_n(A,B,x_0)->pi_n(X,B,x_0)->pi_n(X,A,x_0)->pi_n(A,B,x_0)->...

I guess this is just yet another example of how the abstraction and
generalization tendacies of mathematicians make learning math so
confusing and cryptic...

Anyway, once I figured all that out, the proof was rather simple,
actually... (Once I ignored all the mentions of how this parallels the
long exact sequence in homology and cohomology, which I didn't even look
at at all after realizing the proof of this theorem had nothing to do
with homology.)

### Jason

Aug 6, 2003, 3:36:08 PM8/6/03
to
John Baez wrote:
>Helpful hints: pi_2 of every Lie group is trivial. pi_3 of
>every simple Lie group is Z. pi_n of any Lie group is the
>same as that of any maximal compact subgroup, so you only
>need to know homotopy groups of compact Lie groups.
I'm curious but why is pi_2(G)=0 for any Lie group and pi_3(G)=Z for any
simple Lie group? I think the latter has got to do with the fact that
for any continuous map phi:S^3->G, we can define a locally flat
connection by A=(d phi)phi^-1 with F=0 and since Tr[\int_S^3 F wedge
A-1/3 A wedge A wedge A] is a topological invariant which is additive
under the composition of two homotopies, it defines a homomorphism from
pi_3(G) to R, but I still can't see why the image of the homomorphism is
Z or why its kernel is trivial.

By the same construction, I can see why for any odd number n>=3, the
Chern-Simons invariant, which simply reduces to a constant factor times
Tr[\int_S^n A wedge... A] because F=0 defines a homomorphism from
pi_n(G) to R. Actually, if there are m linearly independent traces, it
would be a map to R^m, but that's just about as far as I can go.

### A.J. Tolland

Aug 7, 2003, 7:19:28 AM8/7/03
to

Which book? Hatcher?

--A.J.

### Michael K Murray

Aug 7, 2003, 7:19:28 AM8/7/03
to
In article <bfjjvt$f6k$1...@news.udel.edu>, Jason <pri...@excite.com>
wrote:

> John Baez wrote:

If you are just looking for lists of groups there are some quite good
ones in the back of the Encyclopedic Dictionary of Mathematics if
you can find one in your library. If you have trouble finding it its

http://www.amazon.com/exec/obidos/tg/detail/-/0262590204/qid=1059386957/s
r=12-1/103-8382105-7214260?v=glance&s=books

For quotient groups as one of other posts recommends try the fact that
there is a long exact sequence of maps and groups as follows. Given G
with subgroup H you have

... -> pi_i(H) -> pi_i(G) -> pi_i(G/H) -> pi_{i-1}(H) -> ...

... -> pi_0(H) -> pi_0(G) -> pi_0(G/H) -> < 1 >

All the maps here are group homomorphisms and the image
of every map is the same as the kernel of the next map along.

Michael

### Jason

Aug 7, 2003, 7:22:04 AM8/7/03
to
John Baez wrote:

>>>Helpful hints: pi_2 of every Lie group is trivial. pi_3 of
>>>every simple Lie group is Z. pi_n of any Lie group is the
>>>same as that of any maximal compact subgroup, so you only
>>>need to know homotopy groups of compact Lie groups.

> Being a mathematician by training, when I said "every Lie group"

> I actually meant "every Lie group" - unlike a physicist, for whom
> it would mean "every Lie group except the one you're interested in".

Sorry to question your "mathematician's every", but isn't SO(2,1) a
simple Lie group and its maximal compact subgroup SO(2)? In that case,
wouldn't Z=pi_3(S0(2,1))=pi_3(SO(2))=0? And is your statement true for
infinite-dimensional Lie groups as well? (I have to work with the
homotopy groups of infinite dimensional Lie groups as well.)

### John Baez

Aug 7, 2003, 6:57:25 AM8/7/03
to
In article <bgjels$4du$1...@news.udel.edu>, Jason <pri...@excite.com> wrote:

>John Baez wrote:

>>>>Helpful hints: pi_2 of every Lie group is trivial. pi_3 of
>>>>every simple Lie group is Z. pi_n of any Lie group is the
>>>>same as that of any maximal compact subgroup, so you only
>>>>need to know homotopy groups of compact Lie groups.

>> Being a mathematician by training, when I said "every Lie group"
>> I actually meant "every Lie group" - unlike a physicist, for whom
>> it would mean "every Lie group except the one you're interested in".

>Sorry to question your "mathematician's every", but isn't SO(2,1) a
>simple Lie group and its maximal compact subgroup SO(2)? In that case,
>wouldn't Z=pi_3(S0(2,1))=pi_3(SO(2))=0?

You're right! When I said "every" I meant "every", but I was
wrong. Mathematicians are precise, but everyone is wrong sometimes,
and here I was precisely wrong.

The false statement on my list was:

"pi_3 of every simple Lie group is Z."

I should have said:

"pi_3 of every compact simple Lie group is Z."

pi_3 of SO(2,1) is indeed 0.

> And is your statement true for
>infinite-dimensional Lie groups as well?

According to mathematical definitions, a Lie group
must be a finite-dimensional smooth manifold.

There are lots of infinite-dimensional groups that are *like* Lie groups -
Banach Lie groups, Hilbert Lie groups and so on -
and they are very important in physics, but mathematicians
don't consider them to be Lie groups. It's difficult to
make quick blanket statements about their topology, and
none of my statements above were intended to apply to them.

### Aaron Bergman

Aug 8, 2003, 8:23:55 AM8/8/03
to

In article <bg7j92$89d$1...@glue.ucr.edu>, John Baez wrote:

>Anyway, taking the universal cover of a space doesn't affect any
>of its homotopy groups except the first - which becomes the trivial
>group.

Which suggests an interesting question: how does this generalize
to higher homotopy? The obvious answer is the universal cover of
loop space. I guess you'd want to take the classifying space of
what came out.

Does this make any sense? Better question: is it all useful?

Aaron
--
Aaron Bergman
<http://www.princeton.edu/~abergman/>

### Jason

Aug 8, 2003, 8:39:45 AM8/8/03
to

Sorry for asking yet another question, but I don't know any "friendly
local algebraic topologist".

It occured to me if you have a map, phi:S^n->G where n is an odd
positive integer and G is a Lie group, then, there is a homomorphism
from pi_n(G) to R given by defining the 1-form connection A to be
(d phi)phi^-1 and taking the trace of the integral of the Chern-Simons
n-form for A since homotopically equivalent mappings (between twice
differentiable mappings, I have to admit, not merely continuous
mappings) give equivalent integrals. This, incidentally, is true for any
compact, boundaryless n-dimensional manifold, not just S^n.

Now is there, in general, for any n-dimensional manifold M and any
manifold X a nontrivial map from the homotopy class of mappings from M
to X to a vector space R^m or C^m given by, for any specific smooth
mapping f:M->X, an integral over an n-form which is defined locally in
terms of f when viewed as a field over M?

I doubt this would be the case in general, but how about particular
cases? For example, since you claim pi_2(G)=0 for any Lie group, there's
no nontrivial homomorphism there, but is there any analogous
construction for pi_4(G), for example? Tr[F^F] wouldn't work because
it's simply zero for a flat connection. Or how about when X is G/H?

Or to take another example, if we have an n+1 dimensional manifold, M
with an n-dimensional boundary and a manifold X with a submanifold Y and
we consider the relative homotopy class of smooth mappings from M to X
such that the boundary of M maps to Y, is there any n+1 form defined
locally in terms of the mapping which gives a homotopy invariant?

For example, if we have an n+1 space, R^(n+1), with a Higgs field taking
values within a representation R, let's say such that set of possible
Higgs VEVs form a submanifold isomorphic to G/H within R, we could
deform R^(n+1) into a disc D^(n+1) (assuming the usual conditions of
fast falloff of field strengths toward vacuum values at infinity) and
now, M would be D^(n+1), X would be R and Y would be isomorphic to G/H.
Now, is there an n+1 form defined locally in terms of the Higgs field
such that its integral gives the topological charge?

The reason I'm asking this is because topological charges are global
things, but yet, surprisingly often, soliton models admit local
topological currents/densities. (It would be a current if we also
include time.)

### John Baez

Aug 8, 2003, 8:23:40 AM8/8/03
to
In article <bg9fja$qbv$1...@news.udel.edu>, Jason <pri...@excite.com> wrote:

>John Baez wrote:

>> you may want to use the long exact sequence of homotopy
>> groups
>>
>> ... -> pi_n(H) -> pi_n(G) -> pi_n(G/H) ->
>>
>> pi_{n-1}(H) -> pi_{n-1}(G) -> pi_{n-1}(G/H) -> ...
>>

>In the version mentioned in the book, they defined the nth relative

>homotopy group, pi_n(X,A,x_0) for a subset A of X and an element x_0 of
>A as the homotopy classes of continuous maps from [0,1]^n to X such that
>the boundary of [0,1]^n maps to A and the set of values with the last
>coordinate equal to 0 maps to {x_0}. After some thought, I realized
>their pi_n(G,H,x_0) is the same thing as pi_n(G/H) if G/H is connected,
>even though this was stated nowhere. And instead of your sequence, they
>gave, for B as a subset of A as a subset of X,
>
>...->pi_n(A,B,x_0)->pi_n(X,B,x_0)->pi_n(X,A,x_0)->pi_n(A,B,x_0)->...

I think there should be an "n-1" or "n+1" somewhere in here.

>I guess this is just yet another example of how the abstraction and
>generalization tendacies of mathematicians make learning math so
>confusing and cryptic...

I actually think it's an example of how mathematicians know more than
one long exact sequence.

The one I mentioned is most easily seen as a special case of "the
long exact sequence of a fibration". A good example of a fibration
is a fiber bundle: one space mapping down onto another in a nice way.
In this case we have G mapping down to G/H, with fiber H.

The one you just mentioned is pretty different, in general.

Your attempt to get this long exact sequence

... -> pi_n(H) -> pi_n(G) -> pi_n(G/H) ->

pi_{n-1}(H) -> pi_{n-1}(G) -> pi_{n-1}(G/H) -> ...

as a special case of this one:

...->pi_n(A,B,x_0)->pi_n(X,B,x_0)->pi_n(X,A,x_0)->....

is noble, but right now I don't see why pi_n(G,H,x_0)
should be the same as pi_n(G/H). Maybe I'm just being
dumb. But anyway, you probably want to learn about the
long exact sequence of a fibration - it makes life easier.

Btw, what is this algebraic topology book that you dislike so much?
There might be better ones.

### Jason

Aug 9, 2003, 1:17:33 AM8/9/03
to
Jason wrote:

> In the version mentioned in the book, they defined the nth relative
> homotopy group, pi_n(X,A,x_0) for a subset A of X and an element x_0 of
> A as the homotopy classes of continuous maps from [0,1]^n to X such that
> the boundary of [0,1]^n maps to A and the set of values with the last
> coordinate equal to 0 maps to {x_0}. After some thought, I realized
> their pi_n(G,H,x_0) is the same thing as pi_n(G/H) if G/H is connected,
> even though this was stated nowhere.

Actually, after even more reflection, that doesn't seem so obvious
anymore. If let's say we have a continuous mapping f:[0,1]^n->G/H such
that f maps the boundary to the identity coset, is there any guarantee
that there exists a continuous function g:[0,1]^n->G such that g(x)
belongs to f(x) for all x and that all such g's for a given f are
homotopically equivalent?

### John Baez

Aug 11, 2003, 4:33:08 AM8/11/03
to
In article <slrnbieoi0....@cardinal2.Stanford.EDU>,
Aaron Bergman <aber...@princeton.edu> wrote:

>In article <bg7j92$89d$1...@glue.ucr.edu>, John Baez wrote:

>>Anyway, taking the universal cover of a space doesn't affect any
>>of its homotopy groups except the first - which becomes the trivial
>>group.

>Which suggests an interesting question: how does this generalize
>to higher homotopy? The obvious answer is the universal cover of
>loop space. I guess you'd want to take the classifying space of
>what came out.

That sounds good.

>Does this make any sense? Better question: is it all useful?

Yes, it makes sense and it's useful.

Suppose someone hands you a simply connected
topological group and you want to kill its
second homotopy group. Here's how:

1) First loop it (= form its loop space)
to push the homotopy groups down one notch.

2) Then kill the 1st homotopy group of the
result by taking the universe cover.

3) Then deloop it (= take the classifying space)
to push the homotopy groups back up one notch.

This is pretty nice.

By looping and delooping more times in steps 1)
and 3) you can also arrange to kill off the bottom
n homotopy groups, for whatever value of n you like.

The thing to beware of is that if your original
group was a nice little Lie group, the result will
usually be a huge infinite-dimensional group.

A classical example involves O(n),
the rotation-reflection group in n dimensions.

If we kill its 0th homotopy group we get SO(n),
the rotation group in n dimensions.

If we then kill its 1st homotopy group we get Spin(n),
the spin group in n dimensions.

The 2nd homotopy group will be trivial so there's no
need to kill it.

The 3rd homotopy group is Z. If we will this we
get String(n), the string group in n dimensions.
This is infinite-dimensional!

