On Symplectic Geometry

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It is said that "the symplectic geomtery is rather loose, like that of rubber sheet which may be stretched by different amounts in different directions and still retain its geomtery".

But the scaling transformation, in general, is not a symplectomorphism. For example, for $(x,\xi)\in \mathbb{R}^2$. Let $T(x,\xi)= (ax,b\xi)$ for some constants $a,b$ such that $ab\neq 1$. Let the symplectic form on $\mathbb{R}^2$ be given by \begin{equation} \sigma\Big((x,\xi),(y,\eta)\Big)=\xi y-\eta x. \end{equation} Clearly $\sigma\Big(T(x,\xi),T(y,\eta)\Big)=ab(\xi y-\eta x)\neq \sigma\Big((x,\xi),(y,\eta)\Big)$.

Can any one help me understand what has been said?

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The symplectic form in $\Bbb R^2$ is the area form, so any scaling done in one direction must be undone in the other direction, as to preserve area. Such a scaling will be a symplectomorphism if and only if $b = 1/a$ (in view of your calculation, this might make more sense than explicitly saying that the condition is $ab=1$, but I suppose this is mainly psychological).

Maybe I should add the following generalization: if $V$ is any (finite-dimensional, say) vector space and $V^*$ is it's dual, then $V\oplus V^*$ has a symplectic structure given by $$\big((x,\xi), (y, \eta)\big) \mapsto``\begin{vmatrix} x & \xi \\ y & \eta\end{vmatrix}"= \eta(x) - \xi(y),$$and the transformation $$V\oplus V^* \ni (x,\xi) \mapsto (ax,b\xi) \in V\oplus V^*$$again will be a symplectomorphism if and only if $b = 1/a$.

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In you example, if you are stretching the two directions, one with $a$ and the other with $1/a$, then the "geometry is preserved". In this sense, it only means that the symplectic form is preserved (which means the area is preserved, in two dimensions).

I think this is just a matter of interpretation. It says that stretching by different amounts will preserve the geometry. This is true because, in general $a\neq 1/a$, so this type of stretching is by different amounts. He didn't say any stretch by different amounts will preserve the geometry.