what is cohomology and what it's physics applications
tedsu...@my-deja.com
Thanks for people's response to my question about fibre bundles.
I have another along the same lines. What is a cohomology (or
cohomology theory) and how is it used in physics?
Thanks,
Ted
Sent via Deja.com http://www.deja.com/
Before you buy.
John Baez
>Thanks for people's response to my question about fibre bundles.
>I have another along the same lines. What is a cohomology (or
>cohomology theory) and how is it used in physics?
I forget how much math you know, so I'll start out by saying some
incredibly elementary stuff and then shoot forwards rapidly to some
more sophisticated stuff. If you know a medium amount of math,
you'll probably be bored at first and then completely confused.
Okay? So: I'll concentrate on what cohomology theory *is* and
leave the physics applications for someone else.
If you have a space, it can have holes of various different
dimensions. A cohomology theory is a way of defining what
you mean by holes, and it lets you add and subtract these holes.
First you gotta understand how we keep track of the dimension of
a hole! It takes a bit of getting used to, so don't argue, just
pay close attention:
Consider a doughnut. We say this has a 1-dimensional hole because
it's possible to put a circle in a doughnut which encircles the
hole of the doughnut, and we can't "pull this circle tight" -
contract it to a point - while keeping it in the doughnut.
Since a circle is 1-dimensional we say the hole is 1-dimensional.
A doughnut has no holes of dimensions other than 1.
Now consider a basketball: just the rubber skin, not the air inside.
We say this has a 2-dimensional hole because it's possible to put a
sphere inside the skin of the basketball which wraps around the
air inside, and we can't "pull this sphere tight" - contract it
to a point - while keeping it in the skin of the basketball.
Since a sphere is 2-dimensional we say the whole is 2-dimensional.
A basketball has no holes of dimensions other than 2.
Now consider an inner tube: just the rubber skin, not the air inside.
This is like a hollow doughnut! It's more complicated than the previous
examples because it has more than one kind of hole. It has a 1-dimensional
hole corresponding to how you can wrap a circle around it the long way,
just like you can with the doughnut. But it also has a 1-dimensional
hole corresponding to how you can wrap a circle around it the short way!
It also has a 2-dimensional hole corresponding to how you can put a
torus inside the skin of the inner tube. This 2-dimensional hole
is a bit like the example of the basketball, except now we're using a
torus to wrap around the air inside the inner tube, instead of a sphere.
If you're smart you can figure out other ways to wrap a circle around
the inner tube: it can sort of *spiral* around the inner tube. But this
kind of 1-dimensional hole is a linear combination of the 1-dimensional
holes I already discussed. If the circle spirals n times around the
short way while it spirals m times around the long way, we say the hole
it represents is nx + my, where x and y form the "basis" of 1-dimensional
holes described in the previous case.
Okay: in general, the cohomology of a space X is a bunch of abelian
groups H^n(X), one for each integer n. The idea is that H^n(X) consists
of all the n-dimensional holes, and we can add and subtract these holes.
Now, so far I've been describing holes by looking at *manifolds* mapped
into our space X. But we could use other things. It's actually very
customary to use *simplices*. Different ways give different cohomology
theories. If we use simplices we get something called "singular cohomology
theory" or "ordinary cohomology theory". If we use manifolds we get
something called "cobordism theory". There are actually lots of different
kinds of cobordism theory depending on what sort of manifolds we use:
there's "oriented cobordism theory" and "unoriented cobordism theory"
and "smooth cobordism theory" and "piecewise-linear cobordism theory"
and so on. And there are lots of other cohomology theories, too, like
K-theory and stable homotopy theory and so on. These cohomology theories
are related in lots of useful ways.
In physics, perhaps the most popular cohomology theory is "deRham
cohomology theory". Here we use differential forms to keep track
of holes in a smooth manifold. Since differential forms are very
important in physics, it's not surprising that deRham theory has lots
of interesting applications. Stokes' theorem, Gauss' theorem, Green's
theorem - they're all just the tip of the iceberg called deRham
theory. You can find an elementary introduction to deRham cohomology
in my book "Gauge Fields, Knots and Gravity", and in lots of other places,
like Flanders' book "Differential Forms with Applications to the Physical
Sciences", or von Westenholz' book "Differential Forms in Mathematical
Physics". If you're a physicist, deRham theory is the place to start!
