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If $E, F$ are real finite dimensional vector spaces and $\mu\colon E \to F$, we can speak of a (total) derivative of $\mu$ in Fréchet sense: $D\mu$, if it exists, is the unique mapping from $E$ to $L(E; F)$, the vector space of linear $E\to F$ mappings, such that for all $x, x_0\in E$ we have

$$\mu(x)=\mu(x_0)+D\mu(x_0)(x-x_0)+o(\lvert x-x_0\rvert). $$

Now since $L(E;F)$ is a vector space itself, the construction can be iterated yielding higher-order derivatives $D^2\mu=D(D\mu), D^3\mu=D(D^2\mu)\ldots$

The concept of first derivative extends to maps $\mu\colon M \to N$ with $M, N$ smooth manifolds, in which case $D\mu\colon TM \to TN$ is defined by

$$D\mu(X_p)(f)=X_p(\mu \circ f), \qquad \forall p \in M,\ \forall X_p \in T_pM,\ \forall f \in C^\infty(N).$$

Question. What about second derivatives? How to generalize the above construction from vector spaces to smooth manifolds?

The obvious way, that of taking $D^2\mu=D(D\mu)$, seems a bit awkward because it involves the complicated tangent-bundle-of-tangent-bundle $T(TM)$. Also, if $\mu=f \colon M \to \mathbb{R}$, I would expect the definition to boil down to

$$D^2f(X_p, Y_p)=(X_pY_p)(f), \qquad \forall p \in M, \forall X_p, Y_p \in T_pM.$$

I would use some reference material on this question and its applications.

Thank you.

share|cite|improve this question
What's awkward about taking the tangent bundle of the tangent bundle? It's a pretty natural thing to do. – Qiaochu Yuan Mar 21 '12 at 19:32
higher derivatives are jets; see e.g. – user8268 Mar 21 '12 at 19:40
Jets are more like Taylor polynomials, rather, no? – Mariano Suárez-Alvarez Mar 21 '12 at 19:48
@QiaochuYuan: It is that I find it hard to visualize a "vector tangent at a vector", while there should be (IMHO) something like "two vectors tangent at the same point". – Giuseppe Negro Mar 21 '12 at 19:49
@GiuseppeNegro: You can kind-of do that in synthetic differential geometry. There, the tangent bundle of a manifold $M$ is defined to be the function space $M^D$, where $D$ is the space of first-order infinitesimals, so the double tangent bundle is just $(M^D)^D$, which everyone knows is naturally isomorphic to $M^{D \times D}$ – which is exactly what you expect. The main thing is that an element of the double tangent bundle shouldn't really be thought of as a pair of tangent vectors, but rather as a kind of infinitesimal square... – Zhen Lin Mar 21 '12 at 21:14

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