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Suppose I know that two vectors $\vec{a}$ and $\vec{b}$ are perpendicular in a given basis spanned by basis vectors $\vec{x}$. Now suppose I transform to another basis $\vec{x'}$ using a symplectic transformation matrix S (i.e. $SJS^{T} = J$ for some skew-symmetric matrix J). Will the transformed vectors a and b still be perpendicular after the transformation? If not, is there a way to figure out a relation between the dot products between the two vectors in the two bases?

Thanks!

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    $\begingroup$ I only can add this part of information: $\langle S a,S b\rangle= a^TS^TSb $ if the vectors are considered as one-column matrices. $S$ need not preserve orthogonality. $\endgroup$ – Berci Mar 10 '13 at 21:05
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Perhaps I misunderstand your question, but here is an example that shows the answer is no to what I think you are asking.

Consider $\mathbb{R}^2$ with the standard symplectic structure given by $\omega( (a,b), (a',b')) = ab' - ba'$. This corresponds to saying $\omega = < J \cdot, \cdot>$ where $J$ is the almost complex structure given by $$ \begin{pmatrix} 0 & -1\\1 & 0 \end{pmatrix}$$

An easy symplectic and orthogonal basis for $\mathbb{R}^2$ is given by $e_1 = (1,0)$ and $e_2 = (0,1)$. A symplectic transformation is given by $e_1 \mapsto e_1 + e_2, e_2 \mapsto e_2$. (i.e. in matrix form $ \begin{pmatrix} 1 & 0 \\ 1 & 1 \end{pmatrix}. )$ The image of this basis is clearly no longer orthogonal.

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The answer is no in general. For a general (nonlinear) symplectic transformation, the vectors will be pushed-forward by the Jacobian of the transformation. We also know that jacobian of a symplectic transformation is a symplectic linear transformation. So in general, if what you propose is true, then it should hold for ALL symplectic transformations, and not just linear ones like you propose.

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