Okay, finally for some more high-powered stuff. For every cohomology
theory H^n, there's a bunch of spaces S(n) called the "classifying spaces"
of that cohomology theory. The cohomology group H^n(X) of any space
X is really just given by [X,S(n)] - the set of homotopy classes of maps
from X to S(n), which turns out to be a group, thanks to some special
stuff about S(n). For example, for ordinary cohomology the classifying
space is called an "Eilenberg-MacLane space" and denoted K(Z,n). In
general, for any cohomology theory, the spaces S(n) fit together into a
gadget called a "spectrum". So in a sense, cohomology theory is really
the study of spectra! This is how the experts think about it. But to
really understand this viewpoint, you gotta understand what's so great
about spectra. Ultimately it turns out that spectra are just a super-duper-
generalization of abelian groups: they are "infinitely categorified,
infinitely stabilized abelian groups". This is why they're so great.
Anyone who wants to learn more about this should read Adams' book "Infinite
Loop Spaces".
Toby Bartels
>Okay: in general, the cohomology of a space X is a bunch of abelian
>groups H^n(X), one for each integer n.
In many cohomology theories,
you can multiply an element of H^m(X) and an element of H^n(X)
to get an element of H^{m+n}(X).
Thus, the cohomology groups together form the cohomology ring.
I just thought I'd mention that.
(In de Rham cohomology, this is related to multiplying an mform and an nform.)
-- Toby
to...@ugcs.caltech.edu
dale hurliman
[Moderator's note: Quoted text deleted. -TB]
> If you have a space, it can have holes of various different
> dimensions. A cohomology theory is a way of defining what
> you mean by holes, and it lets you add and subtract these holes.
>
> First you gotta understand how we keep track of the dimension of
> a hole! It takes a bit of getting used to, so don't argue, just
> pay close attention:
>
> Consider a doughnut. We say this has a 1-dimensional hole because
> it's possible to put a circle in a doughnut which encircles the
> hole of the doughnut, and we can't "pull this circle tight" -
> contract it to a point - while keeping it in the doughnut.
<snip>
All of what you say assumes that your space can be embedded in an
Euclidian space of higher dimension, right? Is this always possible? Is
there a theorem which says that any arbitrary space on N dimensions can
always be embedded in a Euclidian space of M>N dimensions?
John Baez
Toby Bartels <to...@ugcs.caltech.edu> wrote:
>In many cohomology theories,
>you can multiply an element of H^m(X) and an element of H^n(X)
>to get an element of H^{m+n}(X).
>Thus, the cohomology groups together form the cohomology ring.
Right. In general, this happens when the spectrum S(n) defining
your cohomology theory is a "ring spectrum". Ring spectra are to
rings as spectra are to groups: in both cases, we're taking a basic
concept from algebra and then categorifying and stabilizing it
infinitely many times.
For an intro to "categorification" and "stabilization", try:
http://math.ucr.edu/home/baez/week121.html
James Gibbons
> Here's a simple definition of a spectrum: it's a sequence of
> spaces S(n) equipped with basepoints and basepoint-preserving
> homeomorphisms
>
> f: Loops(S(n)) -> S(n-1)
>
> where Loops denotes the space of based loops.
>
> This implies that each space S(n) is a loop space, and thus that
> [X,S(n)] is a group. It also implies that S(n) is a double loop
> space - the loop space of a loop space - and thus that [X,S(n)] is
> an abelian group. So for any spectrum and any pointed space X and
> any n,
>
> H^n(X) = [X,S(n)]
>
> is an abelian group...
In looking over Mark Hovey's book "Model Categories", it seems this
process can be generalized to Model categories (since Top is such)..
yes?
Jim Gibbons
John Baez
dale hurliman <hurl...@nospam.sunlink.net> wrote:
>John Baez wrote:
>> Consider a doughnut. We say this has a 1-dimensional hole because
>> it's possible to put a circle in a doughnut which encircles the
>> hole of the doughnut, and we can't "pull this circle tight" -
>> contract it to a point - while keeping it in the doughnut.
>All of what you say assumes that your space can be embedded in a
>Euclidean space of higher dimension, right?
Actually none of what I said assumes that. It's an important step,
when learning about concepts of space, to stop relying on a picture
of space as embedded in a Euclidean space of higher dimension. It
took people about a century of careful thought to realize this, but
by the mid-1900s it was quite obvious.
>Is this always possible? Is
>there a theorem which says that any arbitrary space of N dimensions can
>always be embedded in a Euclidean space of M>N dimensions?
There are theorems like this, but they don't apply to an "arbitrary
space". For example, lots of spaces are infinite-dimensional, and
these can't be embedded in any Euclidean space at all - but you can
still study their cohomology, and some people spend their whole lives
doing just that.
