Action of the Unitary Group

Question

I am working on the space $V_k (\mathbb{C}^n) = \left\lbrace (v_1, \cdots , v_k ) \in (\mathbb{C}^n)^k | \langle v_i, v_j \rangle = \delta_{ij} \right\rbrace $.

I define the continuous action of $U(n)$ on $V_k (\mathbb{C}^n)$ by $ U \cdot (v_1, \cdots , v_k ) = (U v_1, \cdots , U v_k ) $ .

I know that it's transitive since I can find a matrix $ U \in U(n)$ that takes me from a vector with norm 1 to another vector with norm 1 ( Is this enough? Can this be made more rigorous?)

I want to show that the stabilizer of any element of $V_k (\mathbb{C}^n)$ is isomorphic to $U(n-k)$; but I can't really see it!

score 1 · Answer 1 · answered May 27 '14 at 19:38

1

Here is a hint:

Suppose $v \in (\mathbb{C}^n)^k$, and let $v_{k+1},...,v_n$ be such that $v_1,...,v_n$ is an orthonormal basis of $\mathbb{C}^n$. Let $V = \operatorname{sp} \{v_1,...,v_k \}$.

Suppose $U \in G_v$, and note that $U V = V$ and $U V^\bot = V^\bot$, which means that in the basis $v_1,...v_n$, $U$ has the form $U = \begin{bmatrix} I & 0 \\ 0 & \tilde{U} \end{bmatrix}$, where $\tilde{U}: V^\bot \to V^\bot$.

answered May 27 '14 at 19:38

copper.hat

172,524

You mean $U \in U(n)$? – Mia May 27 '14 at 19:49
1

Well, it is in $U(n)$, but I specifically mean $U \in G_v = { U | U\cdot v = v }$ the stabilizer subgroup of $v$. – copper.hat May 27 '14 at 19:51
My bad... I should have at least read till the end of the statement before asking :) – Mia May 27 '14 at 19:52
No problem, I should have specified what I meant by $G_v$. – copper.hat May 27 '14 at 19:57

Action of the Unitary Group

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