Timeline for An inequality on some pairs of orthogonal vectors
Current License: CC BY-SA 4.0
8 events
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| Oct 27, 2020 at 0:10 | comment | added | fedja | @DavidESpeyer The difficulty is that the extreme case is still the uniform distribution on the 1-dimensional circle, which, even after your parameterization, is not a uniform distribution on the full space, i.e., we cannot just reduce the game to a single "Fourier coefficient" (let's just assume that all vectors have unit length, say, so the second trick with $f,g$ is not needed). But, of course, I'll be happy to be proved wrong in my skeptical view :-) | |
| Oct 26, 2020 at 13:02 | comment | added | David E Speyer | It seems to me that computing the Peter-Weyl expansion of $|x_{ij}|$ is probably doable, at least for $SO(3)$. I am still working on making the rest of fedja's argument abstract enough for me to understand it, though. | |
| Oct 26, 2020 at 13:01 | comment | added | David E Speyer | It seems to me that it is not hopeless to generalize this to higher dimensions. To make this fit in the comments, I'll do $k=3$. Let $a_i = r_i u_i$ and $b_i = s_i v_i$ where $u_i$ and $v_i$ are unit; let $g_i$ be the matrix in $SO(3)$ with columns $(u_i, v_i, u_i \times v_i)$ and let $\mu$ be the counting measure of the $g_i$, with $\mu(SO(3)) =1$. We want to bound $\int_{SO(3)} (r \mu) \ast (s \mu) |x_{12}|$ where the convolution is by $(g_1, g_2) \mapsto g_1^T g_2$ and $x_{ij}$ are the coordinates from $SO(3) \subset GL(3)$. | |
| Oct 3, 2020 at 14:32 | comment | added | fedja | @IvanMeir I'm not sure what references you want: the only thing I use is the identity $\int(\mu*K)d\bar\nu=\sum_n \widehat K(n)\widehat\mu(n)\overline{\widehat\nu(n)}$. The only equality case is that of uniformly distributed on the circle directions and equal lengths, so no finite $n$ configuration can attain it. As to fixed angle, no, the proof doesn't generalize to that case because a) you have two options now, so it is no longer a pure convolution and b) even if you fix the orientation, the Fourier coefficients of the kernel are no longer nice enough to run the argument. | |
| Oct 3, 2020 at 13:11 | comment | added | Ivan Meir | Also can you generalise your proof to the case where the orthogonality condition is replaced by $a_i^Tb_i=|a_i||b_i|\cos(\theta_i)$ so the $a_i$'s and $b_i$'s differ by fixed angles rather than $\pi/2$? I was wondering what might be the equality condition in this case for $k=2$? | |
| Oct 3, 2020 at 10:45 | comment | added | Ivan Meir | Very interesting thank you! Do you have any references that might help one understand the proof. Also would you mind clarifying the case for equality in terms of the vectors $a_i$? | |
| Oct 3, 2020 at 10:27 | comment | added | Giorgio Metafune | Thank you for the very nice proof. It is short, but not really "simple". | |
| Oct 2, 2020 at 19:22 | history | answered | fedja | CC BY-SA 4.0 |