Here's a theorem along the lines you want: a smooth N-dimensional
manifold can always be smoothly embedded in M-dimensional Euclidean
space for some M > N. (In fact there's an explicit upper bound on
the necessary M, but I forget what it is.) Smooth manifolds are
the kind of spaces physicists are most likely to want to know the
cohomology of - for example, general relativity assumes that spacetime
is a smooth manifold. But it turns out to be completely unnecessary,
and usually downright unhelpful, to think of smooth manifolds as
embedded in Euclidean space.
John Baez
James Gibbons <gib...@easyon.com> wrote:
>
[stuff on spectra and cohomology theories deleted]
>In looking over Mark Hovey's book "Model Categories", it seems this
>process can be generalized to Model categories (since Top is such)...
>yes?
Yes, Quillen invented the theory of model categories in large part
to generalize the basic notions of homotopy theory, such as:
fibration
cofibration
weak equivalence
suspension
loop space
cohomology theory
from the category Top to other categories. Of course, homological
algebra can be done in any abelian category, and abelian categories
were invented for precisely for that purpose. But there's more to
homotopy theory than homological algebra. Quillen called this other
stuff "homotopical algebra" and formalized it using model categories.
One reason Quillen wanted to do this was to understand algebraic
K-theory. There's a very nice cohomology theory for topological spaces
called K-theory. This is most easily defined in terms of vector bundles.
Using the analogy between vector bundles over topological spaces and
projective modules over rings, it's easy to define K^0 and K^1 for
rings. For a while it was an open challenge to find the right
definition of K^i of a ring for i > 1. Quillen gained a lot of fame
by defining and proving cool theorems about these "higher algebraic
K-groups". But to fully exploit the analogy between cohomology
theories for topological spaces and cohomology theories for rings
and other algebraic gadgets, he decided to isolate the key ideas
behind homotopy theory in a set of axioms: the axioms for a model
category.
Quillen is a heavy dude. I took some classes with him at MIT.
He always wore jeans, a lumberjack-style shirt and a look of total
concentration. He lectured on whatever he was working on at the
time. Each lecture started out flawlessly, but often they came to
a bumpy halt near the end of class as he reached the limits of what
he had proved so far. The next class he would back up, start over,
and go further. It was very inspiring. At the time he was attempting
a proof of the Atiyah-Singer index theorem using nothing but standard
multivariable calculus. He never quite got it, but his ideas probably
served as the basis of Getzler's proof (which unfortunately used some
more fancy technology). This was about 20 years after his work on
model categories. Unfortunately at the time I was too ignorant to
even know of his earlier work.
Hovey's book "Model Categories" is an excellent treatise if you already
know why you're interested in model categories and you want to master
all the latest high-powered tools. But it's a little bit unsatisfying
if you mainly want to know what model categories are good for. For that,
I'd recommend Quillen's book "Homotopical Algebra" (Lecture Notes in
Mathematics 43, Springer-Verlag, 1967), and also Dwyer and Spalinski's
article "Homotopy theories and model categories" in the "Handbook of
Algebraic Topology" (North-Holland, 1995, pp. 73-126).
I have been thinking about this stuff a lot lately, since I'm writing
a paper with Todd Trimble on an approach to n-categories based on
homotopy theory. I would like to explain some of it in This Week's
Finds. I hope I find time to do so.
James Gibbons
> [..]
> Hovey's book "Model Categories" is an excellent treatise if you already
> know why you're interested in model categories and you want to master
> all the latest high-powered tools. But it's a little bit unsatisfying
I guess I became interested in model categories because I noticed
that the studies of topology (topological spaces), Differential
Geometry (differentiable manifolds), Algebriac Geometry (schemes),
C* Algebras (operator algs.), and Measure theory (measurable spaces)
had similarities between them that might be distilled by the fact
that the above categories could be given a sensible model-theoretic
structure. Hence they might all be termed "Geometric Categories"
in the same sense that the category Top is.
> I have been thinking about this stuff a lot lately, since I'm writing
> a paper with Todd Trimble on an approach to n-categories based on
> homotopy theory. I would like to explain some of it in This Week's
> Finds. I hope I find time to do so.
It is interesting that Homotopical Algebra and Model categories have
connections to n-category theory!
Jim Gibbons
Gary
ba...@galaxy.ucr.edu (John Baez) wrote:
[math and physics deleted]
> Quillen is a heavy dude. I took some classes with him at MIT.
> He always wore jeans, a lumberjack-style shirt and a look of total
> concentration. He lectured on whatever he was working on at the
> time. Each lecture started out flawlessly, but often they came to
> a bumpy halt near the end of class as he reached the limits of what
> he had proved so far. The next class he would back up, start over,
> and go further. It was very inspiring.
