MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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. – Colin Tan Sep 21 '11 at 8:17
up vote 3 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)? – Colin Tan 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.

Not the answer you're looking for? Browse other questions tagged or ask your own question.