The Unapologetic Mathematician

Mathematics for the interested outsider

Nonvanishing Compactly-Supported de Rham Cohomology

Last time we saw that compactly-supported de Rham cohomology is nonvanishing in the top degree for \mathbb{R}^n. I say that this is true for any oriented, connected n-manifold M. Specifically, if \omega\in\Omega_c^n(M), then the integral of \omega over M is zero if and only if \omega=d\eta for some \eta\in\Omega_c^{n-1}(M). That the second statement implies the first should be obvious.

To go the other way takes more work, but it’s really nothing much new. Firstly, if \omega is supported in some connected, parameterizable open subset U\subseteq M then we can pull back along any parameterization and use the result from last time.

Next, we again shift from our original assertion to an equivalent one: \omega_1 and \omega_2 have the same integral over M, if and only if their difference is exact. And again the only question is about proving the “only if” part. A partition of unity argument tells us that we only really need to consider the case where \omega_i is supported in a connected, parameterizable open set U_i\subseteq M; if the integrals are zero we’re already done by using our previous step above, so we assume both integrals are equal to c\neq0. Dividing by c we may assume that each integral is 1.

Now, if p_0\in M is any base-point then we can get from it to any other point p\in M by a sequence of connected, parameterizable open subsets W_i. The proof is basically the same as for the similar assertion about getting from one point to anther by rectangles from last time. We pick some such sequence taking us from U_1 to U_2, and just like last time we pick a sequence of forms \alpha_i supported in W_{i-1}\cap W_i. Again, the differences between \omega_1 and \alpha_1, between \alpha_i and \alpha_{i+1}, and between \alpha_N and \omega_2 are all exact, and so their sum — the difference between \omega_1 and \omega_2 is exact as well.

And so we conclude that the map \Omega_c^n(M)\to\mathbb{R} given by integration is onto, and its kernel is the image of \Omega_c^{n-1}(M) under the exterior derivative. Thus, H_c^n(M)\cong\mathbb{R}, just as for \mathbb{R}^n.

December 8, 2011 Posted by | Differential Topology, Topology | 1 Comment

   

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