Origins, applications, and generalizations of the Q-curvature

by Tom Branson and Rod Gover


Curvature prescription

Everything below will take place in the setting of Riemannian manifolds (or Riemannian conformal manifolds) of even dimension $n$. Of course many statements will also be true for odd-dimensional manifolds and/or pseudo-Riemannian (conformal) manifolds, but our main intent is to make this blurb readable. There will be no reference list here, though there are plans to compile a separate reading list (of real papers) on the topic.

A touchstone in Differential Geometry is the Yamabe equation: for $n>2$,

\begin{displaymath}
\left(\Delta +{\displaystyle{\frac{n-2}{4(n-1)}}}K\right)u={\displaystyle{\frac{n-2}{4(n-1)}}}\hat Ku^{\frac
{n+2}{n-2}}.
\end{displaymath} (1)

This gives the conformal change law for the scalar curvature $K$. That is, if $\omega $ is a smooth function and

\begin{displaymath}
\hat g=e^{2\omega }g,\qquad u:=e^{\frac{n-2}2\omega },
\end{displaymath}

the $\hat K$ given by ([*]) is the scalar curvature of $\hat g$.

Exercise 1. Show that ([*]) implies the conformal change law for the conformal Laplacian

\begin{displaymath}
Y:=\Delta +{\displaystyle{\frac{n-2}{4(n-1)}}}K,
\end{displaymath}

namely

\begin{displaymath}
\hat Y=e^{-\frac{n+2}2\omega }Ye^{\frac{n-2}2\omega }.
\end{displaymath}

Here, and in such formulas below, the function on the very right is to be viewed as a multiplication operator, so the relation really says that for all smooth functions $f$,

\begin{displaymath}
\hat Yf=e^{-\frac{n+2}2\omega }Y\left(e^{\frac{n-2}2\omega }f\right).
\end{displaymath}

When $n=2$, the equation governing the conformal change of $K$ is qualitatively different from a PDE standpoint:

\begin{displaymath}
\Delta \omega +\frac12K=\frac12\hat Ke^{2\omega }.
\end{displaymath} (2)

Like the Yamabe equation, this is quasilinear, but in contrast to Yamabe, it has an exponential (as opposed to power) nonlinearity, and it has an inhomogeneity (the $\frac12K$ term). ([*] is called the Gauss curvature prescription equation), the Gauss curvature in dimension 2 being $\frac12K$.

There is a formal procedure of analytic continuation in dimension (which in fact can be made rigorous) that allows one to guess (or prove) ([*] given ([*])). The Yamabe equation may be rewritten as

\begin{displaymath}
\Delta \left(e^{\frac{n-2}2\omega }-1\right)+
{\displaystyle...
...displaystyle{\frac{n-2}{4(n-1)}}}\hat Ke^{\frac{n+2}2\omega }.
\end{displaymath}

Note that we have slipped in an extra $-\Delta 1=0$ on the left. The advantage of this is that, as a power series in $\frac{n-2}2$, all terms in the equation begin at the first power. Dividing by $\frac{n-2}2$ and then evaluating at $n=2$, we get ([*]).

Exercise 2 (following C.R. Graham). Make the dimensional continuation argument rigorous by looking at stabilisations of the manifold $M$, i.e. the $(n+p)$-dimensional manifolds $M\times T^p$, where $T^p$ is the standard $p$-torus.

There is a generalisation of this whole picture to higher order, in which the role of the pair $(Y,{\sf {J}})$ is played by a pair $(P_m,Q_m)$ consisting of an operator and a local scalar invariant. The $P_m$ are the celebrated Graham-Jenne-Mason-Sparling (GJMS) operators, which by construction have the following properties.

In this last expression, $d$ and $\delta $ are the usual de Rham operators and $\Delta $ is the form Laplacian $\delta d+d\delta $. Note that $P_m$ is unable to detect changes in the $(d\delta )^{m/2-1}$ term of the principal part of $S_m$, sandwiched as it is between a $\delta $ and a $d$.

With these properties, conditions are right to generalise the Yamabe equation to

\begin{displaymath}
\left(P_m+{\displaystyle{\frac{n-m}{2}}}Q_m\right)u={\displa...
...at Q_mu^{\frac{n+m}{n-m}},
\qquad m\notin\{n,n+2,n+4,\ldots\},
\end{displaymath} (3)

where

\begin{displaymath}
u:=e^{\frac{n-m}2\omega }.
\end{displaymath}

Analytic continuation in dimension then yields the following analogue of the Gauss curvature prescription equation. If we denote $P_n$ and $Q_n$ simply by $P$ and $Q$, then
\begin{displaymath}
\fbox{$P\omega +Q=\hat Qe^{n\omega }.$}\end{displaymath} (4)

Though there is much more to be said about the $Q$-curvature, this is probably the central formula of the theory.

Exercise 3. Show that if we have a local invariant $B$ satisfying a conformal change law like ([*], $A\omega +B=\hat Be^{n\omega })$, with $A$ a natural differential operator, then necessarily $A$ is conformally invariant in the sense $\hat A=e^{-n\omega }A$.

