The fundamental group of a topological group
Arkadas Ozakin
Dan Christensen had told that the fundamental group of any topological
group is abelian. I tried to prove this, and I realized that I needed the
intermediate result that if we take the inverses of each point in a given
loop, the resulting loop as an element of the fundamental group is the
inverse of the original one. Put in mathematical notation:
Let X be our topological group, C a loop in X that starts and ends at the
identity, and D another loop with:
D(t) = (C(t))^-1
where ^-1 denotes taking inverses in X. Then,
D = C^-1
where this ^-1 denotes taking inverses in the fundamental group, and "="
means D and C^-1 represent the same element of pi_1(X) (I will continue to
make this abuse of notation below).
I gave a proof for this (hopefully didn't mess up...) in a previous post.
Now by using this result, I'll try to show that the fundamental group of X
has to be commutative. I'll denote the identity elements of X and pi_1(X)
by "e" and "E" respectively, and their group operations by "." and "*".
Let H and K be two loops in X with
K(0) = K(1) = H(0) = H(1) = e
I will start with
/ K(2t) when 0 < t < 1/2
(K * H)(t) = {
\ H(2t - 1) when 1/2 < t < 1
and will try to deform this loop continuously to H * K. I will do this
deformation in three steps. First, I will get rid of K in the first half
of the parameter region and deform the second half while doing that. Then
I will get rid of H in the second half and deform the first half while
doing that. Then I will claim that the resulting curve is deformable to K
* H by using the result about inverses of X and pi_1(X) I cited above.
Step 1:
Let P(s,t) be a one parameter family of curves given by the following
(0<s<1):
/ K(2t).( K(2st) )^-1 when 0 < t < 1/2
P(s,t) = {
\ H(2t-1).( K(2s(1-t)) )^-1 when 1/2 < t < 1
The idea is to start with K*H at s=0, and end up at s=1 with a curve which
is constantly identity in the first half of the parameter region. For each
s, I need to have a continuous curve that starts and ends at the identity,
and I think the form I give above satisfies this constraint. So, I end up
at s=1 with the following curve:
/ e when 0 < t < 1/2
L(t) = {
\ H(2t-1).( K(2(1-t)) )^-1 when 1/2 < t < 1
Step 2:
Now start with L(t), and deform the curve continuously to get rid of the H
in the second part of the parameter region: Let Q(s,t) be a one parameter
family of curves with (0<s<1):
/ ( H(1-2st) )^-1 when 0 < t < 1/2
Q(s,t) = {
\ ( H(2s(t-1)+1) )^-1.H(2t-1).( K(2(1-t)) )^-1 when 1/2 < t < 1
If I didn't mess it up, this should consist of continuous curves that
start and end at e for each s, and at s = 1, it should give:
/ ( H(1-2t) )^-1 when 0 < t <1/2
M(t) = {
\ ( K(2(1-t)) )^-1 when 1/2 < t <1
Step 3:
Now I claim that the curve M can be continuously deformed to H * K. To see
this, note that M = U * V with:
U(p) = ( H(1-p) )^-1
V(p) = ( K(1-p) )^-1
Since plugging in (1-p) as the parameter corresponds to taking inverses in
the fundamental group, by using the result I cited at the very beginning,
we see that U represents the same element of pi_(X) as H does, and
likewise, V = K as elements of pi_1(X). This finishes the proof of K*H =
H*K.
arkadas
P.S. Thanks Dan, for the fun! Initally I thought this was going to be
quick and easy, but the need for an intermediate result made it take a lot
longer than I would like to admit :)
P.S.^2 I realized that the algebraic structure of X takes away some of the
geometric intuition, and one has to think formally about these things -
which was both surprizing and a bit disappointing.. Maybe after playing
with these things for some time, one can gain some sort of pictorial
intuition for the group structure as well.
P.S.^3 A friend of mine had cited a professor saying something like
"topology is like psychology in trying to understand intrinsic properties
of beings, and algebra is like sociology in trying to understand the
structure of their interactions with each other". Hey, I just realized
this makes algebraic topology ~ social psychology...
Michael Weiss
: Dan Christensen had told that the fundamental group of any topological
: group is abelian.
