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If I have an action of a Lie group on a configuration space。

$G\to \text{Diff}(M)$, $g \mapsto \rho_g$, $\rho_g : q \mapsto \rho_g(q)$ (for example a rotation).

Then when we consider the phase spaces $T^*M$, we provide it with the action :

$G\to \text{Diff}(T^*M)$, $g \mapsto \rho^*_{g^{-1}}$, $\rho_{g^{-1}}^* \colon (q,p) \mapsto (\rho_g(q),\rho^*_{g^{-1}}(p))$.

Now I understand that the point $q$ is send to its image under the action, but I don't understand why the moment $p$ is transformed as $\rho^*_{g^{-1}}(p)$ under the action?

Why is the reason we consider this weird action with a pullback on the moment?

Why do we want the moment to transform in this way?

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You want the lifted action to be by exact symplectomorphisms (actually it turns out even better, the lift gives a Hamiltonian action of $G$ on $T^*M$, but that is a bonus).

Suppose you demand that the lifted action commutes with projection and is linear on the fibers. Then on each fiber it is given by a linear map $A_q: T^*_q M\to T^*_{\rho_q} M$. If we demand that the lifted action preserves the tautological 1-form, then since the position part of the tangent vector is transformed by $D\rho_g|_q$, the momentum needs to transform in a way that "undoes" this, i.e. $A_q$ needs to be such that $<p, v>=<A_q p, D\rho_g|_q(v) >$, i.e. $A_q$ is the inverse of the adjoint of $D\rho_g|_q$; since $\rho$ is an action, this is the adjoint of $D\rho_{g^{-1}}$, aka $\rho^*_{g^{-1}}$.

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  • $\begingroup$ Thanks @Max! What do you mean by the position part of the tangent vector? Is it like decomposing a tangent vector $(v,w) \in T_{(q,p)}T^*M$ and saying $v$ is the position part? Why is this position part transformed by $D\rho_g \mid_q$? $\endgroup$
    – roi_saumon
    Jun 11, 2019 at 16:00
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    $\begingroup$ It's a bit more subtle. In the absence of a connection, you can't decompose $T_{(p,q)}T^*M$. But you can still project tangent vector from $T_{(p,q)}T^*M$ to $T_q M$ by $D \pi$. The image of that projection is what I call "the position part". It is transformed by $D\rho_g$ because projection $\pi$ commutes with the lift of $\rho_g$ and $\rho_g$ acts on $T_q M$ by $D\rho_g$. $\endgroup$
    – Max
    Jun 11, 2019 at 17:58
  • $\begingroup$ I see. And when you say the lifted action preserves the tautological 1-form, it means that $A^*\alpha=\alpha$, for $\alpha$ the tautological 1-form? Why does this imply $<p, v>=<A_q p, D\rho_g|_q(v) >$? $\endgroup$
    – roi_saumon
    Jun 11, 2019 at 20:34
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    $\begingroup$ "it means $A^*\alpha=\alpha$." no, that's a type error. A is the restriction of lift of $\rho_g$ (denoted by say $P_g$) to a fiber; we want $P_g^* \alpha=\alpha$. $\endgroup$
    – Max
    Jun 11, 2019 at 21:07
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    $\begingroup$ It implies $<p, v>=<A_q p, D\rho_g|_q(v)>$ because the left hand side is $\alpha(V)$ and right hand side is $(P_g^*\alpha) (V)$ for any $V$ with $D\pi(V)=v$. $\endgroup$
    – Max
    Jun 11, 2019 at 21:10

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