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.)



Ultracoproducts and Cartesian products
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](http://en.wikipedia.org/wiki/Grothendieck_space) as an ultraproduct of $\mathscr{L}\_\infty$-spaces over a countably incomplete ultrafilter. (See Propositon 3.2 [here](http://arxiv.org/pdf/1307.4387v1.pdf).) 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 $c\_0$](http://www.ams.org/journals/proc/1984-091-04/S0002-9939-1984-0746089-2/S0002-9939-1984-0746089-2.pdf), *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](https://academic.oup.com/qjmath/article-abstract/66/4/1063/2362453?redirectedFrom=fulltext) 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.


