Are Clifford groups very *non-commutative*?

Question

Clifford groups seem to be very non-commutative by the relation \begin{equation} \gamma_{i}\gamma_{j}=-\gamma_{j}\gamma_{i}. \end{equation} But is it really so? Can we put this degree of non-commutative in a precise form?

I tried the standard, namely, the commutator\begin{equation} [Cl(n),Cl(n)]=\{\pm1\}, \end{equation} where $Cl(n)$ is the Clifford group of degree $n$. From this it seems Clifford groups are not really that bad for the commutator is rather small.

However the centre $Z(Cl(n))$ is $\{\pm 1\}$ for even $n$ and $\{\pm 1, \pm\gamma_{1}\gamma_{2}\cdots\gamma_{n}\}$ for odd $n$, which seems also small.

So my question consists of two parts: 1. Are Clifford groups very non-commutative? and 2. Are there some convenient/ strong indicators of the degree of non-commutativeness?

I am actually more interested in the second, and have some rather naive thoughts: Maybe the degrees of irreducible representation of $G$, $\{\operatorname{degree}(\alpha):\alpha\in\hat{G}\}$, would contain some hints. Firstly, $\#\{\operatorname{degree}(\alpha)=1\}=\#G/[G,G]$;secondly, $\operatorname{degree}_{\alpha}|\#G/Z(G)$; last but not least, non-commutative can be thought of as a high dimensional phenomena since 1-by-1 matrices are commutative in multiplication.

Again these are just naive thoughts and I would like to hear your opinion on the degree of non-commutativeness. Thanks very much!

(Well, actually this is what truly motivates this question: assume $\#\{\operatorname{degree}(\alpha)=1\}$ and $max_{\alpha}\operatorname{degree}(\alpha)$ or $gcd(\operatorname{degree}(\alpha))$ give some information on the commutativity, then what do things like $\#\{\operatorname{degree}(\alpha)=n\}$ tell us? Higher dimensional commutativity?)

Answer 1

You can always find $g$ such that $ab=gba$. (Namely, $g=aba^{-1}b^{-1}$, the commutator) When $g$ is the identity $1$, you have commutativity, and when your group has a fellow like $-1$, you get anticommutativity.

Really, then, it looks like you would like to measure how "big" $g$ is, where $1$ is considered small. Even $-1$ might not really be considered "small", even if it is in the "wrong direction(?)". Potentially there is some valuation which says when things are "far from 1". Really, commutativity an anticommutativity are both not very pathological.

Answer 2

Some random thoughts: The lack of normal subgroups can be considered a measure of noncommutativity. If you divide $Cl(n)$ by its center, you get $O(n)$. The connected component is the group $SO(n)$ which is simple (has no non-trivial connected normal sungroups) for $n\ne 4$. There is a very nice criteria for when a Lie group is semi-simple (direct product of simple groups). This is the non-degeneracy of the Killing form, which is a bilinear form on the Lie algebra. On the other extreme, the Killing form of an abelian Lie group is zero. So you can think of the number of non-degenerate directions of the Killing form as a measure of noncommutativity.

Answer 3

The center is a measure of commutativity or its lack, namely a commutative group G has Z(G) = G and if Z(G) is a proper subgroup of G , any x not in Z(G) fails to commute with at least one element in G. The commutator is a tougher subgroup to use although [G,G] = (1) iff G is commutative.