Basically I'm interested in operator algebras such as $C^*$ or von Neumann algebras. However I decided to learn a bit about noncommutative geometry (in particular spectral triples). Before doing this it would be nice to learn something about commutative case, namely the construction of the spectral triple associated with a spin manifold. First step involved in the entire construction is the definition of Clifford algebra $Cl(V,g)$ where $V$ is real vector space and $g$ is a bilinear form on $V$ (let us assume that the bilinear form is positive defined and nondegenerate). On the complexification of $Cl(V,g)$, to be denoted by $\mathbb{C}l(V)$ one can define a trace $\tau$ and then a scalar product: $\langle a,b \rangle:=\tau(a^*b)$: the star operation is defined by $v^*=-v$ for $v \in V$ and then extended to be antilinear antyhomomorphism. With respect to this scalar product $A:=\mathbb{C}l(V)$ becomes a Hilbert space. This norm cannot make $A$ a $C^*$-algebra. However I've read in some notes that $A$ indeed can be turned into a $C^*$ algebra. So my questions is the following: is there a canonical norm on $A$ in which $A$ turns out to be a $C^*$-algebra?
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$\begingroup$ When you say you've "read in some notes" -- can you give us the precise statement that you saw? $\endgroup$– Yemon ChoiCommented Jul 14, 2014 at 22:56
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$\begingroup$ Yes, of course: author defines the mapping from complexified Clifford algebra $\mathbb{C}l(V)$ (here $dim V=n$ and $n=2m$ or $n=2m+1$) to $M_{2^m}$ in the even case and in the odd case to $M_{2^m} \oplus M_{2^m}$. This mapping is defined on generators $e_i$ (the basis of $V$) by sending them to $E_i$ where $E_i$ are certain matrices constructed before. Then author claims that this mapping is $C^*$ isomorphism, but nothing was said about $C^*$ structure on $\mathbb{C}l(V)$. $\endgroup$– truebaranCommented Jul 14, 2014 at 23:12
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$\begingroup$ Well, the map you mention is an isomorphism of $\ast$-algebras, and matrix algebras (and direct sums of matrix algebras) have a unique C$^\ast$-norm (the operator norm). Perhaps there is some way to interpret this norm intrinsically on the Clifford algebra; perhaps consider the operator norm of the Clifford algebra acting on itself by left multiplication. $\endgroup$– MTSCommented Jul 15, 2014 at 0:09
1 Answer
There is the following description of the $C^{\ast}$-algebra structure on the CLifford algebra $Cl^{n,n}$. Since any vector space with inner product can be embedded into $R^{n,n}$, this answers your question for any (finite-dimensional) Clifford algebra. Recall that $Cl^{n,n}$ is generated by mutually anticommuting generators $a_i,\ldots, a_n, b_1, \ldots ,b_n$ with $a_{i}^{2}=-1$ and $b_{i}^{2}=1$. Now $Cl^{n,n}$ acts on $\Lambda^{\ast} R^n$ by
$$a_i \mapsto e_i \wedge \_ - \iota_{e_i}; \; b_i \mapsto e_i \wedge \_ - \iota_{e_i}.$$
It is not hard to see that the induced map $\gamma:Cl^{n,n} \to End (\Lambda^{\ast} R^n)$ is an isomorphism. Moreover, when $\ast$ denotes the antihomomorphism of $Cl^{n,n}$ you described in your question, this representation of the Clifford algebra is a $\ast$-homomorphism.
Define the norm on the Clifford algebra so that $\gamma$ is an isometry.
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$\begingroup$ Thank You for your answer: could You explain, how do You understand $Cl^{n,n}$? Since, as I understood, You claim that this is $C^*$-algebra but $C^*$-algebras are complex: but going into complex numbers all possible choices of signature leads to the same $\mathbb{C}l(V)$. Could You also explain how do You understand $e_i \wedge_{-} - \iota_{e_i}$? And finally, which norm do You take on $End(\Lambda R^n)$? Standard matrix norm with respect to euclidean norm on exterior algebra? $\endgroup$ Commented Jul 17, 2014 at 22:56