Clifford module

Remarks

By the universal property of the Clifford algebra, giving $\rho$ is the same as giving a linear map $c \colon V \to \End_{\mathbb{K}}(W)$ with $$c(v)^2 = -q(v) \, \id_W \quad \text{for all } v \in V.$$ Authors who use the convention $v \cdot v = q(v) \cdot 1$ get the same theory with $q$ replaced by $-q$. A Clifford module is irreducible if it has no proper nonzero invariant subspace; when $q$ is nondegenerate, every finite-dimensional Clifford module is a direct sum of irreducible ones.

Example: let $V = \R$ with $q(x) = x^2$, let $W = \C$ as a real vector space and let $c(x)$ be multiplication by $ix$. Then $c(x)^2 = (ix)^2 = -x^2 = -q(x)$, so $W$ is a Clifford module, and in fact $\mathrm{Cl}(V, q) \cong \C$.

Non-example: with the same $V$ and $q$, let $W = \R$ and $c(x) = x \, \id_W$. Then $c(1)^2 = 1 \ne -1$, so this is not a Clifford module. In fact no real vector space of dimension $1$ is a Clifford module here, since no real number squares to $-1$.