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In principle, associative *-algebras can be equipped with multiple norms. Can this be done in such a way that (after closure in the respective norm) they are turned into $C^*$-algebras in multiple, not essentially equivalent ways?

Arnold Neumaier
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1 Answers1

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There is at most one norm making an associative $*$-algebra $\mathcal A$ into a C$^*$-algebra, that is $\mathcal A$ is closed with respect to this C$^*$-norm. This is because for all $a\in \mathcal A$, $$ \|a\|^2 = \|a^*a\| = \textrm{spectral radius of}\: a^*a = \textrm{sup}\{|\mu| : a^*a - \mu I \notin \mathcal A^{-1}\},$$ that is, the norm of any element is determined by an algebraic property. The above is stated for unital $\mathcal A$ but it implies that the same holds true for non-unital as well.

However, if we only require that $\mathcal A$ is not norm closed with respect to this new norm, making $\mathcal A$ into a pre-C$^*$-algebra then there are lots of potential C$^*$-norms. For instance if $\mathcal A$ and $\mathcal B$ are C$^*$-algebras then the algebraic tensor product $\mathcal A\odot \mathcal B$ is a $*$-algebra with many potential norm structures. The most notable are $\mathcal A \otimes_{min} \mathcal B$ and $\mathcal A \otimes_{max} \mathcal B$, the min and max tensor products, of which there are examples where they are not $*$-isomorphic.

Chris Ramsey
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  • ''of which there are examples where they are not ∗-isomorphic'' - I'd appreciate a reference or an example where this is obvious. – Arnold Neumaier May 07 '17 at 13:31
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    Examples of not isomorphic: If $A$ and $B$ are simple C* algebras, then $A\otimes_{\mathrm {min}} B$ is simple. Take $A$ and $B$ to be type II$_1$ factors, and take their maximal C* tensor product---it isn't simple. Another example occurs with free groups, as there is a difference between the regular representation and the universal representation. These should be enough to get you started. – David Handelman May 07 '17 at 13:47