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From: W. Dale Hall on 22 Apr 2010 01:38
Bacle wrote:
> Hi, everyone:
>
> If X,X' are m-dimensional manifolds, it seems clear that if X,X' are
> diffeomorphic, then their intersection forms Q_X and Q_X' are
> isomorphic/equivalent (as quadratic forms.)
>
> It seems clear that the converse is true, i.e., that we must have Q_X
> equivalent to Q_X' (i.e, if Q is a form over a module M .then there
> is a module isomomorphism h:M-->M' with Q_X(a,b)=
>
> Q_X(h(a),h(b))

In your opening statement, X and X' are m-dimensional manifolds.
Now, M is some module. How are you asserting this as any converse
of the former statement?

The converse would be "if Q_X and Q_X' are equivalent, then X and X'
are diffeomorphic. That statement is not true: it is surely possible
for two m-manifolds to have equivalent intersection forms yet be
of different homotopy types. A trivial way to do this is to take
X' = X U (some handles with no effect on middle-dimensional homology).
You can map X' to X by collapsing those handles.

For example, take X = CP^2, X' = CP^2 # S^1 x S^3, the connected sum,
and map X' ---> X by collapsing the S^1 x S^3 to a point. The homology
and cohomology in dimension 2 are preserved, and so you get equivalent
intersection pairings.

>
> Could someone please point me towards an argument to show that we
> must have Q_X~Q_X' for M and M' to be diffeomorphic.?
>
> Thanks.

The intersection pairing for an oriented 2n-dimensional manifold M is a
homomorphism

. : H_n(M) (x) H_n(M) ---> H_0(M)

and can be computed by commutativity of this diagram:
.
H_n(M) (x) H_n(M) -------> H_0(M)
| |
Poincare| |Poincare
Duality | |Duality
| |
v cup v
H^n(M) (x) H^n(M) ---------> H^n(M)

Note that everything here is homotopy-invariant. A homotopy-
equivalence would preserve this diagram, and since a diffeomorphism
is (in particular) a homotopy-equivalence, you've got the
equivalence of pairings you were looking for.

Now, I may have completely misunderstood your question; if so,
perhaps you might restate it.

Dale
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