MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $S_+$ be the cone of psd matrices ($n\times n$ real symmetric positive semidefinite matrices). This cone is a metric space induced from the inner product $\langle A,B\rangle = tr (AB)=tr(BA)$.

The cone $S_+$ seems exceptionally symmetrical, and I am curious to know its symmetry groups. Let me be precise. An $R_+$-linear map $f:S_+\to S_+$ is an additive map such that $f(\alpha A)=\alpha f(A)$ for $\alpha\ge 0$. What is the group of $R_+$-linear automorphisms of $S_+$? What is the subgroup of isometric $R_+$-linear automorphisms of $S_+$?

Apparently, from my geometric intuition, such automorphisms should permute the extremal rays of rank 1 matrices. Since each rank 1 matrix corresponds to a codim 1 plane in $R^n$, my guess is that these automorphism groups should be related to the Grassmanian.

share|cite|improve this question
I am sorry. I have redone the latex. All Rs really should be in blackboard font. – user2529 Sep 21 '11 at 8:17
up vote 2 down vote accepted

The general linear group $GL(n, R)$ acts on $S+$ by $g(x) = g x g^t,$ and this is the full linear automorphism group.

and references therein for more details and related results. I am pretty sure the result goes back to at least Minkowski (for $n=2$ this is the hyperboloid model of hyperbolic space, and the isometry group of $H^2$ (which is identified with $x \in S_+,$ such that $\det x = 1$) is $SL(2, R).$ For higher $n$ the analogous section $\det x = 1$ is a representation of the symmetric space for $SL(n, R)$ (Siegel half-space, I guess; but Siegel defined a metric which makes it into a Riemannian symmetric space).

share|cite|improve this answer
Thank you Igor. Is the isometry group O(n)? – user2529 Sep 23 '11 at 4:15
No, it is much bigger (I think SL(n)) – Igor Rivin Sep 24 '11 at 9:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.