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 ${\bf SO}\_n(\mathbb R)$ ?
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> 
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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 $B$.



What is the "positive part" of the unit ball in $M_n(R)$ ?
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](http://epubs.siam.org/doi/pdf/10.1137/14096339X), SIAM J. Optimization, 25(3), 1314-1343, 2015.


This paper provides explicit spectrahedral representations for conv $SO(n)$ (Theorem 1.3).


