Distinguishing the two irreducible representations of odd-dimensional complex Clifford-Algebras 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 $A$...?
 A: 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
A: By the structure classification theorem, such Clifford algebras are actually two copies of a matrix ring over $\mathbb{C}$.
Being semismple, all of its modules are direct sums of copies of the simple modules.
The irreducible ones would just be the simple ones. (right?)
(When talking about representations as opposed to modules I always feel like there is a subtle difference I haven't mastered...)
