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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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 Posted: Sun May 14, 2006 11:43 pm Post subject:
manifold cohomology isomorphism
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if all the cup products in a n-dimensioanl compact connected manifold
vanish, is the cohomology necessarily isomorphic to that of S^n?
Having a slight difficulty seeing this. Is ita degree of map argument? |
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Lee Rudolph
science forum Guru
Joined: 28 Apr 2005
Posts: 566
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| Posted: Mon May 15, 2006 12:37 am Post subject:
Re: manifold cohomology isomorphism
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oferlock@yahoo.com writes:
| Quote: | if all the cup products in a n-dimensioanl compact connected manifold
vanish, is the cohomology necessarily isomorphic to that of S^n?
Having a slight difficulty seeing this. Is ita degree of map argument?
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I'm not sure what a "degree of map argument" is. I think that
the theorem (supposing it to be one) ought to be provable by a
Poincar\'e duality argument, but algebraic topology is really not
my thing, so that's just a(n educated, but not well-educated) guess.
Lee Rudolph |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Mon May 15, 2006 6:58 pm Post subject:
Re: manifold cohomology isomorphism
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thought about Poincare for a bit, but without any orientability, how
would that apply? |
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W. Dale Hall
science forum Guru
Joined: 29 Apr 2005
Posts: 350
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| Posted: Mon May 15, 2006 11:15 pm Post subject:
Re: manifold cohomology isomorphism
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oferlock@yahoo.com wrote:
| Quote: | thought about Poincare for a bit, but without any orientability, how
would that apply?
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It turns out you don't have to worry about non-orientable
smooth manifolds. More on that later. Further, my nearly total
ignorance of the treatment of PL and topological manifolds
restricts my discussion to smooth manifolds only. From what I
gather, the first Stiefel-Whitney class in H^1(M; Z/2Z) can
be defined as the orientation character in any event, so
perhaps the whole argument goes forward in that case.
BTW, if your manifold is not orientable, you can't get the
cohomology of S^n. H_n(non-orientable n-manifold; Z) = 0,
so H^n(non-orientable n-manifold; Z) is a torsion group. The
best you could possibly do is to get the mod 2 cohomology
of a sphere.
For orientable manifolds, check Allen Hatcher's text which is
available online in .pdf format. Here's the URL:
http://www.math.cornell.edu/~hatcher/AT/ATchapters.html
On p.250 (Chapter 3: Cohomology), he proves this:
Proposition 3.3.8 The cup product pairing is nonsingular
for closed R-orientable manifolds when R is a field, or
when R = Z and torsion in H*(M;Z) is factored out.
and a corollary:
Corollary 3.3.9 If M is a closed connected orientable
n-manifold, then for each element \alpha in H^k(M;Z)
of infinite order that is not a proper multiple of another
element, there exists an element \beta in H^(n-k)(M; Z)
such that \alpha \cup \beta is a generator of H^n(M; Z) = Z.
With coefficients in a field, the same conclusion holds
for any \alpha != 0.
This settles the issue (in the affirmative) for orientable manifolds,
since the corollary could be applied for coefficients in Z/pZ for
all primes p.
For a non-orientable manifold M, if all the products in H*(M;Z/2Z)
vanish, then H*(M;Z/2Z) = H*(S^n; Z/2Z), by this result. However,
H^1(M; Z/2Z) must be nonzero, since a non-orientable manifold has a
nonzero first Stiefel-Whitney class. As a result, for n > 1, your
situation cannot arise for a non-orientable manifold.
Dale. |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Tue May 16, 2006 1:52 am Post subject:
Re: manifold cohomology isomorphism
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that works great! at least one of the cohomologies should have a free Z
summand, or the dual pairing would fail... thanks. |
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W. Dale Hall
science forum Guru
Joined: 29 Apr 2005
Posts: 350
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| Posted: Wed May 17, 2006 7:00 pm Post subject:
Re: manifold cohomology isomorphism
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oferlock@yahoo.com wrote:
| Quote: | here the cup products are zero in Z but not over Z/pZ, and the the
cohomology in Z is not that of a S^2. Excellent example. The original
statement was : "if the cup products vanish in Z, then the cohomology
in Z is that of S^n"... is that statement correct then?
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I'm sorry. I had read your original question literally. I'll grab it
from the article:
if all the cup products in a n-dimensioanl compact
connected manifold vanish, is the cohomology necessarily
isomorphic to that of S^n? Having a slight difficulty
seeing this. Is ita degree of map argument?
