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.



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


