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I am going through the definition of symplectic matrices. While searching through various properties of those class of matrices I found that every symplectic matrix is necessarily invertible and the inverse of such matrices are again symplectic. Then out of curiosity I just ask myself a question $:$ Are the symplectic matrices closed under taking transpose? After several ineffectual attempts I have tried searching it in MSE and here's what I found. I have tried to use the same trick mentioned there earlier but I don't get the point how $\Omega^{-1}$ becomes $\Omega$ in this equation. I know that $\Omega$ is the matrix representation of some skew-symmetric non-degenerate form. So the second equation there clearly suggests that either $\Omega$ is involutary or orthogonal but I can't deduce any of these facts from the given condition. Could anyone please shed some light on it?

Thanks a bunch!

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  • $\begingroup$ I'm not sure what you don't understand in the answer you link. Just write the symplectic condition for $\Omega^{-1}$, as is done in the answer, and then take the inverse on both sides. This will give the symplectic condition for the transpose of $\Omega$. $\endgroup$
    – Captain Lama
    Mar 19 at 16:34
  • $\begingroup$ @CaptainLama$:$ $\Omega$ is not symplectic. Rather, in the answer $S$ is symplectic. The author writes the condition of symplecticity for $S^{-1}$ and then told us to invert the condition. Then finally we ended up with $S \Omega^{-1}S^{\top}=\Omega^{-1}.$ But we want $S \Omega S^{\top}=\Omega.$ When can it happen? I am just trying to explore the possibilities that under what condition on $\Omega$ these two relations represent the same thing and I found that $\Omega=\lambda \Omega^{-1}$ i.e. $\Omega^2 =\lambda I$ for some scalar $\lambda.$ But what is this scalar actually in our case? contd... $\endgroup$
    – Anacardium
    Mar 19 at 18:13
  • $\begingroup$ ...This is what I intended to know. Hopefully my query is clear now. $\endgroup$
    – Anacardium
    Mar 19 at 18:14
  • $\begingroup$ @NicholasTodoroff$:$ Wait! Wait! Wait! Don't be in a hurry or pretend to become super smart. How do you get the second equation from the first? Could you add one more line of argument? $\endgroup$
    – Anacardium
    Mar 19 at 18:34
  • $\begingroup$ @Anacardium I deleted my comment because I was wrong. $\endgroup$
    – Nicholas Todoroff
    Mar 19 at 21:12

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I think there is a small confusion. Let us say that is $\Omega$ is an invertible anti-symmetric matrix, then $S$ is $\Omega$-symplectic if it satifies $S^t\Omega S=\Omega$. Then the $\Omega$-symplectic matrices form a group, so if $S$ is $\Omega$-symplectic then so is $S^{-1}$.

On the other hand, $S^t$ is in general not $\Omega$-symplectic but $\Omega^{-1}$-symplectic, following the computation you did.

But if you use the classical $\Omega=J$, then $J^{-1}=-J$, so being $\Omega^{-1}$-symplectic is the same as being $\Omega$-symplectic.

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  • $\begingroup$ But it should be noted that there is a basis such that $\Omega$ takes on the form $J$, i.e. there is a change of basis matrix $A$ such that $J = A^T\Omega A$. What goes wrong is that $S$ transforms to $S' = A^{-1}SA$ and we know $(S')^T$ is symplectic wrt. $A^T\Omega A$, but transforming back what we get is $A(S')^TA^{-1} = (AA^T)S^T(AA^T)^{-1} \ne S^T$ is symplectic. $\endgroup$
    – Nicholas Todoroff
    Mar 19 at 22:32
  • $\begingroup$ What is $J$ in your answer? If the answer I linked is confusing then how did it get $6$ upvotes and how does one accept it as an answer? $\endgroup$
    – Anacardium
    Mar 20 at 17:32

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