The fact that $P_m$ has an expression with rational dependence on the dimension is crucial to making the analytic continuation rigorous, whether one does it by stablilisation (generalising Exercise 2), or by an algebraic argument using ${\mathbb{R}}(n)$-linear combinations in a dimension-stable basis of invariants.

Very explicit formulas for $P_m$ and $Q_m$ are known up to $m=8$. The $m=4$ case, which was already being discussed in the early 1980's, is particularly appealing as a source of intuition, since the formulas there are still quite manageable. Let $r$ be the Ricci tensor and let

\begin{displaymath}
{\sf {P}}:={\displaystyle{\frac{r-{\sf {J}}g}{n-2}}}.
\end{displaymath}

The Paneitz operator is

\begin{displaymath}
P_4:=\delta S_4d+{\displaystyle{\frac{n-4}{2}}}Q_4,
\end{displaymath}

where

\begin{displaymath}
Q_4:=-2\vert{\sf {P}}\vert^2+\frac{n}2{\sf {J}}^2+\Delta {\sf {J}},
\end{displaymath}

and

\begin{displaymath}
S_4:=d\delta +(n-2){\sf {J}}-4{\sf {P}}\cdot.
\end{displaymath}

Here ${\sf {P}}\cdot$ is the natural action of a symmetric 2-tensor on 1-forms.

The transition from $m=2$ to $m=4$ already points up the fact that the $(P_m,Q_m)$ are not uniquely determined: if $C$ is the Weyl conformal curvature tensor, we could add a suitable multiple of $\vert C\vert^2$ to $P_4$ without destroying any of its defining properties. (The coefficient should be rational in $n$, should have a zero at $n=4$, and should not have poles that create new ``bad'' dimensions.)

The $m=4$ situation is also already big enough to show that the study of $Q$ is not just a disguised study of the conformal properties of the Pfaffian ${\sf {Pff}}$. One of the salient properties of the Pfaffian is that it can be written as a polynomial in $R$, without any explicit occurrences of $\nabla $. For example, in dimension 4,

\begin{displaymath}
32\pi^2{\sf {Pff}}_4=\vert C\vert^2-8\vert{\sf {P}}\vert^2+8{\sf {J}}^2.
\end{displaymath}

But the $\Delta {\sf {J}}$ term in $Q_4$ is an absolutely essential aspect, and it generalises: see Exercise 6 below.

All the conformal change laws we've mentioned are good in odd dimensions, and that for $P_4$ in dimension 3 shows that very strange nonlinearities can occur:

\begin{displaymath}
\left(\delta (d\delta +{\sf {J}}-4{\sf {P}}\cdot)d-\frac12Q_4\right)u=-\frac12\hat Q_4u^{-7},
\end{displaymath}

where $u=e^{\omega /2}$.

Back to the general case, a celebrated property of $Q$ is the conformal invariance of its integral on compact manifolds:

\begin{displaymath}
\int Q\,dv_g\ \mbox{ is conformally invariant.}
\end{displaymath}

Indeed, since $dv_{\hat g}=e^{n\omega }dv_g$,

\begin{displaymath}
\int\hat Q\,dv_{\hat g}=\int(Q+P\omega )dv_g.
\end{displaymath}

But $P\omega $ is an exact divergence, by the $\delta \cdots d$ property of $P$, and thus it integrates to 0, showing invariance. This has an immediate generalisation. If $u$ is a smooth function,

\begin{displaymath}
\int\hat Qu\,dv_{\hat g}=\int(Q+P\omega )u\,dv_g.
\end{displaymath}

Since $P$ is formally self-adjoint, we may move it over to $u$ in the very last term under the integral. If it happens that $u\in{\mathcal{N}}(P)$, there is no contribution from this term. Thus

\begin{displaymath}
u\in{\mathcal{N}}(P)\subset{\mathcal{N}}(d)\;\Rightarrow\;
\int Qu\,dv_g\mbox{ is conformally invariant.}
\end{displaymath}


Relativistic considerations

The Einstein equations are obtained by taking the Einstein-Hilbert action $\int K\,dv$ in dimension 4, and taking the total metric variation. This means we take a compactly supported symmetric tensor $h$ and a curve of metrics $g_\varepsilon $ with $(d/d\varepsilon )\vert _{\varepsilon =0}g_\varepsilon =h$, and compute that

\begin{displaymath}
(d/d\varepsilon)\vert _{\varepsilon=0}\left(\int_{\mathcal{K...
...\right)=\int\langle h,r-{\textstyle{\frac{1}{2}}}Kg\rangle dv,
\end{displaymath}

where ${\mathcal{K}}$ is any compact set containing supp$(h)$. Here we may view $\langle\cdot,\cdot\rangle$ as the pairing of a covariant with a contravariant symmetric tensor, or as the metric ($g_0$) pairing of two covariant tensors.

