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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:

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

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 $$ \mathcal{L}_{v_s} \omega:= \frac{\mathrm d }{\mathrm d t} (\rho_{s,t})^*\omega\big|_{t=s} $$

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?

Epod
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Klaas
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2 Answers2

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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 $s$.

Theo Johnson-Freyd
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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).

Todd
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