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Just a beginner in symplectic geometry, and the definition of symplectic matrix bothers me. A $2n\times 2n$ real matrix $M$ is said to be symplectic if it satisfies the following condition: $$M^T\Omega M=\Omega$$ where $\Omega$ is a fixed $2n\times 2n$ real, invertible and skew-symmetric matrix.

My question is: since $\Omega$ can be arbitrary, so if $\Omega,\Delta$ are both satisfy the condition, then the following statement must be true: $$M^T\Omega M=\Omega \Rightarrow M^T\Delta M=\Delta.$$

But I don't know how to prove this. Can anyone help me? Thanks.

Arc Walker
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2 Answers2

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There are more than one symplectic structure oon a vector space, but they are isomorphic not equal, there exists a linear invertible map such that $f\circ\Delta =\Omega\circ f$ where $\Omega$ and $\Delta$ are the linear map associated to the corresponding matrices.

Tsemo Aristide
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Actually, what you are trying to prove is not strictly true since the definition of the symplectic matrix is basis dependent. To demonstrate this, consider the similarity transformation $M' = D^{-1} M D$ and define $\Delta = D^T \Omega D$. Then the following proposition can be shown with some matrix algebra:

$$M^T\Omega M=\Omega \Rightarrow M'^T\Delta M'=\Delta.$$

In fact, choosing $D$ appropriately allows you to write the symplectic bilinear form in the symplectic basis:

$$\Delta = \begin{bmatrix} 0 & \mathbb{1}_{n \times n} \\ -\mathbb{1}_{n \times n} & 0 \end{bmatrix}. $$

setnoset
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