Actually, the result is true more generally for an H-space, that is, a
topological space with a continuous multiplication that is blessed with a
unit "up to homotopy". That is, there is an element e for which the maps
x |--> ex and x |--> xe are both homotopic to the identity map.
The proof is a fairly short. Let (H,e) be an H-space, let f and g be loops
based at e, and let
F(s,t) = f(s)g(t) (s, t in [0,1])
be a map from the unit square into H. So F(s,0)=F(s,1)=f(s)e is homotopic
to f, and F(0,t)=F(1,t)=eg(t) is homotopic to g. Stated geometrically, the
vertical sides of the square correspond to [f], and the horizontal sides
correspond to [g] (letting [h] stand for the homotopy class of h).
So if we go up and then over, we get [f][g], while over and up gives [g][f].
We can obviously continuously deform the "up and over" path into the "over
and up" path. QED.
Here's another Kool Fact about topological groups: The torus is the only
compact orientable surface that can be made into a topological group.
The proof uses the fact that if a map from a compact surface into itself is
homotopic to the identity, then either it has a fixed point or the surface
has Euler characteristic 0, i.e., is either a torus or a Klein bottle. This
fact is a easy corollary to the Lefschetz fixed point theorem, but probably
is pretty hard to prove from "first principles".
Now, the torus obviously can be given a topological group structure. Can
the Klein bottle? I don't know.
John Baez
same time... it makes me so darn nostalgic I think I could cry...
sniff, sniff.
In article <a46795$imk$1...@bob.news.rcn.net>,
Michael Weiss <mic...@spamfree.net> wrote:
>Here's another Kool Fact about topological groups: The torus is the only
>compact orientable surface that can be made into a topological group.
>
>The proof uses the fact that if a map from a compact surface into itself is
>homotopic to the identity, then either it has a fixed point or the surface
>has Euler characteristic 0, i.e., is either a torus or a Klein bottle. This
>fact is a easy corollary to the Lefschetz fixed point theorem, but probably
>is pretty hard to prove from "first principles".
Yikes! Why not just use what you proved earlier in your post? The
fundamental group of a compact oriented surface is easy to understand,
and clearly it's only abelian for the torus - but you just showed the
fundamental group of any topological group is abelian!
In fact, using what you proved, we know the torus is the only compact
oriented surface that can be made into an "H-space".
>Now, the torus obviously can be given a topological group structure. Can
>the Klein bottle? I don't know.
Ain't its fundamental group nonabelian? I've got one right here on
my desk... it looks pretty darn nonabelian to me!
Michael Weiss
I wrote (in a reply to Arkadas Ozakin's post)
| Now, the torus obviously can be given a
| topological group structure. Can
| the Klein bottle? I don't know.
but as we all know from reading this thread, the fundamental group of a
topological group is abelian. And the fun group of the Klein bottle is NOT
abelian. What was I thinking?!
Anyway, the fixed-point argument I alluded to in my post proves something
different: if a topological group is a compact manifold, then its Euler
characteristic is 0. (Also I think it has to be orientable. Another reason
why the Klein bottle can't be a topological group.)
Michael Weiss
: >Now, the torus obviously can be given a topological group structure. Can
: >the Klein bottle? I don't know.
Despite being overcome with nostalgia, John Baez wrote:
: Ain't its fundamental group nonabelian? I've got one right here on
: my desk... it looks pretty darn nonabelian to me!
Hey, me too! I think I must have been drinking the contents of mine (easy,
with a Klein bottle!) or I would have seen that right away.
But you know, the fixed point argument actually works for any compact
manifold, and tells you something different: the Euler characteristic of a
topological group that is also a compact manifold must be zero.
(Indeed, the fixed point argument shows that the the torus and the Klein
bottle are the only two compact surfaces that can have fixed-point-free
mapping homotopic to the identity -- and they both enjoy such mappings.)
So lessee. Someone shows up at our shop with a compact manifold, and wants
us to install the topological group option. "Is its fundamental group
abelian?" we ask. "Is it orientable? Is its Euler characteristic zero?"