I failed to notice that you meant "with integral coefficients",
probably because you did not state "with integral coefficients".
No matter. If I had known that, I would have shown you the
lens space example at the outset: it has cohomology different
from that of S^3, yet has all nontrivial cup products zero.
According to the results I quoted earlier from Hatcher's book,
for an oriented n-manifold, if there is any (nonzero) free
abelian part in cohomology in any dimension from 1 to n-1, then
there will be a nonzero cup product in integral cohomology.
Thus, in the oriented case, trivial cup products ==> rational
cohomology of a sphere, but the lens space example shows that
torsion groups can exist.
The higher-dimensional lens spaces do have nontrivial cup
products, however, so I don't have a good example there.
In the non-orientable case, consider RP^2, the real projective
plane. Its cohomology is
Z, k = 0
H^k(RP^2) = 0, k = 1
Z/2Z, k = 2
and so RP^2 has only zero cup products (with the universal exception
I mentioned earlier: generator of H^0 cup any other nonzero class).
As with the lens space examples, the higher dimensional examples here
do have nonzero cup products, so your question may still turn out to
have an affirmative answer for n >= 4, with the low dimensional cases
merely being "accidental", to the extent that anything is accidental
in mathematics: the cup products of those spaces need to be zero for
purely dimensional considerations.
However, any non-orientable manifold with all zero cup products
will immediately give a negative answer to your question, since
the free part of top-dimensional cohomology of a non-orientable
manifold is zero.
Dale |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Wed May 17, 2006 10:25 pm Post subject:
Re: manifold cohomology isomorphism
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sorry that i played fast and loose in the original statement. So much
excellent explanation, thanks. I think this issue is clear, and has
given me some good insight at the same time. |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Thu May 18, 2006 2:29 am Post subject:
Re: manifold cohomology isomorphism
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from your earlier one, i sort of thought, if we have oreintability for
Z/k (as in top homology Z/k and Poincare duality holds), then manifold
is orientable in Z. Would that be correct, or would it have to be true
for all prime k for that? |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Thu May 18, 2006 2:32 am Post subject:
Re: manifold cohomology isomorphism
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from your earlier post it seemed that if the manifold is orientable for
Z/k for some k other than 2 (for 2 we know it is), than it is
orientable in Z (as in having top homology Z/k and Poincare holds in
Z/k implying that for Z). Is that correct, or would that require it to
be orientable for all prime k? |
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W. Dale Hall
science forum Guru
Joined: 29 Apr 2005
Posts: 350
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| Posted: Thu May 18, 2006 6:30 am Post subject:
Re: manifold cohomology isomorphism
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oferlock@yahoo.com wrote:
| Quote: | from your earlier post it seemed that if the manifold is orientable for
Z/k for some k other than 2 (for 2 we know it is), than it is
orientable in Z (as in having top homology Z/k and Poincare holds in
Z/k implying that for Z). Is that correct, or would that require it to
be orientable for all prime k?
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Yes. Here's a proof:
Poincare duality wrt Z/pZ coefficients, p odd, implies
H_n(M; Z/pZ) = H^0(M; Z/pZ) = Z/pZ,
assuming M is path-connected. If M were not orientable,
then we would have
H_n(M; Z) = 0,
and from Poincare duality wrt Z/2Z (which always holds
for closed manifolds)
H_n(M; Z/2Z) = H^0(M; Z/2Z) = Z/2Z
so by the Universal Coefficient Theorem (UCT), we
find
H_(n-1)(M) = Z^k (+) Z/2Z
for some non-negative integer k. Thus, UCT also
implies
H_n(M; Z/pZ) = 0,
since UCT gives the short exact sequence
H_n(M)(x)Z/pZ >--> H_n(M;Z/pZ) -->> H_(n-1)(M)*Z/pZ
where H(.) means H(.; Z), and I've used * to denote Tor:
A*B = Tor(A,B)
and >--> is monic, -->> is epic.
Thus, orientable wrt Z/pZ (p odd) implies orientable wrt Z.
Dale. |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Thu May 18, 2006 8:24 am Post subject:
Re: manifold cohomology isomorphism
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H_n(M; Z/2Z) = H^0(M; Z/2Z) = Z/2Z
| Quote: | so by the Universal Coefficient Theorem (UCT), we
find
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| Quote: | H_(n-1)(M) = Z^k (+) Z/2Z
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| Quote: | for some non-negative integer k.
