Friday, December 2, 2016

A Bit of Geometry

Claim: $H^1(\mathbb C^*,\mathbb Z)=\mathbb Z$.

Proof: Let\begin{align*}U_1&=\left\{z\in\mathbb C^*:\Re\left(z\right)>0\right\},\\U_2&=\left\{z\in\mathbb C^*:\Re\left(z\right)<0\right\},\\U_3&=\left\{z\in\mathbb C^*:\Im\left(z\right)>0\right\},\\U_4&=\left\{z\in\mathbb C^*:\Im\left(z\right)<0\right\}\text{, and}\end{align*}$\mathcal{U}=\left\{U_1,U_2,U_3,U_4\right\}$. Then $\mathcal{U}$ is a locally-finite open cover of $\mathbb C^*$. Let $\mathcal{V}\subseteq\mathcal{U}$ Then\begin{equation*}H^q\left(\bigcap_{V\in\mathcal{V}}V,\mathbb Z\right)=0\end{equation*}for all positive integers $q$ since $\bigcap_{V\in\mathcal{V}}V$ is empty or contractible. Therefore, by Leray's theorem, it is the case that $H^1\left(\mathcal{U},\mathbb Z\right)=H^1\left(\mathbb C^*,\mathbb Z\right)$. Let $I=\left\{1,2,3,4\right\}$ and observe that\begin{align*}C^0\left(\mathcal{U},\mathbb Z\right)&=\prod_{\substack{\alpha\in I\\\vphantom{\alpha}}}\mathbb Z\left(U_\alpha\right)=\mathbb Z^4,\\C^1\left(\mathcal{U},\mathbb Z\right)&=\prod_{\substack{\alpha,\beta\in I\\\alpha<\beta}}\mathbb Z\left(U_\alpha\cap U_\beta\right)=\mathbb Z^4\text{, and}\\C^2\left(\mathcal{U},\mathbb Z\right)&=\prod_{\substack{\alpha,\beta,\gamma\in I\\\alpha<\beta<\gamma}}\mathbb Z\left(U_\alpha\cap U_\beta\cap U_\gamma\right)=0.\end{align*}Therefore, the coboundary map $\delta_0:C^0\left(\mathcal{U},\mathbb Z\right)\to C^1\left(\mathcal{U},\mathbb Z\right)$, which is defined by $\sigma\mapsto\delta_0\sigma$ such that\begin{equation*}\left(\delta_0\sigma\right)_{\left(\alpha,\beta\right)}=\sigma_\beta|_{U_\alpha\cap U_\beta}-\sigma_\alpha|_{U_\alpha\cap U_\beta}\text{, where }\left(\alpha,\beta\right)\in I^2\text{ such that }\alpha<\beta,\end{equation*}yields that $\left(a,b,c,d\right)\mapsto\left(c-a,d-a,c-b,d-b\right)$, where $a,b,c,d\in\mathbb Z$. Observe that $d-b=\left(d-a\right)-\left(c-a\right)+\left(c-b\right)$. Therefore, $\text{im }\delta_0=\mathbb Z^3$, which implies that\begin{equation*}H^1\left(\mathbb C^*,\mathbb Z\right)=H^1\left(\mathcal{U},\mathbb Z\right)=\frac{\ker\delta_1}{\text{im }\delta_0}=\frac{\mathbb Z^4}{\mathbb Z^3}=\mathbb Z,\end{equation*}where $\delta_1:C^1\left(\mathcal{U},\mathbb Z\right)\to C^2\left(\mathcal{U},\mathbb Z\right)$ is the zero homomorphism. $\blacksquare$