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



Parametrizing quotient of matrices by the orthogonal group
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.


