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 $\mathcal{U}$.
 A: 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...
A: 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.)
