6
$\begingroup$

In one of Ruelle`s papers "Rotation Numbers for Flows and Diffeomorphisms" Ruelle has the following calculation which I do not understand completely. Assume you have two invertible $2 \times 2$ matrices $A$ and $B$ with polar decompositions $A = U(\theta(A))|A|$ and $B = U(\theta(B))|B|$ where $U(\theta)$ is the planar rotation matrix by $\theta$ and $|B|=\sqrt(BB^T)$ etc. Then he says that $$ |\theta(AB)-\theta(A)-\theta(B)| \leq \pi $$

I don`t quite understand how he gets this result without a constant depending on norms of A and B. One can start by saying $$ AB = U(\theta(AB))|AB| = U(\theta(A))|A| U(\theta(B))|B| $$ $$ = U(\theta(A)+\theta(B))U(-\theta(B))|A|U(\theta(B))|B| $$ so that $$ U(\theta(AB)-\theta(A)-\theta(B)) = U(-\theta(B))|A|U(\theta(B))|B||AB|^{-1}. $$ Somewhere in the paper he gives as a hint $|\theta(PQ)| \leq \pi$ if $P$ and $Q$ are positive but I cant see how to use it.

1 Answers 1

3

Let $P=U(-\theta(B))|A|U(\theta(B))=U(\theta(B))^tAU(\theta(B))$ and $Q=|B|$ then

$AB=U(\theta(A))|A|U(\theta(B))|B|=U(\theta(A))U(\theta(B))PQ=U(\theta(A)+\theta(B))PQ$.

Hence, $PQ=U(-\theta(A)-\theta(B))U(\theta(AB))|AB|=U(\theta(AB)-\theta(A)-\theta(B))|AB|=\theta(PQ)|PQ|$.

By the uniqueness of the polar decomposition of an invertible matrix, we get $\theta(PQ)=\theta(AB)-\theta(A)-\theta(B)$.

Since $P$ and $Q$ are positive definite then $|\theta(PQ)|<\pi$ and the result follows.

  • 0
    I suggest using multiple lines for systems of equalities. Thie last line of math reads completely wrong on mobile because mobile doesn't play well with long equations. – 2017-01-23
  • 0
    Thanks, turned out to be simpler than I imagined just one more polar decomposition on top of the hint he gives :) Not sure why the question itself got so much positive feedback though. Maybe they liked the paper :p – 2017-01-23
  • 0
    @Sina You are welcome – 2017-01-23
  • 0
    @Daniel, A follow up question: Let $g_1,g_2$ be the orthogonal axis scaled by $|B|$ and $e_1,e_2$ be the one scaled by $|A|$. Let $h_1,h_2$ the $e_1,e_2$ rotated by $-\theta(B)$. Then is it true that $\theta(PQ)$ is the smallest angle between the frames $g_1,g_2$ and $h_1,h_2$. This is because $P$ simply acts as scaling on $h_1,h_2$ and $Q$ in $g_1,g_2$. – 2017-01-24
  • 0
    I now think this is not true but if you have any comments its welcome. – 2017-01-24