After all, no point in giving it to the guy in the back room if it fails any
of these tests. He may be a wizard at this stuff, but he also
temperamental.
Will he succeed? Or he hurl fireballs at our head?
Aaron Bergman
wrote:
> Michael Weiss wrote:
> >Now, the torus obviously can be given a topological group structure. Can
> >the Klein bottle? I don't know.
> Ain't its fundamental group nonabelian? I've got one right here on
> my desk... it looks pretty darn nonabelian to me!
Aren't all compact Lie groups orientable?
Aaron
John Baez
Aaron Bergman <aber...@princeton.edu> wrote:
>> Michael Weiss wrote:
Yeah, good point: they have trivial tangent bundles, thanks to
translation symmetry!
I had also classified all 2-dimensional Lie groups not long ago - that's
how we actually got ourselves into this discussion in the first place -
and the Klein bottle was not on the list. So it's clear in a couple of
ways that the Klein bottle cannot be made into a Lie group.
However, that rascal Weiss didn't ask whether the Klein bottle can be
made into a Lie group! He asked whether it can be made into a
*topological* group!
Now, back in 1900 Hilbert posed a bunch of tough math problems, and the
fifth of these asked whether any topological group that's also a
topological manifold can actually be made into a Lie group (in fact not
just smooth but real-analytic). The answer is "yes", so making the
Klein bottle into a topological group would automatically make it into a
Lie group. However, this fact is sufficiently deep that it amounts to
cracking a walnut with a sledgehammer! So, it's nice that merely by
staring at its fundamental group, we can rule out the possibility that
the Klein bottle could be made into a topological group.
Dan Christensen
> Dan Christensen had told that the fundamental group of any topological
> group is abelian. I tried to prove this, and I realized that I needed the
> intermediate result that if we take the inverses of each point in a given
> loop, the resulting loop as an element of the fundamental group is the
> inverse of the original one.
You've discovered the right kind of idea. It's a bit more efficient
if you instead prove that the composition of loops is homotopic to
the pointwise product. So if f and g are loops in X starting and
ending at the identity element e in X, then the composite fg (do f
then g) is homotopic to f*g, where (f*g)(t) = f(t)*g(t) and * is
the multiplication in X. Once you prove this, the same argument
also shows that gf is homotopic to f*g, and so you are done.
Here is a proof that fg ~ f*g, where ~ means "is homotopic to,
relative to the endpoints".
fg ~ (f*c)(c*g) [c denotes the constant path with value e]
~ (fc)*(cg) [(f*h)(k*g) ~ (fk)*hg)]
~ f*g
In fact, this proof works for any H-space: a space X with a map
* : X x X --> X such that the map x |--> x * e is homotopic to
the identity map.
Dan
Marc Nardmann
> So lessee. Someone shows up at our shop with a compact manifold, and wants
> us to install the topological group option. "Is its fundamental group
> abelian?" we ask. "Is it orientable? Is its Euler characteristic zero?"
> After all, no point in giving it to the guy in the back room if it fails any
> of these tests. He may be a wizard at this stuff, but he also
> temperamental.
>
> Will he succeed? Or he hurl fireballs at our head?
The customer, a string theorist, gives us S^7 -- a bit crumpled because
he had to stuff it into some tiny extra dimensions for transport (but we
don't need a smooth structure anyway). "Fundamental group trivial,
(thus) orientable, Euler characteristic 0... Okay, we'll accept it." But
as soon as we hand it over to the back room guy, he yells: "The
topological group option on S^7? What!? Haven't I told you just a few
weeks ago that this is impossible, even for me?" And he mumbles some
spell which only he could understand, something like
"http://www.lns.cornell.edu/spr/2002-01/msg0038424.html". Then he grabs
the S^7, it ignites magically in his hand, and before we can duck, he
throws it at our head.
-- Marc Nardmann
P.S. Dan Christensen told us even more recently:
> 1. Any topology group [...] is an H-space.
[...]
> 4. The only spheres which are H-spaces are S^0, S^1, S^3 and S^7.
So any S^n with odd n>3 would anger the back room guy.
zirkus
>So, it's nice that merely by
>staring at its fundamental group, we can rule out the possibility that
>the Klein bottle could be made into a topological group.
