Clifford modules - How is the grading on $\mathrm{Hom}_{C(V)}(S,E)$ defined?

Question

From the book Heat Kernels and Dirac Operators:

Proposition 3.27. If $V$ is an even-dimensional real Euclidean vector space, then every finite-dimensional $\mathbb{Z}_2$-graded complex module $E$ of the Clifford algebra $C(V)$ is isomorphic to $W\otimes S$, for the $\mathbb{Z}_2$-graded complex vector space $$W = \mathrm{Hom}_{C(V)} (S, E).$$

If we forget about Clifford multiplication, $S$ (the spinor module introduced in proposition 3.19) and $E$ are just graded vector spaces and $L(S,E)$ comes with an obvious grading. Hence each $T\in L(S,E)$ has a unique decomposition $T=A+B$, but $T\in W$ does not imply that $A,B\in W$, does it? So what grading on $W$ is meant in the above proposition?

Answer 1

Warning: The definition of $\mathrm{Hom}_{C(V)} (S, E)$ is not stated in the book and there are different opinions concerning what it should be: As explained here, I think that its elements should be defined to commute with the Clifford action, but in these lecture notes they are defined to super-commute.

But $T\in W$ does not imply that $A,B\in W$, does it?

I think that it actually does: Note that $T\in L(S,E)$ satisfies $T\in W$ if and only if $\gamma_E(a)\circ T=T\circ \gamma_S(a)$ for all $a\in C^0(V)\cup C^1(V)\subset C(V)$. But this equation is equivalent to \begin{equation}\tag{1} \gamma_E(a)\circ A-A\circ \gamma_S(a)=\gamma_E(a)\circ B-B\circ \gamma_S(a) \end{equation} If $a\in C^0(V)\cup C^1(V)$, then one side of $(1)$ is odd and the other side of $(1)$ is even, and hence both sides of $(1)$ are equal to $0$. Since this holds for all $a\in C^0(V)\cup C^1(V)$, we obtain that $A$ and $B$ both commute with the Clifford action.