What is the Center of a Clifford Algebra of Order 2^n?

[sunjin09]

Homework Statement

Show that the center of a Clifford algebra of order 2^n is of order 1 if n is even, and 2 if n is odd

Homework Equations

the center of an algebra is the subalgebra that commutes with all elements
Clifford algebra of 2^n is defined as being spanned by the bases
\gamma_0,\gamma_1,...\gamma_n
where \gamma_0 is the unit element, as well as
\gamma_{\{n_k\}}=\Pi_{n_k}\gamma_{n_k}
where 1\le n_k\le n, and the mutiplication rule is
\gamma_u\gamma_v+\gamma_v\gamma_u=2\delta_{u,v} \gamma_0
where \delta is the Kronecker Delta symbol

The Attempt at a Solution

I was able to show that for n even, only \gamma_0 commutes with all the other bases
and for n odd, only \gamma_0 and \gamma_{\{1,2,...,n\}} commutes with all the other bases, but
how can I know that the center is spanned by the bases that commutes with
all the other bases?

In other words, how do I know that no linear combinations of
non-commuting bases commutes to all bases?

 

[sunjin09]

I finally figured it out. Two things need to be realized:
1.All the basis elements are invertible
2.Any pair of the basis elements either commute with each other or anti-commute with each other;
These may be verified fairly easily from the definition.

Now if x is in the center, x must commute with all the basis elements. For any basis element γ, x may be
written as x=x1+x2, where x1γ=γx1 and x2γ=-γx2; since xγ=γx, we have x2γ=0; since γ is invertible, x2=0;
since x2 which depends on γ equals 0 for all γ, x must contain only components that commutes with all basis,
i.e., γ0 and possibly γ{1,2,...,n}, qed.

fairly straightforward, it feels good to know the answer. Anyone has a simpler proof?