# Cotangent lift of an action and its effect on the moment covector

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?

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 $$=$$, 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 $$\rho^*_{g^{-1}}$$.

• 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 $D\rho_g \mid_q$? Jun 11, 2019 at 16:00
• 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 $D\rho_g$.
– Max
Jun 11, 2019 at 17:58
• 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 $<p, v>=<A_q p, D\rho_g|_q(v) >$? Jun 11, 2019 at 20:34
• "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 $P_g^* \alpha=\alpha$.
– Max
Jun 11, 2019 at 21:07
• 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 $D\pi(V)=v$.
– Max
Jun 11, 2019 at 21:10