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Jan 23, 2018 at 14:12 comment added Eric Reckwerdt @YCor I see, so the quote might not be anything new, just a contextualization of the SO(n) result. I'll check through that Lubotzky reference. If I find a reasonable citation, I'll add it to the wikipedia article. Thanks for the prompt response!
Jan 22, 2018 at 20:58 comment added YCor @ao. I'm not sure what you're asking, but this is a theoretical result. It roughly says that every matrix can be well-approximated by some product, but does not say that there is a procedure to compute efficiently this approximation. I don't know what is the state of art on this question.
Jan 22, 2018 at 19:00 comment added Turbo @YCor what is the closest you have explicitly?
Jan 22, 2018 at 18:58 comment added YCor @ao. hopefully yes but maybe not in 4 lines... I remember nice slides in Sevennec's talk (in Lebesgue measure's 100th birthday, in 2001) with pictures of well and bad distributed points on the sphere, unfortunately they're no in the linked pdf.
Jan 22, 2018 at 18:25 comment added Turbo @ycor might be possible to make it concrete an engineer understands?
Jan 22, 2018 at 17:51 comment added YCor PS 1) I meant convergence in $L^2$-operator norm. 2) This is surveyed in Sevennec's article (in French) umpa.ens-lyon.fr/sevennec/txt-lebesgue.ps
Jan 22, 2018 at 17:02 comment added YCor I think the idea is that if we have a dense subgroup in $G=SO(n)$ with Property T generated by a finite subset $S$, the spectral gap ensures a fast convergence of the $n$-fold convolution $(1_S)^{\ast n}$ to the constant $1_G$, in a suitable sense. Possibly a good reference is Lubotzky's 1994 Birkhäuser book "Discrete groups, expanding graphs, and invariant measures".
Jan 22, 2018 at 16:55 history edited YCor
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Jan 22, 2018 at 16:07 history edited Eric Reckwerdt
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Jan 22, 2018 at 15:20 review First posts
Jan 22, 2018 at 15:24
Jan 22, 2018 at 15:16 history asked Eric Reckwerdt CC BY-SA 3.0