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I'm trying to prove the Grassmann manifold $\mathrm G_k(\mathbb R^n)$ of $k$-dimensional linear subspaces of $\mathbb R^n$ is isotropic and symmetric.

By "isotropic" I specifically that for every point $P \in \mathrm G_k(\mathbb R^n)$, and for every pair of unit tangent vectors $u, v \in T_P \mathrm G_k(\mathbb R^n)$, there is an isometry $\phi : \mathrm G_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ fixing $P$ and with $d\phi_P u = v$. By "symmetric" I mean for every $P \in \mathrm G_k(\mathbb R^n)$, there is an isometry $\phi$ fixing $P$ and so that $d\phi_P = -\mathrm{Id} : T_P \mathrm G_k(\mathbb R^n) \to T_P \mathrm G_k(\mathbb R^n)$. There appear to be a few different definitions for each, and I want to be clear with which one I'm using.

My setting: $\mathrm G_k(\mathbb R^n)$ has a unique smooth structure with respect to which the natural action of $\mathrm{GL}(n,\mathbb R)$ on $\mathrm G_k(\mathbb R^n)$ is smooth. To define the Riemannian metric, we consider the Stiefel manifold $\mathrm V_k(\mathbb R^n)$ of $k$-tuples of orthonormal vectors in $\mathbb R^n$. This may be considered a submanifold of the space $\mathrm M(n \times k, \mathbb R)$ of $n \times k$ real matrices, with the Euclidean metric from identifying $\mathrm M(n \times k, \mathbb R) \approx \mathbb R^{nk}$. The map $\pi : \mathrm V_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ sending a $k$-tuple of orthonormal vectors to its span is a surjective smooth submersion. One can show $\mathrm O(k)$ acts on the right of $\mathrm V_k(\mathbb R^n)$ isometrically, vertically (meaning for $A \in O(k)$, $B \in \mathrm V_k(\mathbb R^n)$, we have $\pi(BA) = \pi(B)$), and transitively on fibers (meaning if $B$ and $B'$ are two orthonormal $k$-tuples with the same span, there's some $A \in \mathrm O(k)$ with $BA = B'$). It follows that there is a unique Riemannian metric on $\mathrm G_k(\mathbb R^n)$ with respect to which $\pi : \mathrm V_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ is a Riemannian submersion, meaning for each $B \in \mathrm V_k(\mathbb R^n)$, $d\pi_B : \ker\left(d\pi_B\right)^\perp \to T_{\pi(B)} \mathrm G_k(\mathbb R^n)$ is a linear isometry. Call this metric $g$ on $\mathrm G_k(\mathbb R^n)$.

My strategy: I know $\mathrm O(n)$ acts on the left transitively and isometrically on $\mathrm V_k(\mathbb R^n)$, and hence acts transitively and isometrically on $\mathrm G_k(\mathbb R^n)$, so $\mathrm G_k(\mathbb R^n)$ is homogeneous. So I only need to prove $\mathrm G_k(\mathbb R^n)$ is isotropic and symmetric at a single point; say the subspace $P = \mathbb R^k \subset \mathbb R^n$, spanned by the first $k$ coordinates. The isotropy group in $\mathrm O(n)$ of this point is $G_P := \mathrm O(k) \oplus \mathrm O(n-k)$. It seems intuitively obvious that with an appropriate choice of matrix $A \in \mathrm O(k) \oplus \mathrm O(n-k)$, the differential of the action map $\theta_A : \mathrm G_k(\mathbb R^n) \to \mathrm G_k(\mathbb R^n)$ may either reflect the tangent space at $P$ or act transitively on unit tangent vectors, but I'm having trouble with the specifics. Plus the coordinates of the Grassmannian seem kind of weird and intimidating. Is there a coordinate-free way to make this argument rigorous?

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    $\begingroup$ That the Grassmannians are symmetric spaces is classic and in many differential geometry books. However, other than the case of projective spaces, what makes you think that they are isotropic? Is $S^2\times S^2$ isotropic, for example? (Generally, you don't want to work with coordinates. You want to work with a Lie algebra decomposition of $\mathfrak o(n)$.) $\endgroup$ – Ted Shifrin Jun 6 at 19:03
  • $\begingroup$ Is $S^2 \times S^2$ isometric to a Grassmannian? I wasn't aware of that. But since $S^2$ is isotropic, it seems like $S^2 \times S^2$ should be as well, should it not? As to why I think Grassmannians are isotropic in the first place, this is an exercise in my book on Riemannian manifolds (Lee, 2nd ed.) $\endgroup$ – D Ford Jun 6 at 19:51
  • $\begingroup$ Yes, the Grassmannian of oriented $2$-planes in $\Bbb R^4$ is isometric to $S^2\times S^2$ (with a shrunken radius). Very few manifolds are isotropic. Has Lee even had you understand how to think about Riemannian symmetric spaces and the tangent bundle of the Grassmannian? At any rate, in the $S^2\times S^2$ case, I doubt there is an isometry carrying $(v,0)$ to $\frac1{\sqrt2}(v,v)$. Can you see why? $\endgroup$ – Ted Shifrin Jun 6 at 21:20
  • $\begingroup$ Uh-oh -- this is a mistake in my book. The problem should just say "homogeneous and symmetric," not isotropic. I've added a correction to my online list. $\endgroup$ – Jack Lee Jun 14 at 23:01

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