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One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none.

On the other hand, your questions seem to really be

If X is a vector field, are the conditions

• $X$ preserves $\omega$ and

• $i_X\omega=\omega(X,-)$ is an exact 1-form

equivalent?"

The answer is "no", but it's "yes" if you replace "exact" by "closed". The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the closedness of $\omega$.

One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none.

On the other hand, your questions seem to really be

If X is a vector field, are the conditions

• $X$ preserves $\omega$ and

• $i_X\omega=\omega(X,-)$ is an exact 1-form

equivalent?"

The answer is "no", but it's "yes" if you replace "exact" by "closed". The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the closedness of $\omega$.

This distinction is important, since lots of symplectic actions don't have moment maps because of the existence of $H^1$. Think about, for example, the 2-torus $T^2$ acting on itself; any invariant volume form can also be thought of as an invariant symplectic form, bu the map $\mathfrak t^2\to H^1(T^2)$ given by $X\mapsto [i_X\omega]$ is an isomorphism. No non-trivial element of the Lie algebra acts by a Hamiltonian vector field, even though they all act by symplectic ones.

One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none (though there always will be if you're willing to replace your group by a central extension).

On the other hand, your questions seem to really be

If X is a vector field, are the conditions

• $X$ preserves $\omega$ and

• $i_X\omega=\omega(X,-)$ is an exact 1-form

equivalent?"

The answer is yes. The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the exactness of $\omega$.

One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none.

On the other hand, your questions seem to really be

If X is a vector field, are the conditions

• $X$ preserves $\omega$ and

• $i_X\omega=\omega(X,-)$ is an exact 1-form

equivalent?"

The answer is "no", but it's "yes" if you replace "exact" by "closed". The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the closedness of $\omega$.

One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none (though there always will be if you're willing to replace your group by a central extension).

On the other hand, your questions seem to really be "if X is a vector field, are the conditions

• $X$ preserves $\omega$ and
• $i_X\omega=\omega(X,-)$ is an exact 1-form

equivalent?"

The answer is yes. The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the exactness of $\omega$.

One thing that seems to be confusing you is that there is no "the moment map." There often several different moment maps for the same action, and as Jose says, sometimes none (though there always will be if you're willing to replace your group by a central extension).

On the other hand, your questions seem to really be

If X is a vector field, are the conditions

• $X$ preserves $\omega$ and

• $i_X\omega=\omega(X,-)$ is an exact 1-form

equivalent?"

The answer is yes. The preservation of $\omega$ by $X$ is the same as saying that the Lie derivative $\mathcal{L}_X\omega=0$. By Cartan's magic formula,

$$\mathcal{L}_X\omega=d(i_X\omega)+i_X(d\omega)=d(i_X\omega)$$ since the second term is zero by the exactness of $\omega$.