Consider the space of $n \times n$ positive definite symmetric matrices and let $\Sigma$ be one such matrix. We make this space into a Riemannian manifold $M$ by means of the metric $$ds^2=tr(\Sigma^{-1}d\Sigma\Sigma^{-1}d\Sigma)$$ Now fix a $p<n$ and consider the collection $\Phi_p$ of all positive definite symmetric matrices which can be written $$B^TB+D$$ for some $B$ and $D$ where $B$ is $p \times n$ and $D$ is a diagonal positive definite matrix. Here $B$ and $D$ are allowed to vary but $p$ must stay fixed. Is $\Phi_p$ an embedded submanifold of $M$ and if so how can I put the induced metric on it?
$\begingroup$
$\endgroup$
21
-
1$\begingroup$ Why is the metric a metric? $\endgroup$– Igor RivinNov 22, 2013 at 19:52
-
2$\begingroup$ In fact, if $X_1 = X_2 = I,$ the distance is not zero, which is a little sad. $\endgroup$– Igor RivinNov 22, 2013 at 19:58
-
1$\begingroup$ So the first thing to test is: say $P$, $Q$ come from this diag+low-rank set. Now, is the geometric mean of $P$ and $Q$ also in this set? If not, then this won't be a submanifold (in the sense that you are searching for)... $\endgroup$– SuvritNov 23, 2013 at 16:36
-
4$\begingroup$ It's not a smooth submanifold in general. Just look at the case $(n,p)=(3,1)$, where $\Phi_1$ contains an open set, but is not open: For $a\in\Phi_1$, you have to have $S := a_{12}a_{23}a_{31}\ge0$. If $S>0$, then you only need that $a_{ij}^2a_{kk} > S$ for all $(i,j,k)$ a permutation of $(1,2,3)$ in order for $a$ to lie in $\Phi_1$. Thus, $\Phi_1$ contains an open set. However, $\Phi_1$ clearly doesn't contain a neighborhood of the identity matrix (which does lie in $\Phi$), since $S$ is not positive at the identity, so it's not an open set. $\endgroup$– Robert BryantNov 24, 2013 at 11:16
-
3$\begingroup$ @IgorRivin: Thanks. Actually, I think that, properly worded, there is a somewhat interesting question here. The OP really should be asking about the structure of this subvariety of 'low-rank' perturbations of a flat submanifold (i.e., the diagonal matrices) of the space of quadratic forms. It's a natural subvariety of this symmetric space, and it's conceivable that knowing something about this space has applications in information geometry. $\endgroup$– Robert BryantNov 25, 2013 at 6:01
|
Show 16 more comments
1 Answer
$\begingroup$
$\endgroup$
This Riemannian metric on the full space of all positive definite matrices turns up in the paper (and others)
There are explicit formulas for geodesics and for curvature. There is only one incomplete geodesic, but the space if far from being geodesically convex.
answered Nov 28, 2013 at 20:36