Let $M$ be a smooth $n$-dimensional Riemannian manifold, $n \ge 3$. Let $C$ denote the Lie algebra of the conformal vector fields on $M$. It is known that $\dim(C) \le \frac{(n+1)(n+2)}{2}$.

Robert Bryant said here that this can be proven using local calculations and the Frobenius theorem. I don't see how to implement this approach.

Any ideas? (Or other elementary arguments?)

I know that a vector field $V$ is conformal if and only if $$ \nabla V+(\nabla V)^T=\frac{2}{n} \text{tr}(\nabla V)\text{Id}_{TM}= \frac{2}{n} \text{div} V \cdot \text{Id}_{TM}.$$

In the case of Killing fields the situation is easier: $\nabla V$ is skew-symmetric, and $V$ is determined by $V|_p,\nabla V|_p $, so the dimension of the Killing algebra is not greater than $n+\frac{n(n-1)}{2}=\frac{n(n+1)}{2}$.

In the conformal case, it is not true that a conformal field is determined by its value and its covariant derivative at a point. (This time the space of possible $\nabla V|_p$ is of dimension $\frac{n(n-1)}{2}+1$).

Also, in the conformal case we really need to use somewhere that $n \ge 3$, since for $n=2$ it can be infinite-dimensional.

  • $\begingroup$ As you surely know, strictly speaking one should say "...smooth $n$-dimensional manifold, $n \geq 3$, equipped with a Riemannian metric $g$" or, more to the point, "...conformal structure $\mathbf c$". $\endgroup$ Jan 30 '18 at 10:10
  • $\begingroup$ This is probably less elementary than you're aiming for in an answer here, but the claim follows from the fact that there is a bijective correspondence (in fact, equivalence of categories) between, e.g., oriented conformal manifolds and Cartan geometries of type $(\operatorname{SO}(n + 1, 1), P)$ satisfying a certain normalization condition, where $P$ is the stabilizer in $\operatorname{SO}(n + 1, 1)$ of a null ray. In particular, for any conformal structure the algebra of conformal Killing fields has dimension $\leq \dim \operatorname{SO}(n + 1, 1) = \frac{1}{2}(n + 1)(n + 2)$. $\endgroup$ Jan 30 '18 at 10:14
  • $\begingroup$ (This perspective and fact are due to Cartan, by the way, though the language is more modern than his.) $\endgroup$ Jan 30 '18 at 10:16
  • $\begingroup$ Thanks, this sounds very interesting. Can you elaborate, or give me a reference where to read about the relevant Cartan geometries? $\endgroup$ Jan 30 '18 at 10:20
  • 1
    $\begingroup$ There is another method that's morally the same as both the Cartan method and to the one Robert Bryant mentions, but which I find more concrete: For a given metric $g$, one can "prolong" the conformal Killing equation to a closed system of linear PDEs that is equivalent to the conformal Killing equation in the sense that solutions of one correspond to solutions of the other. Since the system is closed and linear, it defines a connection (in fact it is natural in that it is constructed invariantly from $(M, g)$ on a particular vector bundle that turns out to have rank $\frac{1}{2}(n + 1)(n+2)$. $\endgroup$ Jan 30 '18 at 10:37

Note that a vector field $X$ is conformal if and only if there is some function $\Lambda$ such that $\nabla X - \Lambda g$ is skew-symmetric. (Throughout this answer I am identifying $TM$ and $T^* M$ by raising and lowering indices implicitly with $g$ - for example here I really mean $\nabla X^\flat - \Lambda g.$) Let $K$ denote this skew tensor, so that we have $$\nabla X = K + \Lambda g.$$

The idea of prolongation is to iterate this process: we now differentiate $K$ and $\Lambda$, introduce new variables (like we did with $K$ and $\Lambda$) for any unknowns, and repeat until the system closes, meaning that we can write the covariant derivative of each of our variables as some "linear combination" of the other variables. Once we have reached this form, we can interpret the system as $D\xi =0$ for some connection $D,$ at which point the dimension of whatever bundle $\xi$ is a section of (which is basically the direct sum of all our variables) gives an upper bound for the dimension of the solution space.