If we take a manifold and reduce the structure
group of its tangent bundle from O(n) to SO(n)
we get an "oriented" manifold. If we reduce the
structure group further to Spin(n) we get a "spin"
manifold. And if we reduce it to String(n) we
get a "string" manifold. Making a manifold into
a string manifold is morally the same as making
its loop space into a spin manifold.

Elliptic cohomology and the Witten genus work well
for string manifolds. The Witten genus is to string
manifolds as the A-hat genus is to spin manifolds.

This is some of the stuff I didn't get around to
including in week197!

The next nontrivial homotopy group of O(n)
(in general) is the 7th one. If we kill that...
well, I'm afraid that's a bit too mysterious for
me to explain well. But it's related to the octonions,
of course.

### Jason

Aug 15, 2003, 5:40:35 PM8/15/03
to
John Baez wrote:
>>...->pi_n(A,B,x_0)->pi_n(X,B,x_0)->pi_n(X,A,x_0)->pi_n(A,B,x_0)->...
> I think there should be an "n-1" or "n+1" somewhere in here.

Right, I accidentally left that out.

> I actually think it's an example of how mathematicians know more than
> one long exact sequence.
>
> The one I mentioned is most easily seen as a special case of "the
> long exact sequence of a fibration". A good example of a fibration
> is a fiber bundle: one space mapping down onto another in a nice way.
> In this case we have G mapping down to G/H, with fiber H.
>
> The one you just mentioned is pretty different, in general.
>
> Your attempt to get this long exact sequence
>
> ... -> pi_n(H) -> pi_n(G) -> pi_n(G/H) ->
>
> pi_{n-1}(H) -> pi_{n-1}(G) -> pi_{n-1}(G/H) -> ...
>
> as a special case of this one:
>
> ...->pi_n(A,B,x_0)->pi_n(X,B,x_0)->pi_n(X,A,x_0)->....
>
> is noble, but right now I don't see why pi_n(G,H,x_0)
> should be the same as pi_n(G/H). Maybe I'm just being
> dumb. But anyway, you probably want to learn about the
> long exact sequence of a fibration - it makes life easier.

Oops, you mean I confused a fibration with relative homotopy groups?
Well, with no one to really guide me, I guess such confusions are
inevitable. Reading the online textbook didn't make that clear at all.
Where can I learn about homotopy groups of fibrations then?

### Matt Reece

Aug 18, 2003, 5:28:53 PM8/18/03
to
Jason <pri...@excite.com> wrote in message news:<bfk0oi$l9g$1...@news.udel.edu>...

Hatcher's book is the only online book I know of; it is used in some
undergraduate classes. (Only the first and fourth chapters are really
relevant to your questions.) My major complaint is that, in some
problems and examples, Hatcher seems to like nasty counterexamples and
expects a lot of ability to visualize funny spaces. The spaces you're
working with are all fairly nice, so you could always skip over
technicalities. The major tool you need to understand is the exact
sequence of a fibration. I like May's "Concise Course in Algebraic
Topology," which deals with these homotopy theory in the first several
chapters (so you wouldn't have to wade through homology as in
Hatcher's book), but all the categorical nonsense might discourage
you. Homotopy groups aren't as nice as homology groups. In particular,
pi_k(X,A) need not equal pi_k(X/A) even for nice spaces X and A with
nice inclusions A -> X. This can be seen from X = RP^n, A = RP^(n-1).
So the exact sequence for a fibration is an important tool.

Here is one result that I don't think anyone has mentioned, which may

pi_1 of a topological group is abelian. (Further, the "product of
loops" determined by traversing one, then the next, is homotopic to
the path given by the pointwise product, i.e. (f*g)(t) = f(t)g(t)).

Since this is true for topological groups, I think it should hold even
for the infinite-dimensional Lie groups you wonder about.

### s.p.research moderator

Aug 18, 2003, 9:30:10 PM8/18/03
to sci-physic...@moderators.isc.org

In article <bgjeav$49a$1...@news.udel.edu>, Jason <pri...@excite.com> wrote:

>Jason wrote:

>> In the version mentioned in the book, they defined the nth relative
>> homotopy group, pi_n(X,A,x_0) for a subset A of X and an element x_0 of
>> A as the homotopy classes of continuous maps from [0,1]^n to X such that
>> the boundary of [0,1]^n maps to A and the set of values with the last
>> coordinate equal to 0 maps to {x_0}. After some thought, I realized
>> their pi_n(G,H,x_0) is the same thing as pi_n(G/H) if G/H is connected,
>> even though this was stated nowhere.

>Actually, after even more reflection, that doesn't seem so obvious
>anymore.

Okay, good - I don't think it's true.

I believe that if A sits inside X nicely (as it does in your example),
the relative homotopy group pi_n(X,A,x_0) is the same as pi_n(X/A,x_0)
where X/A is the result of taking the space X and collapsing all the
points in A to the point x_0.

But this sort of quotient space X/A is *very different* from the
usual definition of G/H when H is a subgroup of G, since in the
latter case we collapse each coset of H down to a different point.

So, there's no reason we should get the same homotopy groups.

What the heck - let's try an example. Let H = U(1) and
G = U(1) x U(1), where H sits in G as the first copy of U(1).

If we take G/H in the group-theoretic sense we get U(1).
In other words, a circle.

If we take G/H in the other sense, we just take a torus
and collapse one of its meridians down to a circle. Imagine
taking a doughnut-shaped balloon, wrapping your thumb
and forefinger around it, and squeezing as hard as you can.

Unless I'm confused these spaces have the same pi_1 but
different pi_2.

pi_2 of a circle is trivial; pi_2 of this other funky space
seems to be Z.

So, use the exact sequence of a fibration - that'll do
what you want.

(Btw, "A sits inside X nicely" is my highly technical jargon
for "(X,A) is an NDR pair", or "A -> X is a cofibration".)

### Squark

Aug 18, 2003, 9:41:23 PM8/18/03
to sci-physic...@moderators.isc.org

ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bh7kc4$ejf$1...@glue.ucr.edu>...

> In article <slrnbieoi0....@cardinal2.Stanford.EDU>,
> Aaron Bergman <aber...@princeton.edu> wrote:
>
> >In article <bg7j92$89d$1...@glue.ucr.edu>, John Baez wrote:
>
> >>Anyway, taking the universal cover of a space doesn't affect any
> >>of its homotopy groups except the first - which becomes the trivial
> >>group.
>
> >Which suggests an interesting question: how does this generalize
> >to higher homotopy? The obvious answer is the universal cover of
> >loop space. I guess you'd want to take the classifying space of
> >what came out.
>
> >...

> >Does this make any sense? Better question: is it all useful?

However, is the classifying space (and thus the space resulting
from this operation) defined precisely or only up to homotopy
equivalence? If so, it makes this operation in a sense "weeker"
than taking the universal cover which results in well-defined
topological space (or even manifold in case the original space
was a manifold).

Best regards,
Squark

------------------------------------------------------------------

Write to me using the following e-mail:
Skvark_N...@excite.exe
(just spell the particle name correctly and change the
extension in the obvious way)

### John Baez

Aug 20, 2003, 4:45:19 AM8/20/03
to
In article <bhgt94$abm$1...@news.udel.edu>, Jason <pri...@excite.com> wrote:

>John Baez wrote:

>> probably want to learn about the
>> long exact sequence of a fibration - it makes life easier.

>Oops, you mean I confused a fibration with relative homotopy groups?

Something like that.

>Well, with no one to really guide me, I guess such confusions are
>inevitable. Reading the online textbook didn't make that clear at all.
>Where can I learn about homotopy groups of fibrations then?

So, if that online book doesn't, it's indecent.
But Hatcher's online book:

http://www.math.cornell.edu/~hatcher/AT/ATpage.html

is quite good. It talks about the exact sequence of relative
homotopy groups on page 344, and the long exact sequence of
a fibration on page 376 (in the section on fiber bundles).
theory.

### John Baez

Aug 20, 2003, 5:24:08 AM8/20/03
to
Jason <pri...@excite.com> wrote:

>I'm curious but why is pi_2(G)=0 for any Lie group and pi_3(G)=Z for any

>[compact] simple Lie group?

That's a good question! I don't understand this as well
as I'd like, so I thought about it for an hour or so on the
bus back from Malacca to Singapore the other day.

Theorem, which says that the nth homotopy group of a space
equals its nth homology group if all the lower homotopy
groups vanish. If we're interested in pi_2 and pi_3, we
can assume without loss of generality that our Lie group
is connected and simply connected - since if it's not, we
can take the universal cover of the connected component
to *make* it so without changing pi_2 or pi_3 (or any of
the higher homotopy groups).

So let's assume our Lie group is connected and simply
connected. This means it has pi_0 = 0 and pi_1 = 0.
Then, to prove pi_2 = 0, it suffices by the Hurewicz theorem
to prove H_2 = 0. And then, to prove pi_3 = Z, it suffices
by the Hurewicz theorem to prove H_3 = Z.

Having reduced these questions about the homotopy groups
pi_2 and pi_3 to questions about the homology groups
H_2 and H_3, I usually declare myself done. :-)

Seriously, at this point I usually feel content with
the fact that even though I don't know why the homology
groups H_2(G) and H_3(G) are, I do know why the corresponding
*real* homology groups are what they are:

H_2(G,R) = 0 (for any connected simply connected Lie group G)
H_3(G,R) = R (for any connected simply connected compact simple Lie group G)

The reason is that I can compute these using de Rham
cohomology. I won't bore you with that now... but it's

Unfortunately, de Rham cohomology doesn't tell us about
the "torsion" part of H_2(G) or H_3(G), which goes away when
tensor with the real numbers. So, this argument doesn't
completely clinch the fact that pi_2 = 0 and pi_3 = Z.

But, I figured out another argument, which you may like,
because it uses the long exact sequence of a fibration.

Perhaps you can try it yourself. I assume you know
pi_2 and pi_3 for the group SU(2), since this group is
just the 3-sphere. So now suppose you want pi_2 and pi_3
of SU(3)! You note that

SU(3)/SU(2) = the unit sphere in C^3
= S^5

so we have a fiber bundle with total space SU(3), base
space S^5 and fiber SU(2). Using the the long exact sequence
of a fibration, we can then read off pi_2 and pi_3 for SU(3)
from the fact that we know them for SU(2).

Give it a shot and tell me if you get stuck!

One more thing - about pi_3(G) = Z:

>I think the latter has got to do with the fact that
>for any continuous map phi:S^3->G, we can define a locally flat
>connection by A=(d phi)phi^-1 with F=0 and since Tr[\int_S^3 F wedge
>A-1/3 A wedge A wedge A] is a topological invariant which is additive
>under the composition of two homotopies, it defines a homomorphism from
>pi_3(G) to R, but I still can't see why the image of the homomorphism is
>Z or why its kernel is trivial.

You're definitely on the right track here! But, since
you are using differential forms, you are not really
making full use of the homotopy group pi_3(G), but instead
the much easier deRham cohomology group H^3(G,R). By
the Hurewicz theorem and some other nonsense, H^3(G,R)
happens to be the same as the space of homomorphisms from
pi_3(G) to R - but as I hinted above, it's easier to see why
H^3(G,R) = R if you think about it using differential forms on
G. It's easy to cook up a closed but non-exact 3-form on any
compact simple Lie group. And this 3-form is what lies behind
the magic of Chern-Simons theory, and a wealth of physics.

### Anon

Aug 21, 2003, 3:28:06 AM8/21/03
to

John Baez wrote:

> Suppose someone hands you a simply connected
> topological group and you want to kill its
> second homotopy group. Here's how:
>
> 1) First loop it (= form its loop space)
> to push the homotopy groups down one notch.
>
> 2) Then kill the 1st homotopy group of the
> result by taking the universe cover.
>
> 3) Then deloop it (= take the classifying space)
> to push the homotopy groups back up one notch.

This probably is irrelavant to whatever I'll ever work on in physics (or
maybe not, who knows?), but what is a classifying space? I guess by loop
space, you simply mean the topological space of all maps from [0,1] to a
topological group with the same initial and endpoint. Does it have to be
a topological group or just any topological space?

Maybe we should crosspost this entire thread to some math newsgroup?

### Jason

Aug 21, 2003, 4:52:01 PM8/21/03
to
John Baez wrote:

> Theorem, which says that the nth homotopy group of a space
> equals its nth homology group if all the lower homotopy
> groups vanish. If we're interested in pi_2 and pi_3, we
> can assume without loss of generality that our Lie group
> is connected and simply connected - since if it's not, we
> can take the universal cover of the connected component
> to *make* it so without changing pi_2 or pi_3 (or any of
> the higher homotopy groups).

>etc...etc...

I guess that means I'd have to learn homology theory in addition to
homotopy theory just to understand what in the world you're talking
about! (And I thought I didn't need to learn anything about homology
just to understand homotopy groups...)

### John Baez

Aug 23, 2003, 2:18:27 AM8/23/03
to
We were talking about killing the 2nd homotopy group of a
connected and simply connected space with basepoint by looping
it to get a topological group, taking the universal cover of that,
and then taking the classifying space of that... and some
people were wondering what the heck all this stuff MEANT.

Squark <fii...@yahoo.com> wrote:

>[...] is the classifying space (and thus the space resulting

>from this operation) defined precisely or only up to homotopy
>equivalence? If so, it makes this operation in a sense "weeker"
>than taking the universal cover which results in well-defined
>topological space (or even manifold in case the original space
>was a manifold).

