I'm trying to prove that the Grassmann manifold $$G_k(\mathbb{R}^n) = \{E = {\rm {\it k} - dimensional\ subspace\ of\ } \mathbb{R}^n\}$$
is equivalent to:
$$G_k(\mathbb{R}^n) = \frac{O(n)}{O(k)\times O(n - k)} \tag1$$
Where $O(n)$ is the orthonormal group of $n\times n$ matrices.
From my research I've seen that Eq. (1) is due to the idea of splitting the original $n$-dimensional subspace into a $k$-dimensional one and its orthonormal complement of $n - k$ dimension; but I don't get how this Grassmann manifold, which is made of vectors ($(k\times 1)$-dimensional matrices - column vectors), is related to $n\times n$ matrices since the quotient in Eq. (1) is the following set, as usual:
$$\frac{O(n)}{O(k)\times O(n - k)} = \{M_n\cdot (O(k)\times O(n - k))\ |\ M_n \in O(n)\} \tag2$$
Can anyone explain me this relation and the prove of Eq. (1)? Thanks in advace! ;)
Let $S_k(\mathbb{R}^n)$ be the Steifel manifold of $k$-frames in $\mathbb{R}^n$: $$ S_k(\mathbb{R}^n) = \{(v_1, \ldots, v_k) \mid \text{the $v_i$ are orthonormal} \}$$ The group $O(n)$ acts transitively on $S_k(\mathbb{R}^n)$ by acting on each vector: $g \cdot (v_1, \ldots, v_k) = (gv_1, \ldots, gv_k)$. The stabiliser of a frame $(v_1, \ldots, v_k)$ will be the subgroup of $O(n)$ which fixes the span of $(v_1, \ldots, v_k)$ and acts nontrivially on its orthogonal complement. This subgroup is isomorphic to $O(n - k)$, and hence we have $$ S_k(\mathbb{R}^n) \cong O(n) / O(n - k)$$ Next, there is a natural map $\phi: S_k(\mathbb{R}^n) \to G_k(\mathbb{R}^n)$ by taking the span of the frame. The fibre of this map over a point $E \in G_k(\mathbb{R}^n)$ will be the set of $k$-frames spanning $E$, which is a set isomorphic to $O(k)$ (an isomorphism of $O(k)$-sets). Hence $$G_k(\mathbb{R}^n) \cong S_k(\mathbb{R}^n) / O(k)$$ These together give the equality you were after.