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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?

Mud
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2 Answers2

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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).

Andreas Cap
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  • Does the cone over a Grassmannian $G(k,n) = SO(n)/S(O(k)\times O(n-k))$ manifold admit an Einstein metric? – Bilateral Jul 31 '15 at 22:12
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    @Bilateral : I don't know the answer to this quesiton. Let me just remark that there is a big difference between question on the $SO(n)$-invariant geometry of $G(k,n)$ (which can esssentially be studied by linear algebra respectively finite dimensional representation theory), and questions about existence of other metrics (or geometric structures) on $G(k,n)$. For the latter, it is much more difficult to exploit the fact that you are dealing with a homgeneous space. – Andreas Cap Aug 02 '15 at 13:11
  • Thanks, I understand. Do you know if someone has checked if if the $SO(n)$-invariant metric on $G(k,n)$ is Einstein on the corresponding cone over $G(k,n)$? (Since as you said the $SO(n)$-invariant metric on $G(k,n)$ is not Einstein). – Bilateral Aug 02 '15 at 18:45
  • @Bilateral: l am pretty sure that the $SO(n)$-invariant metric on $G(k,n)$ is Einstein. What I said is that it does not have constant curvature (which refers to sectional curvature). The argument that it is Einstein should be as follows: The Ricci tensor of the $SO(n)$-invariant metric itself defines a (possibly degenerate) bundle metric on $TG(k,n)$. Such a metric corresponds to a bilinear form on the space of $k\times (n-k)$-Matrices which is invariant under the natural action of $O(k)\times O(n-k)$ (by multiplication from both sides). But there is only one such form up to constant factors. – Andreas Cap Aug 03 '15 at 07:56
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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.

Ali Taghavi
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