Weyl relativity is one proposal for replacing the Einstein-Hilbert action with an action that is invariant under multiplication of the metric by a positive constant: under the variation above,

\begin{displaymath}
(d/d\varepsilon)\vert _{\varepsilon=0}\int_{\mathcal{K}}\vert C\vert^2dv=\int\langle h,{\mathcal{B}}\rangle dv,
\end{displaymath}

where ${\mathcal{B}}$ is the Bach tensor, a trace-free symmetric 2-tensor with the property (when viewed as a contravariant tensor) that under the usual conformal variation,

\begin{displaymath}
\hat{\mathcal{B}}=e^{-2\omega }{\mathcal{B}}.
\end{displaymath}

One aspect of ${\mathcal{B}}$ is that the nonlinear differential operator
\begin{displaymath}
g\mapsto{\mathcal{B}}_g,
\end{displaymath} (5)

carrying a metric to its Bach tensor, is fourth-order quasilinear. Its linearisation is an interesting fourth-order conformally invariant linear differential operator.

In attempting to generalise this to higher dimensions, it's clear that $\int{\sf {Pff}}$ won't help - its total metric variation is 0, since it's a topological invariant. Choices like $\int\vert C\vert^{n/2}$ for $n=8,12,\ldots$ are uninteresting because the linearisations of the analogues of the operators ([*]) have order lower than one might hope for - less than $n$. In fact these linearisations will even vanish when we vary at a conformally flat metric.

Coming to the rescue of the situation is $Q$:

\begin{displaymath}
(d/d\varepsilon)\vert _{\varepsilon=0}\int_{\mathcal{K}}Q\,dv=\int\langle h,{\mathcal{A}}\rangle dv,
\end{displaymath}

where ${\mathcal{A}}$ is the Fefferman-Graham tensor, a symmetric contravariant 2-tensor with the conformal invariance law $\hat{\mathcal{A}}=e^{(2-n)\omega }{\mathcal{A}}$. The linearisation $D$ of the map $g\mapsto{\mathcal{A}}_g$ has order $n$. The operator $D$ is conformally invariant in the sense that $\hat D=e^{-(n+2)\omega }De^{-2\omega }$ when acting on trace-free symmetric contravariant 2-tensors.

Exercise 4. Show that if $\int S\,dv$ is conformally invariant, then its total metric variation tensor ${\mathcal{C}}$ is conformally invariant. Show that if $T$ is any conformally invariant tensor, then the linearisation of the map $g\mapsto T_g$ is conformally invariant on trace-free perturbations of $g$ (and 0 on pure trace perturbations).


Quantum considerations

Let $A$ be a natural differential operator with positive definite leading symbol, and suppose $A$ is a positive power of a conformally invariant operator. For example, $A$ could be one of the GJMS operators, or it could be the square of the Dirac operator. Then in dimensions 2,4,6, and conjecturally in higher even dimensions,

\begin{displaymath}
-\log{\displaystyle{\frac{\det\hat A}{\det A}}}=\alpha \left...
...right\}+\int\left(\overline{F\,dv}-F\,dv\right)+{\mathcal{H}},
\end{displaymath} (6)

where $\alpha $ is a constant, $F$ is a local scalar invariant, and ${\mathcal{H}}$ is a term depending on the null space of $A$. In particular, if the conformally invariant condition ${\mathcal{N}}(A)=0$ is satisfied, then ${\mathcal{H}}=0$. The determinant involved is the zeta-regularised functional determinant of a positively elliptic operator.

Such formulas are the finite variational formulas corresponding to Polyakov formulas, which are infinitesimal variational formulas for the determinant; these take the form

\begin{displaymath}
(-\log\det A)^\bullet=\int\omega (L+h)
\end{displaymath} (7)

where $L$ is a local invariant and $h$ depends on the null space of $A$, and the superscripted bullet denotes the variation $(d/d\varepsilon)\vert _{\varepsilon=0}$. Here the curve along which we vary can be any curve of conformal metrics with $g^\bullet=2\omega g$; it is particularly convenient to consider curves $g_\varepsilon =e^{2\varepsilon \omega }g_0$ for given $g_0$ and $\omega $. As part of the package producing the Polyakov formula, one gets the conformal index property, that

\begin{displaymath}
\int L\,dv\ \mbox{ is conformally invariant.}
\end{displaymath}

In fact, getting from ([*] to ([*])) may be viewed as a process of finding conformal primitives. We say that a functional ${\mathcal{F}}$ on the conformal class $[g_0]$ is a conformal primitive for a local invariant $L$ if $(d/d\varepsilon)\vert _{\varepsilon=0}{\mathcal{F}}=\int\omega L$. Of course this should happen at all possible choices of background metric $g_0$, and all directions of variation $\omega $. This can be said in a more invariant way, following a suggestion of Mike Eastwood. Putting the ``running'' metric $g$ and the background metric on the same footing in a two-metric functional ${\mathcal{F}}(\hat g,g)$ on the conformal class, we require of a conformal primitive that it be

That the $\int(\overline{F\,dv}-F\,dv)$ term in ([*]) should be this way is obvious; for the term involving $P$ and $Q$, it is a subtle point.

