Calculate the Stabilizer of an element in the Grassmannian

Question

I know there is a virtually same question and its answer on this site, but neither the questioner nor the answerer appears to be active at the moment.

Define a group action on Grassmannian $$\begin{aligned} \mathrm{GL}_n(\mathbb{C})\times\mathrm{Gr}_{n,k}&\longrightarrow\mathrm{Gr}_{n,k},\\ (g,W)&\longmapsto g\cdot W=\{gv\mid v\in W\}. \end{aligned}$$ For a standard basis $e_1,...,e_n$ of $\mathbb{C}^n$, let $W$ be a subspace spanned by $e_1,...,e_k$. How to compute the stablizer of $W$?

In this answer, the user set a block matrix to illustrate it. The answer states that $C$ is zero and no restriction on other blocks, but I'm confused how to immediately obtain such a conclusion. I'm not sure if my lack of understanding of Grassmannian or linear algebra is causing this problem. Can anyone help me out with some hints?

Answer 1

Recall that for a matrix $M$, and standard basis vectors $e_i$, we have $Me_i$ is the $i$th column of $M$.

Next $W$ consists in all linear combinations of the $e_i$ of the form $$w=f_1e_1+\dots +f_ne_n$$ with $f_{k+1}=\dots =f_n=0.$

Finally the first $k$ columns of $g$ must have no (zero) contribution in the last $n-k$ components.

And, the last $n-k$ columns are unrestricted, because vectors in $W$ have zeros in the last $n-k$ components anyway.

Answer 2

Hint In terms of the basis $(e_i)$ of $\Bbb C^n$, the condition that $g \in \operatorname{GL}(n, \Bbb C)$ stabilizes $W := \operatorname{span}\{e_1, \ldots, e_k\}$ is precisely that $$g \cdot e_j \in \operatorname{span}\{e_1, \ldots, e_k\}, \qquad j = 1, \ldots , k .$$ Now translate this condition to one on the entries of $g$.