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Let $\newcommand{\CCl}{\mathbb{C}l}\CCl_{p,q}$ be the complex Clifford-Algebra associated to the Minkowski space $\mathbb{R}^{p,q}$ of signature $(p,q)$, where we consider $\mathbb{R}^{p,q}$ as a linear subspace of $\CCl_{p,q}$ in the usual way. Moreover, let $\Phi:\CCl_{p,q}\to\operatorname{End}_\mathbb{C} S$ be a Dirac spinor representation, i.e. a finite-dimensional, irreducible, complex representation of $\CCl_{p,q}$.

Question: Is it possible that $\Phi(v)=1_S$ for some $v\in\mathbb{R}^{p,q}$?

For example, if $n=p+q$ is even, the representation theory for $\CCl_{p,q}$ implies that $\Phi$ is an isomorphism of $\mathbb{C}$-algebras, thus $\Phi(v)=1_S=\Phi(1_{\CCl_{p,q}})$ implies $v=1_{\CCl_{p,q}}$ - a contradiction. So the question is only interesting for the odd-dimensional case.

Robert Rauch
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  • I'm more familiar with using clifford algebras without reference to a particular representation. With that in mind, are you asking if it's possible that $vX = Xv = X$ for some arbitrary $X \in \mathbb{Cl}_{p,q}$? – Muphrid Mar 20 '14 at 23:52
  • Well, I really do not see how your question relates to mine. Do you have problems understanding my question? – Robert Rauch Mar 21 '14 at 00:05
  • Again, I'm pretty ignorant of representation theory, so I tried to make sense of things as best I could. If $\Phi(v)$ is the identity, then $vX$ is represented by $\Phi(v) \Phi(X) = \Phi(X)$, so isn't $vX = X$? – Muphrid Mar 21 '14 at 00:27
  • Ok, now I understand what you mean. The point is, that in our case (where $n$ is odd) $\Phi$ is not injective (in fact, the kernel of $\Phi$ is half of $\mathbb{C}l_{p,q}$) so you cannot conclude $vX=X$. – Robert Rauch Mar 21 '14 at 07:12

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Okay, the answer is No, and the reason is quite simple: It can be shown that the Dirac matrices $\gamma_i:=\Phi(e_i)$ are traceless, so the same is true for all $\Phi(v)$ with $v\in\mathbb{R}^{p,q}$. Now, if $\Phi(v)=1$ we can take traces and find $0=\dim S$, a contradiction.

Robert Rauch
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