# Is the space of positive (semi- or not) definite correlation matrices Polish?

Title basically says it all: Is the space of positive (semi- or not) definite correlation matrices Polish?

Edit: For the sake of completeness, a $p\times p$ positive definite correlation matrix $C$ is a real symmetric matrix with ones on the diagonal and $x'Cx>0$ for any nonzero $x\in \mathbb{R}^p$. I updated the question to include positive semidefinite correlation matrices as well (ie relaxing to $x'Cx\geq0$) because I suppose it's interesting too :)

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This looks relatively straightforward: A subspace of a Polish space is Polish if and only if it is $G_{\delta}$ (by a theorem of Kuratowski, I think). Now the symmetric matrices are closed in all matrices, hence they are Polish. Positive definiteness is an open condition in the symmetric matrices, hence the positive definite matrices are Polish. Having $1$'s on the diagonal is a closed condition, hence the answer to your original question seems to be yes. The positive semi-definite case is even easier (as the space is closed in all matrices). –  t.b. Jul 11 '11 at 0:00
@Theo Buehler Seems to be the case. After your lead and a little digging I found the theorem and proof in Kechris's Classical Descriptive Set Theory (p 17, Thm 3.11 ). Thanks for your help! (I'd accept this comment as an answer too btw) –  JMS Jul 11 '11 at 1:31
Also thanks @amWhy for fixing my sloppy OP :) –  JMS Jul 11 '11 at 1:32
It was just a little tweaking; mainly, I wanted to include your question in the post itself. User's here prefer that the post itself is self-contained. Otherwise, your post was pristine compared to some of the OPs we get! –  amWhy Jul 11 '11 at 1:35
Descriptive set theory saves the day once more! Huzzah! –  Asaf Karagila Jul 11 '11 at 6:23

Let me do it this way:

• The symmetric matrices are closed in all $p \times p$-matrices (obvious), hence they are Polish.
• The positive definite matrices are an open cone in the symmetric matrices, hence they are Polish as well (open subsets of a Polish space are Polish — this is a bit easier to prove than the fact on $G_{\delta}$'s).
• Finally, since the functions $f_{i}:A \mapsto a_{ii}$ are continuous, their simultaneous pre-image of $1$ is closed in the positive definite matrices, hence the positive definite correlation matrices are Polish.

The positive semi-definite case is even simpler, as the conditions are all closed.

Finally, I can only recommend working through the first few sections of Kechris, as most of these arguments become rather simple once one gets used to them.

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Yes, Willie, still on that Kechris riff... –  t.b. Jul 11 '11 at 1:42
Nicely done, thanks again. This stuff is at the boundary of my mathematics but I've checked out the Kechris book, so hopefully it's not a permanent condition :) –  JMS Jul 11 '11 at 15:29
@JMS: Thanks! Maybe Parthasarathy, Probability measures on metric spaces would suit your tastes, interests and needs a bit better than Kechris (should have thought of that earlier). –  t.b. Jul 11 '11 at 15:33
Oooh that does look nice - I've arrived here from studying Bayesian nonparametric statistics, which takes a level of abstract math I never quite got to... –  JMS Jul 11 '11 at 15:40