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this follows from:
Ext(H_(n-1) M; Z/pZ) >--> H^n(M;Z/pZ) -->> Hom(H_n(M);/pZ)
version of UCT, assuming H^n(M;Z/pZ) = 0; but by assumption (correct
me if not) H^n(M;Z/pZ) = Z/pZ by Poincare duality.
Then it would follow that
H_(n-1)(M) = Z^k (+) Z/2Z (+) Z/pZ (or something similar) |
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oferlock@yahoo.com
science forum beginner
Joined: 29 Nov 2005
Posts: 12
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| Posted: Thu May 18, 2006 11:35 pm Post subject:
Re: manifold cohomology isomorphism
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can't seem to get rid of the Z/pZ summand in (as i stated in the post
before this one):
H_(n-1)(M) = Z^k (+) Z/2Z (+) Z/pZ (or something similar),
although your proof says :
H_(n-1)(M) = Z^k (+) Z/2Z
I would really like to know why. |
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W. Dale Hall
science forum Guru
Joined: 29 Apr 2005
Posts: 350
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| Posted: Fri May 19, 2006 5:07 pm Post subject:
Re: manifold cohomology isomorphism
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mskirvin@gmail.com wrote:
| Quote: | oferlock@yahoo.com wrote:
H_n(M; Z/2Z) = H^0(M; Z/2Z) = Z/2Z
so by the Universal Coefficient Theorem (UCT), we
find
H_(n-1)(M) = Z^k (+) Z/2Z
for some non-negative integer k.
this follows from:
Ext(H_(n-1) M; Z/pZ) >--> H^n(M;Z/pZ) -->> Hom(H_n(M);/pZ)
version of UCT, assuming H^n(M;Z/pZ) = 0; but by assumption (correct
me if not) H^n(M;Z/pZ) = Z/pZ by Poincare duality.
Then it would follow that
H_(n-1)(M) = Z^k (+) Z/2Z (+) Z/pZ (or something similar)
I actually slightly disagree with Dale that the UCT implies that
H_(n-1)(M) = Z^k (+) Z/2Z.
I believe that he is using the UCT with tensor product and Tor on the
ends and homology in the middle, not the UCT the OP has listed above.
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That is the version of UCT that I've used. I agree that I neglected
the possibility of odd torsion in H_(n-1)(M; Z), and don't have an
explanation other than a vague recollection of a theorem that says
such can't occur (but I don't have a proof of this theorem, to
which you refer below).
| Quote: | Assuming that M is not orientable, the UCT gives:
Z/2Z = H_n(M, Z/2Z) = Tor(H_(n-1)(M), Z/2Z)
In order for the Tor term to end up as Z/2Z, it doesn't have to contain
a Z/2Z summand, only a Z/mZ summand where m is even. This follows from
the general fact that Tor(Z/nZ, Z/mZ) = Z/dZ where d = gcd(m, n). So,
it seems to me that in order for Z/kZ orientability to imply Z
orientability (at least if the UCT method is to be used to prove it),
we would require that k does not divide m (same m as above).
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Yes. My mistake here.
| Quote: | With that said, I think that it is true that Z/kZ orientability for any
k>2 does imply Z orientability. I've seen this asked on past topology
qualifying exams and I've been working on it for a couple days now.
The solution I've come up with is fairly easy if I use a couple of
results from Bredon's "Topology and Geometry" text.
He gives a very nice characterization of the homology of closed,
connected manifolds with coefficients in any abelian group G. The
result is as follows (see results 7.10 through 7.14 in chapter six of
his text).
H_n(M; G) = G if M is compact and orientable
H_n(M; G) = Tor(G, Z/2Z) if M is compact and not orientable
H_n(M; G) = 0 if M is not compact
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That's the result I had had a vague recollection of, and not having
access to Bredon's book, didn't see a proof of.
| Quote: | With this result, it is possible to prove that the torsion part of
H_(n-1)(M) = Z/2Z or 0 depending on whether M is compact or not. From
here, it is then possible to prove the general result we're looking
for.
Maybe there is a way to see this using the UCT like Dale said in his
post, but I don't see it. If anyone does, I'd certainly like to know.
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Me too. I'll think more on the topic.
Of course, apologies to all for my blunder.
Dale. |
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