# 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 $A$...?

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You are looking for a method to determine whether or not two representations are isomorphic? Did you intend to assume any conditions on them at all? Are we assuming the form is nondegenerate? – rschwieb Sep 10 '12 at 15:31
Oh yes, $q$ is non-degenerate. The representations $\Phi_1,\Phi_2$ under consideration are required to be non-zero, irreducible and finite-dimensional. – Robert Rauch Sep 10 '12 at 15:48
Put all your requirements in the question, please. – rschwieb Sep 10 '12 at 18:20

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

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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...)

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What you write is correct but doesn't answer my question, i guess. Regarding your concerns about representations opposed to modules: whenever you have an associative unital $k$-algebra $A$, then the category of $k$-linear representations of $A$ is equivalent to the category of unital left $A$-modules. Under this equvalence, irreducible corresponds to simple. However, if you, for example, want to classify complex representations of a real algebra up to real isomorphims, things become complicated from the module point of view. – Robert Rauch Sep 10 '12 at 16:49
@RobertRauch Well again, if you want to use real representations only, you need to mention that in your question. From what is currently written, one can only assume you are interested in $\mathbb{C}$ modules. This is mainly because you mentioned complex representations! If you indeed are only interested in complex representations, then what remains to be answered? – rschwieb Sep 10 '12 at 18:19
What I just wrote about real representations in my comment was to illustrate, that there are in fact some traps in identifying representations of an algebra $A$ with modules over $A$, considered as ring. Back to the original question: As you have mentioned, the fact that there are exactly two irreducible (non-zero, complex, finite-dimensional) representations of $Cl(V)$ is a consequence of $Cl(V)\cong End(S)\oplus End(S)$. What I asked above is a different thing: we want to know, how one can decide whether two given irreducible rep. $\Phi_1,\Phi_2$ of $Cl(V)$ lie in the same isomrphism class. – Robert Rauch Sep 10 '12 at 21:07
@RobertRauch Yes, you did write some comments, but they are about real representations which continue to be irrelevant to your question. From what have told me, you agree this algebra has exactly two isoclasses of simple (=irreducible) modules as a complex algebra, which you have also told me correspond exactly to the irreducible complex representations of this algebra. I must be missing something, because to me it looks like the question is answered. :( – rschwieb Sep 11 '12 at 1:30
As I already said, my comments on real representations were only for your pleasure, as you mentioned that you feel unsure about the identifications of representations with modules. And Yes, these comments are "irrelevant" for the original question (sorry for trying to help you with your difficulties), which is now formulated perfectly clear in the original post: given two concrete irreducible (complex, finite-dimensional, nonzero) reps of $Cl(V)$, how can i decide if they are isomorphic?? – Robert Rauch Sep 11 '12 at 11:49