TITLE: Distinguishing the two irreducible representations of odd-dimensional complex Clifford-Algebras
QUESTION [6 upvotes]: The complex Clifford algebra $A$ of a complex, non-degenerate quadratic space $(V,q)$ of odd dimension $2k+1$ admits up to isomorphism exactly two non-trivial, irreducible and finite-dimensional representations on a complex vector space $S$. If $\Phi:A\to \mathrm{End} (S)$ represents one, then $\Phi\circ\chi$ represents the other, where $\chi: A\to A$ is the automorphism defined by $\chi(v)=-v$ for $v\in V$.
My question is now: how can one decide whether two generic (irreducible, non-zero, finite-dimensional complex) representations $\Phi_1,\Phi_2$ of $A$ belong to the same isomorphism class? From what I have seen so far, I suspect that it should have something to do with the volume element of $A$...?

REPLY [5 votes]: I'm sorry for not reacting for a long time. I have to admit my question was a bit vague, but I have found the solution, I was looking for, now. There is a complete isomorphism-invariant $\theta(\Phi)\in\mathbb{Z}_2$ of the irreducible, complex, finite dimensional representations $\Phi$ of the Clifford-Algebra $Cl(V)$ of an odd-dimensional, complex, non-degenerate quadratic space $(V,q)$ defined as follows. Assume we have chosen a preferred orientation in $V$ and consider an oriented orthonormal basis $b_1\dots,b_n$ of $V$. Then we call the element $$\eta=i^{n(n-1)/2}b_1\cdots b_n\in Cl(V)$$ the volume element of $Cl(V)$, and the Clifford Relations imply that $\eta$ does not depend on the chosen basis. The factor is chosen such that $\eta^2=1$, hence $\Gamma=\Gamma(\Phi):=\Phi(\eta)$ also satisfies $\Gamma^2=1$ and decomposes the representation space $S$ of $\Phi$ into $S=S_+\oplus S_-$ with $S_\pm:=\ker \Gamma\mp 1$. Since $n$ is odd, $\eta$ lies in the center of $Cl(V)$ and thus the $S_\pm$ are $Cl(V)$-invariant subspaces of $S$. By irreducibility, we thus have either $S_+=S$ or $S_-=S$, i.e. $\Gamma=\theta(\Phi)\cdot\operatorname{id}_S$ for some $\theta(\Phi)\in\mathbb{Z}_2$. One then easily computes that $\theta(\Phi)$ only depends on the isomorphism class of $\Phi$ and that $\theta(\Phi\circ\chi)=-\theta(\Phi)$, where $\chi\in\operatorname{Aut}(Cl(V))$ is the canonical involution on $Cl(V)$, defining the $\mathbb{Z}_2$-grading. Therefore $\Phi\not\cong\Phi\circ\chi$ and by the structure theorem, these are the only irreducible representations (upto isomorphism).
To summarize, two irreducible (complex, finite-dimensional) representations $\Phi,\Psi$ of $Cl(V)$ are isomorphic if and only if both $\Phi(\eta)$ and $\Psi(\eta)$ both act as either plus or minus the identity on the corresponding representation space.
Cheers,
Robert