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Assume that $M$ is a compact Riemannian manifold whose Laplacian is denoted by $\Delta$. Assume that the Euler characteristic of $M$ is zero. Does $M$ admit a non vanishing vector field $X$ which satisfy $$(*) \qquad \Delta \circ \partial/\partial X=\partial/\partial X \circ \Delta$$

What can be said about the structure of the Lie algebra of all vector fields $X$ with the property $(*)$?

As a second question: Every vector field $X$ on $M$ defines a second order differential operator on $C^{\infty}(M)$ with $$D(f)=\Delta(X.f)-X.\Delta(f)$$

This is a second order operator since the third order terms cancel each others.

What is the principal symbol of this operator , precisely? Can this PDE be an elliptic operator when $M$ is a compact manifold?(I mean:is there an example of this situation in compact case?) Does every compact manifold admit a vector field $X$ for which this PDE would be an elliptic operator?What would be a dynamical interpretation for the index of this PDE. This is a dynamical motivation for the later question..

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    $\begingroup$ In $\mathbb R^n$, the laplacian is invariant under rotations. Did you try X= a killing vector? $\endgroup$ – Marco Farinati Jun 15 at 22:27
  • $\begingroup$ @MarcoFarinati Thank you for your comment. I need to know the remaining part of the expersion $\Delta(X.Y)=\Delta(X).Y+X.\Delta(Y)+2.....$. In fact the equation $*$ in the question defines a second order PDE. I wish to know are there some example of compact manifold for which this PDE is elliptic?and what is its index and the dynamical interpretation of this index. BTW is it obvious every manifold with zero Euler characteristic admit a non vanishing killing vector field? $\endgroup$ – Ali Taghavi Jun 16 at 8:52
  • $\begingroup$ In my previous comment, $X.Y$ is the inner product of two vector fields and by laplacian of a vector field I am considering the Laplacian of the corresponding 1-form. But I wonder what is the remaining part of that formula. i wrote down in local coordinate but i do not know what is the global formula in an abstract manifold? $\endgroup$ – Ali Taghavi Jun 16 at 9:21
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    $\begingroup$ The principal term is $\nabla^{(a} X^{b)} \nabla_{a} \nabla_{b}$, where the parentheses denote symmetrization. So this commutator is as elliptic as the $\nabla_{(a} X_{b)}$ bilinear form is positive definite. Up to a constant, the last expression also happens to be the Lie derivative of the Riemannina metric $g_{ab}$ with respect to $X$. $\endgroup$ – Igor Khavkine Jun 16 at 11:10
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The following formula is known among the experts but hard to find in the literature, so I figure I will document it here. Throughout $(M,g)$ denote an arbitrary pseudo-Riemannian manifold, and $\nabla$ its Levi-Civita connection.

Definition Given a vector field $X$, its corresponding 0th order deformation tensor is defined to be ${}^{(X,0)}\pi := \mathcal{L}_X g$, where $\mathcal{L}_X$ is Lie differentiation with respect to $X$.
The corresponding 1st order deformation tensor is defined using a formula similar to that of Christoffel symbols: $$ {}^{(X,1)}\pi_{ab}{}^c := \frac12 g^{cd} \left[ \nabla_a ( {}^{(X,0)}\pi_{bd}) + \nabla_b ({}^{(X,0)}\pi_{ad}) - \nabla_d ({}^{(X,0)}\pi_{ab}) \right] $$

Lemma Let $\Xi$ be an arbitrary $k$-covariant tensor field. And let $X$ be a vector field. The following formula holds for the commutation: $$ [ \nabla_a, \mathcal{L}_X ] \Xi_{b_1\cdots b_k} = \sum_{j = 1}^k {}^{(X,1)}\pi_{a b_j}{}^c \Xi_{b_1 \cdots b_{j-1} c b_{j+1} \cdots b_k} $$

With the aid of these formulas, we have immediately that, writing $\triangle_g$ for the Laplace-Beltrami operator, first

$$ [ \nabla_X, \triangle_g] f = [\mathcal{L}_X, \triangle_g ] f $$

because Lie derivation and covariant differentiation act identically on scalars, and then

$$ [\mathcal{L}_X, g^{ab}\nabla_a\nabla_b] f = \mathcal{L}_X (g^{ab}) \nabla^a \nabla_b f + g^{ab} [\mathcal{L}_X, \nabla_a] \nabla_b f + g^{ab} \nabla_a [\mathcal{L}_X, \nabla_b ]f $$

The first factor we can compute to get

$$ \mathcal{L}_X(g^{ab}) = - {}^{(X,0)}\pi^{ab} $$

using that $g^{ab} g_{bc} = \delta^a_c$. The third factor vanishes because Lie differentiation commutes with exterior differentiation. And we use our Lemma for the second term. We get, finally

$$ [\nabla_X, \triangle_g] f = - {}^{(X,0)}\pi^{ab} \nabla_a\nabla_b f - g^{ab} ~{}^{(X,1)}\pi_{ab}{}^c \nabla_c f. $$

Remarks:

  • Notice that the first order deformation tensor is defined in terms of the 0th order one. So that when $X$ is Killing, automatically both the ${}^{(X,0)}\pi$ and ${}^{(X,1)}\pi$ vanish, and differentiation with $X$ commutes with the Laplacian.