Some local invariants have other local invariants as conformal primitives. For example, since

\begin{displaymath}
{\sf {J}}^\bullet=-2\omega {\sf {J}}+\Delta \omega ,
\end{displaymath}

we have

\begin{displaymath}
({\sf {J}}^{n/2})^\bullet=-n\omega {\sf {J}}+\frac{n}2{\sf {J}}^{n/2-1}\Delta \omega ,
\end{displaymath}

so

\begin{displaymath}
\left(\int{\sf {J}}^{n/2}dv\right)^\bullet=\frac{n}2\int\omega \Delta ({\sf {J}}^{n/2-1}).
\end{displaymath}

This makes $2J^{n/2}/n$ a conformal primitive for $\Delta ({\sf {J}}^{n/2-1})$. $Q$ is an example of a local invariant that does not have such a local conformal primitive.

In order to handle these objects more cleanly, let's view $P$ and $Q$ as being density-valued objects ${\bf P}$ and ${\bf Q}$, so that a ``weight term'' involving the conformal factor does not appear explicitly. In other words, replace $Pf$ by ${\bf P}f=Pf\,dv_g$, and $Q$ by ${\bf Q}=Q\,dv_g$. (For readers unfamiliar with densities, not much is lost conceptually in assuming our manifold is oriented and talking about $n$-forms instead of scalar densities.) Then

\begin{displaymath}
\hat{\bf Q}={\bf Q}+{\bf P}\omega ,\ \ \ \ \int\hat{\bf Q}=\int{\bf Q}.
\end{displaymath} (9)

Let's also take all the local invariants we consider to be density valued. A local primitive for ${\bf L}$ will be a functional ${\mathcal{F}}$ with ${\mathcal{F}}^\bullet=\int\omega {\bf L}$. The right side of ([*]) (ignoring the term ${\mathcal{H}}$, which may be shown to have a conformal primitive, or which may be eliminated by redefining the determinant) goes over to
\begin{displaymath}
\alpha \left\{\frac12\int\omega {\bf P}\omega +\int\omega {\bf Q}
\right\}+\int\left(\hat{\bf F}-{\bf F}\right).
\end{displaymath} (10)

Recall that when we wrote this, we were thinking about a background metric $g_0$ and a perturbed metric $g_\omega $. But there is an interesting way of rewriting the first term, as

\begin{displaymath}
\frac12\alpha \int\omega ({\bf Q}_\omega +{\bf Q}_0).
\end{displaymath} (11)

To eliminate the appearance of a choice (of $g_0$) being made, we define $c(\hat g,g):=\omega $ for two conformally related metrics $\hat g=e^{2\omega }g$. The two-metric functional $c$ is alternating and cocyclic as above. From this viewpoint, ([*]) is ($\alpha $ times) a two-point functional

\begin{displaymath}
{\mathcal{G}}(\hat g,g):=\frac12\int c(\hat g,g)({\bf Q}_{\hat g}+{\bf Q}_g),
\end{displaymath}

and clearly

\begin{displaymath}
{\mathcal{G}}(\hat g,g)=-{\mathcal{G}}(g,\hat g).
\end{displaymath}

The functional ([*]) is
\begin{displaymath}
\alpha {\mathcal{G}}(\hat g,g)+\int({\bf F}_{\hat g}-{\bf F}_g).
\end{displaymath} (12)

Since the log-determinant functional will obviously satisfy the cocycle condition ([*]), and since the second functional in ([*]) satisfies such a condition, we expect ${\mathcal{G}}$ to behave similarly. One way to see that this expectation is fulfilled is to use the conformal primitive property: for fixed $g$ and $\hat g$, with ${\mathcal{G}}$ for ${\mathcal{F}}$ above, the two sides of ([*]) have the same conformal variation (of $\hat{\hat g}$), and the same value at $\hat{\hat g}=g$.

Exercise 5. Show that if $g_0$, $g_\omega =e^{2\omega }g_0$, $g_\zeta =e^{2\omega }g_0$, and $g_\eta=e^{2\eta}g_0$ are 4 conformally related metrics, then

\begin{displaymath}
\int\omega ({\bf Q}_\zeta -{\bf Q}_\eta)+\int\zeta ({\bf Q}_\eta-{\bf Q}_\omega )+\int\eta({\bf Q}_\omega -{\bf Q}_\zeta )=0.
\end{displaymath}

The following conjecture would be enough to prove the conjecture mentioned at the beginning of the section on the form of the determinant quotient.

Conjecture 1. If ${\bf S}$ is a natural $n$-form and $\int{\bf S}$ is conformally invariant, then

\begin{displaymath}
{\bf S}={\rm const}\cdot{\bf Q}+{\bf L}+{\bf G},
\end{displaymath}

where ${\bf L}$ is a local conformal invariant and ${\bf G}$ has a local conformal primitive. That is, there is a local invariant ${\bf F}$ for which the conformal variation of $\int{\bf F}$ is $\int\omega {\bf G}$.

