Clifford algebra question

sunjin09 · **Joined:** Sat, 18 Feb 2012 22:23:53 UTC **Posts:** 5

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

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

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? Any hints are appreciated.

Shadow · **Posted:** Sat, 18 Feb 2012 22:40:06 UTC

sunjin09 wrote:

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

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

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? Any hints are appreciated.

Just write an arbitrary element out in a basis and assume it commutes with everything, that should lead you directly where you're headed.

sunjin09 · **Joined:** Sat, 18 Feb 2012 22:23:53 UTC **Posts:** 5

Shadow wrote:

sunjin09 wrote:

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

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

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? Any hints are appreciated.

Just write an arbitrary element out in a basis and assume it commutes with everything, that should lead you directly where you're headed.

I realized that any two bases either commute or anti-commute. Suppose x is in center, x= $\sum_{\{n_k\}}a_{\{n_k\}}\gamma_{\{n_k\}}$ , for arbitrary $\gamma_{\{n_i\}}$ , let x=x1+x2, where x1 consists of bases that commutes with $\gamma_{\{n_i\}}$ , and x2 anti-commutes, so that x $\gamma_{\{n_i\}}$ = $\gamma_{\{n_i\}}$ x implies x2 $\gamma_{\{n_i\}}$ =0, but how can I conclude x2=0, since this is not a division algebra? Or is it?

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Clifford algebra question

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