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I am reading ana cannas' lecture notes to try understanding a bit about symplectic geometry.

At page 147, there is this definition :

Definition 24.2 A G-invariant function f : M → R is called an integral of motion of (M, ω, G, μ). If μ is constant on the trajectories of a hamiltonian vector field $v_f$ , then the corresponding one-parameter group of diffeomorphisms $\{exp(tv_f) | t ∈ \mathbb{R}\}$ is called a symmetry of (M, ω, G, μ).

Until now, I actually thought that an integral of motion is a function $f$ which is constant on the trajectories. So how is this definition motivated?

roi_saumon
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    If you take any 1-dimensional (local) subgroup of $G$, you end up with a (local) 1-parameter family of diffeomorphisms. You can think of those as giving you trajectories. Your function $f$ is constant on these. Is this the idea you are looking for? Do you understand the moment map $\mu$? – Sam Lisi Apr 30 '19 at 00:06
  • is this what you mean by local subgroup? Maybe this is what I am looking for but I am not sure to understand. How does it generate a family of diffeomorphism? Why do we need to $f$ be G-invariant? What I understand about the moment map : If we have an element g in the lie algebra of $G$, which I understand as an element that acts infinitesimaly on M, it generates from $X$ a new vector field $X^#$. And $\mu^X$ is kind of the energy of this vector field, i.e. $\mu^X$ is constant on the integral curves of $X^#$. – roi_saumon Apr 30 '19 at 09:49
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    No, that's not what I meant. Consider a simpler case. Let's suppose that $G = S^1$. Then, the Lie algebra is $\mathbb{R}$, and your moment map is the Hamiltonian that generates this $S^1$ action. In this case, the condition that $f$ be $G$-invariant is precisely the one of asking it to be constant on trajectories. – Sam Lisi May 01 '19 at 13:36
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    Consider now a higher dimensional Lie group. A neighbourhood of the identity is the image of a neighbourhood of $0$ in the Lie algebra under the exponential map. If I take a line segment passing through $0$ in the Lie algebra, and look at its image under the exp map, I get something that is like a 1-dimensional subgroup of $G$. The product of two things is defined as long as they are sufficiently small. If the product gets big, they leave the neighbourhood and I don't want to think about them anymore. This is what I meant by a local subgroup. – Sam Lisi May 01 '19 at 13:39
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    Anyway, this point of view allows you to think of the $G$-action as being given by many "directions" of Hamiltonian vector fields with corresponding Hamiltonian functions encoded in $\mu$. Having the function $f$ be $G$-invariant means that it is constant on trajectories of all these different Hamiltonian vector fields. – Sam Lisi May 01 '19 at 13:40
  • Could you please add some details on the phrases : your moment map is the Hamiltonian that generates this S1 action? Also when you say A neighbourhood of the identity is the image of a neighbourhood of 0 in the Lie algebra under the exponential map, it means that there locally a bijection $exp : U \subset g \rightarrow exp(U) \subset G$? How to think about the correspondance $X \leftrightarrow exp(X)$. What do they mean? Is $X$ like the infinitesimal effect of the action while $exp(X)$ the action due to the infinitesimal actions $X$ after $1$ second. I no this may be total nonsense... – roi_saumon May 01 '19 at 17:41

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