Metric on Steifel and Grassmannian manifolds generalizing Fubini-Study If $F$ is $\mathbb{R}, \mathbb{C}$, or $ \mathbb{H}$, the Grassmannian manifold $G_k(\textbf F^n)$ is the space of all $k$ dimensional subspaces of the $n$ dimensional vector space $F^n$. The Stiefel manifold $V_k(\textbf F^n)$ is the set of $k$-tuples representing $k$ orthonormal vectors in $F^n$. In other words $$ V_k(\textbf F^n) = \{A\in\text{Mat}_{n\times k}(\textbf F^n)|A^\ast A = I_{k\times k}\}.$$ 
There is a natural projection $V_k(\textbf F^n)\longrightarrow G_k(\textbf F^n)$ sending a $k$-tuple to the $k$-dimensional subspace that it spans. The fiber of this projection over each point is all $k$-tuples that live in a fixed $k$-dimensional subspace of $\textbf F^n$, which can be thought of $V_k(\textbf F^k) = O(k,\textbf F)$. 
So we have the fibrations $$ O(k,\textbf F)\rightarrow V_k(\textbf F^n)\longrightarrow G_k(\textbf F^n).$$
When $k=1$, these reduce to the Hopf fibrations \begin{eqnarray*}S^0&\rightarrow& S^{n-1}\longrightarrow \mathbb{R}P^{n-1}\\
S^1&\rightarrow& S^{2n-1}\longrightarrow \mathbb{C}P^{n-1}\\
S^3&\rightarrow& S^{4n-1}\longrightarrow \mathbb{H}P^{n-1}\end{eqnarray*}
If each of the spheres $S^{n-1}, S^{2n-1},$ and $S^{4n-1}$ are given the round metric, there are "natural" metrics on $\mathbb{R}P^{n}, \mathbb{C}P^{n},$ and $\mathbb{H}P^{n}$, respectively, defined as the metrics that make these submersions Riemannian submersions. When $\textbf{F} = \mathbb{R}$, this is the constant curvature 1 metric on $\mathbb{R}P^n$ and when $\textbf F=\mathbb{C}$, this is the Fubini-Study metric on $\textbf{C}P^n$. 
Question: Are there "natural" metrics on $V_k(\textbf{F}^n)$ which give generalization of this? That is, is there a generalization of the Fubini-study metric to $G_k(\textbf{F}^n)$. If so, (where) can I find out more information about these metrics? If not, why won't it work?
 A: There is no need to proceed via the Stiefel-Manifolds. You can directly realize the Grassmannians as homogeneous spaces. In the real case, this takes the form $G(k,n)=SO(n)/S(O(k)\times O(n-k))$. This corresponds to a so-called symmetric decomposition of the Lie algebra $\mathfrak{so}(n)$, thus making $G(k,n)$ into a compact symmetric space. (The Riemannian metric is induced from the restriction of the Killing form of $\mathfrak{so}(n)$ to the orthocomplement of $\mathfrak{o}(k)\times\mathfrak{o}(n-k)$). The  resulting metric is relatively simple, but does not have constant curvature. There is a lot known about such spaces, "Riemannian symmetric spaces" is the right key-word to start looking for. The picture for the complex and quaternionic fields is quite similar, you just have to replace $SO$ by $SU$ respectively $Sp$ (quaternionically unitary group).   
A: It is an standard fact that the space of real grassmanian $G(k,n)$ is  homeomorphic to all projections in $M_{n}(\mathbb{R})$ with trace $k$.(A projection is a matrix with $A=A^{tr}=A^{2}$. The inner product of $M_{n}(\mathbb{R})\simeq \mathbb{R}^{n^{2}}$ is $trace(AB^{tr})$
This inner product is invariant under the action of $O(n)$. So actually O(n) is acting on G(k,n) isometricaaly and transitively. The action is transitive since every two projections with the same rank are unitary equivalent, a standard fact in linear algebra. It is also mentioned in "k theory and C* algebra, A freilndly approach" By Wegg Olsen.
 this would implies that the metric has constant curvature.