There's a way to take the classifying space of a topological
group and get a completely specific space.

It goes like this: you stick in one vertex, one edge for
each group element g, one triangle for each pair of group
elements (g,h), and one tetrahedron for each triple (g,h,k),
and so on ad infinitum, putting in one n-simplex for each
n-tuple of group elements. It's supposed to be obvious what
the faces of all these simplices are, so you can see how
they're stuck together. For example, the triangle for (g,h)
looks like this:

o
/ \
g h
/ \
o----gh-----o

where the o is the one vertex. Also, the topology on our
topological group puts a topology on all these simplices, so
you can say (for example) when a point on one triangle is close
to a point on another triangle. So, we get a well-defined
topological space.

If our topological group was called G, this space is called
BG, the "classifying space" of G. It does many marvelous
things, like classify G-bundles, give us characteristic
classes, and serve as a homotopy adjoint to the "looping"
functor from pointed spaces to topological groups. You
may not know what all these things mean, but take my word
In particular, characteristic classes are the bread and butter
of topological physics these days.

BUT, for lots of these things, we only care about what BG is
up to homotopy. So, sometimes people use BG to stand for any
space that's homotopy equivalent to this BG. That gives
you your partially-correct impression that the classifying
space is defined up to homotopy equivalence rather than
"precisely" - i.e. up to a canonical homeomorphism.

Anyway: it's not that it's hard to define BG "precisely"
if that's what you want. It's just that people often don't
care what it is except up to homotopy equivalence. Homotopy
theory encourages this relaxed attitude.

### John Baez

Aug 23, 2003, 2:43:46 AM8/23/03
to
In article <bh95bu$ejl$1...@news.udel.edu>, Anon <an...@anon.com> wrote:

>John Baez wrote:

>> Suppose someone hands you a simply connected
>> topological group and you want to kill its
>> second homotopy group. Here's how:
>>
>> 1) First loop it (= form its loop space)
>> to push the homotopy groups down one notch.
>>
>> 2) Then kill the 1st homotopy group of the
>> result by taking the universe cover.
>>
>> 3) Then deloop it (= take the classifying space)
>> to push the homotopy groups back up one notch.

>This probably is irrelavant to whatever I'll ever work on in physics (or

>maybe not, who knows?), but what is a classifying space?

The classifying space BG of a topological group G is a space
such that

Loops(BG) ~ G

where ~ means "the same up to some equivalence relation I'd
rather not discuss". I explained how to define BG in a
previous post on this thread, but this property is why it's
cool.

But now you'll want to know what "Loops" means:

>I guess by loop
>space, you simply mean the topological space of all maps from [0,1] to a
>topological group with the same initial and endpoint. Does it have to be
>a topological group or just any topological space?

Umm... close, but not quite:

The loop space Loops(X) of a topological space X with a chosen
point x is the topological space of all maps from [0,1] to X
which map 0 and 1 to the point x.

So, we don't need X to be a topological group. But,
it needs to be a little bit better than a topological
space: it needs to be a "pointed space", i.e. one equipped
with a chosen point, usually called the "basepoint".

And when we loop X, all our loops must start and end at
the basepoint - we call them "based loops" if we want to emphasize this.

Loops(X) is not quite a topological group, but it can
be massaged a bit to become one, and homotopy theorists feel
fine about calling that topological group Loops(X), too.
We then have

B(Loops(X)) ~ X

where ~ means "the same up to some other equivalence relation I'd
also rather not discuss".

(Well, *this* one is just homotopy equivalence of pointed spaces,
if that makes any sense to you. The other one was a bit subtler,
but similar.)

Since we have

B(Loops(X)) ~ X

for any pointed space and

Loops(BG) ~ G

for any topological group, there's a sense in which Loops and B
are like "inverses", which is why people also call B "delooping".
There's also a sense in which topological groups and pointed
spaces are just two different ways of looking at the same thing!
This turns out to be incredibly important. Anything you can do
to one, you can do to the other!!!

This is why my construction described above started by looping
and ended by delooping. LOTS of tricks involve doing some process,
then doing something cool you couldn't have done before, and then
undoing the first process. For example, walking around a fence to
get to the other side! Or taking your socks off by first removing
your shoes and then putting them back on. In fact, someone wrote a
whole BOOK of examples of this general strategy for solving problems!

I forget what the book was called.

### John Baez

Aug 23, 2003, 3:01:02 AM8/23/03
to
Jason was wondering why pi_2 of any Lie group is zero,
and why pi_3 of any compact simple Lie group is Z.

For compact simple Lie groups, the answer to both questions
is just that all these groups are just padded-out versions
of SU(2), which is a 3-sphere, and thus has pi_2 = 0 and pi_3 = Z.
Every compact simple Lie group has an SU(2) sitting
inside it, together with a lot of higher-dimensional "padding"
that doesn't affect pi_2 or pi_3.

Unfortunately, before Jason got to discover this for himself,
he seems to have given up, thanks no doubt to the baffling wall
of jargon I erected while explaining some OTHER ways to think

Let's try again.

> In article <bi3a7v$59g$1...@news.udel.edu>, Jason
> <pri...@excite.com> wrote:

>John Baez wrote:

>> Theorem, which says that the nth homotopy group of a space
>> equals its nth homology group if all the lower homotopy
>> groups vanish.

>I guess that means I'd have to learn homology theory in addition to

>homotopy theory just to understand what in the world you're talking

NO! If you read down to the good part, you'll see I found
you a fun way to compute pi_2 and pi_3 of lots of your favorite
Lie groups WITHOUT using anything about homology theory.

In fact, all you need is that "exact sequence of a fibration"
which you've been learning about anyway!

I gave you a nice little warmup example:

>>Perhaps you can try it yourself. I assume you know
>>pi_2 and pi_3 for the group SU(2), since this group is
>>just the 3-sphere. So now suppose you want pi_2 and pi_3
>>of SU(3)! You note that
>>
>>SU(3)/SU(2) = the unit sphere in C^3
>> = S^5
>>
>>so we have a fiber bundle with total space SU(3), base
>>space S^5 and fiber SU(2). Using the the long exact sequence
>>of a fibration, we can then read off pi_2 and pi_3 for SU(3)
>>from the fact that we know them for SU(2).
>>
>>Give it a shot and tell me if you get stuck!

It's fun... it's easy... TRY IT! YOU'LL LIKE IT!

>(And I thought I didn't need to learn anything about homology
>just to understand homotopy groups...)

Well, eventually you need to know about *everything* just to
fully understand *anything*. And certainly you can't get *too*
far in homotopy theory without homology theory - that's one
reason that book you hate so much explained homology for a few
hundred pages before doing homotopy. The other (related) reason
is that homology is a lot easier to compute.

BUT, for THIS problem you can make pretty good progress WITHOUT
using any homology theory.

### A.J. Tolland

Aug 23, 2003, 4:03:28 PM8/23/03
to
On 21 Aug 2003, Anon wrote:
>
> This probably is irrelavant to whatever I'll ever work on in physics (or
> maybe not, who knows?), but what is a classifying space?

It's a geometric space B(G) which carries information about ALL of
the vector bundles with structure group G. As a reminder, the structure
group of a vector bundle tells us what sort of structure we've put on the
bundle. If we've got structure group GL(n,R), we're just studying real
vector bundles. GL(n,C), complex vector bundles. SL(n,R), real vector
bundles with orientations. O(n), real vector bundles with an inner
product. And so on...
Now to get a bit more precise about how classifying spaces carry
information. Suppose we are fixing our attention on vector bundles with
fiber F. (So, real vector bundles have F=R^n for some n, complex have
F=C^n,...) The big idea is that B(G) is the base space of the universal
bundle E_F(G), and that all other vector bundles arise as pullbacks of
this bundle.
What's a pullback? Well, if p: E -> B is a vector bundle, and f:
X -> B is a continuous map, then the pullback of E along f is the vector
bundle p':f^*E -> X with total space f^*E = {(e,x) \in E x X| p(e) =
f(x)}. This definition is not totally enlightening, I suppose. The point
is that the pullback is a natural vector bundle associated to the pair
(p,f), and that it's initial among all such vector bundles.
Another neat thing about pullbacks is that they only depend on the
homotopy class of f. If f and g are homotopic, then f^*E = g^*E. So
every element of [X,B(G)] (the homotopy classes of maps from X to B(G)]
gives us a bundle over X. It turns out that all vector bundles arise this
way.
I'm happy to elaborate on any of this, but I'm bored with theory
so it's time to turn to examples.

Everyone's favorite is CP^\infinity, the set of one-dimensional
complex subspaces of infinite dimensional complex space. This classifies
U(1)-bundles. This generalizes well. If we consider the set of
n-dimensional complex subspaces of infinite-dimensional complex space, the
Grassmanian G(n,infinity;C) we get the classifying space for U(n). We can
also imitate these constructions for the real and quaternionic cases.

> I guess by loop space, you simply mean the topological space of all maps
> from [0,1] to a topological group with the same initial and endpoint.
> Does it have to be a topological group or just any topological space?

The loopspace construction applies to any topological space X.
And you identified it correctly: It's the space of continuous maps from
the circle into X which send some prechosen base point of the circle to
some prechosen basepoint of X. You then make this space of functions into
a topological space by performing some black rites.

> Maybe we should crosspost this entire thread to some math newsgroup?

I'd rather it stay here. The math is relevant to physics.
Classifying spaces are useful in understanding vector bundles, which are
deeply important to modern physics. Likewise, the game of killing
homotopy groups gives us a way of thinking about the relationship between
orthogonal and spin groups.
Furthermore, s.p.r. is a better forum for discussion than the math
newsgroups. We're moderated, so the signal to noise ratio is reasonably
high, and we've got a culture here that encourages nice long threads.

--A.J.

Aug 25, 2003, 5:46:28 PM8/25/03
to
Hi,

ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bh7kc4$ejf$1...@glue.ucr.edu>...

> Suppose someone hands you a simply connected

> topological group and you want to kill its
> second homotopy group. Here's how:

This sounds really cool.

> 1) First loop it (= form its loop space)
> to push the homotopy groups down one notch.

I had some trouble understanding this. I have never read a formal
definition of the loop space of a space, but I am assuming it is "the
space of all loops", given the structure of a topological space in a
natural way: e.g. I would guess loops close to each other in the
original space to be represented by points of the loop space which are
themselves close to each other. In fact I even tend to think that the
loop space of a *manifold* could be given more (natural) structure
than that of a mere topological space -- could it perhaps be made into
an infinite dimensional manifold? I can imagine a "tangent vector" to
the loop space at a given point being something like a *vector field*
on a given loop in the original space... Does this make sense?

Let me move on to my confusion:

I guessed "pushing the homotopy groups down one notch by looping the
space" meant that pi_1 of the loop space would be pi_2 of the original
space. However, an example I try seems to contradict this:

Let T^2 be the two-torus, and take a "base point" in the loop space to
be a contractible loop in T^2. Then, an element of pi_1 of the loop
space would consist of something like taking this contractible loop in
T^2, carrying it around and coming back to it in the end. However, it
seems to me that pi_1 of the loop space *for this particular base
point* will be nothing other than pi_1 of T^2 itself! Is this correct?
Anyway, at least it doesn't seem obvious to me that the fundamental
group of the loop space is equal to pi_2 of the original space.

Now I realize that the loop space of T^2 is not even (path) connected!
One can't start with a contractible loop and end up with a
noncontractible one while moving continuously in the space of loops,
simply due to the definition of contractibility! So a contractible
loop and a noncontractible one live in different connected components
of the loop space. So pi_1 of the loop space can depend on the base
point.

I have some idea about the pi_1 of the loop space of T^2 for a base
point that is one of the generators of pi_1 of T^2 itself, but let me
leave that for now.

Oops! I just I went back and read John Baez's post again, he says:

> Suppose someone hands you a simply connected
> topological group

So I missed "simply connected"... OK, now I am guessing that the
statement is: pi_2 of a simply connected topological group (or
topological space in general?) is the same as pi_1 of its loop space.
It would then seem that pi_1 of the loop space of a *general*
topological space has information about both pi_1 and pi_2 of the
original space. I won't erase my statements above about T^2, in case
anyone feels like pointing out errors, or talking about them to give
some more cool info.

> 3) Then deloop it (= take the classifying space)
> to push the homotopy groups back up one notch.

classifying spaces before I ask anything.

### John Baez

Aug 27, 2003, 3:08:28 AM8/27/03
to

>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<bh7kc4$ejf$1...@glue.ucr.edu>...

>> Suppose someone hands you a simply connected

>> topological group and you want to kill its
>> second homotopy group. Here's how:

>This sounds really cool.

Yeah. Even the phrase "killing the second homotopy group"
sounds cool! And it's a phrase professional homotopy theorists
actually use - so if you say it, you might fool someone into
thinking you're one of those. Which could be dangerous...
but it's never stopped *me*.

>> 1) First loop it (= form its loop space)
>> to push the homotopy groups down one notch.

>I had some trouble understanding this. I have never read a formal

>definition of the loop space of a space, but I am assuming it is "the
>space of all loops", given the structure of a topological space in a
>natural way: e.g. I would guess loops close to each other in the
>original space to be represented by points of the loop space which are
>themselves close to each other.