  • In the case ${}^{(X,0)}\pi = \phi g$ for some scalar function $\phi$ (so $X$ is conformally Killing), one can check that the formula reduces to the one I gave in a comment above.

  • When the function $\phi$ in the previous item is a non-zero constant (which some people refer to as $X$ being a homothetic vector field) one gets the special case $$ [ \nabla_X, \triangle_g] f = \phi \triangle_g f $$

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  • $\begingroup$ Thank you very much and +1 for your answer. I try to understand its details. $\endgroup$ – Ali Taghavi Jun 24 at 16:26
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This is an expansion on Igor's comment and fixes my previous mistake (see also Willie's answer). A direct computation gives $$ D(f) = 2f^{ab}\nabla_{(a}X_{b)} + X_a(\nabla^b\nabla_b\nabla^a-\nabla^a\nabla_b\nabla^b)f + f^a\nabla^b\nabla_bX_a . $$ The middle summand is the same as $R_{ab}f^a$, while the formula $\nabla^b\nabla_aX_b=\nabla_a\nabla^bX_b+R_a{}^bX_b$ yields $$ D(f) = 2f^{ab}\nabla_{(a}X_{b)} + (\nabla^b\nabla_bX_a + \nabla^b\nabla_aX_b - \nabla_a\nabla^bX_b)f^a . $$ This can be rewritten $$ D(f) = \langle L_Xg, \nabla^2f \rangle + \langle \nabla f, \delta L_Xg - d\delta X\rangle , $$ which easily gives the formula for $D$ when $X$ is a conformal Killing field. The interpretations for your questions are as follows:

Case 1: $\nabla_{(a}X_{b)}$ is not identically zero. This is when $D$ is a second-order operator. Note that the trace of $\nabla_{(a}X_{b)}$ is the divergence. By the divergence theorem, the integral of the trace is zero, so the bilinear form $\nabla_{(a}X_{b)}$ cannot be positive definite. Thus there is no example of the type requested in your second question. (Note that on noncompact manifolds there are examples: e.g. Euclidean space with the Euler vector field $X=\sum x^i\partial_{x^i}$, so $D$ is proportional to the Laplacian.)

Case 2: $\nabla_{(a}X_{b)}\equiv0$. (That is, $X$ is Killing.) Since the trace of $L_Xg$ is $2\delta X$, we see that $D\equiv0$. That is, the Lie algebra of vector fields satisfying ($\ast$) is the Lie algebra of Killing vector fields.

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  • $\begingroup$ Thank you very much and +1 for your answe I will come back soon. $\endgroup$ – Ali Taghavi Jun 17 at 19:33
  • $\begingroup$ I would appreciate if you more explain about "If X is parallel and nonvanishing, then (M,g) splits off a flat factor in the direction of X" $\endgroup$ – Ali Taghavi Jun 20 at 21:55
  • $\begingroup$ As another question: is there a vector field $X$ on the plane which possess a closed orbit $\gamma$ and there is a function $f$ such that $D(f)$ is a non vanishing function on $\gamma$ and its interior?($D(f)$ is the same operator as in the question and laplacian is the standard one). what about if we replace the Laplacian by a laplacian associated to a metric compatible to flow of X? $\endgroup$ – Ali Taghavi Jun 20 at 22:03
  • $\begingroup$ I edited the answer to include a reference to a more complete answer to your first question. For your new question in the comments, $D(1)=0$ for any Riemannian manifold and any vector field. $\endgroup$ – Jeffrey Case Jun 24 at 13:31
  • $\begingroup$ The computation doesn't seem entirely correct: $$\nabla^b \nabla_b (Xf) = 2(\nabla^b X^a)( \nabla_b \nabla_a f)+ X^a \nabla^b \nabla_b \nabla_a f + (\nabla^b \nabla_b X^a) \nabla_a f $$ your answer contains the first and third term. The middle term is not $X(\nabla^b \nabla_b f)$! The two differ by a commutation of $[\nabla^b \nabla_b, \nabla_a]$ which introduces another curvature term. For Killing vector fields (and conformal Killing ones also) this term actually completely cancels the Ricci curvature term. $\endgroup$ – Willie Wong Jun 24 at 14:09

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