In symplectic geometry, given a manifold $M$ with closed nondegenerate symplectic 2-form $\omega$, we say that a vector field $X$ is symplectic if $$\mathcal L_X\omega=0\,.$$

I don't know what does it mean to say that $X$ "preserves" the symplectic form, if it is symplectic, how does it preserves the symplectic form ?


1 Answer 1


See sufficient conditions for a flow to be symplectic for the details. In general, suppose we have the following

  • a smooth manifold $M$
  • a smooth vector field $X$ on $M$ with flow $\Phi$ (and for simplicity think of the case where $\Phi$ is defined on $\Bbb{R}\times M$)
  • a tensor field $T$ (of any type) on $M$.

Then, we have that $\mathcal{L}_XT=0$ if and only if for each $s\in\Bbb{R}$ we have $\Phi_s^*T=T$. So, the vanishing of the lie derivative is the "infinitesimal description" (i.e at the level of derivatives) of the statement that the corresponding flow $\Phi$ "always preserves" the tensor field $T$ (here preservation just means is equal even after pulling back).

This definition doesn't require $T$ to be a symplectic form, or anything. This is a completely general observation about how "preservation of tensor fields" is related to Lie derivatives.

In symplectic geometry one calls vector fields $X$ satisfying $\mathcal{L}_X\omega=0$ symplectic vector fields. This is then equivalent to their flows being symplectic-isomorphisms.

In (pseudo)Riemannian geometry, we call a vector field $X$ satisfying $\mathcal{L}_Xg=0$ (where $g$ is the metric tensor field) Killing vector fields. This is then equivalent to saying that the flows of Killing vector fields are isometries. Other standard way of talking about this is that "Killing vector fields are infinitesimal isometries" or "Killing vector fields are infinitesimal generators of isometries" etc.

  • $\begingroup$ Thank you very much @peek-a-boo ! $\endgroup$
    – Mira
    Commented Oct 22, 2021 at 4:20

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