As mentioned in another post on this (new) thread,
the loop space Loops(X) of a topological space X with a chosen
point * consists of all continuous loops in X starting
and ending at *. Professionals call the point * the
"basepoint" and call loops that start and end here "based loops".

We can make Loops(X) into a topological space using a
more-or-less obvious notion of when two loops are close.

(Also, the loop space has a chosen point consisting of the
loop that just sits at * the whole time. So, we can ITERATE
the loop space construction! This turns out to be very important.)

>In fact I even tend to think that the
>loop space of a *manifold* could be given more (natural) structure
>than that of a mere topological space -- could it perhaps be made into
>an infinite dimensional manifold?

Yes! Here's it's nice to redefine the loop space Loops(X) to consist
of all *smooth* based loops in the manifold X. This is an
infinite-dimensional topological manifold of a certain sort -
technically, a "Frechet manifold".

This new space Loops(X) is homotopy equivalent to the old
Loops(X) where we used continuous loops - so for the purposes
of homotopy theory, it's just as good. But for the purposes
of geometry, it's even better.

(There are other versions that are even better, which are
"Hilbert manifolds". But, they're a wee bit trickier to
define.)

>I can imagine a "tangent vector" to
>the loop space at a given point being something like a *vector field*
>on a given loop in the original space... Does this make sense?

Yes, if you fill in the details correctly. You have to
be a bit careful in defining "vector field on a given loop"
when the loop crosses over itself. Also, since we're talking
about *based* loops, the vector field must vanish at the
starting/ending point of the given loop. But if you do all
everything right, you get a theorem describing the tangent
vectors in Loops(X).

>Let me move on to my confusion:
>
>I guessed "pushing the homotopy groups down one notch by looping the
>space" meant that pi_1 of the loop space would be pi_2 of the original
>space.

That's exactly correct.

However, an example I try seems to contradict this:

>Let T^2 be the two-torus, and take a "base point" in the loop space to
>be a contractible loop in T^2. Then, an element of pi_1 of the loop
>space would consist of something like taking this contractible loop in
>T^2, carrying it around and coming back to it in the end. However, it
>seems to me that pi_1 of the loop space *for this particular base
>point* will be nothing other than pi_1 of T^2 itself! Is this correct?

Sounds right.

>Anyway, at least it doesn't seem obvious to me that the fundamental
>group of the loop space is equal to pi_2 of the original space.

The problem arises from the fact that you're not
restricting attention to *based* loops. You get an element
of pi_1 of Loops(T^2) from a loop in Loops(T^2). But when
you read the fine print, this means a *based* loop of *based*
loops in the torus.

Your example above gives a based loop of loops in the
torus, but not a based loop of *based* loops. You
can't take a contractible based loop in T^2 and carry
it all the way around the torus while keeping it *based* at every
stage.

>Now I realize that the loop space of T^2 is not even (path) connected!

Right, and that's still true even if you stick in the
word "based" as you should when defining Loops(X).
But that's precisely as it should be:

pi_0(Loops(X)) = pi_1(X)

just as you'd expect from "pushing homotopy groups down one notch".

>Oops! I just I went back and read John Baez's post again, he says:

>> Suppose someone hands you a simply connected
>> topological group

>So I missed "simply connected"... OK, now I am guessing that the

>statement is: pi_2 of a simply connected topological group (or
>topological space in general?) is the same as pi_1 of its loop space.

Actually, it's true for any topological space with basepoint,
when you work consistently with *based* loops.

(Make sure to use the basepoint in Loops(X) which comes
from the basepoint in X in the manner I described near
the start of this post.)

I'm glad you're taking the trouble to see if I'm really making
sense! I was being a little sketchy at first, not defining
the concept of "loop space" - now I'm trying to be precise.

Unbased loops are also important, especially in the math
of string theory.

### Squark

Aug 28, 2003, 2:52:05 PM8/28/03
to sci-physic...@moderators.isc.org

ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bi70vj$ouh$1...@glue.ucr.edu>...

> There's a way to take the classifying space of a topological
> group and get a completely specific space.
>
> It goes like this: you stick in one vertex, one edge for
> each group element g, one triangle for each pair of group
> elements (g,h), and one tetrahedron for each triple (g,h,k),
> and so on ad infinitum, putting in one n-simplex for each
> n-tuple of group elements.

Fun. Apparently, what you do here is construct BG from its
natural representation as an omega-groupoid? Any group may
be thought of as a groupoid with one element, and it's
natural to try and use that groupoid to get hold of BG.
The first step is replacing the equalities "g times h = gh"
by isomorphisms (the triangles) and so on into higher
dimensions.

> Also, the topology on our
> topological group puts a topology on all these simplices, so
> you can say (for example) when a point on one triangle is close
> to a point on another triangle.

Okay, so now we went and made the omega-groupoid a
"topological omega-groupoid" (in order to throw the topology
of G in), switched to thinking of that as a simplicial
complex and ended up with a specific topological space.
Earlier I thought omega-groupoids only correspond to
homotopy types. Now it seems they correspond to
topological spaces (via the simplices procedure), equivalent
omega-groupoids corresponding to homotopically equivalent
spaces. One can think of this as follows: the objects of the
(omega)category of homotopy types are topological spaces,
homotopically equivalent topological spaces being isomorphic
within this (omega)category.

> If our topological group was called G, this space is called
> BG, the "classifying space" of G. It does many marvelous
> things, like classify G-bundles, give us characteristic
> classes, and serve as a homotopy adjoint to the "looping"
> functor from pointed spaces to topological groups. You
> may not know what all these things mean, but take my word

Hey, I know what these things mean - at least in general
of homotopy types, I suppose?

### John Baez

Aug 28, 2003, 3:19:40 PM8/28/03
to
Squark <fii...@yahoo.com> wrote:

>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<bi70vj$ouh$1...@glue.ucr.edu>...

>> There's a way to take the classifying space of a topological
>> group and get a completely specific space.
>>
>> It goes like this: you stick in one vertex, one edge for
>> each group element g, one triangle for each pair of group
>> elements (g,h), and one tetrahedron for each triple (g,h,k),
>> and so on ad infinitum, putting in one n-simplex for each
>> n-tuple of group elements.

>Fun. Apparently, what you do here is construct BG from its
>natural representation as an omega-groupoid?

Yes - but don't tell anyone *else* that. Most topologists
don't know what an "omega-groupoid is", so they'll think you're
nuts if you say something like this.

>Any group may
>be thought of as a groupoid with one element, and it's
>natural to try and use that groupoid to get hold of BG.

Yes. Indeed, for any category C there is a simplicial
set called its "nerve" and denoted BC. An n-simplex in BC
is just a string of n composable morphisms in C. We can also
take this simplicial set BC and think of it as a topological
space - homotopy theorists don't make a big deal out of this
change of viewpoint.

A group G is a category with one object and all morphisms invertible,
and in this special case the "nerve" BC reduces to the "classifying
space" BG described above... but only the classifying space of the
*discrete* group G. The classifying space of a general topological
group is a bit subtler.

>The first step is replacing the equalities "g times h = gh"
>by isomorphisms (the triangles) and so on into higher
>dimensions.

Right - that's the idea!

>> Also, the topology on our
>> topological group puts a topology on all these simplices, so
>> you can say (for example) when a point on one triangle is close
>> to a point on another triangle.

This is where the classifying space of a *topological* group
gets a bit subtler than that of a discrete group.

>Okay, so now we went and made the omega-groupoid a
>"topological omega-groupoid" (in order to throw the topology

>of G in)...

Right.

>...switched to thinking of that as a simplicial
>complex...

First of all, you mean "simplicial set", not "simplicial complex" -
the former is more general and that's the kind of thing we need
here. A "simplicial complex" is a simplicial set such that two
different n-simplices can't both have exactly the same faces.

Second of all, I was lying: you don't really mean "simplicial set".
While BG of a discrete group naturally begins life as a simplicial set,
BG of a *topological* group naturally begins life as a simplicial *space*.
In other words, it has a topological space of n-simplices for each n,
and the face and degeneracy maps are continuous.

(If you like category theory: a simplicial set is a simplicial object in
Set, while a simplicial space is a simplicial object in Top. We can and
should ponder simplicial objects in *all* our favorite categories.)

>... and ended up with a specific topological space.

Right. There's an easy way to turn any simplicial set into a
topological space, and a slightly trickier way to turn any
simplicial space into a topological space. Both these processes
go by the name of "geometric realization".

>Earlier I thought omega-groupoids only correspond to
>homotopy types. Now it seems they correspond to
>topological spaces (via the simplices procedure), equivalent
>omega-groupoids corresponding to homotopically equivalent
>spaces.

That's true. But the category of "homotopy types" is
really just the category of topological spaces with extra
morphisms thrown in to make all (weak) homotopy equivalences
invertible. The objects are the same. So, it shouldn't
come as a shock that an omega-groupoid can be turned into
a specific space.

>One can think of this as follows: the objects of the
>(omega)category of homotopy types are topological spaces,
>homotopically equivalent topological spaces being isomorphic
>within this (omega)category.

Right. Though if you want to be 100% correct, you should
say that WEAKLY homotopy-equivalent topological spaces are
isomorphic in the category of homotopy types. See that
little word "weak" in parentheses up in my previous paragraph?
This is pretty darn technical, but it's good to know:

"Weak homotopy equivalence" is designed to make locally
nasty spaces (like the rational numbers with their usual topology)
equivalent to locally nice spaces (like the integers with their
usual topology). Homotopy theorists don't care about
locally nasty spaces like the rational numbers or the Cantor
set or the topologist's sine curve. So, they either restrict
attention to locally nice spaces (like those obtained via
geometric realization from simplicial sets), or use "weak
homotopy equivalence" to make locally nasty spaces equivalent
to locally nice ones.

>> If our topological group was called G, this space is called
>> BG, the "classifying space" of G. It does many marvelous
>> things, like classify G-bundles, give us characteristic
>> classes, and serve as a homotopy adjoint to the "looping"
>> functor from pointed spaces to topological groups. You
>> may not know what all these things mean, but take my word

>Hey, I know what these things mean - at least in general
>terms.

Scary, huh? Sometimes you read something, understand it,
and then realize how weird you must be to have understood it. :-)

>of homotopy types, I suppose?

I was being a little lazy there - I don't think this is
a term most people use, so it wasn't designed to have an
ultra-precise meaning - but basically you got my meaning.

More detail:

There's a category

Ho(Top)

of homotopy types, which I described above. I described
of all "weak homotopy equivalences". Unfortunately, I
didn't tell you what a "weak homotopy equivalence" was.
However, you can just use spaces that come from simplicial
sets and throw in inverses to all the homotopy equivalences.
This gives an equivalent category, so it's an equally good
definition of Ho(Top).

If we use spaces with basepoint we get a similar thing
called Ho(Top_*)

And, there's a similar sort of "homotopy category of
topological groups", Ho(TopGrp). This is a bit trickier
to describe, since here the homotopies have to get along
with the group structure, but only weakly.

Anyway, unless I'm making a mistake, we get adjoint functors

B: Ho(TopGrp) -> Ho(Top_*)

and

Loops: Ho(Top_*) -> Ho(TopGrp)

If you understand all this, you may know more math than a
physicist should!

Aug 29, 2003, 3:56:20 AM8/29/03
to
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bihldc$m6g$1...@glue.ucr.edu>...

> As mentioned in another post on this (new) thread,
> the loop space Loops(X) of a topological space X with a chosen
> point * consists of all continuous loops in X starting
> and ending at *. Professionals call the point * the
> "basepoint" and call loops that start and end here "based loops".

So *this* was what I missed. Now it all makes sense. In fact it seems
almost obvious that pi_1 of Loops(X) is equal to pi_2 of X: if I
consider a loop in Loops(X) that starts with the constant loop at *
and also ends with it, what I am doing is like tracing a two-sphere by
loops.

To be more precise, from a homotopy of based loops in X (now I sound
like a professional, eh?) that starts and ends at the constant loop at
*, I can easily get a map from the two-sphere to X -- which of course
is nothing but an element of pi_2(X). Going in the other direction
doesn't seem that difficult, either. In order to show that pi_2(X) =
pi_1(Loops(X)), though, I also need to show that this map between
pi_1(Loops(X)) and pi_2(X) is an isomorphism of groups, but this seems
obvious enough once one recalls the definition of the group operation
in pi_2.

(Hopefully I am not being too careless here and pushing my way through
subtleties by throwing those "obivous enough"'s, "easily"'s, etc.)

Now, can I actually see that that this "pushing the homotopy groups
down one notch" works not only for pi_2 -> pi_1 but for all the higher
homotopy groups? Hey, I think this just calls for an inductive proof.
I've already done the first step, and the inductive step doesn't seem
that difficult, either. Thinking of an element of pi_n(X) as a map
from the cube [0,1]^n to X which sends all the boundary points to the
base point, I can get an element f of pi_n(X) from an element g of
pi_(n-1)(Loops(X)).

To state this again, if g is a map from the cube [0,1]^(n-1) to
Loops(X) that sends all the boundary points of the (n-1)-cube to loops
fixed at *, it doesn't seem hard to treat this as (or turn this into)
a map f from [0,1]^n to X that sends all the boundary points of the
n-cube to *. Drawing a picture for n=3 might help visualize this,
though a formal argument also seems straightforward. The group
operation also seems to work out properly, by sticking the cubes next
to each other.

