The Hessian matrix $\{\partial_i \partial_j f \}$ of a function $f:\mathbb{R}^n \to \mathbb{R}$ depends on the coordinate system you choose. If $x_1,\cdots,x_n$ and $y_1,\cdots,y_n$ are two sets of coordinates (say, in some open neighborhood of a manifold), then $\frac{\partial f(y(x))}{\partial x_i} = \sum_{k} \frac{\partial f}{\partial y_k} \frac{\partial y_k}{\partial x_i}$. Differentiating again, this time with respect to $x_j$, we get $\frac{\partial^2 f(y(x))}{\partial x_i \partial x_j} = \sum_{k} \sum_{l} \frac{\partial^2 f}{\partial y_k \partial y_l} \frac{\partial y_l}{\partial x_j} \frac{\partial y_k}{\partial x_i}+\frac{\partial f(y(x))}{\partial y_k}\frac{\partial^2y}{\partial x_i \partial x_j}$. At a critical point, the second term goes away, so we will consider such a case.

In other words, if the derivative is a differential $1$-form, i.e. $\sum_{i} \frac{\partial f}{\partial x_i} dx_i$, a section of the cotangent bundle, then the second derivative should be $\sum_{k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} dy_k \otimes dy_l$. This makes sense since $dy_k=\sum_{i} \frac{\partial y_k}{\partial x_i} dx_i$, and $dy_l=\sum_{j} \frac{\partial y_l}{\partial x_j} dx_i$, meaning that $\sum_{k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} dy_k \otimes dy_l = \sum_{k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} (\sum_{i} \frac{\partial y_k}{\partial x_i} dx_i) \otimes (\sum_{j} \frac{\partial y_l}{\partial x_j} dx_j) = \sum_{i,j,k,l} \frac{\partial^2 f(y(x))}{\partial y_k \partial x_l} \frac{\partial y_k}{\partial x_i} \frac{\partial y_l}{\partial x_j} dx_i dx_j = \sum_{i,j} \frac{\partial^2 f}{\partial x_i \partial x_j}$, making it coordinate independent. Note that I did not use exterior powers, I used tensor powers, since I wanted to actually find a way to make sense of second derivatives, rather than having $d^2=0$. This means the Hessian should be a rank $2$ tensor ((2,0) or (0,2), I can't remember which, but definitely not (1,1)).

Does this make sense? Can we then express the third, etc, derivative as a tensor? More interestingly, how can this help us make sense of Taylor's formula? Can we come up with a coordinate-free Taylor series of a function at a point on a manifold?

EDIT: An in general, if the first $n$ derivatives vanish, then the $n+1$ derivative should be a rank $n+1$ tensor, right?


No, no, no! You left out a term involving $\frac{df(y(x))}{dy}\frac{d^2y}{dx^2}$. This term vanishes at critical points -- points where $df=0$ -- so that indeed at such a point the Hessian define a tensor -- a symmetric bilinear form on the tangent space at that point -- independent of coordinates. Paying attention to what kind of bilinear form it happens to be is the beginning of Morse theory, but it's only intrinsically defined as a tensor if you're at a critical point.

Notice that even the question of whether the Hessian is zero or not is dependent on coordinates. Even in a one-dimensional manifold.

Taylor polynomials don't live in tensor bundles, but in something more subtle called jet bundles.

  • $\begingroup$ Oops, I didn't notice that. I think I missed it in part because I was so bent on showing that it transformed tensorially. I did notice that the Taylor series in a sense includes tensors of all sorts of types (sort of like the tensor algebra, except where we allow infinite sums). Is this the difference between a tensor bundle and a jet bundle? Or is the jet bundle related to the extra term somehow? $\endgroup$ – David Corwin Jul 15 '10 at 1:16
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    $\begingroup$ Say that two smooth functions, each defined in neighborhood of the point p, have the same $2$-jet if with respect to a coordinate chart they have the same first order and second order partial derivatives. This equivalence relation is independent of coordinates. (If I just said "same second order partial derivatives" it would not be independent.) For each p the vector space of $2$-jets maps onto the vector space of $1$-jets (cotangent space) and the kernel is the space of symmetric bilinear forms on the tangent space. As p varies you have a vector bundle. But it's not any kind of tensor bundle. $\endgroup$ – Tom Goodwillie Jul 15 '10 at 2:54
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    $\begingroup$ You can't make it by applying some functor fiberwise to the tangent bundle, because unlike the tangent bundle its transformation law for coordinate change involves more than the first derivatives $\frac{\partial y}{\partial x}$. $\endgroup$ – Tom Goodwillie Jul 15 '10 at 3:01

Sorry for reviving this question. Everything Tom said is correct, but there is more to say about "coordinate-free Taylor series".

It is true that arbitrary jet bundles $J^k(M,N)$ are subtle. The fibers are vector spaces but the transition maps are not linear. The special case $J^k(M,\mathbb{R})$ is a vector bundle and this is where the Taylor series of a map $f : M \to \mathbb{R}$ lives. The vector bundle $J^k(M,\mathbb{R})$ is not a tensor power of the tangent bundle as Tom pointed out, but it is far easier to understand than an arbitrary jet bundle. Indeed, the structure group of $J^k(M,\mathbb{R})$ is sometimes called the Phylon group.


More generally, for any vector bundle $V \to M$, the vanishing of a section at a point to a given order $k$ is independent of local coordinates and choice of basis of local sections, and when this happens at some point $m \in M$, the $(k+1)$-jet of the local section at that point is a well defined element of $S^kT^*_mM \otimes V_m$.


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