Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I came across a certain generalization of Grassmann manifolds and was wondering what work if any has been done on it. If you take the space of $n\times p$ real matrices, $n>p$, and define an equivalence relationship

$A\sim B \Leftrightarrow \exists M\in GL(p)\mid AM=B$

where $GL(p)$ is the space of invertible $p\times p$ real matrices, then the quotient space defined by this relation is the space of all $p$-dimensional subspaces of $\mathbb{R}^n$, called the Grassmann manifold $Gr(n,p)$.

If you generalize this equivalence relation to also include the addition of a constant to each column:

$A\sim B \Leftrightarrow \exists M\in GL(p), x \in \mathbb{R}^p\mid (A1_n)\left(\begin{array}{c}M\\x^T\end{array}\right)=B$

what is the manifold given by the corresponding quotient space? Does it have a name? Are there analytic solutions for the geodesics on this manifold, like there are for Grassmann manifolds?


This came up while trying to figure out how to maximize Gaussian mutual information under certain constraints. If you have two data matrices $X\in\mathbb{R}^{n\times T}$ and $Y\in\mathbb{R}^{m\times T}$, then the empirical Gaussian mutual information between them can be computed as

$\hat{I}[X;Y] = \frac{1}{2}\mathrm{log}|M_nXX^TM_n^T| - \frac{1}{2}\mathrm{log}|M_nXX^TM_n^T-M_nXY^TM_m^T(M_mYY^TM_m^T)^{-1}M_mYX^TM_n^T|$

where $M_n:=I_n-1_{n\times n}/n$ is the matrix that subtracts the mean from each row of a matrix. This function is invariant under left multiplication of $X$ or $Y$ by an arbitrary matrix in $GL(n)$ or $GL(m)$ respectively, but also under addition of an arbitrary vector in $\mathbb{R}^{n}$ or $\mathbb{R}^{m}$ to each column. I'd like to do optimization of this function directly in the quotient space defined by this equivalence relationship, instead of on $\mathbb{R}^{n\times T} \times \mathbb{R}^{m\times T}$, which is precisely the Cartesian product of the manifold I defined above.


Edit: I realized my previous comment about this manifold being larger than $Gr(n,p)$ is wrong. The equivalence relation is more general than the equivalence relation for $Gr(n,p)$, so the manifold is smaller. It also occurs to me that the construction I described can be given as $GL(n)/(GL(n-p)\times Aff(p)) = GL(n)/(GL(n-p)\times (\mathbb{R}^p \rtimes GL(p)))$ where $Aff(p)$ is the $p$ dimensional affine space. I believe this is equivalent to $O(n)/(O(n-p)\times(\mathbb{R}^p \rtimes O(p))$ where $O(n)$ is the group of $n$ dimensional orthogonal matrices, since $Gr(n,p)$ can be constructed as $O(n)/(O(n-p)\times O(p))$.

share|improve this question

1 Answer 1

This is just the Grassmannian $Gr(n-1,p)$. You can use the freedom to add a constant to each column to make the last entry in each column be $0$, and then you are looking at $(n-1) \times p$ matrices modulo $GL_p$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.