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.



Are there any natural differential operators besides $d$?
I think your question, the way it is stated, makes one want to classify unary and binary (depending how far you generalise the question as written) invariant differential operators on tensor fields. This has been done for unary operators by an awful lot of people, and the statement indeed is that $d$ is the only operator of that sort. More interestingly, there exists a full classification of binary invariant operators, this was done by Grozman around 1980, and is documented in <http://arxiv.org/abs/math/0509562> .


