1
$\begingroup$

enter image description here

This is a question about Example 2.9, in Bott/Tu - Differential Forms in Algebraic Topology.

Consider the decomposition of $S^1=U\cup V$ by two open sets, as in the figure above. Then both $U$ and $V$ are diffeomorphic to $\Bbb R$, and the intersection $U\cap V$ is diffeomorphic to $\Bbb R \coprod \Bbb R$. We know that $H_c^q(\Bbb R)$ (de Rham cohomology with compact support) is $\Bbb R$ if $q=1$ and is $0$ otherwise. Thus to compute $H^q_c(S^1)$, the only interesting portion of the M-V seq. is as follows. $$ 0\to H_c^0(S^1)\to H_c^1 (U\cap V)=\Bbb R^2 \xrightarrow{i} H_c^1(U)\oplus H_c^1(V) =\Bbb R^2 \to H^1_c(S^1)\to 0 $$

It follows that $H_c^0(S^1)$ and $H_c^1(S^1)$ is equal to $\ker (i)$ and $\text{coker}(i)$, respectively. The map $i$ is given by $\omega \mapsto (\omega_U,\omega_V) $, where, for instance, $\omega_U$ is the extension of $\omega $ (which is a $1$-form on $U\cap V$ with compact support) by zero. Then the book says that the image of $i$ is $1$-dimensional so we get $H_c^1(S^1)=\Bbb R=H_c^0(S^1)$. But how do we know that $\text{image}(i)$ is $1$-dimensional?

$\endgroup$
2
  • $\begingroup$ @NoelLundström In the original de Rham cohomology, yes, but with the cohomology with compact support, the arrows are reversed $\endgroup$
    – blancket
    May 24 '20 at 11:56
  • $\begingroup$ Yes ofcourse, my bad. $\endgroup$ May 24 '20 at 12:35
1
$\begingroup$

They show Early in the book that $$H_c^1(\mathbb{R})=\frac{\Omega^1_c(\mathbb{R}^1)}{\ker\int_{\mathbb{R}^1}}$$ Calculate the image of the linear map $(\int,\int)\circ i^*:H^1_c(U \cap V) \to \mathbb{R}^2$ $$\omega \mapsto (\omega_U,\omega_V)\mapsto (\int_{U}\omega_U,\int_V \omega_V)$$ The second maps is an isomorphism. Show that $\int_{U}\omega_U=\int_{V}\omega_V$ to deduce the desired result.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.