An example our professor gave in class for us to work out was that on $M = T^*S^1$, endowed with the canonical symplectic form on the cotangent bundle, the map $\varphi: M \to M$ given by $\varphi(q, p) = (q, p+1)$ is a symplectomorphism, even homotopic to the identity, but not hamiltonian.
By our definition, a symplectomorphism $\varphi$ is hamiltonian if it is of the form $\phi_1^H$, where $\phi_t^H$ is the flow of the hamiltonian vector field corresponding to some (possibly time-dependent) $H \in C^\infty([0,1] \times M)$. The sign convention we use in the definition of hamiltonian vector fields $X_t^H$ is that we require $$ \omega(X_t^H, \cdot) = -dH_t. $$
Idea: I know that an isotopy $\phi_t: M \to M$ is symplectic iff $\imath_{X_t}\omega$ is closed, where $X_t$ is the vector field it defines by $\dot{\phi_t} = X_t(\phi_t)$, and hamiltonian iff it is exact. If I set $\phi_t(q, p) = (q, p + t)$ for $t \in [0,1]$, I get such an isotopy from the identity to $\varphi$, so I have to compute $X_t$ and show $\imath_{X_t}\omega$ defines a nontrivial class in $H^1(M; \mathbb{R})$. My problem lies mainly with the description of the symplectic form on $M$.
Problems:
- Description of $T(T^*S^1)$: We could realize $S^1 \subset \mathbb{C}$ as the unit circle. Then for $q \in S^1$, its tangent space is $T_qS^1 = (\mathbb{R}q)^\perp = \mathbb{R}iq$. The cotangent space thus consists of all linear functionals of the form $p = \langle \beta i q, \cdot \rangle$ for $\beta \in \mathbb{R}$. Then a general path $\gamma$ in $T^*S^1$ should look like $(q(t), p(t))$, where $q$ is a path in $S^1$, so $\langle q(t), q(t) \rangle = 1$, and $p(t)$ is the linear functional $\langle \beta(t)iq(t), \cdot\rangle$ for some real function $\beta$.
Taking the derivative of the constraints on $\gamma$, we get $0 = 2\langle \dot{q}(0), q\rangle$ and $\dot{p}(0) = \langle \dot{\beta}(0)iq + \beta i \dot{q}(0), \cdot \rangle$, where $(q(0), p(0)) = (q, p)$. Hence we should get
$$
T_{(q, p)}(T^*S^1) = \{ (v, \langle w i q + \beta i v, \cdot \rangle) \mid v \in T_qS^1 , w \in \mathbb{R}, p = \langle \beta i q, \cdot \rangle \}.
$$
Identifying $w$ with the functional it defines above, let us just write $((q,p),(v,w)) \in TM$. Then we have $\dot{\phi_t}(q, p) = ((q, p),(0,1))$ for all $t$, so $\phi_t$ is the flow of the vector field $X(q, p) = (0,1)$. Does this make sense to work with/ Is my description correct so far?
- Description of $\omega$: The tautological 1-form is $\lambda_{(q,p)}(v) = p(D\pi(q, p)[v])$ in general, for $\pi: T^*Q \to Q$ the projection. So here, we should have $\lambda_{(q, p)}((v,w)) = p(v) = \langle \beta i q, v \rangle$. To compute the exterior differential, I would have used two paths $\gamma_1, \gamma_2: (-\varepsilon, \varepsilon) \to T^*S^1$ as above, and then used the coordinate free expression of $d$ $$ d\lambda(\dot{\gamma_1}, \dot{\gamma_2}) = \dot{\gamma_1}(\lambda(\dot{\gamma_2})) - \dot{\gamma_2}(\lambda(\dot{\gamma_1})) - \lambda([\dot{\gamma_1}, \dot{\gamma_2}]). $$ If $\gamma_i = (q_i, p_i)$ as above, then $\lambda(\dot{\gamma_i})$ would be the function (in $t$) $p_i(\dot{q_i})$. To compute the expression for $d\lambda$, I need to know how $\dot{\gamma_i}$ acts on this function; I realized that this probably does not make sense as $p_i(\dot{q_i})$ really is a function from $\mathbb{R} \to \mathbb{R}$ and not $M \to \mathbb{R}$. Hence, how do I go about computing this differential? And incidentally, is there a coordinate free version of the exterior differential that lets me do computations with tangent vectors as derivatives of paths?
This is the strategy I came up with so far. Any help would be appreciated, be it with regards to my thoughts above or another approach! I have seen this question, but the explanations are too brief for me to understand.