Let $f:\mathbb{R}\to \mathbb{R}$ be a function, then looking $f$ as a function between manifolds, $df:T\mathbb{R}=\mathbb{R}^2\to \mathbb{R}^2$ and $d^2f:TT\mathbb{R}=\mathbb{R}^4\to \mathbb{R}^4$ are given by

$$df(x,a)=(f(x),\frac{df}{dx}a)$$ $$d^2f((x,a),(b,c))= ((f(x),\frac{df}{dx}a),(\frac{df}{dx}b,\frac{d^2f}{dx^2} ab+ \frac{df}{dx} c))$$

So the second derivative of $f$, as you know it from calculus, is embedded in the differetial geometric definition of $d^2f$ in a wierd way. There is a similar story in higher orders.

Now suppose $f:M\to N$, $f(x)=y$, is a smooth map between two manifolds. As before, for each k, we have $d^kf: TTT\cdots TN \to TTT\cdots TM$.

Question: For $v_1,\cdots,v_k \in T_xM$, is there an object $\frac{d^kf}{dv_1\cdots dv_k}$ similar to the way we define higher derivatives of a function on $\mathbb{R}$? how is it defined? and where does it belong to? (Is $\frac{df}{dv_1\cdots dv_k}\in T_yN$ ?)

Apologies in advance if the question is obvious or absurd.

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    $\begingroup$ en.wikipedia.org/wiki/Jet_bundle $\endgroup$ – Otis Chodosh Jul 14 '12 at 23:10
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    $\begingroup$ Functions whose first derivatives are zero are constant functions. You can identify these topologically. functions whose second derivatives are zero are linear. This is not an intrinsic property of $\mathbb R$. You have to put some extra structure on it. $\endgroup$ – Will Sawin Jul 15 '12 at 2:26
  • $\begingroup$ well, put a metric and connection. not a big deal. $\endgroup$ – Mohammad F. Tehrani Jul 15 '12 at 3:09
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    $\begingroup$ @Mohammad F.Tehrani: to put a metric is a big deal, as the construction will depend on the metric and not be intrinsic to the differentiable manifold. If you accept the object $\partial^k f/\partial \bar v$ you seek to depend upon a choice of metric, you should revise your question accordingly. $\endgroup$ – Benoît Kloeckner Jul 15 '12 at 12:15
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    $\begingroup$ @ Mohammad F. Tehrani: I think what Otis Chodosh means is that the $k$-jet of a function is the appropriate analog, on manifolds, of the system of derivatives up to order $k$. You asked for an object similar to the higher derivatives, and jets are such objects. $\endgroup$ – Andreas Blass Jul 15 '12 at 13:59

It's not very clear to me which properties you expect from that object, but interpreting the question literally the answer is no: there is no natural way to associate to some vectors $v_1,v_2,\ldots,v_k\in T_x M$ and a function $f\in C^\infty(M)$ an object corresponding "only" to the partial derivative $\frac{\partial ^kf}{\partial v_1\cdots\partial v_k}$ $k\geq 2$. As Otis remarks there are jets, but these encode all lower order derivatives too, and there is no natural way to disentangle them. The reason is that all functions with $df(x)\neq 0$ look locally the same, so they would all have "second derivative" zero. This is what Will says in his remark.

On the other hand, if the function $f$ vanishes to order $k-1$ in $x$ (in terms of jets: makes $k-1$-contact with a constant function), then the partial derivative depends only on the vectors $v_1,v_2,\ldots,v_k\in T_x M$ and is an element of $T_{f(x)}Y$.

If you assume a metric on $M$ you should also be able to save the situation, maybe extending the vectors to geoedesics and then by parallel transport to vector fields... but I'm not sure about that.

Edit in response to the comment: a high level answer is this: the fibers of the projection from the space of jets of order k to jets of order k-1 are naturally affine spaces over the vector space $S^k(T^*_x M)\otimes T_y N$. So if two maps have contact of order k-1, then their jets of order k differ by a k-symmetric map $T_x M\times \cdots T_x M \to T_y N$, which in your case is the map you expect in local coordinates $v_1,\ldots ,v_k\to \frac{\partial ^kf}{\partial v_1\cdots\partial v_k}$. There are different proofs for this affine structure in the literature on jets (personally I find non of them particularly enlightening).

In the case that $N=\mathbb{R}$ it is not so hard to show directly, that if you extend your vectors to vector fields $X_1,\ldots,X_2$ then the map $v_1,\ldots ,v_k\mapsto X_k(X_{k-1}(\cdots (X_1(f))\cdots)(x)$ is independent of the choice of extensions when $f$ vanishes to order k-1 in $x$.

  • $\begingroup$ In the paragraph before the last paragraph: " if the function f vanishes to order k−1 ... then the partial derivative depends only on the vectors $v_1,v_2,\cdots,v_k∈T_xM$ and is an element of $T_f(x)Y$." This is sufficient for what I need, How do you show/define it in this case; more specifically, how do you see the result as an element of $T_f(x)Y$. It would be great if you can expand this paragraph more. $\endgroup$ – Mohammad F. Tehrani Jul 15 '12 at 14:47
  • $\begingroup$ I think the book by Saunders, "The geometry of jet bundles" should contain a proof. $\endgroup$ – Michael Bächtold Jul 15 '12 at 18:54

It's true that "jet bundles" is a perfectly correct 2-word answer to the original question, but I feel a Diff Geom 101 discussion could be useful here:

You can differentiate a function in the direction of a tangent vector at a point, but you can't take second derivatives with respect to 2 tangent vectors at a point, because the first derivative has only been defined at a single point. That is, if $f$ is a function on a manifold and $V_p$ and $W_p$ are tangent vectors at $p$, then $V_p f$ makes sense but $W_p (V_p f)$ doesn't make sense.

In order to have it make sense, you have to extend the tangent vector $V_p$ to a vector field. Then $W_p (Vf)$ does make sense, but the answer depends on how you decide to extend $V_p$ (i.e. the expression is not tensorial in $V$). The reason why this is different from Euclidean space is that in Euclidean space there is a default way to extend tangent vectors to vector fields. This is one way to motivate covariant derivatives.

However, in the special case when $df$ vanishes at $p$, the answer does NOT depend on how you extend $V_p$, and hence you can sensibly talk about the second derivatives of a function at a critical point, even without a connection. Furthermore if those second derivatives also vanish at that critical point, then you can have third derivatives, and so on. In this way one can make sense of the concept of "vanishing to order $k$."


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