Kleinbottle.com says that if you order one of their Acme Klein Bottles that
it comes with a topological fundamental group which can be homomorphically
mapped onto the Baumslag-Solitar group:
http://www.kleinbottle.com/why_acme.htm
What does this mean?
John Baez
zirkus <zir...@hotmail.com> wrote:
>Kleinbottle.com says that if you order one of their Acme Klein
>Bottles that it comes with a topological fundamental group which
>can be homomorphically mapped onto the Baumslag-Solitar group:
>http://www.kleinbottle.com/why_acme.htm
What does this mean?
The web is your friend. A quick Google search reveals that the
Baumslag-Solitar group BS(m,n) is the group with two generators a,b
and one relation:
m -1 n
a b a = b
It follows that if Acme Klein Bottles sold you one whose fundamental
group is not actually *isomorphic* to BS(1,-1), it is defective -
and you should demand your money back.
(However, this is purely hypothetical: the Klein bottle I got from
this company definitely has the correct fundamental group.)
zirkus
> The web is your friend. A quick Google search reveals that the
> Baumslag-Solitar group BS(m,n) is the group with two generators a,b
> and one relation:
>
> m -1 n
> a b a = b
Also see:
http://mathworld.wolfram.com/FundamentalGroup.html
> ([...] the Klein bottle I got from
> this company definitely has the correct fundamental group.)
Btw, "Kleinsche Flache" refers to "surface" and not to "flask" or
"bottle" - a minor linguistic nitpick. If you're ever bored you can
calculate the fundamental group of your bottle with the Van Kampen
theorem [1]. What does this have to do with physics you ask? Well,
regarding the geometry of string diagrams, the cross caps can only be
removed with Klein bottle openers (perhaps there will be some readers
who actually believe this :-)
[1] http://www.maths.abdn.ac.uk/~ran/mx4509/mx4509-notes/node22.html
Michael Weiss
: P.S. Dan Christensen told us even more recently:
:
: > 1. Any topology group [...] is an H-space.
: > 4. The only spheres which are H-spaces are S^0, S^1, S^3 and S^7.
Ah, the perils of being away from spr for too long. I thought I was pulling
the thread in a different direction, when actually, by a commodius vicus of
recirculation, I was returning to its headwaters.
Here's a question that popped into my head:
"What groups occur as fundamental groups of compact Lie groups?"
(If the answer to that question is buried somewhere in the thread on
teleparallelism, or the one on G2 holonomy, or the one on Quantum Gravity,
my apologies. Just pretend that Michael accidentally left his machine
unguarded and Oz decided to post from it.)
Obviously the answer has to be "well-known", since compact Lie groups are as
familiar as bagels. And hopefully any question about compact Lie groups is
at least *potentially* relevant to physics, or relevant enough for spr.
So I thought I'd check Dave Rusin's "known math" website (nowadays reached
through www.math-atlas.org). I didn't find the answer, though in the
article "Connections among algebraicity, centers, and the fundamental group
for Lie groups", I did find another puzzle, in fact, one of Greg Kuperberg's
favorite puzzles:
"How many Lie groups (up to isomorphism) are topologically a pair of
circles?"
John Baez
Michael Weiss <mic...@spamfree.net> wrote:
>I thought I was pulling the thread in a different direction, when
>actually, by a commodius vicus of recirculation, I was returning
>to its headwaters.
It's so nice to read posts by people who actually take the time
to write enjoyably!
>Here's a question that popped into my head:
>
> "What groups occur as fundamental groups of compact Lie groups?"
Hmm....
Well, for starters, the only groups we can possibly get are the finitely
generated abelian groups!
Every mathematician learns the classification of finitely generated
abelian groups when they take an introductory course in abstract algebra.
But since some physicists might have skipped that course, let me say
what the result is: a finitely generated abelian group is just a product
of finitely many copies of Z (the integers) together with a finite
collection of groups of the form Z/n (the integers mod n). In fact,
it suffices to let the numbers n be powers of prime numbers! Then
then we have a unique way of factorizing any finitely generated abelian
group. (For example, Z/6 = Z/2 x Z/3, but Z/4 is not Z/2 x Z/2.)
