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Let $\mathrm{M}_{d\times d}\left(\mathbb{C}\right)$ denote the set of all $d\times d$-matrices with complex entries. My goal is to show that the set $\mathcal{M}:= \left\{ \rho\in \mathrm{M}_{d\times d}\left(\mathbb{C}\right) \left|\right. \ \rho\geq0 \ , \ {\rm tr}(\rho) = 1 \right\} $ can be endowed with the structure of a smooth manifold (most likely with boundary).

I have proven that $\mathcal{M}':= \left\{ \rho\in \mathrm{M}_{d\times d}\left(\mathbb{C}\right) \left|\right. \ \rho>0 \ , \ {\rm tr}(\rho) = 1 \right\}$ is indeed a smooth manifold. My argument roughly goes like this:

First we notice that $\mathrm{M}_{d\times d}\left(\mathbb{C}\right)$ is a $2d^{2}$-dimensional real vector space and the set $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$ of all hermitian $d\times d$ matrices with complex entries is a $d^{2}$ dimensional real vector subspace. In particular $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$ trivially forms a $d^{2}$-dimensional manifold since it's globally diffeomorphic to $\mathbb{R}^{d^{2}}$. Next we want to show that the set $\mathrm{P}_{d\times d}\left(\mathbb{C}\right)$ of all positive $d\times d$-matrices (which is a subset of $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$) is open in the relative topology of $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$. Therefor it constitutes a $d^{2}$-dimensional manifold. To see this we can consider the map \begin{align} f: \mathrm{H}_{d\times d}\left(\mathbb{C}\right) &\rightarrow \mathbb{R}^{d^{2}} \nonumber\\ A &\mapsto \left(\lambda_{1},...,\lambda_{d^{2}}\right) \end{align} which maps a hermitian matrix to its eigenvalues. This map is continuous. Since $\mathbb{R}_{>0}^{d^{2}}$ is an open set in $\mathbb{R}^{d^{2}}$ it follows that the preimage $f^{-1}\left(\mathbb{R}_{>0}^{d^{2}}\right)=\mathrm{P}_{d\times d}\left(\mathbb{C}\right)$ is also open in $\mathrm{H}_{d\times d}\left(\mathbb{C}\right)$. Now consider the map \begin{align} g: \mathrm{H}_{d\times d}\left(\mathbb{C}\right) &\rightarrow \mathbb{R} \nonumber\\ \rho &\mapsto {\rm tr}(\rho). \end{align} We need to show that 1 is a regular value of g. To do so we pick a $\rho\in g^{-1}(1)$. Next we choose a curve $\gamma: (-\epsilon,\epsilon) \rightarrow \mathcal{M}$ with $\gamma(0) = \rho$ and $\partial_{t}\gamma(0) = A$ for some $A\in T_{\rho}\mathrm{P}_{d\times d}\left(\mathbb{C}\right)$. We calculate that \begin{equation} dg_{\rho}A := \left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} {\rm tr}(\gamma(t)) ={\rm tr}(A). \end{equation} Certainly $dg_{\rho}$ is surjective. This proofs the result.

I'm wondering whether this proof can be extended to positive-semidefinite matrices since a positive-semidefinite matrix can be approximated by a sequence of positive definite matrices. I'm however not sure whether the "boundary" that we will obtain is still regular in the sense that it contains no "sharp edges".

I would be more than happy for some suggestions and ideas.

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I think you will find that the cone of positive semidefinite matrices is naturally a smooth manifold with boundary at all points except at the origin (the zero matrix), which is the vertex of the cone. Fortunately the hyperplane of matrices with unit trace misses the origin. –  Dan Petersen Jan 13 '13 at 20:42
    
Can you please explain in more detail why this is the case and how I can modify the proof to take positive semi-definite matrices into account? –  MrLee Jan 13 '13 at 20:45
    
In my experience whit BMV conjecture I recomendy arxiv.org/pdf/1107.4875v3.pdf and books.google.com.br/… –  Elias Jan 15 '13 at 18:09
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In case it helps, I just worked out the $d=2$ case by hand. In this case, $\mathcal{M}$ has a boundary diffeomorphic to $S^2$. (In fact, $\mathcal{M}$ can naturally be identified with the ball of radius $\frac{1}{2}$ centered at $(\frac{1}{2},0,0)$.) If you're interested, I can write it up, but there's really not much to it. –  Jason DeVito Jan 15 '13 at 18:29
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@JasonDeVito Actually I did that too prior to offering the bounty. I could not come up with an idea for the general case, though. –  user53153 Jan 15 '13 at 18:49
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1 Answer

Two-by-two semidefinite matrices A (without the trace condition) form a cone with a singularity at the origin. These can be embedded into three-by-three matrices by means of direct sum with the one-by-one matrix [1-tr(A)]. This suggests that your space will have singularities; this is turned into a proof below.

More specifically, the diagonal matrices in your space $M$ form a triangle $DEF$ in $\mathbb R^3$ in the first octant cut out by the relation $x+y+z=1$. The three vertices $D,E,F$ are in a single orbit when one acts by conjugation. This suggests that there might be non-smooth on the boundary of the space, as shown below.

The set of semidefinite matrices $A$ is convex because if $x^* A x\ge0$ and $x^* B x\ge0$ then also $x^* (A+B) x\ge 0$. Thus your space is a convex set and therefore a ball topologically. Using radial scaling you can identify it with a metric ball centered at an interior point, and pull back the smooth structure of the metric ball; in this sense, the space does admit "the structure of a smooth manifold".

However, the boundary of the convex set itself is not smooth. This can be seen as follows. The vertices and the edges of triangle $DEF$ above lie on the boundary of the convex set, since their determinant is zero. Meanwhile, each interior point of the triangle is also an interior point of the convex set, since any perturbation of it is still a positive definite matrix. If the boundary of the convex set were smooth at the vertex $D$, then the tangent plane at $D$ would in particular contain the directions $DE$ and $DF$, and would therefore also contain the entire triangle $DEF$. But we already know that the interior points of the triangle are not on the boundary. Hence the boundary of the convex set cannot be smooth at $D$.

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+1. The argument that you've added in the last paragraph seems to work. –  Jim Belk Jun 13 '13 at 3:53
    
I did not delete the earlier comments because they provide a bit of motivation. –  user72694 Jun 13 '13 at 12:04
    
+1 and here is the same proof reworded. Assume $\mathcal{M}$ has smooth boundary. Since $\frac13 I$ is an interior point of smooth convex body $\mathcal M$, every hyperplane section through $\frac13 I$ is also smooth, by transversality. But the section by diagonal matrices is the triangle with vertices $(1,0,0)$, $(0,1,0)$, $(0,0,1)$. –  ˈjuː.zɚ79365 Jun 14 '13 at 6:15
    
Very nice. It should be mentioned that transversality follows from convexity. –  user72694 Jun 14 '13 at 8:25
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