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?
1 Answer
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 nonunital 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 preC$^*$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.

$\begingroup$ ''of which there are examples where they are not ∗isomorphic''  I'd appreciate a reference or an example where this is obvious. $\endgroup$ May 7, 2017 at 13:31

3$\begingroup$ 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 productit 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. $\endgroup$ May 7, 2017 at 13:47