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Let $G_r(n)$ be the real Grassmannian which is the collection of all $r$ dimensional subspace in $\mathbb{R}^n$ equipped with the usual invariant metric $g$.

Let $U_A\in\mathbb{R}^{n\times r}$ and $U_B\in\mathbb{R}^{n\times r}$ be the orthonormal basis of $A$ and $B$. Let $1\geq\sigma_1\geq...\geq\sigma_r\geq0$ be the singular values of $U_A^TU_B$. It is well known that the geodesic distance under metric $g$ between two elements $A,B\in G_r(n)$ is the following: $$d_g(A,B)=\sqrt{\sum_{i=1}^r\arccos^2\sigma_i}$$ $\arccos\sigma_i,i=1,...,r$ are also called the principal angles between $A$ and $B$.

Now for a given sequence $w_1,...,w_r>0$, define another distance metric on $G_r(n)$, such that for any subspace $A,B$: $$\tilde{d}_W(A,B)=\sqrt{\sum_{i=1}^rw_i\arccos^2\sigma_i}$$ My questions are the following:

  1. Does there exist another Riemannian metric $\tilde{g}$ on $G_r(n)$ such that the geodesic distance is exactly $\tilde{d}_W$? (Due to Alexandrov geometry?)

  2. If $\tilde{g}$ exists, let $\tilde{\mu}$ be the volume measure induced by $\tilde{g}$ and $\mu$ be the volume measure induced by $g$. For a given $A\in G_r(n)$ and $a>0$, What is the relationship between $\mu(\{B\in G_r(n): \tilde{d}_W(A,B)\leq a\})$ and $\tilde{\mu}(\{B\in G_r(n): d_\tilde{g}(A,B)=\tilde{d}_W(A,B)\leq a\})$ (I'm most interested in)

  3. It is known that $G_r(n)$ under metric $g$ has positive Ricci curvature. Does $G_r(n)$ under metric $\tilde{g}$ still have positive Ricci curvature?

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If we multiply $U_A$ and $U_B$ on the left by the same element of $U(n)$, this will preserve $U_A^T U_B$, thus preserve $\sigma_1,\dots, \sigma_r$ and preserve the metric distance between $A$ and $B$. Because the distance is $U(n)$-invariant, if it it arrises from a Riemannian metric, the Riemannian metric must be $U(n)$-invariant (because the metric is the second derivative of the distance squared, say). But there is a unique $U(n)$-invariant Riemannian metric on the Grassmanian, up to constant factor scaling, because the tangent space at a point is an irreducible representation of the stabilizer $U(r) \times U(n-r)$ at this point. So this can only happen if $w_1 = \dots = w_r$.

One can also see this more computationally by verifying that the second derivative of the distance squared to a fixed point, at that fixed point, does not exist.

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