Does $X^T Y=I$ define a manifold?

Question

Consider the set of pairs (X,Y) of real $n\times k$ matrices ($k\le n $) defined by $X^T Y=I$. Is this set a manifold? And if so, what is the tangent space at a point $(X_0,Y_0)$?

Answer 1

Let $E=\{(X,Y)|X^TY=I_k\}$; $X^TY=I_k$ implies that $rank(X)=rank(Y)=k$. Then $(X^T)^+=X(X^TX)^{-1}$ and $X^T(X^T)^+=I_k$.

Moreover $Y=(X^T)^++(I_n-(X^T)^+X^T)Z$ where $Z\in M_{n,k}$ is arbitrary.

Note that $P_X=I_n-(X^T)^+X^T$ is a projector of rank $n-k$. One may write that the set of solutions in the unknown $Y$ is $Y=(X^T)^++\{U\in M_{n,k}|X^TU=0_k\}$, that is $im(U)\subset \ker(X^T)=(im(X))^\perp$; note that $dim((im(X))^\perp)=n-k$ and that $dim(im(U))\leq k$. Thus $U\in L(\mathbb{R}^k,\ker(X^T))$, a vector space of dimension $k(n-k)$. Finally, $E$ is an algebraic set of dimension $nk+k(n-k)=2nk-k^2$; then the $k^2$ algebraic relations associated to $X^TY=I_k$ are algebraically independent. Let $R=\{(X,U)|X^TU=0,rank(X)=k\}$. A pseudo-parametrization of $E$ is $(X,U)\in R\rightarrow (X,(X^T)^++U)\in E$. Since $E$ has no auto-intersection, it is a (not connected) manifold.

I did not study the tangent space. We can use that follows; if $f:X\rightarrow (X^T)^+$, then $Df_X:H\rightarrow -(X^T)^+H^T(X^T)^++(I-(X^T)^+X^T)HX^+(X^T)^+$.