Please indulge a cultural sidebar.
Certain mid-'70s MIT undergraduates took an interest in Quillen's work
soley as a result of an entry in the MIT course catalog. This entry
was on occasion chanted in revered tones, usually in the wee hours of a
Sunday morning, in the distal hours of a party from which all the
woman had already left (that is to say: the set of all parties) (recall
that the drinking age was 18 at the time) (this really was an early
form of performance art).
As with all religious relics, the chant was carefully maintained over
the years by a group of devoted acolytes. It sounded like a secret
language then, and it sounds like a secret language now. I reproduce it
here:
Uhh... before we get to that ... I recall that someone traced back
through the chain of prerequisites and discovered that in order for the
average grad student to take this course, he/she had to have had taken
18.01 when he/she was 13 years old.
------------------------------------------------------------------
18.715 Homological Algebra (A)
Prereq.: 18.706
Year: G (1)
3-0-9
18.716 Homological Algebra (A)
Prereq.: 18.715
Year: G (2)
3-0-9
Abelian catagories, derived functors, spectral sequences, the functors
Tor and Ext, cohomology of sheaves, homological properties of
noetherian rings. Grothendieck topologies, the derived category of an
abelian category, local duality theory, semi-simplicial methods and a
unified cohomology theory for algebraic structures. (Not offered 1973-
74.) D.G.Quillen
--------------------------------------------------------------------
--
------------------------
Gary Pajer
Jim Heckman
ba...@galaxy.ucr.edu (John Baez) wrote:
>
> Here's a theorem along the lines you want: a smooth N-dimensional
> manifold can always be smoothly embedded in M-dimensional
> Euclidean space for some M > N.
Yikes!! Really?! By specifying M-d *Euclidean* space, aren't you
implicitly inducing a metric on the N-d manifold? But I could have sworn
(I find myself saying that a lot, recently :-/) I read it's impossible to
embed, say, a 2-d hyberbolic plane (constant negative Gaussian
curvature) in *any* Euclidean space. Is this not correct?
OTOH, maybe just the *diffeomorphic* structure of the N-d manifold can
be replicated in M-d Euclidean space -- not the entire geometry. Is that
closer to what you mean?
--
~~ Jim Heckman ~~
-- "As I understand it, your actions have ensured that you will never
see Daniel again." -- Larissa, a witch-woman of the Lowlands.
-- "*Everything* is mutable." -- Destruction of the Endless
Toby Bartels
>Here's a theorem along the lines you want: a smooth N-dimensional
>manifold can always be smoothly embedded in M-dimensional Euclidean
>space for some M > N. (In fact there's an explicit upper bound on
>the necessary M, but I forget what it is.)
The upper bound on M is 2N.
It's fairly easy to prove an upper bound of 2N + 1
(although I'm not sure I could do it off the top of my head);
2N, I'm told, is harder.
BTW, you want to say the manifold is *connected*,
at least in the case N = 0
(and possibly other cases if you don't require paracompactness).
-- Toby
to...@ugcs.caltech.edu
John Baez
>> Here's a theorem along the lines you want: a smooth N-dimensional
>> manifold can always be smoothly embedded in M-dimensional
>> Euclidean space for some M > N.
>Yikes!! Really?!
YES!! REALLY!!
(It's nice to see people getting so excited about theorems. For
some reason one rarely sees this on sci.math.research.)
>>By specifying M-d *Euclidean* space, aren't you
>implicitly inducing a metric on the N-d manifold?
Yes - but of course this metric is completely irrelevant to the
point of the theorem, which is a theorem about smooth manifolds,
not Riemannian manifolds.
In other words, I DID NOT SAY THIS:
If you have a smooth N-dimensional manifold with a given Riemannian
metric, there is an isometric embedding of it into M-dimensional
Euclidean space with M > N.
(Here by "isometric" we mean that the given metric on our manifold
is the same as that induced by its embedding in Euclidean space.)
Now, I COULD have said that, because it's true... but it's not
what I actually DID say.
>But I could have sworn
>(I find myself saying that a lot, recently :-/) I read it's impossible to
>embed, say, a 2-d hyberbolic plane (constant negative Gaussian
>curvature) in *any* Euclidean space. Is this not correct?
Yes, I'm pretty sure it's not correct, thanks to the theorem I
just now stated, which is a much deeper and more surprising theorem
than the one I stated before.
>OTOH, maybe just the *diffeomorphic* structure of the N-d manifold can
>be replicated in M-d Euclidean space -- not the entire geometry. Is that
>closer to what you mean?