The point of separating these 3 kinds of terms is that $\int\omega {\bf L}$ will have a very banal conformal primitive, namely itself, or (to write it in a way that makes the properties of a conformal primitive more apparent),

\begin{displaymath}
\frac12\int c(\hat g,g)({\bf L}_{\hat g}+{\bf L}_g).
\end{displaymath}

${\bf Q}$ has an interesting conformal primitive, as discussed above. ${\bf Q}$ is not uniquely defined, but the The ${\bf Q}$ in the statement of the conjecture could be anyone's favorite version of ${\bf Q}$. In fact, ${\bf Q}$ is well-defined up to addition of an ${\bf L}$.

Then there are the following related conjectures:

Conjecture 2. Any ${\bf S}$ as above may be written

\begin{displaymath}
{\rm const}\cdot {\bf Q}+{\bf L}+{\bf V},
\end{displaymath}

where $V$ is an exact divergence.

Conjecture 3. Any ${\bf S}$ as above may be written

\begin{displaymath}
{\rm const}\cdot{\sf Pff}+{\bf L}+{\bf V}.
\end{displaymath}

There are at least 2 filtrations of the local invariants of this type that should be relevant. First, any invariant can be written as a sum of monomial expressions in $R$ and $\nabla $ with $k_\nabla +2k_R=n$, where $k_\nabla $ (resp. $k_R$) is the number of occurrences of $\nabla $ (resp. $R$) in the monomial. If an invariant ${\bf T}$ can be written with $k_\nabla \le p$ for each monomial term, let's say ${\bf T}\in{\mathcal{P}}_p$. Then

\begin{displaymath}
{\mathcal{P}}_0\subset{\mathcal{P}}_2\subset\cdots\subset{\mathcal{P}}_{n-2},\qquad\qquad{\mathcal{P}}_{{\rm odd}}=0.
\end{displaymath}

Pff is in the most elite space, ${\mathcal{P}}_0$. The exact divergences inject into ${\mathcal{P}}_2/{\mathcal{P}}_0$.

Exercise 6. Use the conformal change law for ${\bf Q}$ to show that the class of ${\bf Q}$ in ${\mathcal{P}}_{n-2}/{\mathcal{P}}_{n-4}$ is nontrivial, and agrees with the class of $(\Delta ^{(n-2)/2}{\sf {J}})dv_g$. This establishes that Pff and ${\bf Q}$ are ``at opposite ends'' of the ${\mathcal{P}}$-filtration.

The other filtration is by the degree of the conformal change law. If

\begin{displaymath}
\hat{\bf T}={\bf T}+{\bf T}_1(d\omega )+\cdots+{\bf T}_n(d\omega ),
\end{displaymath}

with ${\bf T}_i$ of homogeneity $i$ with respect to scalar multiples of $\omega $, then say ${\bf T}\in{\mathcal{L}}_p$ if ${\bf T}_q=0$ for $q>p$. Then local conformal invariants are in ${\mathcal{L}}_0$, and ${\bf Q}$ is in ${\mathcal{L}}_1$. Pff on the other hand does not look so great in this filtration.


Other routes to $Q$ and its variants

There is an alternative definition of $Q$ which avoids dimensional continuation. We write ${\mathcal{E}}$ for the space of smooth functions, ${\mathcal{E}}^{1}$ for space of smooth 1-forms and define the special section

\begin{displaymath}
I^g:= \left(\begin{array}{c} 2-n \\ 0 \\ {\sf {J}}\end{array}\right)
\end{displaymath}

of the direct sum bundle ${\mathcal{E}}\oplus {\mathcal{E}}^1 \oplus {\mathcal{E}}$. Let us first set the dimension to be 4, simply present the some results and then explain how this works. Then we get

\begin{displaymath}
\square I^g = \left(\begin{array}{c} 0 \\ 0 \\ Q_4 \end{array}\right) ,
\end{displaymath}

where $\square $ is the coupled conformal Laplacian operator. More precisely $\square = -\nabla^a\nabla_a+(n-2)K/(4n-4)$, which appears to be the usual formula for the conformal Laplacian (cf. $Y$ above), but now $\nabla $ is a connection which couples the usual metric connection with the connection

\begin{displaymath}
\nabla_a \left(\begin{array}{c} \sigma \\ \mu \\ \tau \end{a...
...akebox[0.4em][l]{\tiny $\vert$}}{\sf {P}}\end{array}\right) .
\end{displaymath}

on the sum bundle ${\mathcal{T}}:={\mathcal{E}}\oplus {\mathcal{E}}^1 \oplus {\mathcal{E}}$. This bundle ${\mathcal{T}}$ is called the standard tractor bundle and this connection is usually termed the (normal conformal) tractor connection. It is equivalent to a principal bundle structure known as the (normal conformal) Cartan connection. For those who know about Cartan connections we can say that the tractor bundle and connection is an associated bundle and connection for the Cartan bundle. We have used a metric $g$ to express these objects in terms of a Riemannian structure but in fact the bundle and connection are conformally invariant and so descend to well defined structures on a conformal manifold. In fact, to be more accurate, the decomposition of the standard tractor bundle ${\mathcal{T}}$ is really ${\mathcal{T}}={\mathcal{E}}[1]\oplus {\mathcal{E}}^1\otimes {\mathcal{E}}[1]\oplus {\mathcal{E}}[-1]$ where ${\mathcal{E}}[w]$ indicates the space of conformal densities of weight $w$. The field $I^g$ is a section of ${\mathcal{T}}\otimes {\mathcal{E}}[-1]$. In this section and the next we are allowing tensors and tractor fields to be density valued, to simplify the notation, but partially suppressing the details of weights involved.

