I’m trying to understand solvability of $$ L_X u = f $$ for a vector field $X$ on a manifold, and then compare this with some related linear problems for left-invariant forms on a Lie group (where things look much more algebraic).
0-form case (functions on a manifold)
Let $M$ be a smooth manifold and let $X$ be a complete vector field on $M$ with flow $\phi_t$. Given a smooth function $f \in C^\infty(M)$, I want to understand when there exists a smooth function $u$ with $$ L_X u = X(u) = f. $$
Along each orbit of the flow of $X$, this is just the ODE $$ \frac{d}{dt}\big(u(\phi_t(x))\big) = f(\phi_t(x)). $$
If the orbit through $x$ is periodic of period $T>0$, i.e. $\phi_T(x)=x$, then integrating over one period gives $$ 0 = u(\phi_T(x)) - u(x) = \int_0^T f(\phi_t(x))\,dt, $$ so I get the familiar necessary condition $$ \int_0^T f(\phi_t(x))\,dt = 0 \quad \text{for every periodic orbit of } X. $$
Equivalently:
For every periodic orbit $\gamma$ of $X$, $$ \int_\gamma f\, dt = 0.$$
So for the $0$-form equation $L_X u = f$, the “zero integral on periodic orbits” condition is clearly necessary.
What I am not sure about is whether this condition is also sufficient, or whether there can be more subtle global obstructions.
So my main question is:
Question A.
Let $X$ be a complete vector field on a smooth manifold $M$, and let $f \in C^\infty(M)$. Suppose that for every periodic orbit $\gamma$ of $X$ one has $$\int_\gamma f\,dt = 0.$$ Is this condition sufficient to guarantee the existence of a global smooth solution $u$ to $$L_X u = f \, ?$$ If not, what additional obstructions can occur, or under what extra hypotheses on $X$ or $M$ does this condition become sufficient?
If every orbit is a circle, the answer is Yes. In the circle case, if the equation has no solution, the periodic orbit integrals will explicitly tell you so (by being non-zero). The condition $\int_\gamma f = 0$ is now perfectly perfectly aligned with solvability.
Similar Lie algebra problems on a Lie group
Now let $G$ be a connected Lie group with Lie algebra $\mathfrak{g}$. I identify
- $\mathfrak{g}$ with the left-invariant vector fields on $G$,
- $\mathfrak{g}^*$ with the left-invariant $1$-forms on $G$,
- $\Lambda^2 \mathfrak{g}^*$ with the left-invariant $2$-forms on $G$.
Fix $X \in \mathfrak{g}$. In these left-invariant settings, the equations for $1$-forms and $2$-forms reduce to purely linear problems on finite-dimensional vector spaces, where I think I can describe the solvability conditions quite explicitly.
(1) Left-invariant $1$-forms
If $\beta \in \mathfrak{g}^*$ is left-invariant, then $$ L_X \beta = \operatorname{ad}_X^*(\beta), $$ so solving $$ L_X \beta = \alpha, \qquad \alpha \in \mathfrak{g}^* $$ is equivalent to solving the linear algebra equation $$ \operatorname{ad}_X^*(\beta) = \alpha \quad \text{on } \mathfrak{g}^*. $$
Necessary condition
If $Y \in \mathfrak{g}$ is left-invariant, then $\beta(Y)$ is a constant function on $G$. Using the identity $$ L_X \beta(Y) = X\cdot(\beta(Y)) - \beta([X,Y]) $$ and the fact that $X\cdot(\beta(Y)) = 0$ for left-invariant $X,\beta,Y$, we get $$ (L_X \beta)(Y) = -\beta([X,Y]). $$
If $L_X \beta = \alpha$, then evaluating at $Y$ gives $$ \alpha(Y) = (L_X \beta)(Y) = -\beta([X,Y]). $$
In particular, if $Y$ commutes with $X$, i.e. $[X,Y]=0$, then $$ \alpha(Y) = 0. $$
So a necessary condition is: $$ \alpha(Y) = 0 \quad \text{for all } Y \in \mathfrak{g} \text{ with } [X,Y]=0, $$ i.e. $\alpha$ vanishes on the centralizer $$ \ker(\operatorname{ad}_X) = \{Y \in \mathfrak{g} : [X,Y]=0\}. $$
Sufficiency via linear algebra
Let $A = \operatorname{ad}_X : \mathfrak{g} \to \mathfrak{g}$. Then $A^* = \operatorname{ad}_X^* : \mathfrak{g}^* \to \mathfrak{g}^*$.
In finite-dimensional linear algebra there is the standard identity $$ \operatorname{im}(A^*) = (\ker A)^\circ, $$ where $(\ker A)^\circ \subset \mathfrak{g}^*$ is the annihilator of $\ker A$, i.e. $$ (\ker A)^\circ = \{\lambda \in \mathfrak{g}^* \mid \lambda(v) = 0 \ \text{for all } v \in \ker A\}. $$
Sketch of proof:
If $\lambda \in \operatorname{im}(A^*)$, write $\lambda = A^*(\beta)$. Then for any $v \in \ker A$, $$ \lambda(v) = \beta(A v) = \beta(0) = 0, $$ so $\lambda \in (\ker A)^\circ$. Hence $\operatorname{im}(A^*) \subset (\ker A)^\circ$.
