In "Lectures on Symplectic Geometry" by A. C. da Silva (http://www.math.ist.utl.pt/~acannas/Books/lsg.pdf) the author gives the following definition:

$$ \mathcal{L}_{v_t} := \frac{\mathrm d }{\mathrm d t} (\rho_t)^*\omega\big|_{t=0} $$

where $\rho_t$ satisfies $$ \frac{\mathrm d \rho_t}{\mathrm d t} = v_t\circ\rho_t \qquad \text{and} \qquad \rho_0 = \mathrm{id}. $$

I wonder if this actually makes sense. For time-independent vector fields $v_t=v$ it totally does, but in the time dependent case I have the following objections:

  1. The right-hand side of the definition of $\mathcal{L}_{v_t}$ does not use the parameter $t$. Or is the $t$ in the left-hand side just to denote that we have a time-dependent vector field? But on page 40 the author uses the Cartan formula $$ \mathcal{L}_{v_t}\omega = i_{v_t}\mathrm{d\omega} + \mathrm{d}i_{v_t}\omega $$ where the right-hand side certainly depends on the parameter $t$.
  2. The formula $$ \frac{\mathrm d}{\mathrm dt}\rho_t^*\omega = \rho_t^*\mathcal{L}_{v_t}\omega$$ given on page 36 seems to be wrong when you use the definition of $\mathcal{L}_{v_t}$ given above.

For me everything works when I define instead $$ \mathcal{L}_{v_s} := \frac{\mathrm d }{\mathrm d t} (\rho_{s,t})^*\omega\big|_{t=s} $$

where $\rho_{s,t}$ satisfies $$ \frac{\mathrm d \rho_{s,t}}{\mathrm d t} = v_t\circ\rho_{s,t} \qquad \text{and} \qquad \rho_{s,s} = \mathrm{id}. $$

Does this make sense to you?

  • $\begingroup$ Can anybody help me whith this? $\endgroup$
    – Klaas
    Jul 17, 2013 at 17:34

2 Answers 2


Your corrections are correct. The definition you quoted gives you $\mathcal L_{v_t}|_{t=0}$. Since $\mathbb R$ acts on everything by $t \mapsto t+s$, this definition lets you extract the value of $\mathcal L_{v_s}$ for any other $s$.


I think this works as well. Intuitively, if you are at point $p$ at time $s$, then travelling along the time dependent vector field, you would be at point $\rho_{s,t}(p)$ at time $t$. This interpretation helps to motivate the adapted definition of $\mathcal{L}_{v_s}\omega$ the OP gives. For the proofs, we can write $\rho_{s,t} = \rho_t \circ \rho_s^{-1}$, and this expression allows us to prove the desired formulae with relative ease (compared to the time independent case).


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .