# Comparing two Riemannian metrics on Grassmannian

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?

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 $$w_1 = \dots = w_r$$.