Yes, that's what I meant - because I didn't breathe a word about
"metrics" or "geometry" until you started talking about that. Please
don't mix up your categories - it's a recipe for misunderstanding!
When I say "smooth manifold", I don't mean "Riemannian manifold", and
when I say "embedding", I don't mean "isometric embedding".
John Baez
Toby Bartels <to...@ugcs.caltech.edu> wrote:
>John Baez <ba...@galaxy.ucr.edu> wrote:
>>Here's a theorem along the lines you want: a smooth N-dimensional
>>manifold can always be smoothly embedded in M-dimensional Euclidean
>>space for some M > N. (In fact there's an explicit upper bound on
>>the necessary M, but I forget what it is.)
>The upper bound on M is 2N.
>It's fairly easy to prove an upper bound of 2N + 1
>(although I'm not sure I could do it off the top of my head);
>2N, I'm told, is harder.
Thanks. The trick to remembering this stuff is to remember
that generically, an N-dimensional manifold mapped into an
M-dimensional one intersects itself along a submanifold of
dimension N + N - M. This is just by counting degrees of
freedom. So M = 2N + 1 should certainly work. On the other
hand, M = 2N might give you isolated points of self-intersection.
So in this case, you'll need to do some topology involving
intersection numbers to prove you can get rid of the intersection
points by cleverly cancelling them in pairs. Or something like
that.
>BTW, you want to say the manifold is *connected*,
>at least in the case N = 0
>(and possibly other cases if you don't require paracompactness).
Right, we definitely need paracompactness for these theorems.
Without it, we can always make a "long line", a nonparacompact
1-manifold, that can't be embedded in R^n for the simple reason
that it has more points than R^n does! This is one more reason
people like to tuck paracompactness into the definition of "manifold".
Chris Hillman
> implicitly inducing a metric on the N-d manifold? But I could have sworn
> (I find myself saying that a lot, recently :-/) I read it's impossible to
> embed, say, a 2-d hyberbolic plane (constant negative Gaussian
> curvature) in *any* Euclidean space. Is this not correct?
It's not correct. But there is a subtlety: local versus global plays a
role here! You can embedd in E^3 a whole bunch of surfaces which are
-locally- indistinguishable from H^2 (have constant negative curvature)
but are -globally- quite different. Take a small neighborhood of one of
these: that would be a "local embedding of (a piece of) H^2 in E^2". By
going into higher dimensions you can get a -global- embedding on H^2 in
some E^d. Similar local versus global distinctions arise for embedding
semi-Riemannian maninfolds in higher dimensional E^(p,q).
Chris Hillman
Home Page: http://www.math.washington.edu/~hillman/personal.html
Vesselin G Gueorguiev
> If you have a smooth N-dimensional manifold with a given Riemannian
> metric, there is an isometric embedding of it into M-dimensional
> Euclidean space with M > N.
Is that true for pseudo Riemannian metric spaces? I guess not since
the Riemannian metric is positive definite and so is the Euclidean one,
while the pseudo Riemannian metric is not, but may be there is a suitable
pseudo Euclidean space.
[...]
Penny314
Right. The embedding theorem in the smooth manifold case is due to
Whitney and is called " the Whitney embedding theorem". The isometric embedding
theorem is due to John Nash and is called the Nash embedding theorem
best
pennysmith
Toby Bartels
>Toby Bartels <to...@ugcs.caltech.edu> wrote:
>>John Baez <ba...@galaxy.ucr.edu> wrote:
>>>Here's a theorem along the lines you want: a smooth N-dimensional
>>>manifold can always be smoothly embedded in M-dimensional Euclidean
>>>space for some M > N. (In fact there's an explicit upper bound on
>>>the necessary M, but I forget what it is.)
>>The upper bound on M is 2N.
>>It's fairly easy to prove an upper bound of 2N + 1
>>(although I'm not sure I could do it off the top of my head);
>>2N, I'm told, is harder.
>Thanks. The trick to remembering this stuff is to remember
>that generically, an N-dimensional manifold mapped into an
>M-dimensional one intersects itself along a submanifold of
>dimension N + N - M. This is just by counting degrees of
>freedom. So M = 2N + 1 should certainly work. On the other
>hand, M = 2N might give you isolated points of self-intersection.
>So in this case, you'll need to do some topology involving
>intersection numbers to prove you can get rid of the intersection
>points by cleverly cancelling them in pairs. Or something like
>that.
It would have to be more than just cancelling in pairs.