This construction generalises. In each even dimension $n$ there is a conformally invariant differential operator $\square _{n-2}$ so that for any metric $g$ we have

\begin{displaymath}
\square _{n-2} I^g= \left(\begin{array}{c} 0 \\ 0 \\ Q_{n} \end{array}\right) .
\end{displaymath} (13)

Here $I^g$ is as above, while $\square _{n-2}$ has the form $\Delta ^{n/2-1}
+{\rm LOT}$. The tractor field $I^g$ is not conformally invariant, but it does have an interesting conformal transformation. If $\hat{g}$ is a metric related to $g$ conformally according to $\hat{g}=e^{2\omega }g$ ($\omega $ a smooth function) then
\begin{displaymath}
I^{\hat{g}} =I^g+ D \omega,
\end{displaymath} (14)

where $D$ is a well known second order conformally invariant linear differential operator known as the tractor $D$ operator. From this and ([*] it follows that the $Q$-curvature $\hat{Q})_n$, for $\hat{g}$, differs from $Q_n$ by a linear conformally invariant operator acting on $\omega $. In fact it follows easily from the definition of $\square _{n-2}$ that

\begin{displaymath}
\square _{n-2} D \omega = \left(\begin{array}{c} 0 \\ 0 \\ P_n \omega \end{array}\right)
\end{displaymath}

where $P_n$ is the GJMS operator of order $n$. So we have recovered the now famous property $\hat{Q}_n= Q_n + P_n\omega $ (cf. ([*])).

As a final comment on the above story we should clarify the origins of the tractor field $I^g$ defined and used above. For those who are familiar with tractors a more enlightening alternative definition is

\begin{displaymath}
I^g_A : = -\frac{1}{n}D^A\sigma^{-1}D_{AB}\sigma.
\end{displaymath}

Here $D^A$ is the tractor $D$ operator and $D_{AB}$ is the so-called fundamental $D$ operator. $0\neq\sigma\in \Gamma{\mathcal{E}}[1]$ is the conformal scale corresponding to the metric $g$. The point is that these operators are both conformally invariant and under $g\mapsto \hat{g}=e^{2\omega }g$ we have $\sigma\mapsto e^\omega \sigma$. Since $D_{AB}$ satisfies a Leibniz rule and $\sigma^{-1}D_{AB}\sigma$ is a ``logarithmic derivative'' the conformal transformation law of $I^g$ is no surprise.

The main results above are derived via the ambient metric construction of Fefferman and Graham. Explaining this construction would be a significant detour at this point. Suffice to say that this construction geometrically associates to an $n$-dimensional conformal manifold $M$ an $(n+2)$-dimensional pseudo-Riemannian manifold $\tilde{M}$. The GJMS operators $P_{2\ell}$ arise from powers of the Laplacian $\tilde{\Delta }^\ell$, of $\tilde{M}$, acting on suitably homogeneous functions. The operators $\square _{2\ell}$ arise in a similar way from $\tilde{\Delta }^\ell$ on appropriately homogeneous sections of the tangent bundle $T\tilde{M}$. Such homogeneous sections correspond to tractor fields on the conformal manifold $M$. The results above are given by an easy calculation on the ambient manifold. Thus we can take ([*]) as a definition of the $Q$-curvature; it is simply the natural scalar field that turns up on the right hand side.

While this definition avoids dimensional continuation, there is still the issue of getting a formula for $Q_n$. There is an effective algorithm for re-expressing the ambient results in terms of tractors which then expand easily into formulae in terms of the underlying Riemannian curvature and its covariant derivatives. This solves the problem for small $n$. For example

\begin{displaymath}
Q_6=
8 {\sf {P}}{{}_{
\ifthenelse{\equal{i}{;}}{\vert}{i}}{...
...j}{;}}{\vert}{j}}} {\sf {P}}{{}^{i}{}^{j}} \mbox{{\sf {J}}} +
\end{displaymath}


\begin{displaymath}
8 \mbox{{\sf {J}}}^{3}
- 8
\mbox{{\sf {J}}}{{}_{
\ifthene...
...qual{l}{;}}{\vert}{l}}}
\mbox{C}{{}^{i}{}^{k}{}^{j}{}^{l}}
.
\end{displaymath}

While such formulae shed some light on the nature of the $Q$-curvature it would clearly be ideal to give a general formula or simple inductive formula. From the angle discussed here the missing information is a general formula for the operators $\square _{n-2}$.


Problem 1: Give general formulae or inductive formulae for the operators $\square _{2\ell}$.


