Topology of the space of hermitian positive definite matrices

Question

Let $\mathcal{H}_n \mathbb{C}$ be the set of hermitian $n \times n$ complex matrices. This set carries the structure of a vector space over $\mathbb{R}$ under usual addition. It also inherits the standard euclidean topology from $\mathbb{C}^{n\times n}$. Let $\mathcal{P}$ denote the subset of $\mathcal{H}_n \mathbb{C}$ of positive definite matrices and give it the subspace topology.

My question is: is $\mathcal{P}$ locally compact?

What I understand: $\mathcal{P}$ is closed under linear combination with positive coefficients, in particular it is closed under addition, multiplication by a positive scalar and is convex. In dimension $1$ it is just $]0,+\infty[$, so the answer is yes. My feeling is that the answer should be yes in higher dimensions as well, since $\mathcal{P}$ is a kind of "cone" in a real vector space, but I can't provide a rigorous proof of this.

Answer 1

Edit: I think that the first answer below is more efficient and easier to write. Nevertheless, I have added a more concrete approach which might be closer to what you were after.

By some sort of transitivity: recall that every open or closed subset of a locally compact space is a locally compact space with the induced topology. We will use both, for open and for closed.

I consider every space here equipped with the topology induced by the Euclidean norm of $\mathbb{C}^{n\times n}$. Hence every space is Hausdorff. This is good to know, even though your question does not specifically ask about this aspect.

Like every finite-dimensional vector space over $\mathbb{R}$ or $\mathbb{C}$, $\mathbb{C}^{n\times n}$ is locally compact when equipped with the topology induced by any norm.

Clearly, $\mathcal{H}_n^+\mathbb{C}$, the set of positive semidefinite matrices, is closed in the locally compact $\mathbb{C}^{n\times n}$. So $\mathcal{H}_n^+\mathbb{C}$ is a locally compact space.

Now $\mathcal{P}=\{A\in \mathcal{H}_n^+\mathbb{C}\;;\det A>0\}$. By continuity of the determinant, it follows that $\mathcal P$ is open in the locally compact $\mathcal{H}_n^+\mathbb{C}$. Hence $\mathcal P$ is a locally compact space. QED.

Parametrized alternative: fix $A_0$ hermitian definite positive, and denote $\{t^0_1,\ldots,t_n^0\}$ its (positive) eigenvalues. Now let $\epsilon:=\min t_j^0/2>0$. Then denote $U_n$ the unitary group and $\mathcal{H}_n^{++}$ the cone of positive definte hermitian matrices. Now consider the map $$ \phi:U_n\times \prod_{j=1}^n[t_j^0-\epsilon,t_j^0+\epsilon]\longrightarrow \mathcal{H}_n^{++} $$ which sends $(U,t_1,\ldots,t_n)$ to $U\mbox{diagonal}\{t_1,\ldots,t_n\}U^*$. Since $U_n$ is compact, the domain is compact. And since $\phi$ is continuous, the range of $\phi$ is a compact subset of $\mathcal{H}_n^{++}$ containing $A_0$. So it only remains to check that this is a neighborhood of $A_0$ in $\mathcal{H}_n^{++}$. To that aim, note that it contains $$ \phi(U_n\times \prod_{j=1}^n(t_j^0-\epsilon,t_j^0+\epsilon))\ni A_0 $$ i.e. the set of hermitian definite positive matrices with spectrum $\{t_1,\ldots,t_n\}$ such that, up to a permutation, $|t_j-t_j^0|<\epsilon$ for every $j=1,\ldots,n$. By continuity of polynomial roots over $\mathbb{C}$ applied to the characteristic polynomial, this is open in $\mathcal{H}_n^{++}$. QED.

Answer 2

Edit: First, as $\mathbb{C}^{n\times n}$ is Hausdorff, so is $\mathcal{P}$.

Second, let $B(A,r)$ denotes the ball centred at a matrix $A$ with radius $r$ in $\mathbb{C}^{n\times n}$ and $\bar{B}(A,r)$ its closure. Since eigenvalues are continuous functions of matrix entries, for any $P\in\mathcal{P}$, there exists $r>0$ such that $B(P,r)\cap(\mathcal{H}_n \mathbb{C})\subset\mathcal{P}$. Therefore, for all $\varepsilon\in(0,r)$, the closure of $B(P,\varepsilon)\cap\mathcal{P}$ in $\mathcal{P}$ is simply $\bar{B}(P,\varepsilon)\cap\mathcal{P}=\bar{B}(P,\varepsilon)\cap(\mathcal{H}_n \mathbb{C})$, which is closed and bounded and hence compact in $\mathbb{C}^{n\times n}$. As compactness is an intrinsic property of a set, $\bar{B}(P,\varepsilon)\cap\mathcal{P}$ is also compact in $\mathcal{P}$. Furthermore, $\bar{B}(P,\varepsilon)\cap\mathcal{P}$ has a non-empty interior ($B(P,\varepsilon)\cap\mathcal{P}$). So, it is a compact neighbourhood in $\mathcal{P}$.

In other words, $\mathcal{P}$ is a Hausdorff space such that its every member is contained in a compact neighbourhood. Hence $\mathcal{P}$ is locally compact.