# Is the norm of the Banach space projective tensor product of finite-dimensional C*-algebras a C*-norm?

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.

• 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 – 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 $$\max_{|\alpha|,\beta|\le 1} \|\alpha(1,1)+\beta(1,-1)\|_1= \max_{\alpha,\beta} (|\alpha+\beta|+|\alpha-\beta|)<4$$.
PS: I read Yemon's comment, which practically says the same as mine, only after having typed this$$\dots$$.
• 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 $\ell^\infty(2 \times 2)^*$. – Robert Furber Oct 8 '18 at 20:25