# Understanding the definition of the Paneitz operator

I am given the definition of the Paneitz operator for Riemannian 4-manifolds as follows:

$$P = (-\Delta)^2 + \delta\big(\frac{2}{3}Rg - 2\operatorname{Ric}\big)d$$

where $\Delta$ is the Laplace-Beltrami operator for our fixed metric $g$, $\delta$ is the negative divergence, $R$ is the scalar curvature, $\operatorname{Ric}$ is the Ricci curvature (all with respect to $g$) and $d$ is the differential (acting on functions).

I am struggling to make sense of this: obviously I understand how $(-\Delta)^2$ is an operator acting on functions, but the other part of the RHS I don't understand. If $d$ is to act on a function $f$ to produce a 1-form $df$, and $\delta$ being the negative divergence is to act on a vector field, how can

$$(\frac{2}{3}Rg - 2\operatorname{Ric}\big) \, df$$

be interpreted as a vector field? This term doesn't make sense to me in any case - in the brackets we have a $(0,2)$-tensor, which even if I were to apply a type change to produce a $(1,1)$-tensor, after acting on $df$ this would still give me a $(0,1)$-tensor field rather than a vector field! Should I be considering $g$ and $\mathrm{Ric}$ as their $(2,0)$-tensor types instead? Rather, I imagine it is my understanding of divergence that is wrong here - could someone could please elucidate things? Thank you.

• If you work on this kind of math then you may have many occasions to use LaTeX, which uses almost the same sort of code for mathematical notation as what is used here in MathJax. In correct LaTeX usage one writes \operatorname{Ric} for this sort of thing, and your use of \mathrm{Ric} is why you didn't have proper spacing between the $2$ and $\operatorname{Ric}.$ With \operatorname{}, the spacing depends on the context, with \mathrm the spacing is no more than what appears in things like $2x. \qquad$ – Michael Hardy Jan 6 '18 at 21:43
• @MichaelHardy thanks, I'll do so in the future :) Do you have any pointers on the question? – kt77 Jan 6 '18 at 21:51
• The notation $\delta$ typically refers most directly to the codifferential, i.e. the adjoint of the exterior derivative. It naturally sends $(0,1)$-tensor fields to scalars, as desired. When you compose it with the musical isomorphism to make it act on vector fields it is the negative divergence. – Anthony Carapetis Jan 7 '18 at 0:26

## 1 Answer

There are some implicit musical isomorphisms (i.e., index-raising and index-lowering operations) going on there that are not stated. First, although $df$ is a $1$-form, by means of the metric we can turn it into a vector field $(df)^\sharp = \operatorname{grad} f$.

Next, any covariant $2$-tensor field $T$ can be interpreted as a bundle map from vector fields to covector fields, by sending a vector field $X$ to the $1$-form $T(X,\cdot)$. Thus we can interpret $\big(\frac 2 3 R\, g - 2\operatorname{Ric}\big)df$ as the $1$-form $$\big(\tfrac 2 3 R\, g - 2\operatorname{Ric}\big)(\operatorname{grad} f, \cdot).$$

Finally, apply one more musical isomorphism to turn this $1$-form into a vector field, and then apply the divergence to it.

People who work a lot with Riemannian geometry often get into the habit of just raising and lowering indices automatically without saying anything about it.

• Sorry to bring up an old post, I have recently just come back to this. Whilst I understand your explanation, I don't understand why it wouldn't just be easier to define the operator differently so as to avoid these musical isomorphisms. For instance, why not just replace the $d$ on the far right with a $\operatorname{grad}$? – kt77 Mar 22 '18 at 14:20