1
$\begingroup$

Let $(M,g)$ be a riemannian manifold and $G$ a Lie group acting on $M$ by isometries. Taking a $G$-invariant function $f\colon M \to \mathbb{R}$, we have the riemannian gradient $\operatorname{grad} f$ defined by $$d_xf(Y) = g_x(\operatorname{grad} f (x), Y)$$ Next, we take the flow $\phi_t$ induced by $\operatorname{grad} f$. Using the canonical lift, we get a flow $T^*\phi_t \colon T^*M \to T^*M$ on the cotangent bundle with the corresponding vector field $Z \in \mathfrak{X}(T^*M)$. $T^*M$ is a symplectic manifold with the canonical symplectic 2-form $\omega_0$. Now it is possible to show, that $$d(i_Z\omega_0) = L_Z\omega_0 =0.$$

Fixing $v \in T^*M$ and using the Poincaré lemma, we have a neighborhood $U$ of $v$, and a function $H \colon T^*M \to \mathbb{R}$ with $i_Z\omega_0|_U = dH|_U$.

After all this constructions, is the function $H$ locally invariant under the lifted action of $G$?

$\endgroup$

1 Answer 1

2
$\begingroup$

If $V$ is a vector field on $M$, its lift to $T^*M$ is a Hamiltonian vector field with hamiltonian function $V$ (viewed as a linear function on the fibers). And if the vector field $V$ is $G$-invariant, it is also invariant as a function on $T^*M$ for the lifted action.

This has nothing to do with the fact that $V$ is a gradient.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.