Hi, I am interested in the following question:


Fix $n$.
Let $M\_n$ be matrix algebra over the field of complex numbers with normalized trace $tr\_n$. Let $M\_n^{\omega}$ be an ultrapover of $M\_n$, namely we consider the algebra of all bounded (in norm) sequences in $M\_n$,say $l\_{\infty}(M\_n)$, and take a quotient of this space by sequences $(a\_i)\_{i\in\mathbb{N}}$ with $lim\_{\omega} tr\_n(a\_i^\*a\_i)=0$. Is $M\_n^{\omega}$ is finite-dimensional? What is the structure of $M\_n$?


Also, if $F\_i$ is $n$-dimensional Banach space, what is the dimension of the Banach space ultraproduct of $\{F\_i\}$.



Ultraproduct of n-dimensional Banach spaces and algebras
If the space $F$ is $k$-dimensional with basis $b\_1,\ldots b\_k$ then the ultraproduct with respect to the ultrafilter $U$ will also be $k$-dimensional. To see this let $x\_i\in \ell\_\infty(F\_i)$ and let $q\_{i,j}$ be scalars such that $\sum\_{j=1}^kq\_{i,j}b\_j =x\_i $. By the compactness of the ball in finite dimensional spaces the $q\_{i,j}$ converge to $q\_{j}$ with respect to $U$ and so $\sum\_{j=1}^kq\_{j}b\_j = x$ modulo $U$.


The same argument works of different $k$-dimensional Banach spaces as long as there is a uniform bound on the norm of the identity map from one space to the other to allow the compactness argument to be used --- in other words, there is a single ball containing the unit balls of all the $F\_i$.


