# Largest ideal in bounded linear maps on Schatten-$p$ class

Let $$1\leq p<\infty.$$ Denote $$S_p(\ell_2)$$ be the set of all compact operator $$x$$ on $$\ell_2$$ such that $$Tr(|x|^p)<\infty.$$ Define $$\|x\|_{S_p(\ell_2)}:=Tr(|x|^p)^{\frac{1}{p}}.$$ This makes $$S_p(\ell_2)$$ a Banach space. What is the largest closed two-sided ideal in the Banach algebra of set of all bounded linear maps on $$S_p(\ell_2)$$?

Suppose $$X$$ is a Banach space that has the following property: (A) If $$T$$ and $$S$$ are in the space $$L(X)$$ of bounded linear operators on $$X$$ and the identity $$I_X$$ on $$X$$ factors through $$T+S$$, then either $$I_X$$ factors through $$T$$ or $$I_X$$ factors through $$S$$. It is more or less obvious that if $$X$$ satisfies (A) then $$M_X:=$$ all $$U$$ in $$L(X)$$ s.t. $$I_X$$ does not factor through $$U$$ is the largest ideal in $$L(X)$$. Probably $$S_p$$ satisfies (A); maybe it is even in some paper (maybe by Arazi and Lindenstrauss?).

• Link to an online copy of the Arazy-Lindenstrauss paper, for any readers following this up: numdam.org/item/CM_1975__30_1_81_0 Jun 19 '19 at 17:06
• @Tomek. Why do you think that the fact that $S_1$ is the projective tensor product with $\ell_2$ with itself will be helpful? Jun 28 '19 at 5:01

This is not a complete answer but maybe it will shed some light. Btw. your question is slightly ill-posed as we do not know whether $$B(S_p)$$ has a unique maximal ideal, unless $$p=2$$.

Bill in his answer refers to the set $$\mathscr{M}_X$$ defined here. And indeed, it is very likely that the space $$S_1$$ has the property that $$\mathscr{M}_{S_1}$$ is closed under addition being the projective tensor product of $$\ell_2$$ with itself. The space $$S_1$$ is primary (by a result of Arias and Farmer) and is also isomorphic to the $$\ell_1$$-sum $$\ell_1(S_1)$$.

If I am not mistaken, all known examples of primary spaces $$X$$ isomorphic to their infinite sums have the property that $$\mathscr{M}_X$$ is closed under addition, in which case, their algebras of operators have unique maximal ideals. However, there is no general result that would say it is a theorem, I am afraid.

As for $$p=2$$, the situation is trivial because in that case $$S_2$$ is a separable Hilbert space.

• Thank you everyone. Jun 22 '19 at 6:31
• @ Tomek. Why do you think that the fact $S-$ Jun 28 '19 at 5:01