2
$\begingroup$

Let $V=(V,b)$ be a finite-dimensional vector space equipped with $b$ a symmetric and positive definite bilinear form. And let $\{e_1,\dotsc,e_n\}$ be a orthonormal basis for the subspace $\ker((P_A)^t)$ ($P_A$ is defined below).

For a matrix $A \in \mathrm{O}(V)$, let $\mathrm{O}_*(V)$ be the subset of $\mathrm{O}(V)$ such that the matrix $P_A:=\frac{A-JAJ}{2}$ is invertible, where $J$ is a complex structure (a matrix such that $J^2=-1$).

Let $n=\dim \ker(P_A)$. For every $j \in \{1,\dotsc,n\}$ we define the reflexions $r_j$ such that $r_j(e_j)=-Je_j$, $r_j(Je_j)=-e_j$ and $r_j(v)=v$ for any $v \in V$ such that $b(v,e_j)=b(v,Je_j)=0$. Finally, let $$R:=r_1r_2\dotsm r_n \in \mathrm{O}(V).$$

I need to prove that $$RA \in \mathrm{SO}_*(V),$$ where similarly as $\mathrm{O}(V)$: $\mathrm{SO}_*(V)$ is the subset of $\mathrm{SO}(V)$ such that $P_B:=\frac{B-JBJ}{2}$ is invertible.

I already proved that $RA \in \mathrm{SO}(V)$; the only thing that I haven’t been able to figure out is to prove that $\frac{1}{2}(RA-JRAJ)$ is invertible, since $n$ can be even or odd.

Also, $P_{r_j}$ is not invertible since $\det(r_j)=-1$.

What is a good and optimized approach to deal with the product of reflections $$R=r_1r_2\dotsm r_n?$$

In summary: I’m trying to prove that $RA$ (where $R$ is the product of reflections $r_1r_2\dotsm r_n$) is an orthogonal matrix with $\det(RA)=+1$ and that the matrix $\frac{1}{2}(J-J(RA)J)$ is invertible.

Any help will be greatly appreciated.


UPDATE: Two things:

  1. I made a typo, $\{e_1,\dotsc,e_n\}$ is an orthonormal basis for the subspace $\ker((P_A)^t)$, not $\ker(P_A)$.

  2. On the main reference I'm using, the author establishes the following:

This operators (each reflection $r_j$, with $j \in \{ 1,\dotsc, n \}$) has the identity restricted to the subspace orthogonal to $e_j$ as $p_{r_j}$. Let $R:=r_1\dotsm r_n$, then the operator $R$ is in block form, its lower right corner being the identity on $(\ker((P_A)^t))^{\perp}$. And so $RA \in \mathrm{SO}_*(V)$.

And that's it, I think the author's argument has many gaps or things that I'm not getting. He says he's following Ruijsenaars - On Bogoliubov Transformations. II. The General Case, but I have read it multiple times and I don't see anything like what I'm trying to prove (or at least with this notation).

$\endgroup$
9
  • 2
    $\begingroup$ Related (from 2 days ago): mathoverflow.net/questions/425926/… $\endgroup$ Commented Jul 5, 2022 at 4:28
  • 1
    $\begingroup$ Is $p_A$ the same as $P_A$? $\endgroup$
    – LSpice
    Commented Jul 7, 2022 at 3:42
  • 1
    $\begingroup$ Yes, sorry. I’ll correct it right away. $\endgroup$ Commented Jul 7, 2022 at 3:51
  • 1
    $\begingroup$ @SanHopkins Sadly, $\det(A)$ can be either $-1$ or $1$. :( $\endgroup$ Commented Jul 9, 2022 at 3:52
  • 1
    $\begingroup$ You refer to "the main reference I'm using". What is it? $\endgroup$
    – LSpice
    Commented Jul 9, 2022 at 21:05

0

You must log in to answer this question.