Let $M$ be a closed finite-dimensional smooth manifold (over $\mathbb R$). Let $C^\infty(M) = C^\infty(M,\mathbb C)$ be the algebra of smooth complex-valued functions on $M$, with the natural complex conjugation $f\mapsto \bar f$.

Definition: A (real) density $\mu$ on $M$ is a section of the trivial real line bundle given in local coordinates by the transition maps $\tilde\mu = \bigl| \det \frac{\partial x^i}{\partial \tilde x^j} \bigr| \mu$, where $x^i,\tilde x^j$ are the different systems of coordinates: if $M$ is oriented, then densities are the same as (real) top-forms. A volume form on $M$ is an everywhere-positive density — this condition is invariant under changes of coordinates.

We remark that if $\mu$ is a density, then $\int_M \mu$ is well-defined: cut $M$ into coordinate patches, integrate each patch, and check that the answer doesn't depend on the choice of coordinates.

A choice of volume form $\mu$ determines a Hermitian structure on $C^\infty(M)$, via $\langle f_1,f_2\rangle_\mu = \int_M {\bar f}_1 \, f_2 \, \mu$. A (complex) differential operator $\mathcal D: C^\infty(M) \to C^\infty(M)$ is $\mu$-Hermitian if $\bigl\langle f_1, \mathcal D[f_2] \bigr\rangle_\mu = \bigl\langle \mathcal D[f_1], f_2 \bigr\rangle_\mu$.

Example: Multiplication by a real-valued function $c$ is Hermitian for any measure. Let $a$ be a Riemannian metric on $M$. Then $\sqrt{\left|\det a\right|}$ makes sense as a volume-form on $M$. The Laplace-Beltrami operator on $M$ is given in local coordinates by $\Delta_a = -\left|\det a\right|^{-1/2} \partial_i a^{ij} \left|\det a\right|^{1/2} \partial_j$. It is $\sqrt{\left|\det a\right|}$-Hermitian. More generally, let $b$ be any real one-form on $M$. Then I believe that the operator given in local coordinates by: $$ \frac1{\sqrt{|\det a|}} \bigl(\sqrt{-1}\partial_i + b_i\bigr) a^{ij} \sqrt{|\det a|} \bigl( \sqrt{-1} \partial_j + b_j\bigr) $$ is $\sqrt{\left|\det a\right|}$-Hermitian.

I believe that the following is true (I've checked it by hand for $M=$ the circle):

Propoposition: Let $a$ be Riemannian metric, $\Delta_a$ its Laplace-Beltrami operator, and $\mu$ a volume form. Then $\Delta_a$ is $\mu$-Hermitian if and only if $\mu/\sqrt{\left|\det a\right|}$ is locally constant.

The if part I said above. Thus, by completing the square, it's straightforward to check whether a second-order differential operator can be made Hermitian with the correct choice of volume form (basically, the factors of $\sqrt{-1}$ must be as above). My question is whether you can do this for higher-order differential operators.

Question: Suppose I have a linear differential operator $\mathcal D$, with, say, third- or higher derivatives. How do I determine whether there exists a volume-form $\mu$ so that $\mathcal D$ is $\mu$-Hermitian? If such a $\mu$ exists, how do I find one? How many are there?

For example, on the circle with the usual volume form, the fourth derivative $\partial^4$ is Hermitian, so some of them are. But maybe most of them are not....

Please re-tag as you see fit.

  • $\begingroup$ Incidentally, after talking to some more experts (i.e. grad students in the area) here in Berkeley, I think that everywhere I said "I believe" I'm being optimistic. It seems that the question isn't totally obvious globally even for first-order differential operators. Locally, $\mu$-Hermeticity of a vector field imposes some PDE on $\mu$, which solvable locally the vector field does not vanish, but it wasn't clear to us that even in that case it's solvable globally. $\endgroup$ – Theo Johnson-Freyd Feb 6 '10 at 3:34
  • $\begingroup$ Are you familiar with the module of Kahler differentials? It gives you a completely algebraic way to work with differentials. $\endgroup$ – Harry Gindi Feb 6 '10 at 17:15
  • $\begingroup$ @HG: No. Where can I learn about it? $\endgroup$ – Theo Johnson-Freyd Feb 7 '10 at 1:02
  • $\begingroup$ @Theo Johnson-Frey: Could you please explain, why every second-order differential operator can be made hermitian with the correct choice of a volume form and what the proposition has to do with it? Is there a relation between an arbitrary linear differential operator of second-order and the Laplace-Beltrami operator, or am I misunderstanding something here? Thx $\endgroup$ – unknown Jun 28 '10 at 13:12

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