Grassmannian as a quotient of orthogonal or general linear group 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?
 A: 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.
A: This is a common construction in geometry. If a manifold $M$ admits a transitive action by some Lie Group $G$ then we can identify it with the quotient $G/H$ of the Lie group by the stabiliser of a point. An easy way to see this is as follows. Take a point $x \in M$. Any other point $y\in M$ is equal to $gx$ for some $g\in G$ because the action of $G$ is transitive. If $H_x$ is the stabiliser of our point $x$ then $hx = x$ and thus $ghx = gx$ so we quotient out the action of $H$. Thus we get a bijective map $G/H_x \to M;gH_x \mapsto gx$. The point we chose doesn't really matter as the action of $G$ is transitive and so we usually just write $G/H$. It is not too hard to see that $G/H$ is a manifold and the bijective map is a ($G$-equivariant) diffeomorphism.
The example you're interested in, the Grassmannian, has quite a few permitted transitive Lie group actions. The action of of $GL(V)$ is a natural one (although I'm partial to using $SL(V)$ or even $PSL(V)$ so we have a semisimple Lie group) and the stabiliser for a point (that is some subspace of $V$) is a (maximal) parabolic subgroup. Now instead look at the action of $O(V)$ which is still transitive. We can see that the stabiliser of a subspace consists of the orthogonal transformations which preserve the subspace and its orthocomplement (where the inner product on $V$ is the one we've used to define $O(V)$). Thus it is isomorphic to $O(r)\times O(n-r)$ so that $G(r,n) \cong O(n)/O(r)\times O(n-r)$. Now, in this case there is in fact a canonical metric that is preserved by the $O(n)$ action. So the Grassmannian can be viewed as a Riemmannian manifold (indeed a Riemannian symmetric space) in this way. Note however that the action of $GL(V)$ does not preserve any such metric.
As a side point, we have to be careful calling the general linear group the group of invertible matrices. This relies heavily on a choice of basis. It is more prudent to think of them as invertible linear transformations.
