2
$\begingroup$

I'm trying to understand some different ways to construct the Grassmannian of a real vector space, but I'm having trouble getting some of the notation and definitions.

One definition that I often see given, such as on Wikipedia, is

$$Gr(r,n) = O(n)/(O(r) \times O(n–r))$$

where $Gr(r,n)$ is the Grassmannian of $r$-dimensional subspaces of $\Bbb R^n$.

My first question is: what does this quotient notation mean? $O(n)$ is the group of $n\times n$ orthogonal matrices, whereas $O(r)$ and $O(n-r)$ are the groups of $r \times r$ and $(n-r) \times (n-r)$ orthogonal matrices. How are we viewing the cartesian product of $O(r) \times O(n-r)$ as a group of $n \times n$ matrices?

A different way to construct this is also given by Wikipedia:

First, recall that the general linear group GL(V) acts transitively on the r-dimensional subspaces of V. Therefore, if H is the stabilizer of any of the subspaces under this action, we have

$$Gr(r,V) = GL(V)/H$$

This is also somewhat puzzling to me. You start with the general linear group of invertible $r \times r$ matrices, which acts transitively on subspaces, meaning for any two subspaces $a$ and $b$ there is some element of $GL(V)$ that maps $a$ to $b$.

But then we quotient by what stabilizer, exactly, to construct the Grassmannian as a quotient space?

The only stabilizer that preserves all subspaces are simple scalar transformations; if we mod by those we get the projective linear group $PGL(V)$, not the Grassmannian.

If on the other hand, they mean that we quotient by all such $H$ that are a stabilizer for any subspace, this means that we are quotienting by every element of $GL(V)$ that has any real eigenspace, which is almost the entire group, except perhaps for rotation matrices in even dimensions, where all eigenvalues are complex.

How is this thing supposed to be constructed? And how do these yield any type of metric structure?

$\endgroup$
  • 1
    $\begingroup$ They mean the stabilizer of an $r$-dimensional subspace. For example the stabilizer of $\mathbb{R} \subset \mathbb{R}^2$ is given by the upper triangular matrices $B$ and $\operatorname{GL}_2/B$ is isomorphic to $\mathbb{P}^1$. As for your first question I am pretty sure they consider the map $O(n-r) \times O(r) \to O(n)$ given by $(A,B) \mapsto ((A,0),(0,B))$ where the latter denotes a block matrix. $\endgroup$ – user45878 May 26 at 8:10
2
$\begingroup$

Answer to first question: there is a very often abuse of notation here. The block diagonal subgroup $\begin{pmatrix}A_r\\&A_{n-r}\end{pmatrix}$ is isomorphic to $O(r)\times O(n-r)$, and we are quotienting $O(n)$ by this subgroup. There could be other isomorphic copies of $O(r)\times O(n-r)$ which could give different quotients, but they are of no interest to us.

Let's also settle what quotienting does here. The quotient manifold theorem says

Suppose $G$ is a Lie group acting smoothly, freely and properly on a smooth manifold $M$. Then the orbit space $M/G$ is a topological manifold of dimension $\dim M-\dim G$ and has a unique smooth structure with the property that $M\to M/G$ is a smooth submersion.

In particular, the quotient of a Lie group by a Lie subgroup is a smooth manifold. You don't get a "metric structure" just from this information, although you could define a (noncanonical) Riemannian metric on patches arbitrarily and glue using a partition of unity.

Answer to second question: Note it is "$H$ is the stabilizer of any ...", not "$H$ is the stabilizer of all ...", so $H$ just stabilizes one $r$-subspace of your choice.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.