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For this question I will use the definition that a canonical transformation is a map $T(q,p)$ from the phase space onto itself, which leaves the symplectic 2-form invariant (which is the definition of a symplectomorphism). I know that if the topology of the phase space is allright (which means there are no holes, and any closed path can be drawn to a point), then any vectorfield $V(q,p)$, with a flow $F(q, p, \alpha)$ with $\frac{\partial F}{\partial \alpha} = V(q,p)$ and $F(q,p, 0) = (q, p)$ can be generated by a function on the phase space.

What I'd like to know now is wether every symplectomorphism (not a flow, just a symplectomorphism) $T(q,p)$ can be seen as a flow, that means, for every T, I find a flow $F$ so that $T(q,p) = F(q, p, \alpha)$ for a certain value of $ \alpha $.

For example, take the symplectomorphism $T(q, p) = (p, -q)$. I could represent this symplectomorpohism with the Flow $$ F(q, p, \alpha) = (cos(\alpha) q + sin (\alpha) p , -sin(\alpha) q + cos (\alpha) p) $$ Then it would hold that $F(q, p, \frac{\pi}{2}) = T(q, p)$. Is that true in general? do I find such a flow for any symplectomorphism I want to look at?

I couldn't think of a counter example, but I don't know how to show it either.

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OP asks good questions.

First of all, let us mention that there is a bijective correspondence between 1-parameter symplectic flows and parameter-independent$^1$ symplectic vector fields. The latter are by definition vector fields $X\in \Gamma(TM)$ that preserve the symplectic 2-form ${\cal L}_X\omega=0$.

OP is essentially asking the following (some of it in previous versions of the question).

  1. Is the symplectomorphism group path-connected?

Answer: Not necessarily, there could be topological obstructions.

2D counterexample: Let the phase space $M=\mathbb{S}^2$ be the 2-sphere equipped with the standard symplectic 2-form $\omega$. The second homotopy group $\pi_2(\mathbb{S}^2)\cong\mathbb{Z}$ is non-trivial. $\Box$

  1. Is any symplectomorphism the 1-time of a 1-parameter symplectic flow, i.e. is it the exponential $\exp(X)$ of a symplectic vector field $X$?

Answer: Not necessarily, there could be topological obstructions.

Conjectured 2D counterexample: Consider the phase space $M=\mathbb{R}^2$ with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. Consider the canonical transformation

$$ \begin{bmatrix} Q \cr P \end{bmatrix} ~=~ A\begin{bmatrix} q \cr p \end{bmatrix}, \qquad A~:=~\begin{bmatrix} -2 & 1\cr 3 & -2 \end{bmatrix}~\in~Sp(2,\mathbb{R})~=~SL(2,\mathbb{R}). $$

We claim that this symplectomorphism can not be generated by a symplectic flow. A tell-tale fact is that the matrix $A$ has no square root.

Another clue is that this symplectomorphism only has the origin as a fixed point. This means that a symplectic vector field $X$ (for a flow, if it exists) can at most vanish at the origin.

Interestingly, one can write this symplectomorphism as a canonical transformation with a type-2 generating function

$$F_2(q,P)~=~-\frac{1}{2}qP +\frac{3}{4}q^2 - \frac{1}{4}P^2. $$

However we conjecture that a 1-parameter deformation from the identity $F_2(q,P)=qP$ must always go through a singular point.

See also this related Math.SE post. $\Box$

  1. Is any symplectic vector field $X\in \Gamma(TM)$ a Hamiltonian vector field?

Answer: Not necessarily, there could be topological obstructions. In fact, this is measured by the first Poisson cohomology group.

2D counterexample: Consider the phase space $M=\mathbb{R}^2\backslash\{(0,0)\}$ with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. One may check that the vector field $$X=\frac{q}{q^2+p^2}\frac{\partial}{\partial q} +\frac{p}{q^2+p^2}\frac{\partial}{\partial p} $$ is symplectic but it is not a Hamiltonian vector field. The problem is that the candidate ${\rm arg}(q+ip)$ for the Hamiltonian generator is multi-valued, and hence not globally well-defined.

See also e.g. this & this related Phys.SE posts. $\Box$

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$^1$ Note that there exists a notion of flow corresponding to a parameter-dependent vector field. We shall not consider this in this answer. Such flows are used in e.g. this related MO.SE post.

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    $\begingroup$ Qmechanic gives good answers. $\endgroup$ – Quantumwhisp Mar 11 '18 at 21:54
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    $\begingroup$ Notes for later: (The following is only further speculations/conjectures lacking proofs.) Consider symplectomorphisms of the form $$ Q~:=~ f(q),\qquad P ~:=~p/f^{\prime}(q),$$ on the cotangent bundle $M=T^{\ast}\mathbb{S}^1$with the symplectic 2-form $\omega =\mathrm{d}p\wedge \mathrm{d}q$. Then the exponential map $\exp:{\rm Vect}(M,\omega)\to {\rm Symp}(M,\omega)$ is not even locally surjective arbitrarily close to the identity. $\endgroup$ – Qmechanic Mar 12 '18 at 11:01
  • $\begingroup$ Canonical transformations corresponding to parity transformations in odd spacial dimensions seem to provide rather trivial counterexamples. $\endgroup$ – Dexter Kim Mar 14 '18 at 1:48
  • $\begingroup$ The post you referred to about parameter dependent flows just makes statements on a 2 dimensional phase space. If you consider parameter dependent flows in your answer, would the answer still be no for the question wether one could always find a parameter dependent flow that describes a given canonical transformation? $\endgroup$ – Quantumwhisp Mar 15 '18 at 17:22
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    $\begingroup$ @Dexter Kim: If we identity $z=q+ip$, then a parity transformation $Q=-q$ can be reached via OP's rotation flow $Z(z,\alpha)=e^{i\alpha}z$ with $\alpha=\pi$. $\endgroup$ – Qmechanic Mar 19 '18 at 10:43

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