To prove that the fundamental group of a compact Lie group is
finitely generated and abelian, we just bring out two smallish
sledgehammers and crush the problem. First, the fundamental group
of any topological group is abelian - we've seen the proof of *this*
earlier on this thread! Second, the fundamental group of any
compact manifold is finitely generated - if you don't believe me,
cover the manifold with finitely open many balls in a nice way and
use van Kampen's theorem.
So, the really fun question is "which finitely generated abelian
groups arise as the fundamental group of some compact Lie group?"
The answer is not hard, but let me explain how I blundered my
way to it, because it might be interesting to people who want
to learn stuff about Lie groups.
I started out contemplating a brute-force approach. Compact Lie groups
are a well-understood commodity, and we only care about the connected
ones here, which simplifies things immensely. One can pretty much
just *list* all the compact connected Lie groups, and then see what
all their fundamental groups were!
Here's what compact connected Lie groups are like. A bunch are
of the form G x T where G is a compact simply-connected Lie group
and T is a torus of some dimension or other. All the rest are of
the form (G x T)/N where N is a finite normal subgroup. This group
N may not split up neatly as N1 x N2 where N1 is a normal subgroup
of G and N2 is a normal subgroup of T. However, it's pretty easy
to see that N lies *inside* a finite normal subgroup of the form
N1 x N2.
So, for the brute-force approach, we'd need to figure out
1) what all the compact simply-connected Lie groups G are,
2) what all their normal subgroups N1 are,
3) what all the normal subgroups N2 of all tori T are,
4) what all the subgroups N of these N1 x N2 are,
5) what the fundamental groups of all the resulting (G x T)/N are.
This may sound tough, but it's not really so terrible!
For example, compact simply-connected Lie groups are all products
of *simple* ones. And these simple ones have been completely
tabulated: there's the A series, the B series, the C series, and
the D series, together with five exceptional rascals known as
E6, E7, E8, F4 and G2. They've all been scrutinized under
the microscope by a century's worth of very smart mathematicians,
so practically everything is known about them. In particular,
for each one, the largest possible finite normal subgroup is just
its *center*, which has finitely many elements; all the other normal
subgroups are subgroups of this. You can look up these centers in
your favorite tome on Lie groups, and then work out all their
subgroups.
However, as soon as I got to this point in my deliberations, I
realized that we didn't need to do all this work! Just consider
the A series: these are the groups SU(n). The group SU(n) has
center Z/n. If we form the quotient of SU(n) by Z/n, we get a
compact Lie group called PSU(n), whose fundamental group is Z/n.
Thus, by taking products of PSU(n)'s and tori, we easily see that
EVERY finitely generated abelian group is the fundamental group of
some compact Lie group. We can stop here!
>Obviously the answer has to be "well-known", since compact Lie groups are as
>familiar as bagels. And hopefully any question about compact Lie groups is
>at least *potentially* relevant to physics, or relevant enough for spr.
Sure!
>So I thought I'd check Dave Rusin's "known math" website (nowadays reached
>through www.math-atlas.org). I didn't find the answer, though in the
>article "Connections among algebraicity, centers, and the fundamental group
>for Lie groups", I did find another puzzle, in fact, one of Greg Kuperberg's
>favorite puzzles:
>
>"How many Lie groups (up to isomorphism) are topologically a pair of
>circles?"
Hmm, sounds like a trick question. I can just think of two.
zirkus
> And these simple ones have been completely
> tabulated: there's the A series, the B series, the C series, and
> the D series, together with five exceptional rascals known as
> E6, E7, E8, F4 and G2.
G2 is not really an exceptional group because it is defined by a
generic form. See page 20 of:
arkadas ozakin
> Arkadas Ozakin wrote :
>
> : Dan Christensen had told that the fundamental group of any topological
> : group is abelian.
> Actually, the result is true more generally for an H-space, that is, a
> topological space with a continuous multiplication that is blessed with a
> unit "up to homotopy". That is, there is an element e for which the maps
> x |--> ex and x |--> xe are both homotopic to the identity map.
