# Ultrapowers of operators

Can we prove that for each infinite dimensional Banach space $X$ and any free ultrafilter (possibly over uncountable set of indices) $\mathcal{U}$ the obvious embedding $$({\mathcal{L}(X)})_{\mathcal{U}}\to \mathcal{L} (X_U )$$

is not surjective? Even when $X$ is superreflexive? $X_{\mathcal{U}}$ stands for the Banach space ultrapower of $X$ along $\mathcal{U}$.

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As Andreas suggests, I shall fix $\mathcal U$ to be countably-incomplete. In fact, wlog, $\mathcal U$ will be over $\mathbb N$. If $X$ is not super-reflexive, then you don't even get all the rank-one operators. We know that $(X)_{\mathcal U}^* = (X^*)_{\mathcal U}$ if and only if $X$ is super-reflexive, so there is $\lambda \in (X)_{\mathcal U}^* \setminus (X^*)_{\mathcal U}$. Choose $y=(y_n)\in (X)_{\mathcal U}$. Let $T(x) = \lambda(x)y$ so $T$ is a rank-one map on $(X)_{\mathcal U}$. Suppose $T=(T_n)$. For each $n$ pick $\mu_n\in X^*$ with $\|\mu_n\|\leq 1$ and with $\lim_n \mu_n(y_n)=\lim_n \|y_n\|$ (limits over $\mathcal U$ of course). Set $\mu=(\mu_n)$. Then $$\mu(T(x)) = \lambda(x) \mu(y) = \lambda(x) = \mu((T_n)(x)) = \lim_n \mu_n(T_n(x_n)),$$ which holds for all $x$, so $\lambda = (\mu_n\circ T_n)\in (X^*)_{\mathcal U}$, contradiction.

If $X$ is super-reflexive, then I want to use some "co-ordinate" structure, so I need to think some more...

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Filter the finite subsets of an orthonormal basis for a big Hilbert space to get an ultrafilter and use Dvoretzky's theorem, Matt. Doesn't that do it? –  Bill Johnson Nov 16 '12 at 11:19
Although I agree that ultrafilters over $\mathbb{N}$ is a central case, why can you assume without loss that your ultrafilter is over $\mathbb{N}$? There are (nonisomorphic) countably incomplete ultrafilters on sets of any infinite size. –  Joel David Hamkins Nov 16 '12 at 11:49
Sorry-- sloppy. If $\mathcal U$ is countably incomplete, then let $(A_n)$ be a decreasing sequence of elements of $\mathcal U$ with $\bigcap_n A_n=\emptyset$. Then for example you'd choose $(\mu_i)$ as follows: if $i\in A_n\setminus A_{n+1}$ pick $\mu_i\in X^*$ with $\mu_i(y_i)\geq0$ and with $\mu_i(y_i)> \|y_i\|-1/n$; otherwise choose $\mu_i$ arbitrary (as this won't affect the equivalence class $(\mu_i)$ in $(X^*)_{\mathcal U}$. Hopefully you now see how to make the rest of the argument run... –  Matthew Daws Nov 16 '12 at 12:35
@Bill: Not sure I see this. If $(X)_{\mathcal U}$ is reflexive and has the approximation property then the image of $(\mathcal B(X))_{\mathcal U}$ will contain all compacts. So you need to start building general operators in $\mathcal B((X)_{\mathcal U})$. If e.g. $X$ is Hilbert then I can do this, but I'm don't see how just having a copy of a Hilbert space in $(X)_{\mathcal U}$ is going to help. –  Matthew Daws Nov 16 '12 at 12:38
I wrote quickly, Matt, as I was off to catch a plane. When $X$ is superreflexive and has the approximation property you get uniformly complemented $\ell_2^n$-s in $X$ by Pisier's theorem and thus large complemented Hilbert spaces in ultrapowers. This should reduce the problem to the Hilbert space case by taking another ultrapower, but I am not sure about that. If it works, the remaining case would be superreflexive spaces all of whose complemented subspaces fail the approximation property. I don't think anyone has proved that there are such spaces. –  Bill Johnson Nov 16 '12 at 13:43

If $U$ is a countably complete ultrafilter, then the operation of ultrapower by $U$ does nothing to small (i.e., smaller than the first measurable cardinal) spaces. So to prove non-surjectivity of the map in the question, we wold have to prove in particular that there are no measurable cardinals. We certainly don't know how to do that. (The most likely reason for our inability to do that would be that measurable cardinals are consistent.)

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Thank you. Let me restate my question then. Is there a Banach space $E$ and an ultrafilter $U$ over some cardinal $\lambda>\mathfrak{c}$ such that the above holds and $|E_U|\geqslant \lambda$? –  Slavoj Žižek Nov 15 '12 at 13:48
Andreas, is what you say true for Banach-space theoretic ultrapowers as well as set-theoretic ones? –  Yemon Choi Nov 15 '12 at 20:19
Yemon, if the ultrafilter is countably complete then the two corresponding equivalence relations coincide (since $\mathbb{R}$ is first countable). My guess is that a lot of the Banach-space ultrapower theory is done with the extra assumption that the ultrafilter is countably incomplete. –  Ramiro de la Vega Nov 15 '12 at 22:20
@Yemon: I believe the Banach space ultrapower can be obtained from the set-theoretic ultrapower by first discarding any elements whose norms are infinitely large and then identifying any two elements whose difference has infinitesimal norm. In particular, if you use a countably complete ultrafilter, then the set-theoretic ultrapower of a small structure is the original structure, there are no elements of infinite norm to discard, and the infinitesimal-distance equivalence relation is just equality. So the Banach space ultrapower is also just the original structure. –  Andreas Blass Nov 15 '12 at 23:36
Just out of curiosity, does the following hold for countably complete ultrafilters: $(X\oplus Y)_U \isom X_U \oplus Y_U$, $X,Y$ Banach spaces? –  Bojan Kwitek Dec 29 '12 at 13:20