What is the "positive part" of the unit ball in $M_n(R)$ ? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T18:39:53Z http://mathoverflow.net/feeds/question/89842 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/89842/what-is-the-positive-part-of-the-unit-ball-in-m-nr What is the "positive part" of the unit ball in $M_n(R)$ ? Denis Serre 2012-02-29T08:20:21Z 2012-02-29T09:02:37Z <p>In ${\bf M}_n(\mathbb R)$, let us consider the usual operator norm $$\|A\|=\sup\frac{\|Ax\|}{\|x\|},$$ where $\|x\|$ is the Euclidian norm.</p> <p>The closed unit ball $B$ is the set of <em>contractions</em> (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 ${\bf O}_n(\mathbb R)$. </p> <p>Now, remember that ${\bf O}_n(\mathbb R)$ has <strong>two</strong> connected components, a positive one ${\bf SO}_n(\mathbb R)$ and a negative one ${\bf O}_n^-(\mathbb R)$. </p> <blockquote> <p>What is the convex hull of ${\bf SO}_n(\mathbb R)$ ?</p> </blockquote> <p>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.</p> <p>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.</p> <p>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 $B$.</p>