Short question with long title:

Suppose $T$ is the tangent functor and $T^2:=T\circ T$ is the second order tangent functor.

Are there natural transformations $T\otimes T \Rightarrow T^2$ ?

I consider the 'usual' domain of $T$ and $T^2$ as the category $\mathbf{M}$ of finite dimensional Hausdorff manifolds together with smooth morphisms, but in case there is a difference feel free to restrict to the category $\mathbf{M}_n$ of $n$-dimensional Hausdorff manifolds together with local diffeomorphisms.


1 Answer 1


It sounds very unlikely, other than the trivial composition $T \otimes T \to id \to T^2$

Let $M=\mathbb R^n$. The double tangent bundle is $\mathbb R^{4n}$. The tensor tangent bundle is $\mathbb R^{n+n^2}$. Suppose a group $G$ acts on $M$ through $GL_n$. On the double tangent bundle, $T$ just doubles the representation, so you end up with $4$ copies of the same representation. On the tensor tangent bundle, you end up with one copy of the representation and one copy of the representation tensored with itself. There is not a natural linear map between these other than the one I described, and there is no reason to expect a natural nonlinear map.

  • $\begingroup$ Yes! Very close to my own reasoning. Since $T^2$ is of second order, relative to derivation, a natural transformation must be a Gl(n,2) invariant map (where Gl(n,2) means the second order differential group) between the standard fibers you mentioned. Now since $T\otimes T$ is of first order $Gl(n,2)$ acts on its fibers linear through $Gl(n,1)$ $\endgroup$ Jun 26, 2012 at 1:26

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