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 Post subject: Clifford algebra questionPosted: Sat, 18 Feb 2012 22:31:41 UTC

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

where is the unit element, as well as

where , and the mutiplication rule is

where is the Kronecker Delta symbol

I was able to show that for n even, only commutes with all the other bases
and for n odd, only and 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.

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 Post subject: Re: Clifford algebra questionPosted: Sat, 18 Feb 2012 22:40:06 UTC
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Joined: Wed, 30 Mar 2005 04:25:14 UTC
Posts: 12102
Location: Austin, TX
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

where is the unit element, as well as

where , and the mutiplication rule is

where is the Kronecker Delta symbol

I was able to show that for n even, only commutes with all the other bases
and for n odd, only and 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.

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 Post subject: Re: Clifford algebra questionPosted: Sun, 19 Feb 2012 04:34:51 UTC

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

where is the unit element, as well as

where , and the mutiplication rule is

where is the Kronecker Delta symbol

I was able to show that for n even, only commutes with all the other bases
and for n odd, only and 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=, for arbitrary , let x=x1+x2, where x1 consists of bases that commutes with , and x2 anti-commutes, so that x= x implies x2=0, but how can I conclude x2=0, since this is not a division algebra? Or is it?

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