4

A $2n \times 2n$ matrix $S$ is symplectic, if $SJ_{2n}S^T=J_{2n}$ where \begin{equation} J_{2n} = \begin{bmatrix} 0 & I_n \\ -I_n & 0 \end{bmatrix}. \end{equation} My question is, how to prove that $\det S =1$, without using Pfaffian, as given in the wikipedia article.

RSG
  • 1,293
  • 1
    For future reference, these are related questions without any constraints on the method of proof: http://math.stackexchange.com/questions/242091/why-is-the-determinant-of-a-symplectic-matrix-1 http://math.stackexchange.com/questions/501130/prove-that-det-a-1-with-at-m-a-m-and-m-beginbmatrix-0-i-i http://math.stackexchange.com/questions/930319/sympletic-matrix-must-have-a-determinant-equal-to-one – Chris Culter May 27 '15 at 07:34

3 Answers3

3

Great timing! Check out this recent preprint: Donsub Rim, "An Elementary Proof That Symplectic Matrices Have Unit Determinant", http://arxiv.org/abs/1505.04240. It proceeds by decomposing the inequality $\det(S^TS+I)\geq1$.

Chris Culter
  • 26,806
  • Wow! I missed it. Thanks a lot! Can you suggest some books where symplectic group and it associated mathematical concepts, like Euler decomposition, Iwasawa decomposition, Williamson normal form (for strict positive real matrices) etc. have been studied in details - preferably from a matrix analysis point of view. Can you suggest some books or papers in that direction? – RSG May 27 '15 at 07:31
  • 1
    @RSG Sorry, I've long since forgotten what little I once knew on the topic. :) You might want to raise another question with tag:reference-request for that. – Chris Culter May 27 '15 at 07:37
2

If $S$ is a symplectic matrix, it preserves the standard symplectic form $\omega=\sum_{i=1}^n dx_i\wedge dy_i$, i.e. $S^*\omega=\omega$. Note that $\displaystyle\frac{\omega^n}{n!}$ is the standard volume form of $\mathbb{R}^{2n}$. Now we have \begin{eqnarray} \det(S)\frac{\omega^n}{n!}=S^*\left(\frac{\omega^n}{n!}\right)=\frac{\omega^n}{n!} \end{eqnarray} So $\det(S)=1$.

Alex Fok
  • 4,828
1

After the interruptions and warnings of some moderators, I rewrote my answer in more descent way

In Grove's book, Classical groups and geometric algebra, https://bookstore.ams.org/gsm-39 , He showed that symplectic transvections generate the symplectic group. This means that any symplectic matrix can be written as a product of symplectic transvections. I am explaining more below.

Let $(V,\{ .,. \})$ be a symplectic vector space. Let $a$ be a real number and $u \in V$. Then $T_{a,u}(v)=v+a\{v,u\}u$ is called symplectic transvection. Grove showed that any symplectomorphism can be written as a product of symplectic transvections. We already know that the set of symplectomorphisms is isormorphic to the set of symplectic matrices. So what we do for any symplectomorphism is valid for any symplectic matrix.

My aim is to show determinant of any symplectomorphism is 1. So it is equivalent to say that the determinant of product of symplectic transvections is 1.

Determinant of any matrix is product of its eigenvalues. Let us look at eigenvalues of any transvection. $T_{a,u}(v)=\lambda v \longrightarrow v+a\{v,u\}u = \lambda v \longrightarrow (1-\lambda)v=-a\{v,u\}u$.

We have $(1-\lambda)v=-a\{v,u\}u$. So they are linearly dependent if the coefficients are nonzero. But in this case $\{v,u\}$ vanishes because the form is skew symmetric linear. So $\lambda=1$. Assume they are linearly independent, then the coefficients must be zero, so again $\lambda$ is 1.

Therefore determinant of any symplectomorphism is $1$. Thus determinant of any symplectomorphism is $1$.

For details you can check my notes in the following link https://ferhatkarabatman.github.io/positive_transvections.pdf

Red Phoenix
  • 629