Let $X$ be a non-vanishing real analytic vector field on an open manifold $M$. What kind of obstructions would appear when we search for a Riemannian metric on $M$ such that the space of harmonic functions would be invariant under the derivation operator $f \mapsto X.f$? A harmonic function is a function $f$ which satisfy $\Delta_g (f)=0$ where $\Delta_g$ is the Laplace operator associated to the metric $g$.

  • $\begingroup$ $[X,\Delta]=0.$ $\endgroup$
    – Bazin
    Mar 2 '18 at 15:12
  • 1
    $\begingroup$ @Bazin Yes. But $\Delta_g$ does not exist if we do not have a metric, already. $\endgroup$ Mar 2 '18 at 21:26

(A partial answer; the full picture is likely to be very complicated, with special behaviors in dimensions 1 and 2 compared to higher ones.)

Dimension 1

Given a Riemannian metric on a one dimensional manifold there is, up to a sign, a unique "unit-length" vector field $Y$ tangent to $M$, and in this case you must have $\Delta_g = Y^2$. It is easy to see that $X$ preserves harmonic functions if and only if $[X,Y] = 0$, which implies $X = cY$ for some constant $c$.

Therefore the only obstruction to to the existence of a suitable metric is whether $X$ vanishes: if $X$ is a non-vanishing vector field, then you can define a co-metric by $X\otimes X$. If $X$ vanishes at some point, then such a metric doesn't exist.

General Dimensions

Given a metric $g$, it is well-known that $$ [X, \Delta_g]f = - {}^{(0)}\pi^{ab} \nabla^2_{ab} f - g^{ab} \cdot {}^{(1)}\pi_{ab}{}^c \nabla_c f \tag{*}$$ where $$ -{}^{(0)}\pi^{ab} = \mathcal{L}_X g^{ab} = - \nabla^a X^b - \nabla^b X^a \tag{1}$$ and $${}^{(1)}\pi_{ab}{}^c = \frac12 g^{cd} (\nabla_a {}^{(0)}\pi_{bd} + \nabla_b {}^{(0)}\pi_{ad} - \nabla_d {}^{(0)} \pi_{ab} ) \tag{2} $$

For $X$ to preserve the space of harmonic functions, a sufficient condition (necessary and sufficient if you work only locally) can be found by examining the equations above.

First note that it is not necessary that $[X,\Delta_g] = 0$. If we are only interested in preserving the harmonic functions, it suffices that $[X, \Delta_g] \propto \Delta_g$. Examining (*) this means that:

  1. We want ${}^{(0)}\pi^{ab} = k g^{ab}$ for some function $k$. This means that $X$ is a conformal isometry of $g$.
  2. We want ${}^{(1)}\pi_{ab}{}^c g^{ab}$ to vanish. Together with the previous statement we want $$ 0 = (2-n) \nabla^c k,$$ where $n$ is the number of spatial dimensions. So (as is well known), we conclude that in 2 dimensions a sufficient condition is that $X$ is a conformal isometry of $g$, and in higher dimensions we require that $X$ is a homothety.

Now, homotheties also obey the Jacobi equation, in the form $$ \nabla_a \nabla_b X_c+ R_{bcad} X^d = 0 $$ (for one choice of sign of the Riemann curvature tensor). A particular consequence is that if $X$ and $\nabla X$ both vanish at a point $p$, then $X$ vanish everywhere. This means that we have an obstruction:

Obstruction: Let $n \geq 3$. If there exists a point $p$ such that the vector field $X$ vanishes to second order (in the sense that for any function $f$ and any other vector field $Y$, both $Xf|_p = 0 = YXf|_p$), then there does not exist any metric $g$ on $M$ such that $X$ preserves the harmonic functions.

Remark: In $n = 2$, this is not an obstruction. Consider the vector field $(y^2 - x^2) \partial_x - 2 xy \partial_y$ on $\mathbb{R}^2$. It is simple to check that this vector field preserves the harmonic functions on $\mathbb{R}^2$, but this vector field vanishes to second order at the origin.

  • $\begingroup$ A few more remarks: (a) A sufficient local condition is that $X$ is non-vanishing: locally take a transverse hypersurface and you can build a Riemannian metric for which $X$ is Killing. There can however be global obstructions. (b) I am not sure whether there are further local obstructions when $X$ vanish only to first order. $\endgroup$ Mar 9 '18 at 19:43
  • $\begingroup$ Thanks for your attention to my question and your interesting answer. i try to understand its details. $\endgroup$ Mar 9 '18 at 19:58

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