This seems to be a difficult problem. In another direction there is another exercise to which we already have some answers. One of the features of the $Q$-curvature is that it ``transforms by a linear operator'' within a conformal class. More precisely, it is an example of a natural Riemannian tensor-density field with a transformation law

\begin{displaymath}
N^{\hat{g}}= N^g + L\omega ,
\end{displaymath} (15)

$L$ being some universal linear differential operator. (Here $\omega $ has the usual meaning; $\hat{g}=e^{2\omega }g$.)


Problem 2: Construct other natural tensor-densities which transform according to ([*]). (Note that any solution yields a conformally invariant natural operator $L$.)


From the transformation law for $I^g$ above, we can evidently manufacture solutions to this problem. We have observed already that $I^g$ is a section of the bundle ${\mathcal{T}}[-1]:={\mathcal{T}}\otimes{\mathcal{E}}[-1]$. If $P$ is any scalar (or rather density) valued natural conformally invariant differential operator which acts on ${\mathcal{T}}[-1]$ then $P$ can act on $I^g$, and $P I^g$ has a conformal transformation of the the form ([*]). Using the calculus naturally associated to tractor bundles (or equally effectively, using the ambient metric) it is in fact a simple matter to write down examples, and the possibilities increase with dimension. This is most interesting when the resulting scalar field gives a possible modification to the original $Q$-curvature. For those familiar with densities this means that $P$ should take values in densities of weight $-n$; this is the weight at which densities that can be integrated on a conformal manifold. For example, in any dimension we may take $P$ to be $\iota (D)
\vert C\vert^2$ where $\iota (D)$ indicates a contracted action of the tractor $D$ operator and the square of the Weyl curvature is here viewed as a multiplication operator. In dimension $6$ this takes values in ${\mathcal{E}}[-6]$ and we have

\begin{displaymath}
\iota (D) \vert C\vert^2 I^g = 4 \Delta \vert C\vert^2
\end{displaymath}

and

\begin{displaymath}
\iota (D) \vert C\vert^2 I^{\hat{g}}=
4\Delta \vert C\vert^2 - 16 \delta \vert C\vert^2 d \omega .
\end{displaymath}

Note that in this example the conformally invariant ``$L$-operator'' $\delta \vert C\vert^2 d$ is formally self-adjoint. So for any constant $\alpha $, $Q_6+\alpha \iota (D) \vert C\vert^2 I^g$ is another scalar field with almost the same properties as $Q_6$. It is not so closely related to the GJMS operator $P_6$, but it is related instead to a modification of $P_6$ by $\delta \vert C\vert^2 d$. It is clear that solutions to Problem 2 have a role to play in the problem of characterising the $Q$-curvature and the GJMS operators.


A generalisation: maps like $Q$

So far we have viewed the $Q$-curvature as a natural scalar field. It turns out that if instead we view it as an operator then it fits naturally into a bigger picture. To simplify matters suppose we are working with a compact, oriented, but not necessarily connected, manifold of even dimension $n$. We fix $n$ and so omit $n$ in the notation for $Q$. We can view $Q$ as a multiplication operator from the closed 0-forms ${\mathcal{C}}^0$ (i.e. the locally constant functions) into the space of $n$-forms ${\mathcal{E}}^n$ (which we identify with ${\mathcal{E}}[-n]$ via the conformal Hodge $\star$). With the observations above we have the following properties:

  1. $Q:{\mathcal{C}}^0\to {\mathcal{E}}^n$ is not conformally invariant but $\hat{Q}= Q + P_n\omega $, where $P_n$ is a formally self-adjoint operator from 0-forms to $n$-forms. $P_n$ has the form $d M d$ which implies the next properties.
  2. $Q:{\mathcal{C}}^0\to H^n(M) $ is conformally invariant and non-trivial in general.
  3. If $c\in {\mathcal{C}}^0$ and $u\in {\mathcal{N}}(P_n)$ then $
\int u Q c
$ is conformally invariant.
  4. (See the discussion immediately below.) In each choice of metric $Q:{\mathcal{E}}^0\to {\mathcal{E}}[-n]$ is formally self-adjoint.
  5. (See the discussion immediately below.) $Q 1$ is the $Q$-curvature.
The last properties are trivial since the operator is multiplication by a scalar field and by definition $Q 1=Q$. However we should note that we can add to $Q$ any differential operator that annihilates constants, and properties 1-3 will be unaffected. Property 4 is suggesting that if we do that, then we should insist that the result is formally self-adjoint.

The idea now is to look for analogous operators on other forms. We write ${\mathcal{C}}^k$ for the space of closed $k$-forms. Consider the operator

\begin{displaymath}
M^g=d\delta +2{\sf {J}}-4{\sf {P}}\sharp :{\mathcal{E}}^{n/2-1}\to {\mathcal{E}}^{n/2-1}[-2].
\end{displaymath} (16)

(Note that by the conformal Hodge star ${\mathcal{E}}^{n/2-1}[-2] \cong {\mathcal{E}}^{n/2+1}$, so we can also view this as an operator into $(n/2+1)$-forms.) Then:

Exercise 7. On ${\mathcal{C}}^{n/2-1}$ we have

\begin{displaymath}
M^{\hat{g}}= M^g+ \beta \delta d \omega ,
\end{displaymath}

where $\beta$ is some nonzero constant, $\hat{g}=e^{2\omega }g$, and in the display $\omega $ is viewed as a multiplication operator.

