In this survey of Symplectic Geometry by Arnol'd and Givental in 1985, an example of momentum mappings was given on Page 62 as follows:
The action of the group by left translations on its cotangent bundle is Poisson. The corresponding momentum mapping $P:T^*G \to \mathfrak{g}^*$ coincides with the right translation of covectors to the identity element of the group.
I want to understand this example in a simple and concrete setting. So let's take $G = \mathbb{R}$. Then $T^*G \cong \mathbb{R}^2$, $\mathfrak{g}^* \cong \mathbb{R}$, and the identity element of the group $\mathbb{R}$ is zero. The physical meaning of this example is a particle moving on the real line with coordinate $q$ at a momentum $p$.
How does the momentum mapping $P: \mathbb{R}^2 \to \mathbb{R}$ coincide with the right translation of covectors to zero?
So I'm asking what exactly the map $P$ is and what Arnol'd and Givental meant by "the right translation of covectors to zero".
Best Answer
With your identifications, the mapping would be simply $P(q,p) = p$. Or if you wanted to think of $\mathfrak{g}^*$ as being $T^*_0G$ (the cotangent space at $0\in\mathbb{R}$), it would be $P(q,p) = (0,p)$, i.e. the momentum at $q\in\mathbb{R}$, but translated to the identity $0\in\mathbb{R}$. If your Hamiltonian is translation-invariant, this will be conserved under the Hamiltonian flow.
Also, right translation of a covector to zero would be the map $(q,p)\mapsto (0,p)$. (This is also left translation of a covector to zero - left and right translation agree since the underlying group is Abelian.)