I was scared away by this "H-space" stuff, and thought I would read
this post later. Now I came back and read the below:
> The proof is a fairly short. Let (H,e) be an H-space, let f and g be loops
> based at e, and let
>
> F(s,t) = f(s)g(t) (s, t in [0,1])
>
> be a map from the unit square into H. So F(s,0)=F(s,1)=f(s)e is homotopic
> to f, and F(0,t)=F(1,t)=eg(t) is homotopic to g. Stated geometrically, the
> vertical sides of the square correspond to [f], and the horizontal sides
> correspond to [g] (letting [h] stand for the homotopy class of h).
>
> So if we go up and then over, we get [f][g], while over and up gives [g][f].
> We can obviously continuously deform the "up and over" path into the "over
> and up" path. QED.
Indeed! This is so simple and cool. I've read Dan Christensen's proof,
and learned a new way of thinking, and your proof teaches me yet
another way of thinking. Thanks a lot!
arkadas
Arkadas Ozakin
Dan Christensen wrote in part:
> Here is a proof that fg ~ f*g, where ~ means "is homotopic to,
> relative to the endpoints".
>
> fg ~ (f*c)(c*g) [c denotes the constant path with value e]
This one is simple enough, just multiply each point in the curve by
identity. In fact, (although it's unnecessary) I think a "=" sign would
also work instead of "~".
[Moderator's note: no, you need to reparametrize the path to get
from f to f*c, or from g to c*g. - jb]
> ~ (fc)*(cg) [(f*h)(k*g) ~ (fk)*(hg)]
[minor typo corrected]
I stared at the paranthetical remark totally puzzled for a moment, than it
became totally obvious... Actually, once again, it seems to me that a "="
sign would also work in place of the "~".
[Moderator's note: again, some reparametrization is needed. - jb]
> ~ f*g
I can see that the constant "c" parts of the curves (fc) and (cg) can be
got rid of easily (while keeping the endpoints of both curves fixed at
identity) to give this result. And of course, a "=" would not have worked
at this step.
Every steps seems so obvious, but when I look at what we showed finally,
it seems like magic.
Dan had also said:
> Once you prove this, the same argument also shows that gf is homotopic
> to f*g, and so you are done.
I will try to do this part in order to gain familiarity with this stuff (I
will even try to rewrite the paranthetical remark):
(gf) ~ (c*g)(f*c)
~ (cf)*(gc) [because: (a*b)(p*q) ~ (ap)*(bq)]
~ f*g
Ok, I guess I did it, but it still seems like magic.
I think what is a new way of thinking for me is to think of a curve as a
"*" of two curves, and to deform the two curves that go into this *
product. I was more comfortable with contionuously deforming parts of the
curve in the parameter space, but not deforming the "parts" that go into a
product (the *, that is).
Thanks a lot!
arkadas
Michael Weiss
: I was scared away by this "H-space" stuff, and thought I would read
: this post later. Now I came back and read the below:
Yeah, I should have done the proof first for a topological group, then
explain how it can be generalized, instead of starting with the more general
case.
: Indeed! This is so simple and cool. I've read Dan Christensen's proof,
: and learned a new way of thinking, and your proof teaches me yet
: another way of thinking. Thanks a lot!
I'd be surprised if Dan Christensen's proof wasn't really the same thing,
just with a different jacket and tie on, but I'm too lazy to check.
If you go to
http://www.math.cornell.edu/~hatcher
and download AT2001.pdf, and go to the beginning of Chapter 1, you'll find a
delightful pair of examples, with an abelian and a non-abelian fundamental
group. Just to whet your appetite, the examples involve the famous
"Borromean rings" (or Ballantine ale rings) --- the three rings where any
two by themselves are unlinked, but all three together cannot be separated.
Turns out that if you modify the rings so two of them are now linked, then
the third magically becomes *unlinked*. And the fundamental groups lie at
the root of this strange behavior.
arkadas ozakin
> arkadas ozakin
[...]
> : Indeed! This is so simple and cool. I've read Dan Christensen's proof,
> : and learned a new way of thinking, and your proof teaches me yet
> : another way of thinking. Thanks a lot!
