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.
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.