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While reading about the structure of Clifford algebra, there were two facts listed as bullet points about the center of Clifford algebra based on the parity of the dimension of the underlying vector space that I'm now curious about.

Say you have a finite dimensional vector space $V$ over a field $K$, such that $\operatorname{char} K\neq 2$, and $G$ is a symmetric bilinear form. Let $\mathrm{Cl}_G(V)$ denote the corresponding Clifford algebra.

Apparently, if $\dim V$ is even, then the center of $\mathrm{Cl}_G(V)$ coincides with $K$. However, if $\dim V$ is odd, then the center is actually a $2$-dimensional vector space over $K$.

These seem like good interesting facts to know, but I couldn't find an authoritative reference containing a proof of either of them. Does anybody here have a nice proof of either one/both I could read over? I'd appreciate it very much, thanks.

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The parity issue is discussed in detail in E. Meinrenken's Notes math.toronto.edu/mein/teaching/clif_main.pdf . See Proposition 2.6 in particular, which gives the result over $\mathbb{C}$. Another good reference is Chevalley's book. –  Jonathan Apr 8 '12 at 2:48
Thanks for the helpful notes @Jonathan! –  Jacqueline Pauwels Apr 9 '12 at 21:16
The "Apparently" and "However" statements hold only if the form $G$ is nondegenerate. –  darij grinberg Apr 11 at 1:19

1 Answer 1

up vote 7 down vote accepted

If $e_1, e_2, \ldots, e_n$ are an orthogonal basis in $V$, then the relations $$e_i e_j = - e_j e_i$$ hold for $i \ne j$ in the corresponding Clifford algebra. From this, we can derive the relation $$e_A e_B = e_B e_A (-1)^{|A| \cdot |B| - |A \cap B|}$$ where $A$ and $B$ are subsets of $\{1,2,\ldots,n\}$ and $e_A = \prod_{i \in A} e_i$, the product taken in increasing order of indices. (To see this, note that every time you move an "element" $e_i$ of $e_B$ "past" all of $e_A$ you introduce a minus sign for each element of $A$ if $i \not \in A$, but one less minus sign than elements of $A$, if $i \in A$.)

From this, we can prove that if $n$ is even, the only element of the form $e_A$ in the center of the Clifford algebra is 1, and if $n$ is odd, we also have $e_1 e_2 \ldots e_n$. (To rule out all other elements, let $A$ be a nonempty proper subset of indices; let $B = \{i, j\}$ where $i \in A$, $j \not \in A$, then the above formula shows that $e_A e_B = -e_B e_A$. Then show that $e_1 e_2 \ldots e_n$ works precisely when $n$ is odd.)

Then, it's straightforward to consider the case of a general element $\sum c_A e_A$, since multiplying such a sum by a fixed $e_B$ acts "independently" on each $e_A$.

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Thanks Ted. I've been reading over this, but I don't quite follow the second paragraph addressing the cases where $n$ is even and odd. Do you mind explaining that paragraph in a bit more detail if you have the time? Thank you. –  Jacqueline Pauwels Apr 9 '12 at 21:26
If $A$ is a nonempty proper subset of $\{1,2,\ldots,n\}$, and $B=\{i,j\}$ where $i \in A, j \not \in A$, we have $e_A e_B = -e_B e_A$. This shows that $e_A$ doesn't commute with $e_B$; therefore, $e_A$ is not central. So the only possible elements of the form $e_A$ in the center are 1 and $e_1 e_2 \ldots e_n$. Of course 1 is always central, and the displayed formula above relating $e_A e_B$ with $e_B e_A$ shows that $e_1 e_2 \ldots e_n$ is central precisely when $n$ is odd. –  Ted Apr 10 '12 at 5:57
Thanks again Ted! –  Jacqueline Pauwels Apr 14 '12 at 5:34
The statements beginning with "From this" hold only when the form $G$ is assumed to be nondegenerate. In the general case (when the form $G$ is allowed to be singular), the center of the Clifford algebra is spanned by all $e_A$ such that either ($A = \left\{1,2,...,n\right\}$ and either $G = 0$ or $n$ is odd) or ($\left|A\right|$ is even and $e_i$ is singular for every $i \in A$). –  darij grinberg Apr 11 at 1:18
And in a similar vein, it can be shown that the supercenter of the Clifford algebra (i.e., the set of all elements whose supercommutator with everything is $0$) is spanned by all $e_A$ such that $e_i$ is singular for every $i\in A$. –  darij grinberg Apr 11 at 1:23

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