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!
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.