Can I identify $S_k(V)$ with an homogeneous space?

Question

I'm in trouble with a question: Let $V$ be an $n$-dimensional vector space over $\mathbb R$. Can I identify the manifold, $$S_k(V):=\{(X_1, \ldots, X_k): X_1, \ldots, X_k\in V\ \textrm{are linearly independent}\}$$ with an homogeneous space? Thanks..

Answer 1

Yes, you can.

The general approach is the following:

Suppose $G$ is a Lie group which acts on a manifold $M$ transitively. Let $p\in M$ and let $H = \{g\in G: gp = p\}$. The map $G/H\rightarrow M$ given by mapping $gH$ to $gp$ is a well defined diffeomorphism between $G/H$ and $M$.

To apply this to your setting, we must show that $S_k(V)$ is a manifold, and then show that $G = Gl(n,\mathbb{R})$ acts transitively on it and then compute the stabilizer subgroup. The stabilizer subgroup is not $Gl(n-k,\mathbb{R})$, but does contain it as a subgroup. The stabilizer subgroup is (up to conjugacy) all matrices of the block form $$\begin{bmatrix} I_{k\times k} & v_{k\times n-k}\\ 0_{n-k\times k} & B_{n-k\times n-k}\end{bmatrix}$$ where $B$ is invertible and $v$ is an arbitrary $k\times n-k$ matrix. (Since $S_k(V)$ has dimension $kn$ and $\dim G/H = \dim G - \dim H$, $H$ cannot possibly be $Gl(n-k,\ \mathbb{R})$).

Since $\det$ is continuous, and vectors $\{X_1,...,X_k\}$ are independent iff some $k\times k$ minor in them is nonzero, this shows $S_k(V)$ is an open subset of the manifold/vector space $V^k$. In particular, $S_k(V)$ is naturally a manifold, and multiplication by matrices is a smooth map.

Now, given $A\in Gl(V) = Gl(n,\mathbb{R})$, define $A\ast(X_1,..., X_k) = (AX_1, \ldots, AX_k)$. Since $A$ is invertible, the set $\{AX_1, ..., AX_k)\}$ is independent, so this really does define a smooth action of $G=Gl(V)$ on $S_k(V)$.

To see this action is transitive, let $e_i$ denote the vector in $V\cong \mathbb{R}^n$ with a $1$ in the $i$th slot and $0$s else where. It's enough to show that given any $(X_1,\ldots , X_k)\in S_k(V)$, that there is an $A\in G$ with $A(e_1,\ldots , e_k) = (X_1,\ldots, X_k)$. But, given $\{X_1, \ldots X_k\}$, we can complete it a basis $\{X_1,\ldots , X_n\}$ of V. Letting $A$ be the matrix whose columns are the $X_i$s, we see $A$ is invertible and $Ae_i = X_i$.

Finally, we compute $H$ the point $p = (e_1,\ldots, e_k)$. But it's easy to see that the conditions $Ae_1 = e_1$,..., $Ae_k = e_k$ force the $I$ and $0$ block in the above description. The fact that $B$ must be invertible follows from the fact that $\det(A)\neq 0$.

Together, this shows $S_k(V) = Gl(n,\mathbb{R})/H$ where $H$ is also matrices of the above form.

Answer 2

I could be wrong, but $S_k(V)\cong Gr(k,n)\times GL(k,\mathbb{R})$? A point in $S_k(V)$ can be given as a $k$-dimensional linear subspace in $V$ (the space spanned by $\{X_1,\ldots X_k\}$), together with a $k$-tuple of $k$ linearly independent vectors in $\mathbb{R}^k$, ie a non-singular $k\times k$ matrix over $\mathbb{R}$.