Let $\mathcal{M}$ be a real, compact, orientable manifold and let $X$ be a vector field on $\mathcal{M}$. Consider the functional

$$E(f) = \int_{\mathcal{M}} X_p(f)^2 dV$$

where $X_p(f)$ is the directional (Lie) derivative of $f$ along $X$ at the point $p$ and $dV$ is a volume form on $\mathcal{M}$ -- this functional essentially measures the total amount of change in $f$ along $X$ over all of $\mathcal{M}$ in the $L^2$ sense. Then $\delta E(f)$ is a differential operator whose eigenspectrum

$$\delta E(f) = \lambda f$$

(for $\lambda \in \mathbb{R}$) yields the critical points of $E$ over the set of functions with unit norm. Is there an established name for this operator (or functional) and/or its corresponding eigenspectrum?

The prototype for this operator is Dirichlet's energy

$$\int_{\mathcal{M}} ||\nabla f||^2 dV$$

which has as its (unit-norm) critical points the Laplacian eigenspectrum

$$\nabla^2 f = \lambda f,$$

the main difference being that Dirichlet's energy measures the total gradient, i.e., the change in all directions, rather than just the change along a particular direction at each point.

  • $\begingroup$ One small comment is that compact orientable manifolds do not come with a preferred volume form. As your prototype suggests, you are perhaps working on a compact, orientable riemannian manifold. $\endgroup$ Jan 26, 2010 at 9:41
  • $\begingroup$ Maybe I'm missing something, but I would have thought that critical points of E would correspond to zeros of dE. I.e., elements in the kernel of the corresponding differential operator, rather than just eigenvectors. $\endgroup$
    – Joel Fine
    Jan 26, 2010 at 11:30
  • $\begingroup$ José: I nearly added that condition, but decided to write "a volume form" rather than "the volume form" instead. (Hopefully that's sufficient, though now I'm uncertain whether a choice of volume form induces a Riemannian metric.) $\endgroup$ Jan 26, 2010 at 15:51
  • $\begingroup$ Joel: you are correct -- the spectrum comes from critical points of the functional over the set of functions with unit norm (I was so used to working with this setup that I forgot to mention it!). Fixed above. $\endgroup$ Jan 26, 2010 at 15:53
  • $\begingroup$ @fuzzytron: 1st, I agree, it looks much better now you've mentioned the unit norm condition. 2nd, your problem makes perfect sense with just a volume form. This is enough to give you an L^2 inner product on functions and for your differential operator to make sense. You don't need a metric for your question to be meaningful. However, for future reference, it's important to note that a volume form is strictly less information than that given by a metric - there are many different metrics which have the same volume form. (It's enough to see this is true for inner-products in linear algebra.) $\endgroup$
    – Joel Fine
    Jan 26, 2010 at 18:27

1 Answer 1


There is an simple explicit formula for your operator in terms of known operators. To see this, note that $\delta E_f(g)$ (the differential of $E$ at the function $f$ in the direction $g$) is equal to $$ 2 \int X(f) X(g) dV = 2\int \left[L_X(g X(f) dV )-g\left(X(X(f)) dV+X(f)L_X(dV)\right)\right] $$ where $L_X$ is the Lie deriviative with respect to $X$. Now, $\int L_X(\alpha) = 0$ for any top degree form $\alpha$, so we get $$ \delta E_f(g) = - 2 \int_X g \left[X(X(f)) + X(f) div(X)\right]dV $$ where the divergence of $X$ is defined by $div(X)= L_X(dV)/dV$. So, using the $L^2$-inner product on functions, we can interpret the 1-form $\delta E$ as the differential operator $$ D :f \mapsto - X(X(f)) - X(f) div(X). $$ (This is the same way that the differential of Dirichlet energy is seen as the Laplacian.)

Note that the leading order part of D is just $X^2$ and so, in particular, $D$ is not elliptic. You'd expect this, of course, because $E$ only sees the change of $f$ in the $X$-direction.


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