I hope that I am just confused, but I don't see why a sympletic map must have a determinant equal to one and not minus one?-Could anybody help me with that?- I am referring to the group
$$Sp = \{T \in GL(\mathbb{R}); TJT^T = J\},$$
where $J$ is the canonical matrix, see e.g. here mathworld
If anything is unclear, please let me know.
This answer is similar to Charlie Frohman's, but may also be helpful. The symplectic form determined by $J$ is the skew-symmetric bilinear map $\omega : V \times V \to \mathbb{R}$, defined as $\omega(v,u) = v^{T}Ju$, and $(T^{*} \omega)(v,u) = \omega(Tv,Tu) = v^{T} T^{T}J Tu.$ So the condition you wrote above is exactly that $T^{*} \omega = \omega$. The symplectic form $\omega$ determines an orientation via its top exterior power $\omega^{n} = \omega \wedge ... \wedge \omega$, and $T^{*} \omega^{n} = (T^{*}\omega)^{n} = \omega^n$. So $T^{*} \omega$ and $\omega$ determine the same orientation. Which is to say that if a collection $v_{1},...,v_{n}$ is positively oriented, then so is $Tv_{1}, ..., Tv_{n}$. Therefore, the symplectic transformations must preserve orientation. This is equivalent to saying that $\det(T) = 1$.
