---
tag: G155QN3Z4Y
kind: definition
name: Spencer cohomology
msc: [58A20, 35N10, 58J10]
defines: BKKAA5JKT0
uses: [05EGHEQZAW, 0BF2DE2SYK, 0MDMZ965VE, 2C3PAQ7H50, WQ82C0M3QB, YZY11W08SD, GC5BDZ6QBC]
sources:
  - url: https://en.wikipedia.org/wiki/Spencer_cohomology
---

## Definition

Let $V$ and $W$ be finite-dimensional real [[vector spaces|05EGHEQZAW]], with $\dim V = n$.
Let $S^k = S^k V^* \otimes W$ be the space of $W$-valued [[homogeneous polynomials|2C3PAQ7H50]] of degree $k$ on $V$, with $S^k = 0$ for $k < 0$, and let $\Lambda^j = \Lambda^j V^*$ be the [[exterior power|0MDMZ965VE]].
Choose a basis $e_1, \dots, e_n$ of $V$ with dual basis $e^1, \dots, e^n$, and let $\partial_i$ be the derivative of polynomials in the direction $e_i$.
The *Spencer operator* is the linear map
$$\delta \colon S^k \otimes \Lambda^j \to S^{k-1} \otimes \Lambda^{j+1}, \qquad \delta(P \otimes \omega) = \sum_{i=1}^n \partial_i P \otimes (e^i \wedge \omega).$$
It does not depend on the basis, and $\delta \circ \delta = 0$.

Let $q \ge 1$ and let $g_q \subseteq S^q$ be a linear subspace, the symbol of a system of order $q$.
Put $g_k = S^k$ for $k < q$, and for $k \ge q$ define the prolongations inductively by
$$g_{k+1} = \set{ P \in S^{k+1} : \partial_v P \in g_k \text{ for all } v \in V }.$$
Then $\delta(g_k \otimes \Lambda^j) \subseteq g_{k-1} \otimes \Lambda^{j+1}$, so the spaces $g_k \otimes \Lambda^j$ form [[chain complexes|WQ82C0M3QB]], the *Spencer complexes* of $g_q$.
The *Spencer cohomology* of $g_q$ is
$$H^{k,j}(g_q) = \frac{\ker\big(\delta \colon g_k \otimes \Lambda^j \to g_{k-1} \otimes \Lambda^{j+1}\big)}{\delta(g_{k+1} \otimes \Lambda^{j-1})}.$$

## Remarks

$\delta \circ \delta = 0$ because partial derivatives commute while $e^i \wedge e^l = -e^l \wedge e^i$.
Indexing conventions differ between authors, for instance by shifting $k$.

Example: take $g_q = S^q$, the symbol of the empty system.
Then $g_k = S^k$ for all $k$, and for each $m \ge 1$ the sequence $0 \to S^m \to S^{m-1} \otimes \Lambda^1 \to \dots \to S^0 \otimes \Lambda^m \to 0$ is exact.
It is the degree $m$ part of the de Rham complex of polynomial forms, and it is dual to the [[Koszul complex|YZY11W08SD]].
So $H^{k,j}(S^q) = 0$ whenever $k + j \ge 1$.

Example: for the Laplace equation on $\R^2$, $q = 2$, $W = \R$ and $g_2$ is the space of quadratic forms $a x^2 + b xy + c y^2$ with $a + c = 0$.
Its prolongation $g_k$ is the space of harmonic homogeneous polynomials of degree $k$, which is $2$-dimensional for each $k \ge 1$.

Consider a system of [[partial differential equations|GC5BDZ6QBC]] whose symbol has nonzero Spencer cohomology in the relevant degrees.
Such a system may have hidden integrability conditions, and its power series solutions cannot always be built degree by degree.