Since $\mathfrak{g}$ is finite-dimensional, $$ \dim(\operatorname{im}(A^*)) = \dim(\mathfrak{g}^*) - \dim(\ker A^*) = \dim\mathfrak{g} - \dim(\ker A), $$ while $$ \dim((\ker A)^\circ) = \dim\mathfrak{g}^* - \dim(\ker A) = \dim\mathfrak{g} - \dim(\ker A). $$ Thus the two subspaces $\operatorname{im}(A^*)$ and $(\ker A)^\circ$ have the same dimension, and with the inclusion above this gives equality: $$ \operatorname{im}(A^*) = (\ker A)^\circ. $$
Applied to $A = \operatorname{ad}_X$, this says: $$ \operatorname{im}(\operatorname{ad}_X^*) = \{\alpha \in \mathfrak{g}^* : \alpha(Y) = 0 \text{ for all } Y \in \ker(\operatorname{ad}_X)\}. $$
Therefore the equation $$ \operatorname{ad}_X^*(\beta) = \alpha $$ is solvable in $\beta \in \mathfrak{g}^*$ if and only if $$ \alpha(Y) = 0 \quad \text{for all } Y \in \mathfrak{g} \text{ with } [X,Y]=0. $$
So in this left-invariant $1$-form setting, the condition “$\alpha$ vanishes on all commuting directions” is both necessary and sufficient.
(2) Left-invariant $2$-forms
Now let me look at the analogous problem for left-invariant $2$-forms.
Left-invariant $2$-forms on $G$ correspond to elements of $\Lambda^2 \mathfrak{g}^*$. If $\beta \in \Lambda^2 \mathfrak{g}^*$ is left-invariant, then $L_X$ acts as the dual of the induced map $$ \operatorname{ad}_X : \Lambda^2 \mathfrak{g} \to \Lambda^2 \mathfrak{g}, $$ defined on decomposable bivectors by $$ \operatorname{ad}_X (Y \wedge Z) = [X,Y] \wedge Z + Y \wedge [X,Z], $$ and extended linearly.
Thus the equation $$ L_X \beta = \alpha, \qquad \alpha \in \Lambda^2 \mathfrak{g}^* $$ is equivalent to the linear algebra equation $$ \operatorname{ad}_X^*(\beta) = \alpha \quad \text{on } \Lambda^2 \mathfrak{g}^*, $$ where now $\operatorname{ad}_X^*$ is the dual operator to $$ \operatorname{ad}_X : \Lambda^2 \mathfrak{g} \to \Lambda^2 \mathfrak{g}. $$
Necessary condition
Let $B = \operatorname{ad}_X : \Lambda^2 \mathfrak{g} \to \Lambda^2 \mathfrak{g}$. If $W \in \Lambda^2 \mathfrak{g}$ lies in $\ker B$, then for any $\beta$ we have $$ (B^*\beta)(W) = \beta(BW) = \beta(0) = 0. $$
In particular, for a decomposable element $W = Y \wedge Z$, the condition $W \in \ker B$ reads $$ \operatorname{ad}_X (Y \wedge Z) = [X,Y] \wedge Z + Y \wedge [X,Z] = 0. $$
So if $L_X \beta = \alpha$ (equivalently $B^* \beta = \alpha$), then for all pairs $(Y,Z)$ with $$ 0 = [X,Y] \wedge Z + Y \wedge [X,Z], $$ we must have $$ \alpha(Y,Z) = \alpha(Y \wedge Z) = 0. $$
That is, a necessary condition is:
For all pairs $(Y,Z) \in \mathfrak{g} \times \mathfrak{g}$ such that $$ 0 = [X,Y] \wedge Z + Y \wedge [X,Z],$$ we have $$\alpha(Y,Z) = 0. $$
Sufficiency via the same linear algebra identity
Again, $B$ is a linear map on a finite-dimensional vector space $\Lambda^2 \mathfrak{g}$, so the same linear algebra argument as before gives $$ \operatorname{im}(B^*) = (\ker B)^\circ, $$ where $(\ker B)^\circ \subset (\Lambda^2 \mathfrak{g})^* \cong \Lambda^2 \mathfrak{g}^*$ is the annihilator of $\ker B$.
Thus the equation $$ \operatorname{ad}_X^*(\beta) = \alpha $$ on $\Lambda^2 \mathfrak{g}^*$ is solvable if and only if $$ \alpha(W) = 0 \quad \text{for all } W \in \ker\big(\operatorname{ad}_X : \Lambda^2 \mathfrak{g} \to \Lambda^2 \mathfrak{g}\big), $$ i.e. iff $$ \alpha(Y,Z) = 0 \ \text{whenever } [X,Y] \wedge Z + Y \wedge [X,Z] = 0. $$
So in this left-invariant $2$-form setting, the condition “$\alpha$ vanishes on all pairs $(Y,Z)$ with $[X,Y] \wedge Z + Y \wedge [X,Z] = 0$” is again both necessary and sufficient.
Back to the $0$-form case
In the left-invariant $1$- and $2$-form problems on a Lie group, the solvability conditions for $L_X \beta = \alpha$ reduce cleanly to linear algebra: $\alpha$ must lie in the image of the appropriate dual operator, which can be expressed concretely as annihilating the kernel of $\operatorname{ad}_X$ (on $\mathfrak{g}$ or on $\Lambda^2 \mathfrak{g}$, respectively).
In the original $0$-form problem on a general manifold, the natural necessary condition is the “zero integral over each periodic orbit” condition. My question is whether, in that functional setting, $$ \int_\gamma f\,dt = 0 \ \ \text{for every periodic orbit } \gamma $$ already gives a full characterization of solvability of $L_X u = f$, or whether additional global obstructions can arise beyond the periodic orbits.