What is the center of a Clifford algebra

Question

I have for short interess at Clifford algebras, and I've read on the german wikipedia that the center, that the set $$\{x\in CL| x\cdot y= y\cdot x \;\forall y\in CL\}$$ was $$\mathbb{R}\cdot 1_{CL}$$ But I couldn't figure out why. Does someone have an idea? Thanks in advance!

Answer 1

$ \newcommand\Cl{\mathrm{Cl}} \newcommand\Ext{{\bigwedge}} \newcommand\mult\mathbf \newcommand\K{\mathbb K} \newcommand\rad{\mathrm{rad}} $

Your statement is only correct for even-dimensional vector spaces together with a non-degenerate quadratic form. Below I lay out the correct statement for characteristic $\not=2$.

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The Clifford algebra $\Cl(V, Q)$ is constructed over a vector space $V$ with quadratic form $Q$; I will assume the field of scalars $\K$ of $V$ has characteristic $\not=2$ since things change considerably in that case. Assume $V$ has dimension $n$.

There is a canonical vector space isomorphism $\Ext V \cong \Cl(V, Q)$, where $\Ext V$ is the exterior algebra (which I will note is exactly the Clifford algebra with $Q = 0$). Perhaps the most straightforward way to see this is to consider any orthogonal basis $\{e_i\}$ for $V$. We can write $$ \Cl(V, Q) = \K + V + V^2 + V^3 + \cdots + V^n, $$ where a power $V^k$ should be interpreted as the set of all linear combinations of products of $k$ elements of $V$, and this gives a basis for $\Cl(V, Q)$ generated by $\{e_i\}$ using the fact that $e_ie_j = -e_je_i$ when $i\not= j$. Now let $\mult i, \mult j$ be any multi-indices, i.e. tuples of any length (including zero) with distinct positive integer entries; we define $e_{()} = 1$ and $e_{\mult i} = e_{i_1}e_{i_2}\cdots e_{i_k}$ if $\mult i = (i_1, i_2, \dotsc, i_k)$, and we define $\mult i\cap \mult j$ as the intersection of the associated sets. Then we define $$ e_{\mult i}\wedge e_{\mult j} = \begin{cases} e_{\mult i}e_{\mult j} &\text{if }\mult i\cap\mult j = \varnothing, \\ 0 &\text{otherwise}, \end{cases} $$ which extends by linearity to the entirety of $\Cl(V, Q)$. (Of course, we would have to show that this construction is independent of the orthogonal basis chosen. There are nicer basis-free ways of constructing this isomorphism, but that would be too much of a digression.)

Now let $\Cl^k(V, Q)$ be the image of $\Ext^{\!k} V$ under the isomorphism $\Ext V\cong\Cl(V, Q)$. If $V$ has dimension $n$ and field of scalars $\K$, then the center of $\Cl(V, Q)$ is exactly $$ Z(\Cl(V, Q)) = \begin{cases} \K\oplus\mathfrak R_+ &\text{if }n\text{ is even}, \\ \K\oplus\mathfrak R_+\oplus\Cl^n(V, Q) &\text{if }n\text{ is odd}. \end{cases} \tag{$*$} $$ The set $\K\oplus\mathfrak R_+$ is the even subalgebra of $\Cl(V, Q)$ generated by the radical $V^\perp$ of $V$ under $Q$; that is to say that $$ V^\perp = \{v \in V \;:\; \forall w \in V.\:B(v, w) = 0\}, $$$$ \mathfrak R = \sum_{k=1}^n (V^\perp)^k,\quad \mathfrak R_+ = \mathfrak R\cap\Cl_+(V, Q) = (V^\perp)^2 + (V^\perp)^4 + \cdots, $$ where $B$ is the bilinear form associated with $Q$ and $\Cl_+(V, Q)$ is the even subalgebra of $\Cl(V, Q)$: $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)), $$$$ \Cl_+(V, Q) = \K + V^2 + V^4 + \cdots = \K \oplus \Cl^2(V) \oplus \Cl^4(V) \oplus\cdots. $$ Note that if $Q$ is non-degenerate then $V^\perp = \{0\} = \mathfrak R$.

The equation ($*$) is straightforward though perhaps tedious to confirm by again considering an orthogonal basis $\{e_i\}$ on $V$, where we then have $$ \Cl(V, Q) = \bigoplus_{k=0}^n\Cl^k(V, Q),\quad \Cl^k(V, Q) = \{\sum_{|\mult i|=k}\alpha_{(\mult i)}e_{\mult i} \;:\; \alpha_{(\mult i)} \in \K\}. $$ $|\mult i|$ is the length of $\mult i$ and $\alpha_{(\mult i)}$ is meant to indicate that there is a distinct variable for each multi-index. For example, when $n = 3$ $$ \Cl^2(V, Q) = \{\alpha e_1e_2 + \beta e_1e_3 + \gamma e_2e_3 \;:\; \alpha,\beta,\gamma \in \K\}. $$ We would also want to make use of the fact that $$ RX = R\wedge X,\quad XR = X\wedge R $$ for any $R \in \mathfrak R$ and $X \in \Cl(V, Q)$.