Second differential of a function

While it may not be as general, a scalar field [itex]\psi[/itex] obeys

[tex]\nabla \wedge \nabla \psi = 0[/tex]

In 2d, this reduces to

[tex]\left( e^x \partial_x + e^y \partial_y\right) \wedge \left(e^x \partial_x \psi + e^y \partial_y \psi \right) = e^x \wedge e^y \left( \partial_x \partial_y \psi - \partial_y \partial_x \psi \right) = 0[/tex]

This is trivially satisfied by the equality of mixed partial derivatives. In one dimension, you don't have two basis vectors, and there is no wedge product to speak of at all--I've heard of the wedge product referred to as an antisymmetric quadratic form, but that's started to get out of my area of familiarity.

 

I dont' know this book, but they are probably referring to the second order part of the Taylor expansion. This is a quadratic form that is composed of second derivatives, and it is the next best local information about a function once you know its value and its derivatives at a point. In other words, they are talking about the Hessian matrix.