4
$\begingroup$

Is there an explicit formula to parameterize the Grassmannian $P$ (which is a $(a+b)\times (a+b)$ dimensional matrix) $$P\in \frac{U(a+b)}{U(a)\times U(b)}$$ by $a\times b$ complex independent parameters?

(One may normalize $P$ such that $P^{\dagger}=P, P^2=I_{N}$ where $I_N$ is the $N\times N$ identity matrix. )

$\endgroup$
3
  • $\begingroup$ The wikipedia article en.wikipedia.org/wiki/Grassmannian explains this. $\endgroup$
    – Alan Muniz
    Sep 8, 2018 at 2:38
  • $\begingroup$ @AlanMuniz Thanks! I don't see the wiki page contains a square matrix which has extra unitary structure. Could you point it out? I appreciate your help. $\endgroup$
    – user34104
    Sep 8, 2018 at 20:13
  • $\begingroup$ It is in the section "The Grassmannian as a homogeneous space". The section above it explains how to find the coordinate charts. $\endgroup$
    – Alan Muniz
    Sep 8, 2018 at 20:28

1 Answer 1

4
$\begingroup$

There is a nice local parametrization of the Grassmannian around a fixed subspace. Fix an $a$-dimensional subspace $V\subset\mathbb C^{a+b}$ and consider its orthocomplement $V^\perp$. The parametrization covers $U:=\{W:W\cap V^\perp=\{0\}\}$, which is an open subset in the Grassmannian. Any subspace in this set is the graph of a unique linear map $V\to V^\perp$ (with $V$ itself corresponding to the zero map). To see this, consider the two projections corresponding to $V\oplus V^\perp\cong\mathbb C^n$. By definition of $U$, for any $W\in U$, the restriction of the first projection to $W$ is injective and thus a linear isomoprhism. The map corresponding to $W$ then is the composition of the inverse of the first projection with the restriction of the second projection.

To get an explicit parametrizations, chose orthonormal bases $\{v_1,\dots,v_a\}$ for $V$ and $\{\tilde v_1,\dots,\tilde v_b\}$ for $V^\perp$. Then for a subspace $W\in U$, and each $i=1,\dots,a$, there are unique complex numbers $c_{ij}$ for $j=1,\dots,b$ such that $v_i+\sum_{j=1}^bc_{ij}\tilde v_j$ lies in $W$. Conversely, you associate to a matrix $(c_{ij})$ the subspace spanned by the vectors $v_i+\sum_{j=1}^bc_{ij}\tilde v_j$ for $i=1,\dots,a$, which are linearly independent by construction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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