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Let $A$ and $B$ denote $C^{\ast}$-algebras. Let $\lVert\cdot\rVert_h$ and $\lVert\cdot\rVert_{\text {max}}$ denote the Haagerup norm and max $C^*$-norms on $ A \otimes B$, respectively. I am looking for a reference/proof for the following result:

$\lVert\cdot\rVert_{\text{max}} \leq \lVert\cdot\rVert_h$.

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  • $\begingroup$ @NikWeaver: Maximum cross norm is Banach space projective, no? $\endgroup$
    – Math Lover
    Commented Feb 2, 2022 at 5:51
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    $\begingroup$ I was a little confused reading the comments and the question. Does $\max$ refer to the max $C^*$-norm? Then indeed that's not the greatest cross norm (which is the Banach space projective tensor norm). $\endgroup$ Commented Feb 2, 2022 at 9:23
  • $\begingroup$ @MatthewDaws: Indeed! Edited, thanks. $\endgroup$
    – Math Lover
    Commented Feb 2, 2022 at 10:09

1 Answer 1

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Let $v \in A \otimes B$, where $A\otimes B$ is the algebraic tensor product.

The Haagerup norm of $v$ is the infimum of the expressions $\sqrt{\|\sum_{i=1}^{r} x_{i} x_{i}^{\ast}\|}\cdot \sqrt{\|\sum_{i=1}^{r} y_{i}^{\ast} y_{i} \|}$, taken over all decompositions $v=\sum_{i=1}^{r} x_{i} \otimes y_{i}$ ($r$ can be arbitrarily large).

On the other hand, the maximal norm is the supremum of the norm $\|\sum_{i=1}^{r} \pi(x_i) \sigma(y_{i})\|_{B(H)}$ over all pairs of representations $\pi: A \to B(H)$, $\sigma: B \to B(H)$ with commuting ranges. Note that $\sum_{i=1}^{r} \pi(x_i) \sigma(y_{i})$ can be interpreted as the product of the row $(\pi(x_1),\dots, \pi(x_r))$ and the column $(\sigma(y_1),\dots, \sigma(y_r))^{T}$, so we get can estimate $\|\sum_{i=1}^{r} \pi(x_i) \sigma(y_{i})\|_{B(H)}$ by

$ \sqrt{\|\sum_{i=1}^{r} \pi(x_{i} x_{i}^{\ast})\|_{B(H)}} \cdot \sqrt{\|\sum_{i=1}^{r} \sigma(y_{i}^{\ast} y_{i})\|_{B(H)}}. $

As representations of $C^{\ast}$-algebras are contractive, this is clearly not greater than the quantity defining the Haagerup norm.

It might be useful to note that we did not even have to use the fact the representations $\pi$ and $\sigma$ had commuting ranges.

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    $\begingroup$ Nice answer. I guess your final comment is related to the connection between the Haagerup tensor product and free products of $C^*$-algebras, as explored e.g. in Pisier's book. $\endgroup$ Commented Feb 2, 2022 at 9:28
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    $\begingroup$ @MatthewDaws, that's correct, but I really didn't want to elaborate on that, in order to keep the answer cleaner. Moreover, I am definitely underqualified to say something more about this. :) $\endgroup$ Commented Feb 2, 2022 at 9:52
  • $\begingroup$ @MateuszWasilewski: Thank you :) $\endgroup$
    – Math Lover
    Commented Feb 2, 2022 at 10:43

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