I am trying to parametrize the collection of $d\times m$ real matrices quotient $d\times d$ orthogonal matrices. Formally, define $\sim$ on $\mathbb{R}^{d\times m}$ by $X\sim Y$ if there exists an orthogonal $d\times d$ matrix $Q$ such that $QX = Y$. Then $\sim$ is an equivalence relation on $\mathbb{R}^{d\times m}$. I want to parametrize $\mathbb{R}^{d\times m}/{\sim}$. I think this quotient set is a kind of manifold, just cannot figure it out. Please help. I want to know what the dimension of this manifold is and how to parametrize it. Thanks
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$\begingroup$ It suffices to just work with the Gram matrix of a $d\times m$ matrix, since this is all one cares about (as $Y^{T}Y=X^{T}Q^{T}QX=X^{T}X$). When the columns of $X$ are unit norm, Gram matrices are identified with the name 'Elliptope' in various contexts, (see manopt.org/reference/manopt/manifolds/symfixedrank/… for example). $\endgroup$– Josiah ParkCommented Oct 14, 2019 at 4:47
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1$\begingroup$ Also, if you require the matrices to be exactly of rank $d$, then the space is a manifold. $\endgroup$– Josiah ParkCommented Oct 14, 2019 at 14:46
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$\begingroup$ Thanks for providing so many useful materials to look at. Yes, for my purpose, the matrices are of rank d. I am trying to optimize a function on this quotient space. Do you know whether this manifold (under rank d assumption) is compact or not? $\endgroup$– Min WuCommented Oct 14, 2019 at 15:32
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$\begingroup$ Also, how do I parametrize this manifold? (Since I still need the parametrization to deal with the optimization problem.) $\endgroup$– Min WuCommented Oct 14, 2019 at 15:41
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1$\begingroup$ For the rank $d$ matrices, the matrices can be parametrized as they are here: manopt.org/reference/manopt/manifolds/fixedrank/…. This follows the embedded geometry described in Bart Vandereycken's 2013 paper: "Low-rank matrix completion by Riemannian optimization". $\endgroup$– Josiah ParkCommented Oct 14, 2019 at 16:04
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1 Answer
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It is not a manifold, but there are still ways to talk about the dimension of the quotient space in terms of principal orbits. Chapter 4 of G.E. Bredon's Introduction to compact transformation groups, Academic Press, 1972 is a useful reference on this topic, specifically Theorem 3.8 there (re-phrased below).
3.8. Theorem$\ \ $Let $G$ be a compact group acting locally smoothly on the $n$-manifold $M$ with $M^{*}=M/G$ connected. If $d$ is the dimension of a principal orbit (i.e., the maximal dimension among all orbits), then $\dim M^{*}=n-d$.
This theorem gives that the dimension of the quotient space $\mathbb{R}^{d\times m}/{\sim}$ is given by the highest dimension of an element of the orbit space (that is, the quotient space has dimension equal to the dimension of the principal orbits). So the dimension of the quotient space (for $m\geq d$) is $dm-\binom{d}{2}$, as one would expect.
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$\begingroup$ I made the same typographical correction in this answer as in the question. $\endgroup$ Commented Mar 27, 2021 at 20:12