>
> I'd be surprised if Dan Christensen's proof wasn't really the same thing,
> just with a different jacket and tie on, but I'm too lazy to check.
I'll give a cartoon version of Dan's proof, both to make myself digest
it better, and to explain what kind of different way of thinking from
yours I thought it involved. After I explain the notation, the proof
is just going to be a "movie".
Here is the notation: This:
f f f f f f f f
|---------------|
denotes the curve f(t). The left end is t=0, the right end is t=1. So,
both ends are actually mapped to "e", the identity element.
This:
f f f f c c c c
|---------------|
denotes the curve fc - I will use c to denote the constant curve at
"e", the identity element of the group.
And finally this:
a a a a a a a a
|---------------|
b b b b b b b b
denotes the curve (a*b), given by (a*b)(t) = a(t)*b(t) where "*"
denotes the multiplication in the topological group. (Obviously, a and
b are two curves.)
So, here goes: I'll start with fg and deform it to gf. (Again, c
denotes the constant curve c(t) = e, "e" being the identity of the
group.)
f f f f g g g g
|---------------|
=
f f f f c c c c
|---------------|
c c c c g g g g
~
c f f f f c c c
|---------------|
c c c g g g g c
~
c c f f f f c c
|---------------|
c c g g g g c c
~
c c c f f f f c
|---------------|
c g g g g c c c
~
c c c c f f f f
|---------------|
g g g g c c c c
=
g g g g f f f f
|---------------|
Voila! (Hey, I had said that Dan's proof seemed like magic, but not
anymore! Now I feel comfortable with it. Long live picture proofs!)
Now, are this proof and yours the same thing in disguise? That's not
very obvious to me. Even if you can find a suitable deformation of one
of the proofs to the other, I would say they have different "pants" as
well as jackets and ties!
Michale Weiss also said:
> Just to whet your appetite, the examples involve the famous
> "Borromean rings" (or Ballantine ale rings) --- the three rings where any
> two by themselves are unlinked, but all three together cannot be separated.
> Turns out that if you modify the rings so two of them are now linked, then
> the third magically becomes *unlinked*. And the fundamental groups lie at
> the root of this strange behavior.
I'll check this out, thanks!
arkadas
arkadas ozakin
> > fg ~ (f*c)(c*g) [c denotes the constant path with value e]
>
> This one is simple enough, just multiply each point in the curve by
> identity. In fact, (although it's unnecessary) I think a "=" sign would
> also work instead of "~".
and the moderator noted:
> [Moderator's note: no, you need to reparametrize the path to get
> from f to f*c, or from g to c*g. - jb]
Dan had defined the * operation on loops as (f*g)(t) := f(t)*g(t)
where the second * is the group multiplication in the topological
group. So,
(f*c)(t) = f(t)*c(t)
= f(t)* e
= f(t)
It seems to me that no reparametrization is needed.
I had also said:
> once again, it seems to me that a "=" sign would also work in place of the
> "~"
where I was referring to the equation
> (f*h)(k*g) ~ (fk)*(hg)
The moderator said:
> [Moderator's note: again, some reparametrization is needed. - jb]
Writing out the definitions will make things look complicated, so I'll
use words:
(fk)*(hg) means:
- "squeeze" all four curves into parameter regions of length 1/2
- stick k to the end of f, and g to the end of h
- multiply the corresponding elements of the two resulting curves
But instead of this, I can consider (f*h)(g*k):
- multiply the corresponding elements in f and h (end up with f*h)
- multiply the corresponding elements in g and k (end up with g*k)
- squeeze the two resulting curves into parameter regions of lengths
1/2
- and stick the second one to the end of the first one.
The result of this, it seems to me, is equivalent (without a
reparametrization) to that of the above.
Using the notation I made up in a post in which I gave a pictorial
version of Dan's proof, both of these curves can be denoted by:
f f f f g g g g
|---------------|
h h h h k k k k
arkadas
Chris Hillman
> If you go to
> http://www.math.cornell.edu/~hatcher
> and download AT2001.pdf, and go to the beginning of Chapter 1, you'll find a
> delightful pair of examples, with an abelian and a non-abelian fundamental
> group.
