[Nakahara GTP] 12.3 The de Rham complex

    This article is one of Manifold, Differential Geometry, Fibre Bundle.

Let $M$ be an $m$-dimensional compact orientable manifold with no boundary. By now we are familiar with the de Rham complex, $$\begin{align}\cdots\stackrel{d}{\rightarrow}\Omega^{r-1}(M)^\mathbb{C}\stackrel{d}{\rightarrow}\Omega^r(M)^\mathbb{C}\stackrel{d}{\rightarrow}\Omega^{r+1}(M)^\mathbb{C}\stackrel{d}{\rightarrow}\cdots\end{align}$$ where $\Omega^r(M)^\mathbb{C}=\Gamma(M,\wedge^rT^*M^\mathbb{C})$. We complexified the forms so that we may apply the AS index theorem. The exterior derivative satisfies $d^2=0$. To show that (1) is an elliptic complex, we have to show that $d$ is elliptic. To find the symbol for $d$, we note that $$\sigma(d,\xi)\omega=d(f\tilde{s})|_p=df\wedge\tilde{s}+fd\tilde{s}|_p=\xi\wedge\omega$$ where $p\in M$, $\omega\in\Omega^r_p(M)^\mathbb{C}$, $f(p)=0$, $df(p)=\xi$, $\tilde{s}\in\Omega^r(M)^\mathbb{C}$ and $\tilde{s}(p)=\omega$; see (12.1.4). We find $$\begin{align}\sigma(d,\xi)=\xi\wedge.\end{align}$$

This defines a map $\Omega^r(M)^\mathbb{C}\rightarrow \Omega^{r+1}(M)^\mathbb{C}$ and is non-singular if $\xi\ne0$. Thus, we have proved that $d:\Omega^r(M)^\mathbb{C}\rightarrow \Omega^{r+1}(M)^\mathbb{C}$ is elliptic and, hence, (1) is an elliptic complex. Note, however, that the operator $d:\Omega^k(M)\rightarrow \Omega^{k+1}(M)$ is not Fredholm since $\ker d$ is infinite dimensional. To apply the index theorem to this complex, we have to consider the de Rham cohomology group $H^r(M)$ instead. The operator $d$ is certainly Fredholm on this space/

Let us find the index theorem for this complex. We note that $\dim_\mathbb{C}H^r(M;\mathbb{C})=\dim_\mathbb{R}H^r(M;\mathbb{R})$. Hence, the analytical index is $$\begin{align}\mbox{ind}d=\sum^m_{r=0}(-1)^r\dim_\mathbb{C}H^r(M;\mathbb{C})\nonumber\\ =\sum (-1)^r\dim_\mathbb{R}H^r(M;\mathbb{R})=\chi(M)\end{align}$$ where $\chi(M)$ is the Euler characteristic of $M$. Suppose $M$ is even dimensional, $m=2l$. The RHS of (12.2.1) gives the topological index $$\begin{align}(-1)^{l(2l+1)}\int_M\mbox{ch}\left(\oplus^m_{r=0}(-1)^r\wedge^rT^*M^\mathbb{C}\right)\left. \frac{\mbox{Td}(TM^\mathbb{C})}{e(TM)}\right|_{vol}.\end{align}$$ The splitting principle yields $$\mbox{ch}\left(\oplus^m_{r=0}(-1)^r\wedge^rT^*M^\mathbb{C}\right)\\ =1-\mbox{ch}(T^*M^\mathbb{C})+\mbox{ch}(\wedge^2T^*M^\mathbb{C})+\cdots+(-1)^m\mbox{ch}(\wedge^mT^*M^\mathbb{C})\\ =1-\sum^m_{i=1}e^{-x_i}(TM^\mathbb{C})+\sum_{i<j}e^{-x_i}e^{-x_j}(TM^\mathbb{C})+\cdots\\ +(-1)^me^{-x_1}e^{-x_2}\cdots e^{-x_m}(TM^\mathbb{C})\\ =\prod^m_{i=1}(1-e^{-x_i})(TM^\mathbb{C})$$ where we have noted that $x_i(T^*M^\mathbb{C})=-x_i(TM^\mathbb{C})$. (Let $L$ be a complex line bundle and $L^*$ be its dual bundle. $L\otimes L^*$ is a bundle whose section is a map $\mathbb{C}\rightarrow\mathbb{C}$ at each fibre of $L$. $L\otimes L^*$ has a global section which vanishes nowhere (the identity map, for example) from which we can show $L\otimes L^*$ is a trivial bundle. We have $c_1(L\otimes L^*)=c_1(L)+c_1(L^*)=0$, hence $x(L^*)=-x(L)$. The splitting principle yields $x_i(T^*M^\mathbb{C})=-x_i(TM^\mathbb{C})$.) We also have $$\mbox{Td}(TM^\mathbb{C})=\prod^m_{i=1}\frac{x_i}{1-e^{-x_i}}(TM^\mathbb{C})\\ e(TM)=\prod^l_{i=1}x_i(TM^\mathbb{C}).$$ Substituting these in (4), we have $$\begin{align}\mbox{ind}d=\int_M(-1)^{l(2l+1)}(-1)^l\left( \prod^l_{i=1}x_i(TM^\mathbb{C})\right)=\int_Me(TM).\end{align}$$ 

If $m$ is odd, it can be shown that $$\begin{align}\mbox{ind}d=0\end{align}$$ which is in harmony with the fact that $e(TM)=0$ if $\dim M$ is odd. In any case, the index theorem for the de Rham complex is $$\begin{align}\chi(M)=\int_Me(TM).\end{align}$$

Example 12.3. Let $M$ be a two-dimentional orientable manifold without boundary. Equation (7) reads $$\begin{align}\chi(M)=\frac{1}{4\pi}\int_M\epsilon^{\alpha\beta}\mathcal{R}_{\alpha\beta}=\frac{1}{2\pi}\int_M\mathcal{R}_{12}\end{align}$$ which is the celebrated Gauss-Bonnet theorem. For $\dim M=4$, it reads as $$\begin{align}\chi(M)=\frac{1}{32\pi^2}\int_M\epsilon^{\alpha\beta\gamma\delta}\mathcal{R}_{\alpha\beta}\wedge\mathcal{R}_{\gamma\delta}.\end{align}$$