Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
    
1  
Answer to EDIT: Right. –  Deane Yang Jul 15 '10 at 3:24
    
You may also be interested in the book Natural Operations in Differential Geometry by Kolar, Michor, and Slovak. (My apologies for not typing in the proper diacritics on the letters.) –  Willie Wong Jul 15 '10 at 10:32
add comment

1 Answer

up vote 10 down vote accepted

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.

share|improve this answer
    
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? –  David Corwin Jul 15 '10 at 1:16
2  
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. –  Tom Goodwillie Jul 15 '10 at 2:54
1  
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 $\frac{\partial y}{\partial x}$. –  Tom Goodwillie Jul 15 '10 at 3:01
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.