Do all these seem to make sense?

> >Now I realize that the loop space of T^2 is not even (path) connected!

[...]

> But that's precisely as it should be:
>
> pi_0(Loops(X)) = pi_1(X)
> just as you'd expect from "pushing homotopy groups down one notch".

That's neat!

> Unbased loops are also important

How about pi_1(unbased loops(X))? (Now *that* doesn't look like
professional notation, does it?). Thinking a bit about S^2, T^2, and a
circle attached to S^2, I am still guessing that this group has
information about both pi_1(X) and pi_2(X). Is there a known relation
between these groups (which, if exists, would presumably depend on the
base loop one chooses)?

Once I read the posts about classifying spaces, I bet I'll find more

### A.J. Tolland

Aug 29, 2003, 3:59:05 PM8/29/03
to sci-physic...@moderators.isc.org

On Mon, 25 Aug 2003, arkadas ozakin wrote:

> I have never read a formal definition of the loop space of a space, but
> I am assuming it is "the space of all loops", given the structure of a
> topological space in a natural way: e.g. I would guess loops close to
> each other in the original space to be represented by points of the loop
> space which are themselves close to each other.

astray. The crucial thing you've forgotten in the definition of L(X) is
that X is a pointed space; that is, it comes with a chosen basepoint x.
The loop space L(X) is the set of maps from S^1 to X which send 1 to x.
This matters when we topologize L(X). The normal topology on L(X)
is the one where a map f: Y -> L(X) is continuous if and only if the
associated map F: Y * S^1 -> X is continuous. The * product is what you
get when you take the cartesian product of Y and S^1 and then quotient out
by the subspaces {y} x S^1 and Y x {1}. The * product is like the
cartesian product, but it actually gives you back a pointed space.
This topology is called the compact-open topology, for reasons
that you will have to read about somewhere else. The definition is not
particularly enlightening.

[aside: Sometimes we have to fix the compact-open topology a little bit
so that its compact subspaces are closed. Usually we surpress mention of
this little technical problem.]

> I guessed "pushing the homotopy groups down one notch by looping the
> space" meant that pi_1 of the loop space would be pi_2 of the original
> space.

Yeah, this is a general fact about topology

pi_n(L(X)) = pi_{n+1}(X)

It follows from the fact that S^n * S^1 = S^{n+1}. (Convince
yourself; I'm too lazy!) If we use this fact and the crucial property of
the compact-open topology we can calculate
pi_n(L(X))
= [S^n,L(X)] (by definition)
= [S^n * S^1 ,X] (property of compact-open topology)
= [S^{n+1},X] (fact above)
= pi_{n+1}(X) (by definition)

[ , ] denotes homotopy classes of based maps.

> However, an example I try seems to contradict this: Let T^2 be the
> two-torus, and take a "base point" in the loop space to be a
> contractible loop in T^2.

[snip]

> Then, an element of pi_1 of the loop space would consist of something
> like taking this contractible loop in T^2, carrying it around and coming
> back to it in the end. However, it seems to me that pi_1 of the loop
> space *for this particular base point* will be nothing other than pi_1
> of T^2 itself! Is this correct?

Nope. Just for the record:
pi_1(T^2) = the free group on two generators.
pi_2(T^2) = pi_2(S^1 x S^1) = pi_2(S^1) x pi_2(S^1) = 0

with it. I think the problem is in your notion of what a loop in L(X)
is. A loop S^1 -> L(X) is really a function S^1 * S^1 -> X, because of
the way we defined the topology on loop space. In other words, a loop in
L(X) is a base-point fixing map S^2 -> X. This doesn't correspond well to
What you've described above looks more like a map S^1 x S^1 -> X.
The problem with this is that S^1 x S^1 is not a space with basepoint, so
we're already on the wrong track. You got even more confused when you
tried to use the contractibility of L(X)'s basepoint to kill one of the
S^1's. You can't actually do this when we're dealing with pointed spaces.

I hope that helped. I'm not very good at deciphering incorrect
reasoning, especially when it's mostly good.

--A.J.

### Aaron Bergman

Aug 30, 2003, 5:15:52 PM8/30/03
to
In article <bh7kc4$ejf$1...@glue.ucr.edu>, John Baez wrote:
>
>The next nontrivial homotopy group of O(n)
>(in general) is the 7th one. If we kill that...
>well, I'm afraid that's a bit too mysterious for
>me to explain well. But it's related to the octonions,
>of course.

Out of curiosity:

What's the obstruction to lifting to the group you get when you
kill pi_7. IIRC, p_1/2 is the obstruction to lifting to a
String(n) structure.

### Iain

Aug 31, 2003, 5:52:31 PM8/31/03
to
ba...@galaxy.ucr.edu (John Baez) wrote in message
news:<bilkkc$oh3$1...@glue.ucr.edu>...

> Squark <fii...@yahoo.com> wrote:

> >ba...@galaxy.ucr.edu (John Baez) wrote in message
> >news:<bi70vj$ouh$1...@glue.ucr.edu>...

> >> There's a way to take the classifying space of a topological
> >> group and get a completely specific space.
> >>
> >> It goes like this: you stick in one vertex, one edge for
> >> each group element g, one triangle for each pair of group
> >> elements (g,h), and one tetrahedron for each triple (g,h,k),
> >> and so on ad infinitum, putting in one n-simplex for each
> >> n-tuple of group elements.

> >Fun. Apparently, what you do here is construct BG from its
> >natural representation as an omega-groupoid?
>
> Yes - but don't tell anyone *else* that. Most topologists
> don't know what an "omega-groupoid is", so they'll think you're
> nuts if you say something like this.

Is an omega-groupoid a category with an infinite number of objects,
all of whose morphisms are isomorphisms? What happens if you try to
define a category with more than a countable number of objects?

> >Any group may
> >be thought of as a groupoid with one element, and it's
> >natural to try and use that groupoid to get hold of BG.

> Yes. Indeed, for any category C there is a simplicial
> set called its "nerve" and denoted BC. An n-simplex in BC
> is just a string of n composable morphisms in C. We can also
> take this simplicial set BC and think of it as a topological
> space - homotopy theorists don't make a big deal out of this
> change of viewpoint.

And this can be applied to the category Top, forming a topological
space BTop? Bizarre.

> A group G is a category with one object and all morphisms invertible,
> and in this special case the "nerve" BC reduces to the "classifying
> space" BG described above... but only the classifying space of the
> *discrete* group G. The classifying space of a general topological
> group is a bit subtler.

> >The first step is replacing the equalities "g times h = gh"
> >by isomorphisms (the triangles) and so on into higher
> >dimensions.

> Right - that's the idea!

That makes sense.

[wads of unnecessary quoted text deleted by moderator]

> (If you like category theory: a simplicial set is a simplicial object in
> Set, while a simplicial space is a simplicial object in Top. We can and
> should ponder simplicial objects in *all* our favorite categories.)

I really should learn some more category theory - it seems to pop up
all over the place.

> >Earlier I thought omega-groupoids only correspond to
> >homotopy types. Now it seems they correspond to
> >topological spaces (via the simplices procedure), equivalent
> >omega-groupoids corresponding to homotopically equivalent
> >spaces.

> That's true. But the category of "homotopy types" is
> really just the category of topological spaces with extra
> morphisms thrown in to make all (weak) homotopy equivalences
> invertible. The objects are the same. So, it shouldn't
> come as a shock that an omega-groupoid can be turned into
> a specific space.

Is there a similar notion for topological spaces with more structure,
e.g. manifolds? Could there be applications to GR?

> There's a category
>
> Ho(Top)
>
> of homotopy types, which I described above. I described
> it by saying you start with Top and throw in inverses
> of all "weak homotopy equivalences". Unfortunately, I
> didn't tell you what a "weak homotopy equivalence" was.

Any good references for this?

> However, you can just use spaces that come from simplicial
> sets and throw in inverses to all the homotopy equivalences.
> This gives an equivalent category, so it's an equally good
> definition of Ho(Top).
>
> If we use spaces with basepoint we get a similar thing
> called Ho(Top_*)
>
> And, there's a similar sort of "homotopy category of
> topological groups", Ho(TopGrp). This is a bit trickier
> to describe, since here the homotopies have to get along
> with the group structure, but only weakly.
>
> Anyway, unless I'm making a mistake, we get adjoint functors
>
> B: Ho(TopGrp) -> Ho(Top_*)
>
> and
>
> Loops: Ho(Top_*) -> Ho(TopGrp)
>
> If you understand all this, you may know more math than a
> physicist should!

No danger of that :-(

-I

Aug 31, 2003, 6:18:37 PM8/31/03
to
"A.J. Tolland" <a...@math.berkeley.edu> wrote in message
news:<Pine.SOL.4.44.0308262130520.16634-100000@blue1>...

> On Mon, 25 Aug 2003, arkadas ozakin wrote:
>
> > I have never read a formal definition of the loop space of a space, but
> > I am assuming it is "the space of all loops", given the structure of a
> > topological space in a natural way: e.g. I would guess loops close to
> > each other in the original space to be represented by points of the loop
> > space which are themselves close to each other.

> astray. The crucial thing you've forgotten in the definition of L(X) is
> that X is a pointed space; that is, it comes with a chosen basepoint x.

"Forgotten"? I was actually just *guessing* a definition! John Baez
also pointed out that the picture I had in mind for the loop space was
not the standard one which involved fixing a base point for the loops.
As I explained in my post in reply to his, after restricting attention
to loops with a given base point, what he said about "pushing the
"classifying spaces".)

(I don't think my problem was due to my vagueness, by the way,
although I won't claim I was stating things precisely. I was just
looking at pi_1 of the space of *all* loops in X, not just the ones
with a given base point.)

> The * product is what you get when you take the cartesian product of
> Y and S^1 and then quotient out by the subspaces {y} x S^1 and Y x
> {1}.

Ok, I had no idea what "quotienting by a subspace" meant, so I
committed a huge sin and (gasp!) opened a book... I think what you are
calling the * product is what Allen Hatcher calls the smash product in
his book (http://www.math.cornell.edu/~hatcher/AT/ATpage.html see p.
14), and denotes by /\, rather than * (he uses * for something else).
I haven't actually gone through all the stuff that leads to it, but I
think I roughly understand the definition.

[Moderator's note: yes, Tolland is talking about the smash
product, often denoted /\ but sometimes *. - jb]

> > I guessed "pushing the homotopy groups down one notch by looping the
> > space" meant that pi_1 of the loop space would be pi_2 of the original
> > space.

> Yeah, this is a general fact about topology
>
> pi_n(L(X)) = pi_{n+1}(X)
>
> It follows from the fact that S^n * S^1 = S^{n+1}. (Convince
> yourself; I'm too lazy!) If we use this fact and the crucial property of
> the compact-open topology we can calculate
> pi_n(L(X))
> = [S^n,L(X)] (by definition)
> = [S^n * S^1 ,X] (property of compact-open topology)
> = [S^{n+1},X] (fact above)
> = pi_{n+1}(X) (by definition)
>
> [ , ] denotes homotopy classes of based maps.

pi_{n+1}(X) is more than just the homotopy classes of maps, there is
also the group operation, so in order to show pi_n(L(X)) =
pi_{n+1}(X), one also needs to think about the group operations on
both sides. I don't think this is much harder, though; as I mentioned
above, I have already convinced myself in another post that the group
isomorphism thing also works out properly.

Thanks for writing things formally like this, though (gotta learn some
professional notation!). I am not yet comfortable with this
"compact-open topology" (I've just seen the smash product thing, I am
yet to think about the definition of the topology you gave in your
post), even though I am guessing it just states formally the intuitive
picture I mentioned before.

> > However, an example I try seems to contradict this: Let T^2 be the
> > two-torus, and take a "base point" in the loop space to be a
> > contractible loop in T^2.
> [snip]
> > Then, an element of pi_1 of the loop space would consist of something
> > like taking this contractible loop in T^2, carrying it around and coming
> > back to it in the end. However, it seems to me that pi_1 of the loop
> > space *for this particular base point* will be nothing other than pi_1
> > of T^2 itself! Is this correct?

> Nope. Just for the record:
> pi_1(T^2) = the free group on two generators.
> pi_2(T^2) = pi_2(S^1 x S^1) = pi_2(S^1) x pi_2(S^1) = 0
>

I don't think it does. As I mentioned above (and as you also observed
in your post), when I said "loop space", I was thinking about the
space of all loops in T^2, without the requirement of fixing the base
point. Apparently one can't "push down the homotopy groups one notch"
by taking this space of "unbased loops" in X.

> I hope that helped.

Thanks!

### John Baez

Aug 31, 2003, 8:23:48 PM8/31/03
to
Iain <iainm...@yahoo.com> wrote:

>ba...@galaxy.ucr.edu (John Baez) wrote in message

>news:<bilkkc$oh3$1...@glue.ucr.edu>...

>Is an omega-groupoid a category with an infinite number of objects,
>all of whose morphisms are isomorphisms?

No - the "omega" doesn't refer to the number of objects.
No matter how many objects it has, a category all of whose
morphisms are isomorphisms is called just a "groupoid".

