how to calculate the quotient group of $U/DU$ where $U$ is an arbitrary unitary matrices, and $DU$ is diagonal unitary matrices?

Question

Question is the calculation of $U/DU$ where $U$ is a unitary matrices with non-degenerate eigenvalues, and $DU$ are diagonal unitary matrices with non-degenerate eigenvalues.

The answer is $UA_{0}U^{-1}\cong U/DU$ where $A_0$ is diagonal matrix with eigenvalues of $1,\ 1/2,\ 1/3\ ...\ 1/n$.

How ever I can see how one can calculate this the quetient is defined as follows: $G/H$ is a group its elements are the equivalence clases $[g]$ where the equivalence is defined by $g\sim g '$ if $g'=gh$. I cant find the above expression by using this definition. The original source that I encountered this problem is this;

Answer 1

[Reference] Wikipedia: Homogeneous space and Wikipedia: Group action

Note. I will use a general notation for the unitary groups.

Let $U(n)$ be the set of unitary matrices, and let $DU(n)$ be the set of unitary diagonal matrices.

Note. $DU(n)$ is a closed subgroup of the Lie group $U(n)$ so that the quotient group $U(n)/DU(n)$ is a smooth manifold.

Let $\mathfrak{M}\equiv\{ UA_0U^{-1} : U\in U(n) \}$ where $A_0$ is the diagonal matrix with entries $1,\frac{1}{2},\dotsc,\frac{1}{n}$. Notice that $\mathfrak{M}$ is not a group, since $\mathfrak{M}$ does not contain the identity matrix $I$.

The unitary group $U(n)$ acts on $\mathfrak{M}$ (by conjugation) $$ U(n) \times \mathfrak{M} \to \mathfrak{M} \quad\text{defined by}\quad (V, UA_0U^{-1}) \mapsto V(UA_0U^{-1})V^{-1} $$ for $V\in U(n)$ and $UA_0U^{-1}\in\mathfrak{M}$. In general the action of $V$ on $UA_0U^{-1}$ is denoted by $V\cdot(UA_0U^{-1})$.

(1) The action is well-defined since $$ V\cdot(UA_0U^{-1}) \equiv V(UA_0U^{-1})V^{-1} = (VU)A_0(VU)^{-1} \in \mathfrak{M} $$ (2) (Identity) For the identity matrix $I$, we have $$ I\cdot(UA_0U^{-1}) \equiv I(UA_0U^{-1})I^{-1} = UA_0U^{-1} $$ (3) (Compatibility) For $V,W\in U(n)$, we have $$ \begin{align*} (VW)\cdot(UA_0U^{-1}) &\equiv (VW)(UA_0U^{-1})(VW)^{-1} \\ &= V(W(UA_0U^{-1})W^{-1})V^{-1} = V\cdot(W\cdot(UA_0U^{-1})) \end{align*} $$ (4) Moreover the action is differentiable because it consists of matrix multiplications.

Claim. The homogeneous space $\mathfrak{M}$ is diffeomorphic to the quotient group $U(n)/DU(n)$.

1. $\mathfrak{M}$ is a homogeneous space.

A homogeneous space for a (Lie) group $G$ is a non-empty (smooth manifold) $X$ on which $G$ acts transitively (from Wikipedia: Homogeneous space).

In our case, the action of $U(n)$ on $\mathfrak{M}$ is faithful so that "$\mathfrak{M}$ is a homogeneous space for $U(n)$" is equivalent to "$\mathfrak{M}$ is a single $U(n)$-orbit". But this is trivial by the definition of $$ \mathfrak{M}=\{ U\cdot A_0\equiv UA_0U^{-1} : U\in U(n) \} = \text{$U(n)\cdot A_0$ (the $U(n)$-orbit of $A_0$) } $$

2. $\mathfrak{M}\cong U(n)/DU(n)$

For a fixed $x\in X$, consider the map from $G$ to $X$ given by $g\mapsto g\cdot x$ for all $g\in G$. The image of this map is $G\cdot x$, the $G$-orbit of $x$. The standard quotient theorem of set theory then gives a natural bijection between $G/G_x$ and $G\cdot x$. Here $G_x=\{g\in G : g\cdot x=x\}$ denotes the stabilizer subgroup of $G$ with respect to $x$ (from Wikipedia: Group action).

The statement above is very basic fact. Moreover, the bijection becomes a diffeomorphism when $G$ is a Lie group. In our case, since $\mathfrak{M}$ is a single $U(n)$-orbit of $A_0$, we have $$ \mathfrak{M} \cong U(n)/U(n)_{A_0} $$ It remains to show that the stabilizer $U(n)_{A_0}=\{V\in U(n) : V\cdot A_0=A_0\}$ is nothing but $DU(n)$. It is easy to check using matrix multiplication and leave it to an exercise.