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.

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    $\begingroup$ I think the answer is no unless either $A$ or $B$ is one-dimensional, by looking at the diagonal copies of $D=\ell_\infty^2$ inside $A$ and $B$. The point is that the natural map $D\hat\otimes_\gamma D \to D \otimes_{\rm min} D$ is not isometric. If I have time later I'll try to check this and flesh out the details $\endgroup$ – Yemon Choi Oct 8 '18 at 18:59

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$\dots$.

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    $\begingroup$ I saw your answer while typing essentially the same thing -- the norm of the $2 \times 2$ Hadamard matrix is 2 as a bilinear form on $\ell^\infty(2) \times \ell^\infty(2)$ (and hence in the dual space of $\ell^\infty(2) \hat{\otimes} \ell^\infty(2)$) but 4 as an element of $\ell^\infty(2 \times 2)^*$. $\endgroup$ – Robert Furber Oct 8 '18 at 20:25

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