Just saw the PUBLISHED version of this beautiful ebook! Yaay! Topology
fans who haven't seen this book should definitely check it out.
author = {Hatcher, Allen},
title = {Algebraic Topology},
publisher = {Cambridge University Press},
year = 1992},
Also, contrary to something I mentioned earlier, if BIP is correct, it
seems that CUP -is- planning a paperback edition of the long awaited
second edition of Exact Solutions of Einstein's Field Equation. Double
yaay!
Chris Hillman
Home page: http://www.math.washington.edu/~hillman/personal.html
arkadas ozakin
> But instead of this, I can consider (f*h)(g*k):
Oops. I should have said (f*h)(k*g).. Sorry for the confusion...
> - multiply the corresponding elements in f and h (end up with f*h)
> - multiply the corresponding elements in g and k (end up with g*k)
Here, I should have said:
multiply the corresponding elements of k and g, end up with k*g.
> - squeeze the two resulting curves into parameter regions of lengths 1/2
> - and stick the second one to the end of the first one.
>
> The result of this, it seems to me, is equivalent (without a
> reparametrization) to that of the above.
>
> Using the notation I made up in a post in which I gave a pictorial
> version of Dan's proof, both of these curves can be denoted by:
>
> f f f f g g g g
> |---------------|
> h h h h k k k k
I should have said:
f f f f k k k k
|---------------|
h h h h g g g g
Michael Weiss
quote the key movie frames:
f f f f c c c c
|---------------| Frame 1
c c c c g g g g
c f f f f c c c
|---------------| Frame 2
c c c g g g g c
c c f f f f c c
|---------------| Frame 3
c c g g g g c c
c c c f f f f c
|---------------| Frame 4
c g g g g c c c
c c c c f f f f
|---------------| Frame 5
g g g g c c c c
Then he says:
: Now, are this proof and yours the same thing in disguise? That's not
: very obvious to me. Even if you can find a suitable deformation of one
: of the proofs to the other, I would say they have different "pants" as
: well as jackets and ties!
A homotopy of proofs, eh? I don't know if I can exactly *deform* one proof
into the other, but I think they still are basically the same proof.
Remember that I started with a map of the unit square into the topological
group via the formula F(s,t) = f(s)g(t). So if we have a path in the unit
square, composing it with F gives us a path in the topological group.
Each of the frames in your movie corresponds to a path in the unit square,
and the movie as a whole gives a homotopy from the "up-and-over-path" to the
"over-and-up" path. (All the paths go from the (0,0) corner to the (1,1)
corner. Or bottom-left to top-right, the way I think of it.)
Frame 1 is just the "up-and-over" path.
Frame 2 starts by going up the left edge *almost* to the top-left corner,
but then cuts across diagonally to the top edge and finishes by going along
it to the top-right corner.
Frame 3 goes along the diagonal, bottom-right to top-left.
Frame 4 starts by going along the bottom edge, but then cuts diagonally
across to the right edge and finishes by going up it to the top-right
corner.
Frame 5, finally, is just the "over-and-up" path.
Maybe I should that between your version and the version I gave, we have a
complete suit, jacket, tie, and pants!
arkadas ozakin
thing, which sounds correct to me, except for one unimportant
technicality:
> c f f f f c c c
> |---------------| Frame 2
> c c c g g g g c
[...]
> Frame 2 starts by going up the left edge *almost* to the top-left corner,
> but then cuts across diagonally to the top edge and finishes by going along
> it to the top-right corner.
I think it also spends some time at the corners. Actually, had I
followed what Dan said more closely in the cartoon proof, what you
said would be precisely correct, without any time spent at the
corners.
So, they _were_ the same proof after all! But I still think I learned
two different ways of thinking from them - in a way somewhat similar
to what William Thurston tells us about different ways of looking at
the derivative in his wonderful paper "On Proof and Progress in
Mathematics" (http://xxx.lanl.gov/abs/math.HO/9404236).
arkadas
[Who was scared because of the similarity of the words "nifty" and
"filthy", went and checked a dictionary, and spent the rest of the day
with a happy smile on his face.]