Is the notion of symplectic matrix independent of the choice of $J$?

Question

A $2n\times 2n$ matrix $A$ is called symplectic if $A^T J A = J$, where $J$ is a fixed invertible, skew symmetric matrix. Generally, $J$ is taken to be the block matrix $J = \begin{pmatrix} 0 & I_n \\ -I_n & 0\end{pmatrix}$.

Is the notion of symplectic matrix independent of the choice of $J$?

Answer 1

As mentioned above, the groups are all isomorphic, however they are not all the same subset of $M_{2n}(\mathbb R)$.

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Example. From MathWorld we have the following nice examples of symplectic matrices:

$$I_4, X = \begin{pmatrix}1&0&0&1\\0&1&1&0\\0&0&1&0\\ 0&0&0&1\end{pmatrix}, Y = \begin{pmatrix}0&1&0&1\\1&0&1&0\\0&0&0&1\\0&0&1&0\end{pmatrix}$$

Let's now try a different skew-symmetric nondegenerate matrix like $$J = \begin{pmatrix}0&0&2&0\\0&0&0&1\\-2&0&0&0\\0&-1&0&0\end{pmatrix}$$

A computation shows $X^TJX \neq J$. You can copy this following into WolframAlpha to check it:

[[1,0,0,1],[0,1,1,0],[0,0,1,0],[0,0,0,1]]^T*[[0,0,2,0],[0,0,0,1],[-2,0,0,0],[0,-1,0,0]]*[[1,0,0,1],[0,1,1,0],[0,0,1,0],[0,0,0,1]]

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Please note that one should not be that surprised by this! The same thing happens for instance with orthogonal groups. If one changes to a basis that is not orthonormal, the orthogonal matrices expressed in the new basis will no longer belong be in the standard subset $O_n \subset M_n(\mathbb R)$.

There is a little disparity between Wikipedia's definitions of symplectic matrix and orthogonal matrix in this sense. Orthogonal groups are of the form $A^TIA = I$ where $I$ is the identity matrix, but for a general nondegenerate symmetric bilinear form one replaces $I$ with any invertible symmetric matrix $S$. The group $\{A \in GL_n(\mathbb R) | A^TSA = S\} \cong O_n$ is certainly not equal to $O_n$ for all the same reasons as above.

Answer 2

Let $W^{2n}$ be a real vector space of dimension $2n$ and let $\omega$ be an alternating bilinear form which is non-degenerate, then according to the linear symplectic Darboux's theorem, there exists $(e_1,\ldots,e_n,f_1,\ldots,f_n)$ a basis of $W$ such that: $$\omega=\sum_{k=1}^n{e_k}^*\wedge{f_k}^*,$$ said differently the matrix of $\omega$ in this basis is equal to $J$, indeed for all $(i,j)\in\{1,\ldots,n\}^2$, one has: $$\omega(e_i,e_j)=\omega(f_i,f_j)=0,\omega(e_i,f_j)=\delta_{i,j}.$$ Whence, the group of linear endomorphisms of $W$ that preserves $\omega$: $$\{\ell\in\operatorname{End}(W);\forall(v,w)\in W^2,\omega(\ell(v),\ell(w))=\omega(v,w)\}$$ is isomorphic to the usual set of symplectic matrices (taking their matrices in the Darboux basis), which is now an intrinsic definition for $\operatorname{Sp}(2n)$.

Remainders.

• A bilinear form $\omega$ is non-degenerate if and only if for all $v\in W$, $\omega(v,\cdot)\colon W\to W^*$ is a linear isomorphism.

• The proof of the linear symplectic Darboux's theorem is done by induction on the half dimension of $W$.