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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 $B$.

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  • $\begingroup$ If $B$ is open, then $B$ has no extremal points, and so is not the convex hull of the set of its extremal points. $\hspace{.2 in}$ If $B$ is closed, then $B$ has members that are not contractions. $\;\;$ $\endgroup$
    – user5810
    Commented Feb 29, 2012 at 8:44
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    $\begingroup$ Ricky, it is customary in many places/books/texts/whatever in analysis to use the word "contraction" to mean "distance non-increasing". This is how almost all practising operator theorists and most functional analysts I've met use the word, for instance $\endgroup$
    – Yemon Choi
    Commented Feb 29, 2012 at 8:44
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    $\begingroup$ The answer for $n=3$ is given in $\S 4.1$ of arxiv.org/abs/0911.5436. In $\S 4.4$ of ibid, there's a discussion of some properties of the convex hull of $SO(n)$ for larger $n$. $\endgroup$
    – Laurent Berger
    Commented Feb 29, 2012 at 15:31
  • $\begingroup$ Merci, Laurent ! $\endgroup$
    – Denis Serre
    Commented Feb 29, 2012 at 15:47
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    $\begingroup$ A few basic remarks: (a) the set of extremal points of the convex hull of $SO_n$ is exactly $SO_n$ (idem for $O^-_n$); (b) for $n$ even these sets are symmetric; (c) for all $n\ge 2$ the convex hull of $SO_n$ and $O^-_n$ have 0 in their intersection, as we see looking just at the convex hull of diagonal $\pm 1$-matrices (d) for $n\ge 3$ the convex hull of $SO_n$ and $O^-_n$ have nonempty interior (hence their intersection is a neighborhood of 0). (This is because it contains 0 and using [I skip details] that the representation in $\mathbf{R}^n$ is absolutely irreducible for $n\ge 3$.) $\endgroup$
    – YCor
    Commented Nov 10, 2016 at 5:06

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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).

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  • $\begingroup$ Thanks a lot ! Apparently, the authors ignored my MO question. $\endgroup$
    – Denis Serre
    Commented Nov 9, 2016 at 17:25

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