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Consider the Grassmannian manifold $G(M, N)$ of $M$-dimensional subspaces in $R^N$. I want to approximate (stochastically) an integral of the form $$ \int_{G(M, N)} f(v) \, dv, $$ where $f : G(M, N) \to R$ is some function and $dv$ is the Haar measure on the Grassmannian. I want to approximate the integral with sampling, and therefore I need a method to uniformly draw samples with respect to the measure dv.

I'm happy about hints / references on how to do that.

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  • $\begingroup$ What do you mean by Haar measure here? $\endgroup$
    – hunter
    Nov 26, 2018 at 16:09
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    $\begingroup$ It's the pushforward measure of the Haar measure on the orthogonal group $O(N)$ under the map $f_H(g) = g H, g \in O(N)$, $f_H : O(N) \to G(N, M)$. I found this paper, arxiv.org/abs/math-ph/0609050 which provides a method to uniformly sample from $O(N)$ which by the above solves my question I think. $\endgroup$
    – yon
    Nov 26, 2018 at 16:16
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    $\begingroup$ Fill an $M\times N$ matrix with independent standard Gaussian random variables. The row space will have the desired Haar distribution. $\endgroup$ Nov 26, 2018 at 16:18
  • $\begingroup$ @kimchilover Do you have a reference for that? $\endgroup$
    – Călin
    Oct 7, 2019 at 16:20
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    $\begingroup$ @Călin A standard reference is A. T. James, "Normal multivariate analysis and the orthogonal group". Ann. Math. Statistics 25 (1954), 40–75. But there might be newer & more accessible explanations. If I find a good one I'll post it here. $\endgroup$ Oct 7, 2019 at 17:23

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To expand on one of the comments to your question. Indeed, sampling each component of an $M \times N$ matrix independently from the standard normal distribution yields a random $M$-dimensional subspace of $\mathbb{R}^N$.

Using the same reference as in this (related) question, i.e., Chikuse, Y. (2003). Statistics on Special Manifolds, we have from Theorem 2.2.2. that if the elements of $Z \in \mathbb{R}^{M \times N}$ are i.i.d. from $\mathcal{N}(0, 1)$, then $P = Z (Z^\top Z)^{-1} Z^\top$ is uniformly distributed on the subset of $\mathbb{R}^{N \times N}$ containing rank-$M$ projection matrices $P = X X^\top$, which is just another representation of the Grassmann manifold.

Thus, if you instead represent points via the column space of rank-$M$ matrices from $\mathbb{R}^{N \times M}$, you can use the matrix $Z$ directly. Note that $Z^\top Z$ has almost surely full rank, so its inverse above only amounts to a change of basis which should keep you in the same equivalence class (i.e., same Grassmann point).

Similarly, if you use orthonormal $M$-frames, then Gram-Schmidt should give you yet another representation of the same subspace.

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