Maximal ideals of ultraproducts of full matrix algebras

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.

• 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? Oct 18 '20 at 3:52
• @NikWeaver, thanks but isn't this ideal contained in ideals that are analogously defined with p-Schatten norm convergence for p less than 2? Oct 18 '20 at 7:49
• @TomaszKania: That doesn't work, as your sequence also needs to be norm bounded, as (surely?) your traces are normalised. Oct 18 '20 at 9:06
• 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.) Oct 18 '20 at 15:19

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 $$\sum w_ixw_i^*\geq \frac{\varepsilon}{2}I.$$
• Nice! It's basically the $II_1$ factor proof, with a little extra care because this isn't a von Neumann algebra. Oct 18 '20 at 16:46
• 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 $\prod M_n$). Oct 19 '20 at 2:23