An omega-groupoid is something very different. The rough
idea that an omega-category is a generalization of a category
which has:

objects,
morphisms between objects,
2-morphisms between morphisms,
3-morphisms between 2-morphisms,
4-morphisms between 3-morphisms,
.
.
.

and so on, together with various ways of composing n-morphisms -
here's where all the subtlety lies. An omega-groupoid is an
omega-category where all the n-morphisms are invertible in some
sort of weak sense.

of N-Categories", starting here:

http://math.ucr.edu/home/baez/week73.html

It's supposed to be sort of fun, goofy and nontechnical...
see what you think.

>What happens if you try to
>define a category with more than a countable number of objects?

That's no big deal. In fact, many of the categories mathematicians
study are so big that the objects don't even form a *set* - instead,
they form a PROPER CLASS! A proper class is like a set that's too
big to be a set without Russell's paradox kicking in. You know:
"the set of all sets that don't have themselves as members" -
if it's a member of itself it's not, but if it's not, it is.

For example, when we consider the category Set, whose objects are
sets and whose morphisms are functions, it's not good to talk about
its *set* of objects. There's no set of all sets. There's just a
proper class.

Other examples include Top, or Vect. There's no set of all topological
spaces, or vector spaces.

This stuff shouldn't scare you; there are ways to handle it,
and for the most part it's not really a big deal.

However, you have somehow managed to head straight for an
example where you can get into trouble if you're not careful:

>> >Any group may
>> >be thought of as a groupoid with one element, and it's
>> >natural to try and use that groupoid to get hold of BG.

>> Yes. Indeed, for any category C there is a simplicial
>> set called its "nerve" and denoted BC. An n-simplex in BC
>> is just a string of n composable morphisms in C. We can also
>> take this simplicial set BC and think of it as a topological
>> space - homotopy theorists don't make a big deal out of this
>> change of viewpoint.

>And this can be applied to the category Top, forming a topological
>space BTop? Bizarre.

Hey, cool! You found an interesting example. But it's
an example where Russell's Paradox will bite you in the butt
if you're not careful! This space BTop is like the "space
of all spaces", since it has one vertex for each topological
space. So, we get in trouble if we try to treat BTop as a
topological space just like any other, because then it will
be one of the vertices of ITSELF. That's the sort of freaky
situation that's not allowed in ordinary set theory. Furthermore,
Top doesn't have a set of objects, just a proper class.

There's a way around this, though, and the result is actually
*omega-groupoid* version of Top which has

objects being topological spaces,
morphisms being homotopy equivalences,
2-morphisms being homotopies between these morphisms,
3-morphisms being homotopies between these 2-morphisms,
.
.
.

This omega-groupoid has a proper class of objects but using
various sorts of sneaky tricks you can squash it down so it
does. Let me call this squashed-down version TOP. Then
we can form the classifying space BTOP, which deserves to be
called "the space of all (small) spaces". And this space
BTOP turns out to be quite useful in topology... though
I must admit, I've never heard anyone but my friend James Dolan
talk about it. Most people are too scared of it, so they use
certain cheap imitations.

Btw, I should warn everyone that there's a certain topological
group called Top, and when normal people talk about BTop they
always mean the classifying space of that group - not what we're

>I really should learn some more category theory - it seems to pop up
>all over the place.

Of course it does! It's basically just a systematic study
of every aspect of mathematics that pops up all over the place.

>> That's true. But the category of "homotopy types" is
>> really just the category of topological spaces with extra
>> morphisms thrown in to make all (weak) homotopy equivalences
>> invertible. The objects are the same. So, it shouldn't
>> come as a shock that an omega-groupoid can be turned into
>> a specific space.

>Is there a similar notion for topological spaces with more structure,
>e.g. manifolds?

Dunno.

> Could there be applications to GR?

Dunno. All the math I'm talking about shows up a lot in
gauge theory and loop quantum gravity, though.

>> There's a category
>>
>> Ho(Top)
>>
>> of homotopy types, which I described above. I described
>> it by saying you start with Top and throw in inverses
>> of all "weak homotopy equivalences". Unfortunately, I
>> didn't tell you what a "weak homotopy equivalence" was.

>Any good references for this?

Any decent intro to homotopy theory will do. Personally
I have a fondness for G. W. Whitehead's "Introduction to
Homotopy Theory". Even if you don't read it, you'll
develop muscles from carrying this tome around, so
you'll simultaneously impress your friends as a genius
and a body-builder! Seriously, it's good; don't be
intimidated by its bulk. Or if you want something free,
try Allen Hatcher's online book entitled "Algebraic Topology":

http://www.math.cornell.edu/~hatcher/

### John Baez

Aug 31, 2003, 8:51:31 PM8/31/03
to

>How about pi_1(unbased loops(X))? (Now *that* doesn't look like
>professional notation, does it?)

When I want to pretend I'm a real professional, I write LX for
the space of unbased loops in the space X, and Omega(X) for the
space of based loops in the pointed space X. But since capital
Greek letters look a bit goofy in ASCII, and I'm trying to be
easily understood by nonexperts, I've been writing Loops(X)

>Thinking a bit about S^2, T^2, and a
>circle attached to S^2, I am still guessing that this group has
>information about both pi_1(X) and pi_2(X). Is there a known relation
>between these groups (which, if exists, would presumably depend on the
>base loop one chooses)?

Everything you said in this post seems right to me, including
your guess that pi_1 of LX is a kind of "blend" of pi_1 and
pi_2 of X. To make this conjecture make sense, we can pick
some basepoint in X, so we can define pi_n(X), and then use
the constant loop sitting at this point to serve as the basepoint
of LX, so we can define pi_n(LX).

If I really wanted to work this out, I would start by studying the
fibration

LX
|
|p
v
X

where the projection p maps each loop in X to its basepoint.
The "fiber" of p over a given point * of X consists of the
loops based at that point, so we can call this fiber Omega(X).
By using the long exact sequence of a fibration (which we've
been discussing a lot here lately), we can relate the homotopy
groups of LX to those of X and Omega(X). But, the homotopy
groups of Omega(X) are just those of X shifted down one notch,
so we'll get something like what you said - with some extra
subtleties thrown in.

People have studied this stuff already. It's actually more
complicated and exciting to think about the *homology* groups
of LX; these are related to something called "cyclic homology",
by Loday.

### John Baez

Aug 31, 2003, 9:16:02 PM8/31/03
to
In article <slrnbkvgh1....@cardinal4.Stanford.EDU>,
Aaron Bergman <aber...@princeton.edu> wrote:

>In article <bh7kc4$ejf$1...@glue.ucr.edu>, John Baez wrote:

>>The next nontrivial homotopy group of O(n)
>>(in general) is the 7th one. If we kill that...
>>well, I'm afraid that's a bit too mysterious for
>>me to explain well. But it's related to the octonions,
>>of course.

>Out of curiosity:
>
>What's the obstruction to lifting to the group you get when you

>kill pi_7? IIRC, p_1/2 is the obstruction to lifting to a
>String(n) structure.

I think it's something sorta like the 2nd Pontryagin class, p_2.
After all, the nth Pontryagin class comes from an element of the
4nth cohomology group of BO(infinity), and the 7th homotopy group
of O(infinity) is the 8th homotopy group of BO(infinity), which,
ahem, should have *something* to do with its 8th cohomology group.

end of Charles Thomas' book "Elliptic Cohomology". I believe
he uses something like O<8> to stand for the topological
group obtained by killing the first 7 homotopy groups of O(infinity),
so you could poke around looking for that.

I find this book quite mysterious, by the way. After talking
a bit about elliptic cohomology, he immediately proceeds to spend
a lot of time calculating it for BG where G is some funky "Mathieu
group" like M_{22} - one of the sporadic finite simple groups.
This is not the space I would have first wanted to know the
elliptic cohomology of! Presumably he has some reason for doing
what he does, probably related to Monstrous Moonshine. But, I've
never managed to figure out what this reason is.

Wild guess of the day: If killing the first 3 homotopy groups of O(n)
gives a group String(n) related to strings (= 1-branes),
maybe killing the first 7 gives a group related to 5-branes!
Do you have any desire to equip manifolds with "5-brane structures"?

### Iain

Sep 3, 2003, 4:03:08 PM9/3/03
to
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<biu3ik$44i$1...@glue.ucr.edu>...

> Iain <iainm...@yahoo.com> wrote:
>
> >ba...@galaxy.ucr.edu (John Baez) wrote in message
> >news:<bilkkc$oh3$1...@glue.ucr.edu>...
>
> >Is an omega-groupoid a category with an infinite number of objects,
> >all of whose morphisms are isomorphisms?
>
> No - the "omega" doesn't refer to the number of objects.
> No matter how many objects it has, a category all of whose
> morphisms are isomorphisms is called just a "groupoid".

That confusion is now sorted - the special case occurs at the other
end, so to speak, with a groupoid with one object being a group (and a
monoid).

> An omega-groupoid is something very different. The rough
> idea that an omega-category is a generalization of a category
> which has:
>
> objects,
> morphisms between objects,
> 2-morphisms between morphisms,
> 3-morphisms between 2-morphisms,
> 4-morphisms between 3-morphisms,
> .
> .
> .
>
> and so on, together with various ways of composing n-morphisms -
> here's where all the subtlety lies. An omega-groupoid is an
> omega-category where all the n-morphisms are invertible in some
> sort of weak sense.

When you say weak sense, do you mean that if there are two morphisms
f:C->D g:D->C, then instead of fg = 1_C, fg ~ 1_C, where the ~
represents some kind of isomorphism?

> of N-Categories", starting here:
>
> http://math.ucr.edu/home/baez/week73.html
>
> It's supposed to be sort of fun, goofy and nontechnical...
> see what you think.

I haven't finished it yet, but it's been fun so far. I must check out
the hint you left about applying category theory (specifically
groupoids) to SR.

> >What happens if you try to
> >define a category with more than a countable number of objects?
>
> That's no big deal. In fact, many of the categories mathematicians
> study are so big that the objects don't even form a *set* - instead,
> they form a PROPER CLASS! A proper class is like a set that's too
> big to be a set without Russell's paradox kicking in. You know:
> "the set of all sets that don't have themselves as members" -
> if it's a member of itself it's not, but if it's not, it is.

<snip>

> >> Yes. Indeed, for any category C there is a simplicial
> >> set called its "nerve" and denoted BC. An n-simplex in BC
> >> is just a string of n composable morphisms in C. We can also
> >> take this simplicial set BC and think of it as a topological
> >> space - homotopy theorists don't make a big deal out of this
> >> change of viewpoint.
>
> >And this can be applied to the category Top, forming a topological
> >space BTop? Bizarre.
>
> Hey, cool! You found an interesting example. But it's
> an example where Russell's Paradox will bite you in the butt
> if you're not careful! This space BTop is like the "space
> of all spaces", since it has one vertex for each topological
> space. So, we get in trouble if we try to treat BTop as a
> topological space just like any other, because then it will
> be one of the vertices of ITSELF. That's the sort of freaky
> situation that's not allowed in ordinary set theory. Furthermore,
> Top doesn't have a set of objects, just a proper class.

So if you have BVect, and try to treat it as a vector space, will you
have itself as one of its own vertices?

> There's a way around this, though, and the result is actually
> quite cool. Actually, it's even better if you start with an
> *omega-groupoid* version of Top which has
>
> objects being topological spaces,
> morphisms being homotopy equivalences,
> 2-morphisms being homotopies between these morphisms,
> 3-morphisms being homotopies between these 2-morphisms,
> .
> .
> .
>
> and so on ad infinitum.

What happens if one tries to form a *Z*-groupoid of Top? Has anyone
tried to make sense of n-morphisms for negative n? Indeed, what about
for non-integral n?

> This omega-groupoid has a proper class of objects but using
> various sorts of sneaky tricks you can squash it down so it
> does. Let me call this squashed-down version TOP. Then
> we can form the classifying space BTOP, which deserves to be
> called "the space of all (small) spaces". And this space
> BTOP turns out to be quite useful in topology... though
> I must admit, I've never heard anyone but my friend James Dolan
> talk about it. Most people are too scared of it, so they use
> certain cheap imitations.

Curious. I would have thought that if it was useful, that it would
therefore be quite commonly used?

> Btw, I should warn everyone that there's a certain topological
> group called Top, and when normal people talk about BTop they
> always mean the classifying space of that group - not what we're

> >I really should learn some more category theory - it seems to pop up
> >all over the place.
>
> Of course it does! It's basically just a systematic study
> of every aspect of mathematics that pops up all over the place.

OK - so that was a bit of a silly thing to say. I suppose it was a bit
like being surprised at the ubiquity of set theory.

> >> That's true. But the category of "homotopy types" is
> >> really just the category of topological spaces with extra
> >> morphisms thrown in to make all (weak) homotopy equivalences
> >> invertible. The objects are the same. So, it shouldn't
> >> come as a shock that an omega-groupoid can be turned into
> >> a specific space.
>
> >Is there a similar notion for topological spaces with more structure,
> >e.g. manifolds?
>
> Dunno.
>
> > Could there be applications to GR?
>
> Dunno. All the math I'm talking about shows up a lot in
> gauge theory and loop quantum gravity, though.

So maybe then - something to look into.

