Let $\lambda = (\lambda_1, \ldots, \lambda_r)$ and $\mu = (\mu_1, \ldots, \mu_r)$ be partitions such that $\mu_j = \lambda_j +1$ for one index $j$ and $\mu_i = \lambda_i$ for all other $i$. Then there is a natural transformation $\alpha_{\mu/\lambda}: \mathbb{S}_{\lambda}(V) \otimes V \to \mathbb{S}_{\mu}(V)$, where $\mathbb{S}_{\kappa}$ denotes the $\kappa$-Schur functor; $\alpha_{\mu/\lambda}$ is unique up to scaling.

For a smooth manifold $X$, let's define a $\mu/\lambda$-differential operator to be a map $\delta$ from sections of $\mathbb{S}_{\lambda} T^{\ast} X$ to sections of $\mathbb{S}_{\mu} T^{\ast} X$ such that, for any smooth function $f$ and section $v$, we have the Leibniz rule $$\delta(f v) = \alpha_{\mu/\lambda}(df \otimes v) + f \delta(v).$$

Let's define a natural $\mu/\lambda$-differential operator to be a choice $\delta_X$ of a $\mu/\lambda$-differential operator on each manifold $X$ such that, if $\phi: X \to Y$ is a smooth map, then $\phi^{\ast} \circ \delta_Y = \delta_X \circ \phi^{\ast}$.

Are the only natural differential operators scalar multiples of the exterior derivative $d$ with $\mu= 1^{k+1}$ and $\lambda = 1^k$?

**Motivation:** Just curiosity. I've been trying to make $d$ sound natural this term, and one thing that I've said a lot is that it is the only thing that commutes with pullback, so I'm curious if this formalization of that this is true.

Natural Operations in Differential Geometryis dedicated to answering questions of this form: emis.de/monographs/KSM/kmsbookh.pdf $\endgroup$ – Qiaochu Yuan Mar 22 '15 at 19:13isa nonlinear differential operator on $S^2_n(T^*M)$, the bundle ofnondegeneratequadratic forms on $M$ that is natural for diffeomorphisms, namely the operator that assigns to each nondegenerate quadratic tensor its Levi-Civita connection, which is a section of the naturalaffinebundle whose sections are the torsion-free connections on $M$. $\endgroup$ – Robert Bryant Mar 23 '15 at 9:006more comments