extreme points of the unit ball of the Schatten classes?

Question

Suppose $1<p<\infty$. What are the extreme points of the unit ball of the Schatten classes $S^p$?

See below for the definition of $S^p$:

http://en.wikipedia.org/wiki/Schatten_norm

Answer 1

In general, every noncommutative $L^p$ space (every Schatten class $S^p$ in particular) is uniformly convex and uniformly smooth for $1<p<\infty$, as follows from the noncommutative Clarkson inequalities. As noted by Norbert, uniform convexity implies that every element of the unit sphere is an extreme point of the unit ball. The converse is clear. So the set of extreme points of the unit ball of a noncommutative $L^p$ space is its unit sphere, for every $1<p<\infty$.

Noncommutative $L^p$ spaces: given a von Neumann algebra $M$ and a normal (semi)-finite faithful trace on $M$, the noncommutative $L^p$ space $L^p(M)$ is the completion of the span of positive elements of $M$ with $\tau$-finite support for the norm $$ \|x\|_p=(\tau(|x|^p))^\frac{1}{p} $$ where $|x|=\sqrt{x^*x}$. Then for conjugate exponents, $\frac{1}{p}+\frac{1}{q}=1$, the dual of $L^p(M)$ is naturally isometric to $L^q(M)$ for $1\leq p<\infty$, since in particular the noncommutative Hölder inequality holds. In particular, for $p=1$, we recover with $L^1(M)$ the unique predual of $M\simeq L^\infty(M)$.

Schatten classes: in the case where $M=B(H)$ and $\tau$ is the usual trace, finite for trace-class operators $S^1$, we get $L^1(B(H))=S^1$ and, more generally, $L^p(B(H))=S^p$.

Noncommutative Clarkson inequalities: the following is true for every noncommutative $L^p$ spaces, in particular for Schatten classes.

If $1\leq p\leq 2$ and $q$ is the conjugate exponent, then $$ \left(\frac{1}{2} (\| x+y\|_p^q+\|x-y\|_p^q) \right)^\frac{1}{q}\leq (\|x\|_p^p+\|y\|_p^p)^\frac{1}{p} \qquad \forall x, y \in L^p(M). $$ If $2\leq p\leq \infty$ and $q$ is the conjugate exponent, then $$ \left(\frac{1}{2} (\| x+y\|_p^p+\|x-y\|_p^p) \right)^\frac{1}{p}\leq (\|x\|_p^q+\|y\|_p^q)^\frac{1}{q} \qquad \forall x, y \in L^p(M). $$

Sketch: this is first seen to be true for $p=1,2$ in the former case, and for $p=2,\infty$ in the latter case. Then it follows by interpolation.

It is not hard to show that these inequalities imply uniform convexity for $1<p<\infty$. By reflexivity, this implies uniform smoothness, as $X$ is uniformly convex/smooth if and only if $X^*$ is uniformly smooth/convex.

Answer 2

In Nonpositive curvature in p-Schatten class, C. Conde it is stated that $S^p$ is uniformly convex and as the consequence all points of unit sphere are extreme points.