Timeline for Is it true $\left\|\log(RS)\right\|≤\left\|\log(R)+\log(S)\right\|$ for all $R,S \in \mathrm{SO}(3)$, where $\|\cdot\|$ is the Frobenius norm?
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
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| May 9, 2023 at 0:58 | comment | added | Spencer Kraisler | Let us continue this discussion in chat. | |
| May 9, 2023 at 0:18 | comment | added | Robert Bryant | About your question about the textbook, I don't have it in front of me, so I'll have to wait until tomorrow to answer that. | |
| May 9, 2023 at 0:15 | comment | added | Robert Bryant | (cont....) For example, consider the exponential map from the tangent space at the north pole $n=(0,0,1)$ of the unit $2$-sphere to the $2$-sphere, so that $\exp_n(r\cos\theta,r\sin\theta,0) = (\sin r\cos\theta,\sin r\sin\theta,\cos r) = (x,y,z)$, then $\exp_n^*(dx^2+dy^2+dz^2) = dr^2+\sin^2r\,d\theta^2$, whereas the flat metric on the tangent plane is $dr^2 + r^2\,d\theta^2$. So the difference is $(r^2-\sin^2r)d\theta^2\ge0$. | |
| May 9, 2023 at 0:09 | comment | added | Robert Bryant | Well. $\exp:T_xM\to M$ is the $g$-geodesic exponential map, and $\exp^*(g)$ is the pullback of $g$ under this smooth map, so its quadratic form (with variable coefficients on the vector space $T_xM$ thought of as a smooth manifold. We can think of $g_x$ which is a quadratic form on $T_xM$ as a Riemannian metric with constant coefficients (so it's determined by its value at the origin $0_x\in T_xM$. The statement $exp^*(g)\le g_0$ is the statement that, at every point $v\in T_xM$, the quadratic differential $g_0 - \bigl(exp^*(g)\bigr)_v$ is nonnegative. (cont...) | |
| May 8, 2023 at 23:46 | comment | added | Spencer Kraisler | I have the textbook in front of me, seems to be something related to differential invariants, of which I lack a bit of understanding. | |
| May 8, 2023 at 23:25 | comment | added | Spencer Kraisler | Could you explain what $\exp^*(g) \leq g_0$ means? What is $\exp^*$ and how can it be "less than" an inner product? Also, would you know where in Helgason I could find this proof? | |
| May 8, 2023 at 17:34 | comment | added | Robert Bryant | I guess the generalization to complete Riemannian manifolds $(M,g)$ with nonnegative sectional curvature is that, for any $x\in M$, we have $\exp_x^*(g)\le g_x$ as quadratic forms on $T_xM$, which would then imply that $d\bigl(\exp_x(a),\exp_x(b)\bigr) \le |a-b|$ for $a,b\in T_x$ for which $\alpha(t) = \exp_x(ta)$ and $\beta(t) = \exp_x(tb)$ for $0\le t\le 1$ are $g$-length minimizing geodesics between their endpoints. | |
| May 8, 2023 at 16:44 | comment | added | Spencer Kraisler | Wow! What a remarkable inequality. Although I am not an expert in Lie theory, I have read quite a few papers and sections on the topic in textbooks. Though I never came across anything like this. I wonder if this could be generalized to compact Riemannian manifolds with non-negative sectional curvature. | |
| May 8, 2023 at 16:43 | vote | accept | Spencer Kraisler | ||
| May 8, 2023 at 12:35 | history | edited | Robert Bryant | CC BY-SA 4.0 |
Cleaned up some typos
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| May 8, 2023 at 12:30 | history | answered | Robert Bryant | CC BY-SA 4.0 |