Why is this a smooth manifold?

Question

I read that the space of quantum states, i.e. the space of density operators

$\mathcal{S}_n = \{ \rho \in H_n : \rho \geq 0, \, \, Tr[\rho]=1\}$

is a smooth manifold of dimension $n^2-1$, without further explanation. Could anyone help me understand why this is the case?

Here $H_n$ denotes the space of hermitian matrices: $H_n = \{ A \in \mathbb{C}^{n \times n} : \, A^* =A \}$, and $\rho \geq 0$ means that the matrix $\rho$ is positive semi-definite.

Answer 1

Preimage theorem:

Theorem. Let $f:X\to Y$ be a smooth map, and let $y\in Y$ be a regular value of $f$. Then $f^{-1}(y)$ is a submanifold of $X$. If $y\in {\text{im}}(f)$, then the codimension of $f^{-1}(y)$ is equal to the dimension of $Y$. Also, the tangent space of $f^{-1}(y)$ at $x$ is equal to $\ker(df_{x})$.

Answer 2

This is false, since the set has a boundary, but is not a even a manifold with boundary, since the boundary (the set of singular matrices with trace equal to one) is very singular. However, the set of positive definite hermitian matrices with trace equal to one is, indeed, an open manifold.