Suppose you're trying to embed S^1 in R^2
and you foolishly end up with a figure "8".
This is generic in that all intersections are transversal,
which I believe is the setting for the formula N + N - M,
so you have the 0D intersection as you should,
but you can't straighten it out by cancelling in pairs.
>Right, we definitely need paracompactness for these theorems.
>Without it, we can always make a "long line", a nonparacompact
>1-manifold, that can't be embedded in R^n for the simple reason
>that it has more points than R^n does! This is one more reason
>people like to tuck paracompactness into the definition of "manifold".
I forgot about the long line, one of my favorite counterexamples.
For those who don't know it, let A be the smallest ordinal
larger than the cardinality of the continuum,
and put A copies of the interval [0,1[ end to end.
-- Toby
to...@ugcs.caltech.edu
Toby Bartels
>Ah! Now there's a term I've see bandied about here that I don't fully
>grok: what is "semi-Riemannian" (vs. "Riemannian"?) vs. ...? I've
>always thought in terms of "metric" vs. "affine" (="projective"?) vs. ...
>What's the correspondence?
It's not that. Since this is s.p.r, let me offer this analogy:
Riemannian : semiRiemannian :: space : spacetime.
On a Riemannian manifold, the signature must equal the dimension,
the metric must be positive definite, all vectors must be spacelike.
These restrictions are removed in the semiRiemannian case.
In fact, a better analogy than the above would be
Riemannian : Lorentzian :: space : spacetime.
Riemannian implies 0 timelike dimensions; Lorentzian implies 1.
SemiRiemannian implies any number you like.
-- Toby
to...@ugcs.caltech.edu
Chris Hillman
On 14 Dec 1999, Jim Heckman wrote:
> As I said, I find this surprising -- even *more* than surprising. Is there
> a constructive method of finding such an E^d and embedding, or is it
> more by way of an 'existence' proof?
I think its pretty constructive, especially if you don't try for the
optimal dimension. I think this is explained in great detail in the book
by Spivak.
> Can you,
Not off the top of my head, but if I thought about long enough, probably
yes.
> or anyone, exhibit an explicit, isometric, global embedding of
> H^2 in some E^d, where H^2 is diffeomorphically R^2 and admits a metric:
>
> ds^2 = dr^2 + sinh(r)^2.d{phi}^2, for 0 <= r < +\inf, {phi} mod 2{pi}
or in the UHP model of Poincare
ds^2 = (dx^2 + dy^2)/y^2, y > 0
>From which, with aid of an LFT, you can obtain Poincare's disk model, not
be confused with Klein's disk model, which has yet another metric; for all
of these see the "coordinate tutorial" on my collection of posts on my
relativity pages (see url below).
Any of these might be easier than the "polar" coordinates you gave.
> and E^d is diffeomorphically R^d and admits a metric:
>
> ds^2 = dx_1^2 + dx_2^2 + ... + dx_d^2, for -\inf < x_n < +\inf
I am sure the answer is "yes"; indeed this was probably known to Klein and
his students. Probably people who have published on "embedding classes"
of solutions in gtr would know an explicit global embedding.
> > Similar local versus global distinctions arise for embedding
> > semi-Riemannian
>
> Ah! Now there's a term I've see bandied about here that I don't fully
> grok: what is "semi-Riemannian" (vs. "Riemannian"?) vs. ...?
Riemannian metric positive definite quadratic form
semi-Riemannian metric semi-definite quadratic form
(similar reference; or see the textbook by O'Neill).
> By E^(p,q) you mean generalizations of flat Minkowski space-time,
> right? If so, I already know how to globally embed H^2 isometrically in
> E^(2,1). :-)
Right and right. Which is one reason right there why semi-Riemannian
metrics are worthwhile even if you are really more interested in (global)
Riemannian geometry.
Miguel Carrion Alvarez
> Toby Bartels <to...@ugcs.caltech.edu> wrote in message news:82c7mm$6...@gap.cco.caltech.edu...
>
> > I forgot about the long line, one of my favorite counterexamples.
> > For those who don't know it, let A be the smallest ordinal
> > larger than the cardinality of the continuum,
> > and put A copies of the interval [0,1[ end to end.
I thought that the long line used the first noncountable
ordinal... But of course that's the same as your long line if you
assume the continuum hypothesis.
> Very interesting. Could you explain this a bit more? I can't
> figure out what "n copies of an interval end to end" might mean
> if n is uncountable. Doesn't the very idea of nonzero-length
> intervals end to end imply that the set of intervals is
> countable, i.e., we can speak of the first interval, the second
> interval, etc.?