> >> There's a category
> >>
> >> Ho(Top)
> >>
> >> of homotopy types, which I described above. I described
> >> it by saying you start with Top and throw in inverses
> >> of all "weak homotopy equivalences". Unfortunately, I
> >> didn't tell you what a "weak homotopy equivalence" was.
>
> >Any good references for this?
>
> Any decent intro to homotopy theory will do. Personally
> I have a fondness for G. W. Whitehead's "Introduction to
> Homotopy Theory". Even if you don't read it, you'll
> develop muscles from carrying this tome around, so
> you'll simultaneously impress your friends as a genius
> and a body-builder! Seriously, it's good; don't be
> intimidated by its bulk. Or if you want something free,
> try Allen Hatcher's online book entitled "Algebraic Topology":
>
> http://www.math.cornell.edu/~hatcher/

Hatcher's book seems good (and a lot cheaper :-)

-I

### Aaron Bergman

Sep 3, 2003, 5:35:52 PM9/3/03
to sci-physic...@moderators.isc.org

In article <biu6ki$4rq$1...@glue.ucr.edu>, ba...@galaxy.ucr.edu (John Baez)
wrote:

> In article <slrnbkvgh1....@cardinal4.Stanford.EDU>,
> Aaron Bergman <aber...@princeton.edu> wrote:
>
> end of Charles Thomas' book "Elliptic Cohomology". I believe
> he uses something like O<8> to stand for the topological
> group obtained by killing the first 7 homotopy groups of O(infinity),
> so you could poke around looking for that.

I don't suppose there's any quick and easy explanation as to why p_1/2
is the obstruction, is there? From what I've been told, it should be
equivalent or at least related to the existence of a spin structure on
the free loop space.

> I find this book quite mysterious, by the way. After talking
> a bit about elliptic cohomology, he immediately proceeds to spend
> a lot of time calculating it for BG where G is some funky "Mathieu
> group" like M_{22} - one of the sporadic finite simple groups.
> This is not the space I would have first wanted to know the
> elliptic cohomology of! Presumably he has some reason for doing
> what he does, probably related to Monstrous Moonshine. But, I've
> never managed to figure out what this reason is.
>
> Wild guess of the day: If killing the first 3 homotopy groups of O(n)
> gives a group String(n) related to strings (= 1-branes),
> maybe killing the first 7 gives a group related to 5-branes!

Steal my wild guesses, will you? No fair.

> Do you have any desire to equip manifolds with "5-brane structures"?

Anything that might help us learn about five-branes is interesting.
First, I need to learn what string structures have to do with strings,
though. The math papers are somewhat lacking in physical intuition.

Just in general, isn't the obstruction in C^2(M,Ker(String(n)->Spin(n))
(a Cech cocycle). (I want to write H^2, but I'm scared that the kernal
isn't abelian and then I'm not really sure about things).

Aaro

### John Baez

Sep 4, 2003, 2:42:18 AM9/4/03
to
Iain <iainm...@yahoo.com> wrote:

>John Baez wrote:

>> An omega-groupoid is an
>> omega-category where all the n-morphisms are invertible in some
>> sort of weak sense.

>When you say weak sense, do you mean that if there are two morphisms
>f:C->D g:D->C, then instead of fg = 1_C, fg ~ 1_C, where the ~
>represents some kind of isomorphism?

Yeah. More precisely, we say an n-morphism f: C -> D is
"weakly invertible" if there exists an n-morphism g: D -> C
such that there exist (n+1)-morphisms A: fg -> 1_C and B: gf -> 1_D
which are... weakly invertible!

Get it? The definition of "weakly invertible" is recursive.
And the recursion only stops if our omega-category is an
N-category, meaning that all n-morphisms with n > N are identities.
In this case an N-morphism is weakly invertible iff it's
invertible, but the concept gets subtler for n-morphisms with
n < N.

>> [...] we get in trouble if we try to treat BTop as a

>> topological space just like any other, because then it will
>> be one of the vertices of ITSELF. That's the sort of freaky
>> situation that's not allowed in ordinary set theory. Furthermore,
>> Top doesn't have a set of objects, just a proper class.

>So if you have BVect, and try to treat it as a vector space, will you
>have itself as one of its own vertices?

Well, you can "try" to treat BVect as a vector space, but
I don't see how you're going to try it. It's a topological
space, not a vector space in any natural way. And worse,
you seem to be wanting to pretend it's a vector space just
so you can get in trouble with Russell's paradox! I won't
help you do that - it'd be like giving a juvenile delinquent
bottle rockets.

>What happens if one tries to form a *Z*-groupoid of Top?

It is a Z-groupoid in a natural way, but most people
don't know about Z-groupoids so they say it's a "symmetric
monoidal topological groupoid", which is more or less the
same thing as a Z-groupoid with only identity n-morphisms
for n negative.

>Has anyone tried to make sense of n-morphisms for negative n?

Yup, but not too many people - mainly just James Dolan and I.
Topologists work with such things but mostly without realizing
they talk about "spectra". They do know how to make Top into
a spectrum, using something called the Segal construction. But
in fact, later this month I'm going to give a talk at a homotopy

>Indeed, what about for non-integral n?

No, that seems a bit too goofy even for me! At least right now.

>> This omega-groupoid has a proper class of objects but using
>> various sorts of sneaky tricks you can squash it down so it
>> does. Let me call this squashed-down version TOP. Then
>> we can form the classifying space BTOP, which deserves to be
>> called "the space of all (small) spaces". And this space
>> BTOP turns out to be quite useful in topology... though
>> I must admit, I've never heard anyone but my friend James Dolan
>> talk about it. Most people are too scared of it, so they use
>> certain cheap imitations.

>Curious. I would have thought that if it was useful, that it would
>therefore be quite commonly used?

No. Since mathematics is pure thought, mathematicians can
invent new tools at a rapid rate - so fast that nobody can
keep up with learning about all these tools, even all the
tools they'd find useful if only they knew about 'em! So,
there are lots of mathematical tools floating around that
mathematicians and physicists would find useful, but haven't

### Aaron Bergman

Sep 4, 2003, 1:07:39 PM9/4/03
to sci-physic...@moderators.isc.org

In article <abergman-7C1B89.23191331082003@localhost>,
Aaron Bergman <aber...@physics.utexas.edu> wrote:

> In article <biu6ki$4rq$1...@glue.ucr.edu>, ba...@galaxy.ucr.edu (John Baez)
> wrote:
>
> > In article <slrnbkvgh1....@cardinal4.Stanford.EDU>,
> > Aaron Bergman <aber...@princeton.edu> wrote:
> >
> > end of Charles Thomas' book "Elliptic Cohomology". I believe
> > he uses something like O<8> to stand for the topological
> > group obtained by killing the first 7 homotopy groups of O(infinity),
> > so you could poke around looking for that.
>
> I don't suppose there's any quick and easy explanation as to why p_1/2
> is the obstruction, is there? From what I've been told, it should be
> equivalent or at least related to the existence of a spin structure on
> the free loop space.

Having done a little reading, there's another definition of a string
structure, too.

Given the Spin(n) bundle, TM -> M, one can loop everything giving the
LSpin(n) bundle LTM -> LM. LSpin(n) has a central U(1) extension and the
obstruction to the lift is the Dixmier-Douady class in H^3(LM). This
transgresses down to p_1/2.

I don't understand the relation of this definition to what we've been

Aaron

### Iain

Sep 6, 2003, 4:17:57 PM9/6/03
to
ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bj6msa$eip$1...@glue.ucr.edu>...

> Iain <iainm...@yahoo.com> wrote:
>
> >John Baez wrote:
>
> >> An omega-groupoid is an
> >> omega-category where all the n-morphisms are invertible in some
> >> sort of weak sense.
>
> >When you say weak sense, do you mean that if there are two morphisms
> >f:C->D g:D->C, then instead of fg = 1_C, fg ~ 1_C, where the ~
> >represents some kind of isomorphism?
>
> Yeah. More precisely, we say an n-morphism f: C -> D is
> "weakly invertible" if there exists an n-morphism g: D -> C
> such that there exist (n+1)-morphisms A: fg -> 1_C and B: gf -> 1_D
> which are... weakly invertible!
>
> Get it? The definition of "weakly invertible" is recursive.
> And the recursion only stops if our omega-category is an
> N-category, meaning that all n-morphisms with n > N are identities.
> In this case an N-morphism is weakly invertible iff it's
> invertible, but the concept gets subtler for n-morphisms with
> n < N.

OK, so we have a collection of morphisms. Take an n-morphism f. Then
it is weakly invertible if there is an n-morphism g, and there are
(n+1)-morphisms A: fg -> 1_C and B: gf -> 1_D which are weakly
invertible. For A to be weakly invertible, we want (n+2)-morphisms
that will generalize the concept of inverses for A. So we have -
A':1_C -> fg such that there are A1:AA' -> fg, A2: A'A -> 1_C which
are themselves weakly invertible, etc. So if the morphisms with n > N
are the identity map, we may have (for instance) A1:AA' -> 1_C, A2:
A'A -> 1_C.

> >> [...] we get in trouble if we try to treat BTop as a
> >> topological space just like any other, because then it will
> >> be one of the vertices of ITSELF. That's the sort of freaky
> >> situation that's not allowed in ordinary set theory. Furthermore,
> >> Top doesn't have a set of objects, just a proper class.
>
> >So if you have BVect, and try to treat it as a vector space, will you
> >have itself as one of its own vertices?
>
> Well, you can "try" to treat BVect as a vector space, but
> I don't see how you're going to try it.

Yes, you are correct, it is a toplogical space. I wonder though how to
add extra structure to it, for instance a vector space. Of course,
that would require a field (or ring, if we are looking at modules
instead) of scalars, and also some abelian group for the vectors.
Perhaps use the vector spaces that we used to construct the
topological space BVect?

It's a topological
> space, not a vector space in any natural way. And worse,
> you seem to be wanting to pretend it's a vector space just
> so you can get in trouble with Russell's paradox!

My intention was to find another example of where Russell's paradox
kicks in - and it seems it wasn't particularly well chosen one (OK, so
it was rubbish)

I won't
> help you do that - it'd be like giving a juvenile delinquent
> bottle rockets.
> >What happens if one tries to form a *Z*-groupoid of Top?
>
> It is a Z-groupoid in a natural way, but most people
> don't know about Z-groupoids so they say it's a "symmetric
> monoidal topological groupoid", which is more or less the
> same thing as a Z-groupoid with only identity n-morphisms
> for n negative.
>
> >Has anyone tried to make sense of n-morphisms for negative n?
>
> Yup, but not too many people - mainly just James Dolan and I.
> Topologists work with such things but mostly without realizing
> that's what they're doing.

I suppose this ties in with your comment below, about the abundance of
(possibly) useful mathematical tools lying around, but that need a bit
of "advertising" to point out their usefulness. (And I don't mean by

> they talk about "spectra". They do know how to make Top into
> a spectrum, using something called the Segal construction. But
> in fact, later this month I'm going to give a talk at a homotopy
> theory conference in Canada and ask people why they don't talk

I googled the "Segal construction", the words in the SPR post (from
yourself) were familiar, but... Maybe, someday, when I have time, I
will attempt to learn what all those words mean. Just for fun.

> >Indeed, what about for non-integral n?
>
> No, that seems a bit too goofy even for me! At least right now.

I was thinking of perhaps representing the Z-groupoids by an integer
valued function, and extending its definition to the complex plane,
and seeing what the corresponding groupoid looked like (if anything).

<snip>

> >Curious. I would have thought that if it was useful, that it would
> >therefore be quite commonly used?
>
> No. Since mathematics is pure thought, mathematicians can
> invent new tools at a rapid rate - so fast that nobody can
> keep up with learning about all these tools, even all the
> tools they'd find useful if only they knew about 'em! So,
> there are lots of mathematical tools floating around that
> mathematicians and physicists would find useful, but haven't

A pity - but then, life is short. Also see my above remark about ads.

-I

### Jason

Sep 8, 2003, 1:53:48 AM9/8/03
to sci-physic...@moderators.isc.org

ba...@galaxy.ucr.edu (John Baez) wrote in message news:<bhveno$fm$1...@glue.ucr.edu>...

> Perhaps you can try it yourself. I assume you know
> pi_2 and pi_3 for the group SU(2), since this group is
> just the 3-sphere. So now suppose you want pi_2 and pi_3
> of SU(3)! You note that
>
> SU(3)/SU(2) = the unit sphere in C^3
> = S^5
>
> so we have a fiber bundle with total space SU(3), base
> space S^5 and fiber SU(2). Using the the long exact sequence
> of a fibration, we can then read off pi_2 and pi_3 for SU(3)
> from the fact that we know them for SU(2).
>
> Give it a shot and tell me if you get stuck!

I didn't know you expected me to reply. Yes, that works for SU(3), but
apparently, it depends on a property known as the homotopy lifting
property. Does that also work for infinite dimensional Lie groups?

### John Baez

Sep 11, 2003, 3:32:42 PM9/11/03
to
Jason <pri...@excite.com> wrote:

>ba...@galaxy.ucr.edu (John Baez) wrote in message
>news:<bhveno$fm$1...@glue.ucr.edu>...

