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).
I asked previously whether $X^{\mathcal U} \times Y^{\mathcal U} \cong (X \times Y)^{\mathcal U}$, and Tomek Kania showed that the answer is no.
Let us now define a unital $C(X)$-algebra. A unital C*-algebra $B$ is a $C(X)$-algebra if it is equipped with a unital *-homomorphism $\theta$ from $C(X)$ to the centre of $B$. This allows one to view $B$ as an algebra of sections of a bundle over $X$; the section at $x\in X$ is defined to be the quotient of $B$ by the ideal generated by $\theta(C(X\setminus \{x\}))$.
There is a canonical map $C(X)_\omega \to C(X\times Y)_\omega$, so that the codomain is a $C(X^{\omega})$-algebra. The answer to my previous question shows that it is not the trivial $C(X^\omega)$-algebra with fibres $C(Y^\omega)$. My new question is: what are the fibres?
I can answer this question for a dense set of points, namely those elements of $X^\omega$ that correspond to a character on $C(X)_\omega$ of the form \begin{equation} (f_n)_n \mapsto \lim_{n \to \omega} f_n(x_n), \end{equation} for a sequence $(x_n)_n$ in $X$. The fibre at such a point is $C(Y)_\omega$. There is nothing subtle about proving this.
