Trace of the chiral matrix of a subspace Let $(V,Q)$ be a pair consisting of a $\mathbb{C}$-vector space $V$ together with a nondegenerate bilinear form $Q$ and let $V_0\subseteq V$ be a linear subspace such that $Q\vert_{V_0}$ is nondegenerate. The inclusion $V_0\hookrightarrow V$ induces a morphism of algebras $Cl(V_0,Q) \to Cl(V,Q)$, where $Cl$ is the Clifford algebra, so that if $M$ is a $Cl(V,Q)$-module, then $M$ is also automatically a $Cl(V_0,Q)$-module. This is in particular true in case $M=S_V$ is a irreducible spinor representation of $Cl(V,Q)$. Now let $\Gamma_{chir,0}$ be the chiral matrix of $(V_0,Q)$, i.e., the product of the elements from a $Q$-othonormal basis of $V_0$ multiplied with a suitable power of $i$ so that $\Gamma_{chir,0}^2=1$. By the above, the element $\Gamma_{chir,0}$ acts on $S_V$. My question is: what is the trace of $\Gamma_{chir,0}$ as an operator on $S_V$? A couple of cases are classical: if $V_0=V$ then this trace is $0$ or $\dim S_V$ (with the correct normalization of the chiral matrix) depending whether $\dim V$ is even or odd. Is there something that can be said in general for a subspace $V_0$ of $V$?
The motivation for this question is to try to understand whether it is possible that the 1-eigenspace of $\Gamma_{chir,0}$ on $S_V$ with $\dim V=2n$ can be identified with the irreducible spinor representation $S_{V_0}$ of $Cl(V_0,Q)$ (when $\dim V_0=2m$) or with the sum of the two irreducible spinor representations $S_{V_0}^+$ and $S_{V_0}^-$ of $Cl(V_0,Q)$ (when $\dim V_0=2m+1$) in some nontrivial cases. It seems to me that this is true for $(\dim V,\dim V_0)=(4,2)$ and $(6,3)$ and I'm wondering about other possibilities. 
 A: A complete answer can be found in Appendix E to de Wit-Laenen-Smith book on Field Theory in Particle Physics (available here: http://www.nikhef.nl/~t45/ftip/AppendixE.pdf). Namely, in a suitable orthonormal basis for $V$, the endomorphism of $S_V$ (and more generally of an arbitrary $Cl(V,Q)$-module $M$) corresponding the chiral element $\Gamma_{chir,0}$ of $Cl(V_0,Q)$ is a matrix of the form $\Gamma^A$ in the notation of Appendix E, and so as soon as $V_0\neq 0$ and $V_0\neq V$ one has $\mathrm{tr}(\Gamma^A)=0$. This is equation (E.21) in the book.
As a consequence, there are infinitely many possibilities for having a possible identification of the 1-eigenspace of $\Gamma_{chir,0}$ on $S_V$ with $S_{V_0}$ (or with $S_{V_0}^+\oplus S_{V_0}^-$ depending on the parity of the dimension of $V_0$), and these can be completely described. Namely, since the trace of the chiral element $\Gamma_{chir,0}$ acting on $S_V$ is zero and its eigenvalues are $\pm1$ we see that the dimension of the 1-eigenspace is $1/2 \dim S_V$. Therefore for $\dim V_0$ even we find $2^{\frac{1}{2}\dim V_0}=2^{\frac{1}{2}\dim V -1}$ and so $\dim V_0=\dim V-2$. The $(4,2)$ case is one of these cases. For $\dim V_0$ odd we find $2^{\frac{1}{2}(\dim V_0-1)+1)}=2^{\frac{1}{2}\dim V -1}$ and so $\dim V_0=\dim V-3$. The $(6,3)$-case is one of these. What singles out the two cases $(4,2)$ and $(6,3)$, if one wants to single out them by some reason, is that they are the only possible cases with $\dim V_0 = \frac{1}{2}\dim V$.