That is only true in the usual real line. The long line is called
"long" for a reason :-)
Anyway, you have to be a little more careful that to say "put A copies
of the [0,1) interval end to end". The way I have seen the long line
defined is: let Omega be the set of all countable ordinals plus the
first uncountable ordinal (this is an uncountable set), and insert
between each pair of consecutive ordinals a copy of (0,1).
Now, what does the set of all countable ordinals look like? Something
like this:
Start with the natural numbers:
1,2,3,...
now call w the first ordinal larger than any natural number:
1,2,3,...,w
Note that w does _not_ have an immediate predecessor, w-1. if w-1 were
an ordinal, w-1 could not be a natural number (for then w would be a
successor to a natural, and hence a natural) but then it would be
larger than any natural, thus contradicting the definition of w. This
is a curious property of the ordinal set, that all elements have
immediate successor but the "limit ordinals" do not have immediate
predecessors. Now stick another copy of the naturals after w:
1,2,3,...,w,w+1,w+2,...,2w
We are still not done, as you might guess.
1,2,3,...,w,w+1,w+2,...,2w,2w+1,2w+2,...,nw,nw+1,nw+2,...
...,w^2,w^2+1,w^2+2,...,w^2+w,w^2+w+1,...
And so on. These are all still countable ordinals. You can still go
past w^n for all n and still be in the realm of countable
ordinals. You get the idea.
I am just now confused about w^w, because that's not countable, but
the smallest ordinal larger than all the w^n (which I would call w^w)
is still countable. Can anyone come to the rescue?
Anyway, now call Q the smallest noncountable ordinal, add that at the
end of all the countable ordinals and insert a copy of the (0,1)
interval between each two _consecutive_ ordinals. Now each non-limit
ordinal looks like the 0 in (-1,1), but for limit ordinals (like w)
there is no predecessor, so you can only insert an interval between,
say, w and w+1. So you get something like
(the reals)...,w,(0,1),w+1,(0,1),...
Miguel
Chris Hillman
> Toby Bartels <to...@ugcs.caltech.edu> wrote in message
> news:82c7mm$6...@gap.cco.caltech.edu...
>
> > I forgot about the long line, one of my favorite counterexamples.
> > For those who don't know it, let A be the smallest ordinal
> > larger than the cardinality of the continuum,
> > and put A copies of the interval [0,1[ end to end.
>
> Very interesting. Could you explain this a bit more?
In lieu of an explanation (which I don't have the energy to attempt right
now) here is a quick reference: Munkres, Elements of Topology has a very
readable discussion of this example.
Marc Nardmann
> Toby Bartels <to...@ugcs.caltech.edu> wrote in message news:82c7mm$6...@gap.cco.caltech.edu...
>
> > I forgot about the long line, one of my favorite counterexamples.
> > For those who don't know it, let A be the smallest ordinal
> > larger than the cardinality of the continuum,
> > and put A copies of the interval [0,1[ end to end.
>
> Very interesting. Could you explain this a bit more? I can't
> figure out what "n copies of an interval end to end" might mean
> if n is uncountable. Doesn't the very idea of nonzero-length
> intervals end to end imply that the set of intervals is
> countable, i.e., we can speak of the first interval, the second
> interval, etc.?
You can find the long line in almost every book on general topology. Try
Steen/Seebach: "Counterexamples in topology", for example. The precise
definition is as follows: Form the cartesian product of the sets \aleph_1
and [0,1[, where \aleph_1 is the smallest uncountable cardinal number. (The
famous continuum hypothesis -- which is independent of the ZFC axioms [the
Zermelo-Fraenkel axioms of set theory including the axiom of choice] --
says that this cardinal number is also the cardinal number of the set of
reals [which equals the cardinal number of the power set of the natural
numbers]. But the continuum hypothesis plays no role here.) We identify the
class of cardinal numbers with a subclass of the class of ordinal numbers.
(Consult a book on set theory if you don't understand this statement.) Then
\aleph_1 is an ordinal number, and thus it is the (uncountable) set of all
ordinal numbers which are smaller than itself. Now put the lexicographic
order on this set: If x0,x1 are in \aleph_1 and y0,y1 are in [0,1[, then we
define (x0,y0) < (x1,y1) to hold if and only if x0 < x1 or (x0=x1 and
y0<y1). Finally, equip this ordered set with the order topology, i.e. the
topology generated by all sets of the form {a | b<a} and {a | a<b} (where b
is an element of our cartesian product). You get a topological space, in
fact, a topological manifold (connected Hausdorff, but not metrizable).
