TITLE: What is the "positive part" of the unit ball in $M_n(R)$ ?
QUESTION [12 upvotes]: In ${\bf M}_n(\mathbb R)$, let us consider the usual operator norm
$$\|A\|=\sup\frac{\|Ax\|}{\|x\|},$$
where $\|x\|$ is the Euclidian norm.
The closed unit ball $B$ is the set of contractions (in the terminology used by operator theorists). It is a convex compact subset of ${\bf M}_n(\mathbb R)$. By Krein-Milman (finite dimensional case), it is the convex hull of its subset ${\rm ext}(B)$ of extremal points. It turns out that ${\rm ext}(B)$ is the orthogonal group ${\bf O}_n(\mathbb R)$. 
Now, remember that  ${\bf O}_n(\mathbb R)$ has two connected components, a positive one  ${\bf SO}_n(\mathbb R)$ and a negative one ${\bf O}_n^-(\mathbb R)$. 

What is the convex hull of  ${\bf SO}_n(\mathbb R)$ ?

Clearly, it is a compact convex subset, included in $B$. It is a strict subset of $B$, because it does not meet  ${\bf O}_n^-(\mathbb R)$. The title refers to the "positive part" of $B$, but this could be inappropriate, in the sense that it could meet the convex hull of ${\bf O}_n^-(\mathbb R)$ non-trivially.
Remark also that this convex hull is invariant under multiplication at right or left by an element of ${\bf SO}_n(\mathbb R)$. Therefore it would be enough to decide which diagonal matrices ${\rm diag}(a_1,\ldots,a_n)$ with $|a_1|\le a_2\le\cdots\le a_n$ it contains.
When $n=2$,  ${\bf SO}_n(\mathbb R)$ is a circle and its convex hull is a disk, obviously a much smaller set (even from the dimensional point of view) than $B$.

REPLY [10 votes]: I'm a bit late in answering this. But in case there is still interest, please have a look at:
Saunderson, Parrilo, Willsky. Semidefinite descriptions of the convex hull of rotation matrices, SIAM J. Optimization, 25(3), 1314-1343, 2015.
This paper provides explicit spectrahedral representations for conv $SO(n)$ (Theorem 1.3).