Definitions
Consider a Clifford module, i.e. suppose that we are given the following data:
- An even-dimensional real vector space $V$ together with a symmetric bilinear form $Q$.
- A super-vector space $E=E^0\oplus E^1$ together with a function $V\times E\to E$ (of course this function is assumed to have some particular properties, see the comments).
Recall that the multiplication extends to a morphism $$\gamma:\mathrm{Cl}(V,Q)\to\mathrm{End}(E)$$ of $\mathbb Z/2$-graded super-algebras. The chirality operator $\Gamma\in L(E,E)$ is defined as follows: \begin{equation} \forall v\in E:\Gamma v=\begin{cases} v&\text{if }v\in E^0\\-v&\text{if }v\in E^1 \end{cases} \end{equation} Let $e_1,\ldots,e_n$ be an orthonormal basis of $V$, then $\omega\equiv e_1\cdots e_n$ is called a volume element - there are two volume elements, one for each orientation of $V$. Note that we have $\omega^2=\pm 1$ (the sign only depends on the dimension of $V$).
Question
For simplicity, suppose that $\omega^2=1$. Hence $\gamma(\omega)\in\mathrm{End}(E)$ is an involution and defines a decomposition $E=E^+\oplus E^-$ such that we can consider the supertrace w.r.t. to this decomposition. Another possibility is to consider the supertrace w.r.t. to the already given decomposition $E=E^0\oplus E^1$. Are they related to each other? For example, can we prove that $\Gamma=\gamma(\omega)$ for one of the two volume elements?
No because $\omega^2 = \pm1$ but $\Gamma^2 = 1$ and $\gamma$ is an algebra morphism: $$ \gamma(\omega)^2 = \gamma(\omega^2) = \gamma(\pm1) = \pm\gamma(1) = \pm1. $$ You would need to restrict $Q$ and/or the dimension of $V$ for this to be possible. I don't know what a "spinor module" is, but this can't change the dependence on $Q$ and $\dim V$.