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I was going through some lectures, then I realised that they compute the DeRham Cohomology of $\mathbb{R}^0$ and when they wanted to compute the compactly supported DeRham Cohomology then they used the fact that since $\mathbb{R}^0$ is a compact space, the DeRham Cohomology groups of usual and compactly supported coincide.

Is it true in general? That both is DeRham Cohomology coincide for any compact manifold $M$.

I was thinking about concluding something using poincare duality. That is since

$$H^q(M)=H^{n-q}_c(M)$$ where $M$ is an n-manifold.

If I assume it to be compact then I get $$H^q_c(M)=H^q(M)=H^{n-q}_c(M)$$

So $$H^q_c(M)=H^{n-q}_c(M)$$ for a compact manifold $M$. Does this look correct. Any nice conclusion using this?

permutation_matrix
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  • Do you know the difference between the definitions of de Rham cohomology and compact supported de Rham cohomology? – Michael Albanese Nov 04 '22 at 10:40
  • @MichaelAlbanese, Yes. I am using that equality because of poincare duality lemma – permutation_matrix Nov 04 '22 at 10:51
  • My comment refers to your first question (is this true in general?). You don't need Poincaré duality for that. What is the difference in the definitions of the two cohomology groups? – Michael Albanese Nov 04 '22 at 10:58
  • @MichaelAlbanese, the difference comes from the space of $k-$differential forms. The differential forms are taken to be of form $\omega=f dx_{i_1}dx_{i_2}\dots dx_{i_k}$ where the function $f$ is compact. Further the maps are defined accordingly. Now when the manifold $M$ is compact then I will get its support to be closed subset of compact space which seem to be compact. Hence both will coincide in case of compact manifolds. – permutation_matrix Nov 04 '22 at 12:47
  • Exactly! You've answered your own question. – Michael Albanese Nov 04 '22 at 13:20
  • @MichaelAlbanese, yes that cool but I am wondering if we can conclude something nice about the Cohomology of a compact manifold using Poincare's Duality? That is I think for example for a manifold of dimension $6$, I will get that $H^0(M)=H^6(M)$ ; $H^1(M)=H^5(M)$ ; $H^2(M)=H^4(M)$ and so on. I think this is very interesting result but I am not quite sure. – permutation_matrix Nov 04 '22 at 16:10
  • note that it depends upon the dimension of the manifold, and that the manifold is compact. For example for $M=S^n$, we have $H^0(M)=H^n(M)=\mathbb{R}$. – permutation_matrix Nov 04 '22 at 16:14
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    Yes. For example, the Euler characteristic of a closed odd-dimensional manifold is zero. See Corollary 3.37 of Hatcher's Algebraic Topology. – Michael Albanese Nov 05 '22 at 00:15

1 Answers1

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As established in the comments, if $M$ is compact then $H^k(M) = H^k_c(M)$ because every form has compact support. In particular, for a closed orientable $n$-manifold, we have

$$H^k_c(M) = H^k(M) \cong H^{n-k}(M) = H^{n-k}_c(M).$$

If the manifold is not compact, then these two groups may not be isomorphic.

Example: Consider the non-compact manifold $M = S^2\times\mathbb{R}$.

For a (potentially non-compact) orientable $n$-manifold, we always have $H^k(M) \cong (H^{n-k}_c(M))^*$, see Remark $5.7$ of Differential Forms in Algebraic Topology by Bott and Tu. So $0 = H^1(M) \cong (H^2_c(M))^*$ and therefore $H^2_c(M) = 0$. On the other hand, $\mathbb{R} \cong H^2(M) \cong (H^1_c(M))^*$, so $H^1_c(M) \cong \mathbb{R} \not\cong 0 = H^2_c(M)$.

Michael Albanese
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