It turns out we only need one more variable in this case. Following Appendix A2 of these notes by Rod Gover, if we introduce $Q_i = \nabla_i \Lambda + P_{ij} X^j$ where $P$ is the Schouten curvature tensor, then after commuting a bunch of derivatives the system can be written as \begin{align} \nabla_i X_j &= K_{ij} + \Lambda g_{ij} \\ \nabla_i \Lambda &= Q_i - P_{ij} X^j\\ \nabla_i K_{jk} &= - P_{ij} X_k - P_{ik} X_j - g_{ij} Q_k - g_{ik} Q_j + W_{lijk}X^l \\ \nabla_i Q_j &= -P^k_i K_{jk} - P_{ij} \Lambda - C_{kij} X^k. \end{align} Here $W,C$ are the Weyl and Cotton curvature tensors respectively - in particular note that $g,P,W,C$ are all fixed tensor fields, so if we think of $\xi =(X,\Lambda,K,Q)$ as a section of $$E := TM \oplus \mathbb R \oplus \Lambda^2 TM \oplus TM$$ then the system can be written $\nabla \xi + L(\xi)=0$ for some $L\in \Gamma(T^*M \otimes \operatorname{End}(E)).$ Thus conformal Killing vectors $X$ are in $1-1$ correspondence with the sections of $E$ that are parallel with respect to the linear connection $D = \nabla + L;$ so the space of solutions has dimension at most $$\dim E = n + 1 + \binom n 2+n=\frac{(n+1)(n+2)}2.$$

The assumption $n>2$ is used somewhere in the calculation for $\nabla Q$: you can see the easy version (for a flat metric) in these slides by Michael Eastwood. (I wussed out and don't feel like doing the hard version of the calculation myself - hopefully you can work it out.) In the case $n=2$, I believe you would find that the system never closes, no matter how long you prolong the prolongation.

Edit: I recalled receiving a nice handout at a talk of Michael's a few years ago, so I dug it out of the closet. I think its introduction is a better reference than either of my links above, and it can be found online here. It includes some well-written motivation as well as the details of the calculation I omitted.

  • $\begingroup$ This is a very nice exposition! I'll add that we can regard the "flat model" of conformal geometry in dimension $n \geq 3$ to be an $\operatorname{SO}(n + 1, 1)$-invariant conformal structure on $\operatorname{SO}(n + 1, 1) / P \cong \Bbb S^n$, where $P$ is the stabilizer in $\operatorname{SO}(n + 1, 1)$ of a null ray in $\Bbb R^{n + 1, 1}$. Then, we can identify $E$ with the associated bundle $\operatorname{SO}(n + 1, 1) \times_P \mathfrak{so}(n + 1, 1)$ and the connection on $E$ with the connection induced on that bundle by the Maurer-Cartan form on $\operatorname{SO}(n + 1, 1)$. $\endgroup$ Jan 30 '18 at 13:50
  • $\begingroup$ (cont.) In that setting, any choice of metric $g$ in the conformal class determines an identification $E = TM \oplus \Bbb R \oplus \Lambda^2 TM \oplus TM$ as in the answer, and this corresponds exactly to the $\mathfrak{so}(n, \Bbb R)$-module decomposition $\mathfrak{so}(n + 1, 1) \cong \Bbb R^n \oplus \Bbb R \oplus \mathfrak{so}(n, \Bbb R) \oplus (\Bbb R^n)^*$. The idea of the Cartan approach to conformal geometry is to generalize this entire picture suitably to "nonflat" geometry. $\endgroup$ Jan 30 '18 at 13:53
  • 1
    $\begingroup$ Thanks! I really need to put the effort in and learn Klein/Cartan geometry properly sometime, it all seems so beautiful. $\endgroup$ Jan 30 '18 at 14:17
  • $\begingroup$ @AnthonyCarapetis Thanks. That last reference is indeed fantastic. BTW, if I am not mistaken the proof there (Example 1.2.2 ) shows also that a conformal vector field is uniquely determined by its $2$-jet at a single point (if the manifold is connected). This is nice; I wonder if there is an analogous result for conformal maps. $\endgroup$ Feb 1 '18 at 14:09
  • $\begingroup$ @AsafShachar: right - this is simply because parallel sections are determined by their value at a point, and $\xi(p)$ is determined by $j^2_pX.$ (I guess we were already using this fact implicitly when we bounded the solution space by the fiber dimension.) The analogous 2-rigidity for conformal maps is indeed true, though I'm not sure if it can be proved by similar means. $\endgroup$ Feb 2 '18 at 1:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.