10
$\begingroup$

Let $\mathscr U$ be a non-principal ultrafilter over the natural numbers. Let $M_{\mathscr U}$ be the ultraproduct of all full matrix algebras $M_n$ along $\mathscr U$. This is a C*-algebra that is not simple as it contains a non-zero proper ideal, for example $\{[(x_n)]\colon \lim_{n, \mathscr U} \|x_n\|_{\rm HS} = 0\}$, where $\|\cdot\|_{\rm HS}$ stands for the Hilbert–Schmidt norm.

  1. Is the cardinality of the set of maximal ideals of $M_{\mathscr U}$ known?
  2. Does $M_{\mathscr U}$ have an ideal of finite-codimension?

I anticipate that for Q1 the answer should be $2^{\mathfrak{c}}$ and for Q2 it should be no but I am somehow stuck.

$\endgroup$
  • 6
    $\begingroup$ Isn't the ideal you mention itself maximal? Because the tracial ultraproduct of matrix algebras is a $II_1$ factor and therefore simple as a C*-algebra (Theorem III.1.7.11 of Blackadar's Operator Algebras). Could this be the only maximal ideal? $\endgroup$ – Nik Weaver Oct 18 at 3:52
  • 1
    $\begingroup$ @NikWeaver, thanks but isn't this ideal contained in ideals that are analogously defined with p-Schatten norm convergence for p less than 2? $\endgroup$ – Tomasz Kania Oct 18 at 7:49
  • 1
    $\begingroup$ @TomaszKania: That doesn't work, as your sequence also needs to be norm bounded, as (surely?) your traces are normalised. $\endgroup$ – Matthew Daws Oct 18 at 9:06
  • 2
    $\begingroup$ Yes, I think that's right. If you have a normalised trace, then restricted to the unit ball (for the operator norm, i.e. $\infty$-norm), the Shatten norms are all equivalent. (That's certainly true in the commutative case.) $\endgroup$ – Matthew Daws Oct 18 at 15:19
14
$\begingroup$

I think Nik Weaver is right that the ideal mentioned is the unique maximal ideal. This simultaneously answers both questions (since the quotient is clearly infinite dimensional). Let $\tau$ be the trace on $M_\mathcal{U}$ defined as $\tau(x_n)=\lim_{n\rightarrow \mathcal{U}}\tau_n(x_n)$ where $\tau_n$ is the normalized trace on $M_n.$ As Nik Weaver mentioned in the comments, the ideal $\{ x\in M_\mathcal{U}:\tau(x^*x)=0 \}$ is maximal. I claim this is the only maximal ideal. First we need a lemma from linear algebra

Claim: Let $a\in M_N$ be positive norm 1 and set $\varepsilon=\tau_N(A)>0.$ Then there are $k=\frac{2}{\varepsilon}$ partial isometries $v_1,...,v_k\in M_N$ such that $\sum v_i^*av_i\geq \frac{\varepsilon}{2}I.$

Proof of Claim: Order the eigenvalues of $a$ as $a_1\geq a_2\geq\cdots \geq a_N$ and if $a_{\frac{N\varepsilon}{2}+1}<\frac{\varepsilon}{2}$ the trace is strictly less than $\varepsilon$ hence $a_i\geq \frac{\varepsilon}{2}$ for $1\leq i\leq \frac{N\varepsilon}{2}.$ Let $v_1$ project onto the $\frac{N\varepsilon}{2}$-dimensional subspace spanned by the first $\frac{N\varepsilon}{2}$ eigenvectors (corresponding to the ordering of the eigenvalues $a_i$). Then twist this projection down the line with appropriate partial isometries to obtain the claim.

Back to the Answer: Let $I$ be an ideal that contains a positive, norm 1 element $x=(x_n)$ such that $\tau(x)>0.$ We will show that the ideal generated by $x$ contains the identity. By replacing $(x_n)$ with an equivalent sequence we can assume each $x_n$ is positive, norm 1 and has $\tau_n(x_n)\geq\varepsilon$ for some $\varepsilon>0.$ Now just apply the above claim coordinate wise to produce $k=\frac{2}{\varepsilon}$ partial isometries $w_1,...,w_k\in M_\mathcal{U}$ so $\sum w_ixw_i^*\geq \frac{\varepsilon}{2}I.$

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Nice! It's basically the $II_1$ factor proof, with a little extra care because this isn't a von Neumann algebra. $\endgroup$ – Nik Weaver Oct 18 at 16:46
  • 5
    $\begingroup$ This is essentially proved by Wright [Annals of Math 1954] (mathscinet.ams.org/mathscinet-getitem?mr=65037) who classified the maximal ideals of a finite AW*-algebra (such as the $\ell_\infty$-sum $\prod M_n$). $\endgroup$ – Narutaka OZAWA Oct 19 at 2:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.