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Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X} \omega=\omega$. The existence of a Liouville vector field implies $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$? (No it is not).

Added: are there some sufficient conditions for the existence of Liouville vector fields? About symplectic surfaces with boundaries?

Thanks!

Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X} \omega=\omega$. The existence of a Liouville vector field implies that $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$?

Thanks!

Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X} \omega=\omega$. The existence of a Liouville vector field implies $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$?

Added: are there some sufficient conditions for the existence of Liouville vector fields?

Thanks!

Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X} \omega=\omega$. The existence of a Liouville vector field implies $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$? (No it is not).

Added: are there some sufficient conditions for the existence of Liouville vector fields? About symplectic surfaces with boundaries?

Thanks!

Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X} \omega=\omega$. The existence of a Liouville vector field implies $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$?

Thanks!

Let $(M, \omega)$ be a symplectic manifold. A vector field $V: M \to TM$ is Liouville if $L_{X} \omega=\omega$. The existence of a Liouville vector field implies $(M, \omega)$ is exact: the one-form $\lambda = i_V \omega$ satisfies $d\lambda=d\circ i_V\omega = L_V\omega=\omega$. In particular, there is no Liouville vector field on any closed (compact and boundaryless) symplectic manifold.

My question is about the existence of Liouville vector fields. Is it a sufficient condition that $\partial M\neq \varnothing$?

Added: are there some sufficient conditions for the existence of Liouville vector fields?

Thanks!