Explicit example of an equivariant embedding of $U(n)/( U(k) \times U(n-k))$ into a finite dimensional $U(n)$-representation

Question

We know that if $H$ is a closed subgroup of a compact Lie group $G$ one can find a finite dimensional $G$-representation $V$ and an element $v_0 \in V$ such that $\textrm{Stab}(v_0)= H$. This gives a $G$-equivariant embedding \begin{equation} G/H \rightarrow V \\ gH \mapsto gv_0. \end{equation}

A possible construction is consider the Hilbert space $L^2(G)$, construct an element $f \in L^2(G)$ such that $\textrm{Stab}(f)=H$. Using Peter-Weyl theorem we obtain that $L^2(G) \cong \bigoplus_i X_i$. Letting $f_i$ be the projection of $f$ into $X_i$ we have $H= \cap_i\textrm{Stab}(f_i)$. Now the descending chain condition for closed subgroup of compact Lie groups provides a description of $H$ as a finite intersection of the form $\cap_{j=1,..., n} \textrm{Stab}(f_{i_j})$. Letting $v_0 =f_{i_1} + \cdots + f_{i_n}$ and $V=X_{i_1} \oplus \cdots \oplus X_{i_n}$ we obtain $\textrm{Stab}(v_0)= H$ as wanted. More details can be found in :

1. R.S. Palais, "Classification of $G$-spaces" Proposition 1.4.1,

and different approaches can be found in:

2. G.E. Bredon,"Introduction to compact transformation groups" Ch $0$ Theorem 5.2,
3. T. Brocker,T. tom Dieck, "Representations of compact Lie groups" III Theorem 4.6.

All the construction in the references apply to the case I am interested of, i.e., $G= U(n)$ and $H=U(k) \times U(n-k)$ for $k \le n$ as block matrix subgroup. The main issue is that this construction is not enough concrete for what I need and I was trying to find a more explicit construction of such representation this case. Since I am not an expert in representation theory of compact Lie groups any suggestion or insight it might be helpful.

Answer 1

You can just use the embedding $f\colon G/H=U(n)/(U(k)\times U(n-k))\to M_n(\mathbb{C})$ given by $f(gH)=gpg^{-1}$, where $p=1_k\oplus 0_{n-k}$. This gives a homeomorphism from $G/H$ to the space $$ P = \{q\in M_n(\mathbb{C}) : q^2=q^\dagger=q,\;\text{trace}(q)=k\}. $$ The map $f$ is equivariant if we let $G$ act on $M_n(\mathbb{C})$ by conjugation.