>> Perhaps you can try it yourself. I assume you know
>> pi_2 and pi_3 for the group SU(2), since this group is
>> just the 3-sphere. So now suppose you want pi_2 and pi_3
>> of SU(3)! You note that
>>
>> SU(3)/SU(2) = the unit sphere in C^3
>> = S^5
>>
>> so we have a fiber bundle with total space SU(3), base
>> space S^5 and fiber SU(2). Using the the long exact sequence
>> of a fibration, we can then read off pi_2 and pi_3 for SU(3)
>> from the fact that we know them for SU(2).
>>
>> Give it a shot and tell me if you get stuck!

>I didn't know you expected me to reply.

With enthusiasm and delight, naturally!

>Yes, that works for SU(3), [...]

Good. You know that pi_i(S^j) is 0 if i < j and Z if i = j.
So, I assume you used this portion of the long exact sequence:

pi_2(SU(2)) -> pi_2(SU(3)) -> pi_2(SU(3)/SU(2))

and the facts that

pi_2(SU(2)) = pi_2(S^3) = 0 (since SU(2) = S^3)

and

pi_2(SU(3)/SU(2)) = pi_2(S^5) = 0 (since SU(3)/SU(2) = S^5)

to get the exact sequence

0 -> pi_2(SU(3)) -> 0

which implies

pi_2(SU(3)) = 0.

Similarly, I assume you used this portion of the long exact sequence:

pi_4(SU(3)/SU(2)) -> pi_3(SU(2)) -> pi_3(SU(3)) -> pi_3(SU(3)/SU(2))

and the facts that

pi_3(SU(2)) = pi_3(S^3) = Z

and

pi_3(SU(3)/SU(2)) = pi_3(S^5) = 0
pi_4(SU(3)/SU(2)) = pi_4(S^5) = 0

to get the exact sequence

0 -> Z -> pi_3(SU(3)) -> 0

which implies

pi_3(SU(3)) = Z.

> but how about the other Lie groups?

Well, how about SU(4)? Now that you've done SU(3), that's
the obvious one to tackle next.

You probably can see that

SU(4)/SU(3) = the unit sphere in C^4
= S^7

so we have a fiber bundle with total space SU(4), base
space S^7 and fiber SU(3). Using the the long exact sequence
of a fibration, can you figure out pi_2 and pi_3 of SU(4),
using the fact that you know them for S^7 and SU(3)?

And then, can you see how to exploit the same trick to calculate
pi_2 and pi_3 for a whole wad of compact simple Lie groups?

>apparently, it depends on a property known as the homotopy lifting
>property.

Right, that's the property a "fibration" must have. Luckily,
any fiber bundle built from finite-dimensional manifolds has it.

>Does that also work for infinite dimensional Lie groups?

As I mentioned before, there's no precise definition of
"infinite dimensional Lie group", so there aren't really
theorems about them. The problem is that there isn't really
a precise definition of "infinite dimensional manifold".

Instead, there are manifolds that look locally like a Hilbert
space, called "Hilbert manifolds", and more general manifolds
that look locally like a complete normed vector space, called
"Banach manifolds", and even more general ones called "Frechet
manifolds" and so on. Correspondingly there are "Hilbert Lie
groups" and "Banach Lie groups" and "Frechet Lie groups" and
so on.

All these various flavors of infinite-dimensional Lie group
come up in practice. If for example you consider a loop group,
i.e. a group of functions from the circle to some (finite-dimensional!)
Lie group G, to make things precise you have to say whether the
functions are continuous or infinitely differentiable or... whatever!

If you work with continuous functions from the circle to G,
you get a BANACH Lie group. If you use infinitely differentiable
ones, you get a FRECHET Lie group. If you use functions whose
derivatives up to the nth derivative are all square-integrable,
you get a HILBERT Lie group (if n is big enough - n = 1/2 will
do). The last of these three choices looks the most technical,
especially if you're not comfortable with fractional derivatives,
but Hilbert Lie groups are so nice that this choice is quite
popular - you can read all about it in Pressley and Segal's
book "Loop Groups".

Anyway, you're probably reeling in shock at this point, but
you can't really have theorems without definitions. Here's
the good news: all these various flavors of infinite-dimensional
Lie groups are topological groups. And, I believe that whenever
G is a topological group and H is a closed subgroup, there is
a fibration with total space G, fiber H and base space G/H.

(If I'm wrong, surely some topologist out there will correct me!)

So, the short answer to your question is "yes, as long as you
use one of the flavors of infinite-dimensional Lie group that I've
discussed". There are also other flavors that work too.

But, you should be aware that all these nitpicky nuances I've
just discussed are not just things that mathematicians have
created to make life miserable for people like you. The
details can sometimes AFFECT THE ANSWER when you start computing
homotopy groups. For example, there are different flavors of the
infinite-dimensional Lie group you might naively call U(infinity).
One of these is contractible so all its homotopy groups vanish.
Others, which most people prefer, have nonvanishing homotopy groups.

..........................................................................

Next year's estimated US federal spending:

Military deployment and reconstruction in Afghanistan and Iraq: $87 billion Department of Education:$45.2 billion
Environmental Protection Agency: $7.6 billion Mike Leavitt, nominated to head the Enviromental Protection Agency, has invented the word "enlibra" to mean "bring balance to enviromental policy" - a euphemism for "allow companies to pollute more". Expect to start hearing this word. ### John Baez unread, Sep 11, 2003, 8:18:50 PM9/11/03 to >ba...@galaxy.ucr.edu (John Baez) wrote in message >news:<bj6msa$eip\$1...@glue.ucr.edu>...

>> More precisely, we say an n-morphism f: C -> D is

>> "weakly invertible" if there exists an n-morphism g: D -> C
>> such that there exist (n+1)-morphisms A: fg -> 1_C and B: gf -> 1_D
>> which are... weakly invertible!
>>
>> Get it? The definition of "weakly invertible" is recursive.

>OK, so we have a collection of morphisms. Take an n-morphism f. Then

>it is weakly invertible if there is an n-morphism g, and there are
>(n+1)-morphisms A: fg -> 1_C and B: gf -> 1_D which are weakly
>invertible. For A to be weakly invertible, we want (n+2)-morphisms
>that will generalize the concept of inverses for A. So we have -
>A':1_C -> fg such that there are A1:AA' -> fg, A2: A'A -> 1_C which
>are themselves weakly invertible, etc.

Right.

>So if the morphisms with n > N
>are the identity map, we may have (for instance) A1:AA' -> 1_C, A2:
>A'A -> 1_C.

Hmm. Well, the real point is that in an N-category, for an
N-morphism to be weakly invertible is precisely for it to be
invertible! This follows straight from the fact that all
n-morphisms with n > N are identity morphisms.

>> They do know how to make Top into
>> a spectrum, using something called the Segal construction.

>I googled the "Segal construction", the words in the SPR post (from

>yourself) were familiar, but... Maybe, someday, when I have time, I
>will attempt to learn what all those words mean. Just for fun.

There I was talking about something completely different:
the "GNS construction" due to Gelfand, Naimark, and
my thesis advisor Irving Segal. That's a trick for
getting a representation of a C*-algebra on a Hilbert
space starting from a state on this algebra.

In the text you quote above, I was talking about
a construction due to Graeme Segal for getting a
spectrum from a symmetric monoidal category. The
_Infinite Loop Spaces_. See Section 2.5, "Segal's machine".

Graeme Segal did some great work on homotopy theory
before moving on to lay the foundations for string theory
with his work on loop groups and conformal field theory.
Only a few people seem to realize how related his old
work and his new work are.

### Jason

Sep 16, 2003, 1:56:07 AM9/16/03
to sci-physic...@moderators.isc.org

John Baez wrote:
> Instead, there are manifolds that look locally like a Hilbert
> space, called "Hilbert manifolds", and more general manifolds
> that look locally like a complete normed vector space, called
> "Banach manifolds", and even more general ones called "Frechet
> manifolds" and so on. Correspondingly there are "Hilbert Lie
> groups" and "Banach Lie groups" and "Frechet Lie groups" and
> so on.

> But, you should be aware that all these nitpicky nuances I've

> just discussed are not just things that mathematicians have
> created to make life miserable for people like you. The
> details can sometimes AFFECT THE ANSWER when you start computing
> homotopy groups. For example, there are different flavors of the
> infinite-dimensional Lie group you might naively call U(infinity).
> One of these is contractible so all its homotopy groups vanish.
> Others, which most people prefer, have nonvanishing homotopy groups.

This made me think. Since the measure of functional integrals supposedly
has its support almost everywhere on proper distributions (I'm just
quoting hearsay here...), if we have let's say a nonlinear sigma model
where the target manifold, M has a nontrivial pi_3 and there's one
particular point on the manifold, phi_0 with the least potential energy,
then we can have topological solitions (in 3+1 dimensions). But, as you
mentioned, if the topology depends on whether we're working with
continuous spaces, differentiable spaces or distribution spaces (as well
as asymptotic boundary conditions which somehow have to be reflected in
the topology), then we could have a condition where topological solitons
can "tunnel" through and disappear!

### John Baez

Sep 23, 2003, 6:47:25 PM9/23/03
to
In article <l03102801bb90e7b17b52@[65.141.92.24]>,
Tony Smith <tsm...@spamfree.net> cited:

>http://xxx.lanl.gov/abs/hep-th/0102183
>
>in which C. I. Lazaroiu says, in a footnote on page 10:
>
>"... As a matter of fact, there are good reasons
>(coming from the analysis of the A-twisted topological string
>theory) to believe that A-type D-branes in a type II compactification
>give a nonassociative category
>(likely an Aoo category), due to worldsheet instanton effects.

I don't like writing "A_infinity" as "Aoo", since I pronounce
the latter as "Aooh!".

So, henceforth I will speak of "A_infinity categories" rather
than "Aoo categories" - except if I'm discussing mathematics
while walking barefoot across hot asphalt.

Anyway:

It's misleading to call an A_infinity category a "nonassociative"
category. It's really more like a category where composition is
associative UP TO HOMOTOPY.

A wee bit more precisely, an A_infinity category is a gadget
where for each pair of objects x,y we have a topological space
of morphisms f: x -> y, and composition is associative up
to a specified homotopy, which satisfies something called the
"pentagon identity" up to specified homotopy, which satisfies some
other equation up to a specified homotopy, and so on ad infinitum.

If we get tired and quit after the nth stage of this definition,
we have just an "A_n category".

A_infinity categories are not at all exotic. Here's a simple
example. Take any topological space X. Take as objects the
points of X, and take as morphisms f: x -> y the parametrized paths

f: [0,1] -> X

that start at x and end at y. If the path

f: [0,1] -> X

ends where the path

g: [0,1] -> X

starts, we can compose them and get a path

fg: [0,1] -> X

that spends the first half of its time going along the
path f at double speed, and the second half of its time
going along g at double speed.

This composition is not associative!!!

If you don't believe me, check it. The point is that
(fg)h and f(gh) are not parametrized the same way.

Nonetheless, there's a more or less obvious homotopy from
(fg)h to f(gh), which does the reparametrization.

This shows we have an A_1 category. Construction of all
the higher homotopies necessary for an A_infinity category
is left as an exercise for the reader. :-)

This stuff is related to the operation of gluing together two
open strings and getting a new one. So, it should show up in
open string field theory. But, I don't pretend to understand
what Lazaroiu is talking about in the passage you quote.

might imagine, an A_infinity category is a special sort of
weak infinity-category. However, I'll just say 2 things

1) Often when people talk about A_infinity categories they
mean a gadget that has a CHAIN COMPLEX of morphisms between
any two objects, instead of a topological space of morphisms.
Here we need to use chain homotopies instead of homotopies.

2) Stasheff was the first to invent this A_infinity business,
but he talked about "A_infinity spaces", where you've got a
topological space equipped with a product that's associative
up to a specified homotopy, which satisfies the pentagon identity
up to a specified homotopy, etcetera ad infinitum. Since you
like the nonassociative product on S^7, you might like to peek
at his book

H-spaces from a homotopy point of view, Lecture Notes in Mathematics
161, Springer-Verlag, Berlin (1970), ii-95.

in which he gives a nice explanation of A_infinity and A_n spaces,
and discusses the subject of which spheres can be made into A_n spaces.

Any Lie group is an A_infinity space, since it has a *strictly*
associative product, so we can take all the homotopies to be trivial.
Thus, S^0, S^1 and S^3 can be made into A_infinity spaces. S^7
cannot be made into a Lie group, but you might wonder if it could
still be made into an A_infinity or A_n space. I think the best
you can do is make it an A_1 space, meaning (if I have the numbering
system right) that you can give it a product, but this product cannot
be associative up to homotopy.

If I'm right, this means the fascinating subject of A_n spaces is
of limited relevance to the octonions, except in the following way:

3) There's a cool construction of the "classifying space of an
A_n space", and this explains why you can define OP^1 and OP^2
but no higher octonionic projective spaces - but I said I'd

By the way - I'm at a homotopy theory conference at the University
of Western Ontario right now, and Jesper Grodal just presented
an example of a finite CW complex which is homotopy equivalent
to a loop space but not homotopy equivalent, or even "rationally
equivalent", to a Lie group. The simplest known example has
dimension 1254. Check it out:

http://www.arxiv.org/abs/math.AT/0306234

Since any loop space is an A_infinity space, this is an example
of an A_infinity space that manages to mimic a Lie group fairly
well without actually being one. Whoops! I said I'd stop talking