# Ultraproduct of n-dimensional Banach spaces and algebras

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 $\{F_i\}$.

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A good reference for this sort of stuff (which would also have answered your question) is Heinrich's article, now available for free: gdz.sub.uni-goettingen.de/en/dms/load/img/… –  Matthew Daws Jul 6 '11 at 12:41
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 $F_i$.
The condition on your second paragraph is always true, as a k-dim Banach space is always at most $\sqrt{k}$ distant from $\ell^2_k$ in the Banach-Mazur distance-- see en.wikipedia.org/wiki/Banach%E2%80%93Mazur_compactum –  Matthew Daws Jul 6 '11 at 12:37