Show that the Grassmanian is a manifold using Godement's theorem on $O(n)$

Question

I am trying to prove that the Grassmannian $G_k(\mathbb R^n)$ is a manifold using Godement's theorem on $O(n)$. I know that $G_k(\mathbb R^n) = O(n) / (O(k) \times O(n-k))$, but I am not sure how I can use Godement's theorem, reproduced below, to derive the result.

Godement's thoerem: If $\sim$ is an equivalence relation on a smooth manifold $M$, then $M / \sim$ has a manifold structure such that $\pi : M \to M / \sim$ becomes a submersion, provided that the diagonal $R \subset M \times M$ is a properly embedded submanifold and the restriction $\pi_1|_R : R \to M \times \to M$ where the second arrow is a projection, is a submersion.

The difficulty seems to be that $O(k) \times O(n-k)$ is not an equivalence relation. I am so confused, as well as how to show that the corresponding $R$ is a properly embedded submanifold. Can anyone help?

Answer 1

Here’s what we can say.

Theorem (Godement’s Crtierion).

Let $M$ be a smooth manifold and $\mathcal{R}\subset M\times M$ an equivalence relation on $M$. The equivalence relation $\mathcal{R}$ is regular (i.e $M/\mathcal{R}$ equipped with the quotient topology is such that the topology is Hausdorff and there exists a smooth structure for which $\pi:M\to M/\mathcal{R}$ is a smooth submersion) if and only if $\mathcal{R}$ is a closed subset and a smooth embedded submanifold of $M\times M$ and the projection $\text{pr}_2: \mathcal{R}\to M$ (or equivalently $\text{pr}_1: \mathcal{R}\to M$) is a smooth submersion.

Furthermore, $\dim (M/\mathcal{R})=\operatorname{codim}(\mathcal{R})=2\dim M-\dim \mathcal{R} $.

Now, you can specialize this theorem to the case where the equivalence relation $\mathcal{R}$ is actually the orbit relation arising from a smooth action of a Lie group $G$ on $M$.

Corollary (Orbit Manifold Theorem).

Let $M$ be a smooth manifold, and $G$ a Lie group acting smoothly on $M$ (for concreteness, say from the right), and let $\mathcal{R}=\{(x,y)\in M\times M\,:\text{there is a $g\in G$ such that $x=y\cdot g$}\}$ be the orbit relation. The equivalence relation $\mathcal{R}$ is regular if and only if it is a closed subset and an embedded submanifold of $M\times M$.

In this case, the orbit manifold $M/G$ (this is the usual notation rather than $M/\mathcal{R}$ or $M/\sim$) has dimension equal to $\dim M-\dim G+\dim S_x$ where $S_x\subset G$ is the stabilizer subgroup of any point $x\in M$.

So the point is that when the equivalence relation comes from a group, the fact that the projection $\text{pr}_2: \mathcal{R} \to M$ (or $\text{pr}_1$) is a submersion is automatic. With this, show that quotients of Lie groups by a closed subgroup are again Lie groups (below, $G$ plays the role of $M$ and $H$ that of $G$).

Corollary.

Let $G$ be a Lie group and $H$ a closed subgroup (so it is automatically an embedded submanifold… or for now you can just assume you have an embedded submanifold). Consider the right-action of $H$ on $G$ given by multiplication $(g,h)\mapsto gh$. Then, the orbit manifold $G/H$ exists, and has dimension $\dim G-\dim H$.

To prove this, consider the map $f:G\times G\to G$ given by $f(x,y)=x^{-1}y$. This is a $C^{\infty}$ map and it is a submersion (it’s the composition of the inversion diffeomorphism $G\to G$ with the multiplication submersion $G\times G\to G$). Notice that the orbit relation is \begin{align} \mathcal{R}&=\{(x,y)\in G\times G\,:\,\text{there is an $h\in H$ such that $y=xh$}\}\\ &=\{(x,y)\in G\times G\,:\, x^{-1}y\in H\}\\ &=f^{-1}(H). \end{align} Since $H$ is a closed subset, the preimage is closed. Also, since $H$ is an embedded submanifold and $f$ is a submersion, it follows the preimage is an embedded submanifold. Lastly, the dimension claim is obvious once you realize that this action is free, i.e the stabilizers are all $\{e\}$, so have dimension $0$.

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The usual definition of the Grassmannian $G_k(\Bbb{R}^n)$ is as the set of all $k$-dimensional subspaces in $\Bbb{R}^n$. We now define a left-action of the Lie group $O(n,\Bbb{R})$ on this set $(R,E)\mapsto R[E]$, sending each subspace $E$ to its image $R[E]$ under the orthogonal transformation. This action is certainly transitive (basic linear algebra). Fixing a subspace $E\in G_k(\Bbb{R}^n)$, the (rough) orbit-stabilizer theorem tells us that the mapping $R\mapsto R[E]$ descends to a bijection $O(n,\Bbb{R})/\text{Stab}(E)\to \text{Orbit}(E)=G_k(\Bbb{R}^n)$.

To get a more concrete description of the stabilizer, let us simplify our lives and choose the subspace $E=\Bbb{R}^k\times\{0_{\Bbb{R}^{n-k}}\}$. Note that if $R$ is an orthogonal transformation of $\Bbb{R}^n$ which stabilizes $E$, then it also stabilizes $E^{\perp}$. So, any such $R$ must have a block form \begin{align} R= \begin{pmatrix} A&0\\ 0& B \end{pmatrix}, \end{align} where $A\in O(k,\Bbb{R})$ and $B\in O(n-k,\Bbb{R})$. Hence, the stabilizer $\text{Stab}(E)$ is an embedded Lie subgroup of $O(n,\Bbb{R})$ which is very naturally diffeomorphic to $O(k,\Bbb{R})\times O(n-k,\Bbb{R})$. Hence, by applying our corollary we see that $\frac{O(n,\Bbb{R})}{O(k,\Bbb{R})\times O(n-k,\Bbb{R})}\equiv O(n,\Bbb{R})/\text{Stab}(E)$ has a smooth structure (the latter is the formal meaning of the former, which is the quotient you see).

We can now transport this smooth structure of this quotient onto $G_k(\Bbb{R}^n)$ via the bijection described above, and so we have a smooth structure on the Grassmannian set. It turns out that this smooth structure is actually independent of the specific subspace $E=\Bbb{R}^k\times \{0_{\Bbb{R}^{n-k}}\}$ used.