# Confusion on de Rham cohomology of manifolds

1. Sep 30, 2011

### kakarotyjn

Consider the infinite disjoint union $$M = \coprod\limits_{i = 1}^\infty {M_i }$$,where $$M_i 's$$ are all manifolds of finite type of the same dimension n.Then the de Rham cohomology is a direct product $$H^q (M) = \prod\limits_i {H^q (M_i )}$$(why?),but the compact cohomology is a direct sum $$H_c^q (M) = \mathop \oplus \limits_i H_c^q (M_i )$$(why?).

Taking the dual of the compact cohomology is a direct sum $$H^{q}_{c}(M)=\oplus_{i}H^{q}_{c}(M_i)$$(why?).

The question is indicated by red word,by the way I need to prove Kunneth Formula for compact cohomology using Poincare duality and the Kunneth formula for de Rham cohomology.But I don't know how to deal with the Dual space $$(H^{n-q}_{c}(m))^{*}$$.Any ideas?

Thank you very much!

Last edited: Sep 30, 2011
2. Oct 1, 2011

### quasar987

This seems natural enough no? I mean when you think about it at the level of chains... a q-form on M is entirely determined by its restriction to each component M_i, so this sets up an isomorphism between C^q(M) and the product of ther C^q(M_i). And this identification identifies Z^q(M) with the product of the Z^q(M_i) and B^q(M() with the product of the B^q(M_i). So that's that.

As for compact cohomology, the restriction homomorphism at the chain level maps Ccq(M) not unto the whole product of the Ccq(M_i)'s but onto the direct sum only, simply because if you take a q-form w on M, then it must be 0 onall but finitely many iof the M_i's since it has compact support!

Show quite generally that for a collection V_i of vector spaces, the dual of their direct sum is naturally isomorphic to the direct sum of their dual.

So using PD and Kunneth for the Rham, you end up with a natural iso

$$(H_c^{n-q}(M))^*\cong \oplus_{i+j=q}(H_c^{n-i}(M))^*\otimes (H_c^{n-j}(M))^*$$

First use that for any two vector spaces, $V^*\otimes W^*$ is naturally isomorphic to $(V\otimes W)^*$.

Then use the natural ismorphism about duals of direct sum proven above. Then you have

$$(H_c^{n-q}(M))^*\cong \left(\oplus_{i+j=q}H_c^{n-i}(M)\otimes H_c^{n-j}(M)\right)^*$$

Then use that if V,W are naturallyu isomorphic, then so are V* and W*, and finally, use the natural iso btw a v. space V and its double dual V** to obtain the natural iso

$$H_c^{n-q}(M)\cong \oplus_{i+j=q}H_c^{n-i}(M)\otimes H_c^{n-j}(M)$$

So in the end, the problem is one big exercice in finding natural isomorphisms!

3. Oct 3, 2011

### kakarotyjn

Thank you very much quasar987!

But I'm sorry that I still can't understand why $$\omega$$ with compact support is not 0 only on finitely many of M_i,then the compact cohomology is a direct sum ?Could you recommend me some books deal with direct sum and direct product.

And I'm sorry that the third formula is not $$H_c^q (M) = \oplus _i H_c^q (M_i )$$,but is $$(H_c^q (M))^* = \prod _i H_c^q (M_i )$$,the dual space is direct product.

And these days I product another question that:Could all the forms in $$H^* (M \times N)$$ be represented by $$\omega \wedge \eta$$ for $$\omega and \eta$$ in H^*(M) and H^*(N) in respect? If it's yes,how to represent (x+y)dxdy in H(R2)by
$$\omega \wedge \eta$$ for omega is in H(R1) and eta is in H(R1)?

Thank you!

4. Oct 3, 2011

### kakarotyjn

another question:why the pairing of poincare duality is from H^*(M) and H^*_c(M)?
why not H^*(M) and H^*(M) or H^*_c(M)and H^*_c(M)?

Thank you !

5. Oct 3, 2011

### quasar987

Given a family V_i of vector spaces indexed by some set I, the direct product of the V_i is the vector space of all the "I-tuples" or "I-indexed sequences" $(v_i)_{i\in I}$ (formally, this idea is made precise by identifying $(v_i)_{i\in I}$ with the maps $f:I\rightarrow \bigcup_iV_i$ such that f(i) is in V_i.)

The direct sum of the V_i is the linear subspace $\bigoplus_iV_i$ of $\prod_iV_i$ consisting of all the I-tuples (v_i) for which all but finitely many of the v_i are 0.

Last edited: Oct 3, 2011
6. Oct 3, 2011

### quasar987

With the definition above, it should be clear. We have a map of chain complex $C^q(M)\rightarrow \prod_iC^q(M_i)$ defined by $\omega\mapsto (\omega|_{M_i})_{i\in I}$. But omega has compact support, so it is zero on all but finitely many of the M_i (otherwise, we could construct a sequence in supp(omega) by taking no more than 1 point in each M_i, and so this sequence would have no convergent subsequence, which contradicts compactness). So we see that $(\omega|_{M_i})_{i\in I}$ actually lives in the smaller space $\bigoplus_iC^q(M_i)$! Show that $\omega\mapsto (\omega|_{M_i})_{i\in I}$ is an isomorphism onto $\bigoplus_iC^q(M_i)$(easy).

Maybe just Lee's Intro to smooth manifold (the appendix) if what I said just above does not quench your thirst.

Ah, you're right, the correct formula to prove is

$$\left(\bigoplus_iV_i\right)^*\cong \prod_iV_i^*$$

But this has no impact on the steps I suggested to solve your problem about kunneth for compact support. Why? Because notice that in the event that the indexing set I is finite (as is the case for the derahm cohomology of a compact manifold), then $$\prod_iV_i = \bigoplus_iV_i$$!

Last edited: Oct 3, 2011
7. Oct 3, 2011

### quasar987

Ah, but what I wrote in post #2 about Kunneth is pretty much nonsense notation wise. For instance, for the first formula, I meant to write

$$(H_c^{n-q}(M\times N))^*\cong \oplus_{i+j=q}(H_c^{n-i}(M))^*\otimes (H_c^{n-j}(N))^*$$

etc. I apologize.

8. Oct 3, 2011

### quasar987

Notice that it does not make sense to talk about $$\omega \wedge \eta$$ if omega and eta do not live on the same manifold... But actually when you think about it a bit it does make sense, if by $$\omega \wedge \eta$$ we actually mean $$pr_M^*\omega \wedge pr_N^*\eta$$. And this has a name in algebraic topology.. it is called the cohomology cross product of omega and eta and we write

$$\omega\times \eta := pr_M^*\omega \wedge pr_N^*\eta$$

This is defined at the level of chains, but it persist at the cohomology level and it is precisely this map that gives the isomorphism in the Kunneth theorem.

9. Oct 3, 2011

### quasar987

Mmmh.. Well, first notice that in the case M is compact, all the options you wrote above are the same because H*(M)=H*_c(M) in that case. In case M is not compact, we at least need one of the 2 forms omega and eta to be of compact support if their exterior product is to be of compact support. And this we need if

$$\int_M\omega\wedge \eta$$

is to be well-defined.

Last edited: Oct 3, 2011
10. Oct 5, 2011

### kakarotyjn

Thank you very much quasar987!Now I know it better!