Let $A$ and $B$ be two finite-dimensional C\*-algebras. 
Let $\gamma$ denote the projective Banach space tensor product norm on the algebraic tensor product $A\odot B$, so $\gamma(t)=\inf\{\sum\_{i}\|a\_i\|\|b\_i\|:t=\sum\_i a\_i\otimes b\_i\}$. Then $\gamma(ts)\leq\gamma(t)\gamma(s)$ and $\gamma(t^\*)=\gamma(t)$, but does $\gamma$ also satisfy $\gamma(t^\*t)=\gamma(t)^2$? 


I know that $\gamma$ should be equivalent to the C\*-algebra norm on $A\odot B$, since all norms on finite-dimensional spaces are equivalent, but I would like to have equality, since this would imply that the embedding of finite-dimensional C\*-algebras with subunital completely positive maps in the category of Banach spaces with contractions and with the projective tensor product as a monoidal product is a strong monoidal functor. 



Is the norm of the Banach space projective tensor product of finite-dimensional C*-algebras a C*-norm?
I think there are various reasons from abstract tensor norm theory why this is impossible, but I'll try to give a concrete example. First, a $C^\*$-norm on an algebra is uniquely determined; therefore, if $A=\ell\_\infty(2)=\mathbf C^2$ with the max-norm, then the tensor norm on $A\otimes A$ is the injective tensor norm $\otimes\_\varepsilon$, identifying $A\otimes A$ isometrically with $\mathbf{C}^4$ in the max-norm. (If $A=C(K)$, then $A\otimes\_\varepsilon A = C(K\times K)$.) So we have to argue that $\otimes\_\gamma$ is not $\otimes \_\varepsilon$ in this example. 


For this let us compare the duals; $(\ell\_\infty(2)\otimes\_\gamma \ell\_\infty(2))^\* = L(\ell\_\infty(2), \ell\_1(2)) =: X$ and $(\ell\_\infty(2)\otimes\_\varepsilon \ell\_\infty(2))^\* = (\ell\_\infty(2; \ell\_\infty(2))^\* = \ell\_1(2; \ell\_1(2)) := Y$. Specifically, consider the element of the dual represented by $(\alpha,\beta)\mapsto \alpha (1 , 1) + \beta (1,-1)$. The norm of this element in $Y$ is $\|(1,1)\|\_1 + \|(1,-1)\|\_1 = 4$, but the norm in $X$ is $\max\_{|\alpha|,\beta|\le 1} \|\alpha(1,1)+\beta(1,-1)\|\_1= \max\_{\alpha,\beta} (|\alpha+\beta|+|\alpha-\beta|)<4$. 


So the norms are isometrically different. 


PS: I read Yemon's comment, which practically says the same as mine, only after having typed this$\dots$. 


