Elliptic complex

Remarks

For $N = 1$ the complex is a single operator $P_0$, and it is elliptic exactly when $P_0$ is an elliptic operator. When $M$ is compact without boundary, the cohomology groups $\ker P_k / \operatorname{im} P_{k-1}$ are finite-dimensional, and the index of the complex is $\sum_k (-1)^k \dim(\ker P_k / \operatorname{im} P_{k-1})$. Suppose that all $P_k$ have the same order and that $M$ and the bundles carry metrics, and write $P = \sum_k P_k$ and $P^* = \sum_k P_k^*$ for the formal adjoint. Then $P + P^* \colon \Gamma(\bigoplus_k E_{2k}) \to \Gamma(\bigoplus_k E_{2k+1})$ is elliptic and has the same index as the complex.

Example: the de Rham complex $0 \to \Omega^0(M) \xrightarrow{d} \Omega^1(M) \xrightarrow{d} \cdots \xrightarrow{d} \Omega^n(M) \to 0$ of a manifold of dimension $n$, with the exterior derivative. It satisfies $d \circ d = 0$, and the symbol of $d$ at $\xi$ is $\omega \mapsto \xi \wedge \omega$. For $\xi \ne 0$, complete $\xi$ to a basis $\xi = e_1, e_2, \ldots, e_n$ of $T^*_x M$. A form $\omega$ with $\xi \wedge \omega = 0$ is then a sum of terms containing $e_1$. So it equals $\xi \wedge \eta$ for some $\eta$, which proves exactness. Its cohomology is the de Rham cohomology, and its index is the Euler characteristic of $M$.

Non-example: on a manifold of dimension $n \ge 2$, the two-term complex $0 \to \Omega^0(M) \xrightarrow{d} \Omega^1(M) \to 0$ satisfies clause 1 but not clause 2. The symbol sequence $0 \to \R \to T^*_x M \to 0$ is not exact at $T^*_x M$, because the image of $c \mapsto c \, \xi$ is a line in a space of dimension $n$.