The concept "length" has nothing to do with it. (You said "nonzero-length
intervals".) If you insist that a "length" (in some canonical sense) should
be assigned to the long line then it would not be a real number [perhaps a
so called Conway number]. (I mentioned that the topology of the long line
is not induced by any metric!) Maybe some remarks about ordinal numbers are
helpful, I've got the impression that that's your problem. The ordinal
numbers extend the idea of counting. You begin counting 0,1,2,.. . When you
have counted all the natural numbers, you continue: \omega, \omega + 1,
\omega + 2, ... . After that, 2 * \omega, 2 * \omega + 1, .... Some day
you'll arrive at \omega ^2, \omega ^2 +1, ... . But even then, you've
counted only countably many numbers! You can find a bijection from the set
of natural numbers to the set of ordinal numbers you have counted so far.
But some day, when you've counted for a really long time, there's no longer
a bijection: You have reached the ordinal \aleph_1.
Hope it helps. Otherwise ask again.
Marc Nardmann
[Moderator's note: It's always hard for me to tell when a discussion has
ventured too far into pure mathematics to be reasonably relevant to
physics. However, I think that this particular thread should continue
elsewhere, such as sci.math. -MM]
John Baez
Toby Bartels <to...@ugcs.caltech.edu> wrote:
>I was confused because John said the long line had
>more points than the continuum, but I think this is wrong.
I know this thread is drifting dangerously far from physics,
but just before it drifts away from this newsgroup I'd like
to apologize for this mistake. If I was thinking anything
when I said this, I guess I was thinking about a "really long
line", formed by stitching together a lot of copies of (0,1] -
where by "a lot" I mean more than the cardinality of the
real numbers. Now I am confused about whether such a thing
would qualify as a (nonparacompact) manifold or not. But
anyway, it isn't the usual "long line", which still has
the same cardinality as the real numbers.
I will set followups to sci.math....
Robert Israel
"Yaakov Eisenberg" <y...@peoplepc.com> writes:
> Toby Bartels <to...@ugcs.caltech.edu> wrote in message news:82c7mm$6...@gap.cco.caltech.edu...
>
> > I forgot about the long line, one of my favorite counterexamples.
> > For those who don't know it, let A be the smallest ordinal
> > larger than the cardinality of the continuum,
> > and put A copies of the interval [0,1[ end to end.
> Very interesting. Could you explain this a bit more? I can't
> figure out what "n copies of an interval end to end" might mean
> if n is uncountable. Doesn't the very idea of nonzero-length
> intervals end to end imply that the set of intervals is
> countable, i.e., we can speak of the first interval, the second
> interval, etc.?
Except that it's an uncountable "etc.".
To define the line a bit more formally, for each ordinal b <= A
you have a copy I_b of the interval, and the long line L is the
disjoint union of these copies. An order relation is defined in
L: x <= y if either x is in I_b and y in I_c with b < c, or
x and y are in the same I_b and x <= y in the usual order on I_b.
The topology on L is the order topology: basic open sets are of
the form (u,v) = {x: u < x < v} for u < v in L. Thus each nonzero
point in I_b has a neighbourhood in I_b, while each neighbourhood of
the 0 in I_b contains points of other copies (in the case of a limit
ordinal, it contains infinitely many copies).
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Michael Weiss
| I thought that the long line used the first noncountable
| ordinal...
I thought so too.
| But of course that's the same as your long line if you
| assume the continuum hypothesis.
Not necessarily, because we are talking about order-isomorphism, and not just
equal cardinality. In fact the long line has a cofinal subset that is
order-isomorphic to the first uncountable ordinal, while the "very long line"
(let's call it that) does not.
Here's how I'd define it. Let W be the set of all countable ordinals; so in
fact, the order-type of W is the first uncountable ordinal. (Indeed, if you
construct ordinals in the slick von Neumann way, W *is* the first uncountable
ordinal.)
Now let LL be the cartesian product W x [0,1), i.e., the set of all ordered
pairs <a,x> where a is a countable ordinal and x is a real number in [0,1).
Order LL lexicographically, i.e., <a,x> is less than <b,y> if a is less than
b, or if a=b and x is less than y. LL is the long line. (At least, it's the
"right-half" of the long-line; gluing on a "left-half" is left as an exercise
for the reader ;-)
| I am just now confused about w^w, because that's not countable, but
| the smallest ordinal larger than all the w^n (which I would call w^w)
| is still countable. Can anyone come to the rescue?
Sure. You can define ordinal exponentition and cardinal exponentiation, and
they don't agree.
I better post this quick before the moderator catches on that we're doing math
and not physics....