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I am trying to constrain the space of matrices used for the layers of a neural network to those in 𝑆𝑂(𝑛). It is proven that 𝑆𝑂(𝑛) is a manifold. I'm trying to find a way to smoothly traverse this manifold.

Note this is a bit different than just sampling 𝑆𝑂(𝑛) using something like Graham-Schmidt. The function used to construct a matrix would need to be, and ideally run in linear time rather than the cubic time needed for Graham-Schmidt. The sampling of 𝑆𝑂(𝑛) would not need to be uniform.

I believe I read a post here before that had a function for constructing said matrices, but I haven't been able to find it again.

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  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    May 3, 2022 at 15:30
  • $\begingroup$ Not requiring a uniform or other distribution leaves too much freedom, such as selecting the same matrix all the time. $\endgroup$
    – Ruy
    May 3, 2022 at 17:35
  • $\begingroup$ @Ruy it's okay to always start at the same initial point, so long as I can traverse the manifold from the starting point. $\endgroup$
    – jbbj94
    May 3, 2022 at 17:48
  • $\begingroup$ What exactly do you mean by "traversing". $\endgroup$
    – Ruy
    May 3, 2022 at 17:52

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If I understand your question correctly you are looking for a parametrization of $SO(n)$ that you can then perhaps also use for optimization procedures. Such a method has, for example, been discussed here. I am not an expert on this, but essentially we can use the exponential map $$\exp(A) := I + A + \frac{A^2}{2} + \dots $$ to translate an optimization problem over a compact group such as $SO(n)$ into a problem over the space of skew-symmetric matrices. The latter can be parametrized more easily (see the paper for details).

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    $\begingroup$ Small comment: The exponential map of a Lie group is not in general surjective, but it is surjective for the group $SO(n)$. $\endgroup$
    – Levent
    May 3, 2022 at 19:32
  • $\begingroup$ Thank you! The exponent of a skew-symmetric matrix was indeed what I was looking for. Thanks as well for helping bridge some of the terminology. $\endgroup$
    – jbbj94
    May 4, 2022 at 16:37

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