Proving that the Grassmannian is a smooth manifold

Question

I am just trying to come up with a simple proof showing that the Grassmannian ($Gr(k,\mathbb{R}^n)$) is a manifold.

I was referring to Lee's Introduction to Smooth Manifolds but I can't seem to follow his construction for the charts of the Grassmanian. I follow till the point where he argues that $Gr(k, \mathbb{R}^n)$ should be a $k(n-k)$-dimensional manifold.

It appears to me that he makes an appeal to $Hom(P, Q)$ (where $P$ is a $k-dimensional$ subspace of $\mathbb{R}^n$ and $P\oplus Q = \mathbb{R}^n$), when endowed with a basis becomes identical to $M((n-k)\times k, \mathbb{R})$ which is in turn identical to $\mathbb{R}^{(n-k)k}$. The route he takes beyond this point confuses me.

Is there a simpler way of showing that $Gr(k, \mathbb{R}^n)$ is a manifold, after knowing that it is $k(n-k)$-dimensional?

I appreciate any and all help!

Answer 1

Let me offer a more geometric point of view, assuming you are convinced that $O(n)$ is a compact Lie group. Consider $O(n)$ as the space of orthonormal $n$-frames in $\mathbb{R}^n$. We can define an action of $O(n-k)\times O(k)$ on $O(n)$ as follows: for an element $(\rho,\eta)\in O(n-k)\times O(k)$ and an orthonormal $n$-frame $(v_1,\dots,v_n)$, we let $\rho$ act on the last $n-k$ vectors and $\eta$ act on the first $k$ vectors of $(v_1,\dots,v_n)$ in the usual way. If you think about it for a second, you will see that every orbit of this action defines a $k$-dimensional subspace of $\mathbb{R}^n$. You can see the Grassmannian as $$ Gr_k(\mathbb{R}^n) = O(n)\big/O(n-k)\times O(k)$$ The orbit space of a free action of a compact Lie group on a manifold is a smooth manifold. Therefore, $Gr_k(\mathbb{R}^n)$ is a smooth manifold.