Degrees of freedom of a $d$ level system and the dimension of its Bloch manifold A qubit is described by two independent degrees of freedom that parametrize the Bloch sphere.
Question:
For a level $d$ level system, i.e., a qu$d$it, what are the corresponding degrees of freedom? And what is the dimension of its Bloch manifold?

Naively, we know that given a qu$d$it state $|d\rangle$ we can easily formulate its $d\times d$ density matrix as $|d\rangle \langle d| := \rho_d$.
Given it is a Hermitian matrix, an element of $SU(d),$  we have $d^2-1$ degrees of freedom. This matches the dimension of the adjoint representation of the corresponding Lie algebra and makes sense since we should be able to write $\rho_d$ as a linear superposition of the $d^2-1$ generators.
However, not all angles in the $SU(d) \subset SO(d^2-1)$ manifold are allowed. For example, just like for a qubit, the trace constraint ${\rm Tr}(\rho_d)=1$ will remove a degree of freedom. Furthermore, $\rho_d$ is a Positive Semidefinite matrix (PSD) and I am not sure if further degrees are removed because of this.
Thoughts?
 A: *

*The space of pure states on a $d$-dimensional system is the complex projective space $\mathbb{CP}^{d-1}$, which is a manifold with real dimension $2d-2$ (and also a $(d-1)$-dimensional complex manifold).


*The space of all $d$-dimensional states, that is, the set of unit-trace positive semidefinite $d\times d$ Hermitian matrices, is a convex (compact) subset of the $(d^2-1)$-dimensional affine subspace of unit-trace Hermitian matrices. Note that neither of these are vector spaces. The former is a compact subset, and the latter is an an affine space. This $(d^2-1)$-dimensional affine subspace is embedded in the $d^2$-dimensional (real) vector space of Hermitian $d\times d$ matrices.


*Absolutely all unitary operations (of suitable dimension) are allowed. The trace constraint is automatically maintained in all cases, because $\operatorname{Tr}(U\rho U^\dagger)=\operatorname{Tr}(\rho)$ for any (special or not) unitary operation $U\in\mathbf{U}(d)$. In fact, if you consider the space of all states, unitary operations will also only connect specific subsets of states; more precisely, they will create disjoint orbits of states sharing the same purity.


*I mostly already implicitly said this above, but let me stress that $\mathbf{SU}(d)$ is the (Lie) group of special unitary matrices, not the space of Hermitian matrices. Its Lie algebra $\mathfrak{su}(d)$ is kinda the same as the space of Hermitian matrices, but with the notable difference that while these are isomorphic as real vector spaces, only one of them has a Lie algebra structure.


*The PSD constraint doesn't actually remove degrees of freedom, but rather imposes further constraints that make the set of states compact. See e.g. Why does the space of pure qudit states have dimension $2(D-1)$, rather than $D^2-2$? and Characterisation of the generalised Bloch space in spherical coordinates.
In the case of a single qubit, the PSD constraint is what brings you from a full vector space to a finite-volume sphere.