Note that the conformal variation term $\delta d$ is the Maxwell operator and is formally self-adjoint. So $M^g$ satisfies the analogue of property 1 above. The analogue of property 3 is an immediate consequence, i.e., $\int \langle u, M c\rangle$ is conformally invariant where now $c$ is a closed $(n/2-1)$-form and $u\in {\mathcal{N}}(\delta d)$ (so in fact by compactness $u,c$ are both closed). Next observe, by inspection, that $M^g$ is formally self-adjoint. So we have analogues for 1,3,4. There is also a bonus property, which is clear from the transformation law displayed:

\begin{displaymath}
\delta M^g d :{\mathcal{E}}^{n/2-2}\to {\mathcal{E}}^{n/2-2}[-4]
\end{displaymath}

is a non-trivial conformally invariant operator. In dimension 4 this is the Paneitz operator.

So finally we need an analogue for property 2. It is clear that $M^g$ is conformally invariant as a map ${\mathcal{C}}^{n/2-1}\to {\mathcal{E}}^{n/2-1}[-2]/{\mathcal{R}}(\delta )$, so this is an analogue. But we can do more. There is no reason to suppose the image is co-closed. On the other hand note that $\delta M^g$ is conformally invariant on ${\mathcal{C}}^{n/2-1}$ and so we have the following:

\begin{displaymath}
\begin{array}{l}
M^g: {\mathcal{H}}^{n/2-1}\to H^{n/2+1}(M) ...
...{\mathcal{C}}^{n/2-1}\to {\mathcal{E}}^{n/2-2}[-4])
\end{array}\end{displaymath} (17)

is conformally invariant. The space ${\mathcal{H}}^{n/2-1}$ may be viewed as the space of ``conformal harmonics''. Evidently $\dim ({\mathcal{H}}^{n/2-1})$ is not always the same as the Betti number $b^{n/2-1}$, but the elliptic coercivity of the pair $(d, \delta M)$ gives it a good chance of returning the Betti number off a set of conformal structures that is somehow small. One should also check that the map ([*]) is non-trivial.


Fact: Let $M=S^p\times S^q$, where $p=n/2-1$, $q=n/2+1$, with the standard Riemannian structure. Then $\phi\in {\mathcal{H}}^p$ if and only if $\phi$ is harmonic. Furthermore, the map ([*]) is non-trivial.


In some recent work the authors have used the ambient metric, and its relationship to tractors, to show that the above construction generalises along the following lines: There are operators $M_k^g:{\mathcal{E}}^k\to {\mathcal{E}}^{n-k}$ ($k\leq n/2-1$), given by a uniform construction, with the following properties:

  1. $M_k^g:{\mathcal{C}}^k\to {\mathcal{E}}^{n-k}$ has the conformal transformation law $M^{\hat{g}}_k= M^g_k + L_{k}\omega $, where $L_{k}$ is a formally self-adjoint operator from $k$-forms to $(n-k)$-forms, and is a constant multiple of $d M_{k+1}^g d$.
  2. ${\mathcal{H}}^k:= {\mathcal{N}}(d M^g_k: {\mathcal{C}}^k\to {\mathcal{E}}^{n-k+1} )$ is a conformally invariant subspace of ${\mathcal{C}}^k$ and $M_k:{\mathcal{H}}^k\to H^{n-k}(M) $ is conformally invariant. There are conformal manifolds on which $M_k$ is non-trivial.
  3. If $c\in {\mathcal{C}}^k$ and $u\in {\mathcal{N}}(L_k)$ then

    \begin{displaymath}
\int \langle u, M_k^g c \rangle
\end{displaymath}

    is conformally invariant.
  4. For each choice of metric $g$, $M_k^g:{\mathcal{E}}^k\to {\mathcal{E}}^{n-k}$ is formally self-adjoint.
  5. $M_0^g 1$ is the $Q$-curvature.
From the uniqueness of the Maxwell operator at leading order (as a conformally invariant operator ${\mathcal{E}}^{n/2-1}\to {\mathcal{E}}^{n/2-1}[-2]$), and the explicit formula ([*] it is clear $M_{n/2-1})^g$ is not the difference between any conformally invariant differential operator and a divergence (even as an operator on closed forms). A similar argument applies to the $M_k$ generally. Thus, from the point of view that the $Q$-curvature is a non-conformally invariant object that in a deep sense cannot be made conformally invariant, but one which nevertheless determines a global conformal invariant, the operators $M_k^g$ give a genuine generalisation of the $Q$-curvature to an operator on closed forms.

Problems $k$: There are analogues for the operators $M_k^g$ of most of the conundrums and problems for the $Q$-curvature.




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