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Let $X$ be a metrizable compact topological space, let $\mathcal U$ be an ultrafilter, and denote by $X^{\mathcal U}$ the ultracopower of $X$ with respect to $\mathcal U$. As a C$^*$-algebraist, I prefer to define $X^{\mathcal U}$ as the compact Hausdorff space satisfying \begin{align*} C(X^{\mathcal U},\mathbb C) &\cong \prod_{\mathcal U} C(X,\mathbb C) \\ &:= \ell^\infty(\mathbb N, C(X, \mathbb C))/\{(f_n)_n \mid \lim_{n\to\mathcal U} \|f_n\|_{\infty} = 0\}. \end{align*} It can alternatively be described, directly, as ultrafilters in $\mathbb N \times X$ which respect $\mathcal U$ (see Bankston's "Reduced coproducts of compact Hausdorff spaces" for a more precise version of this).

My question is: if $Y$ is another metrizable compact topological space, is it the case that $(X \times Y)^{\mathcal U} \cong X^{\mathcal U} \times Y^{\mathcal U}$? Equivalently, in terms of C$^*$-algebras, is it the case that $$ \prod_{\mathcal U} C(X \times Y, \mathbb C) \cong \left(\prod_{\mathcal U}C(X, \mathbb C)\right) \otimes \left(\prod_{\mathcal U} C(Y, \mathbb C)\right)? $$

My intuition, looking at the C$^*$-algebra version, is that this isn't true, although I've found it difficult to prove this.

(Note: we could do an ultracoproduct with respect to a sequence of spaces instead, but this is notationally messier and, well, let's first see the answer to the question I stated before going there.)

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  • $\begingroup$ Anything in the Ge-Hadwin paper? $\endgroup$ – Yemon Choi Aug 13 '14 at 8:36
  • $\begingroup$ You might want to clarify for the non-FA readers that your direct product is (I assume) the one in the Banach space category, i.e. bounded sequences and not all sequences, and that your tensor product is a completion of the algebraic tensor product :) $\endgroup$ – Yemon Choi Aug 13 '14 at 8:38
  • $\begingroup$ Yemon: No, nothing in this direction seems to be addressed in the Ge-Hadwin paper. And thank you for your second comment; changes have been made. $\endgroup$ – Aaron Tikuisis Aug 13 '14 at 10:46
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Aaron, the answer is big no if $X$ and $Y$ are infinite (no metrasibility assumptions required to show that the answer is always negative).

Indeed, $C((X\times Y)^{\mathcal{U}})$ is a Grothendieck space as an ultraproduct of $\mathscr{L}_\infty$-spaces over a countably incomplete ultrafilter. (See Propositon 3.2 here.) On the other hand $$C(X^{\mathcal{U}}\times Y^{\mathcal{U}})\cong C(X^{\mathcal{U}})\otimes C(Y^{\mathcal{U}})\cong C(X^{\mathcal{U}}, C(Y^{\mathcal{U}})) $$ contains a complemented (Banach-space) copy of $c_0$ by the main result of

P. Cembranos, $C(K, E)$ contains a complemented copy of $c_0$, Proc. Amer. Math. Soc. 91 (1984), 556-558

so it is manifestly not Grothendieck. The conclusion is that $C((X\times Y)^{\mathcal{U}})$ and $C(X^{\mathcal{U}}\times X^{\mathcal{U}})$ are not even Banach-space isomorphic.

Remark. One can tweak Cembranos' proof to obtain the following result:

If $A$ is a C*-algebra, $E$ is a Banach space (both $A$ and $E$ are infinite dimensional) and $\gamma$ is any Banach-space tensor cross-norm on $A\odot E$, then $A\otimes_\gamma E$ is not a Grothendieck space.

As a corollary to this one obtains a negative answer to Simon Wasserman's quesiton whether the Calkin algebra can be written as a non-trivial tensor product of two C*-algebras. I might write this down at some point.

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Here's another way to see that the answer is no for infinite $X$ and $Y$, from a more model-theoretic point of view. As an ultrapower over a countably incomplete ultrafilter, $C(X\times Y)^{\mathcal{U}}$ is countably saturated. It follows from this (actually, from the weaker property of countable degree-$1$ saturation) that $C(X \times Y)^{\mathcal{U}}$ is $SAW^*$ (see Farah and Hart, "Countable saturation of corona algebras"), and hence cannot be the tensor product of two infinite-dimensional algebras (see Saeed Ghasemi, "$SAW^*$-algebras are essentially non-factorizable).

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  • $\begingroup$ Just to complement your answer, my remark actually covers Ghasemi's result as SAW*-algebras are Grothendieck. However, one can easily prove that $K=(X\times Y)^U$ is an F-space in which case $C(K)$ is a Grothendieck space; in other words no C*-machinery is required here. $\endgroup$ – Tomasz Kania Sep 1 '14 at 21:30

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