What does commutativity of partial derivatives imply about geometry about the surface of a function?

Question

Suppose we have a function $f(x,y)$ and it has the property that:

$$ \partial_x \partial_y f(x,y) = \partial_y \partial_x f(x,y)$$

What does this imply about the geometry of the surface described the function?

Answer 1

The second order differential operator commutes for $ f\in \mathcal{C}^2$. Which means $\partial_x\partial_y f$ is defined in the neighborhood of $(x_0, y_0)$ and continuous at $(x_0, y_0)$ for all points in the domain of $f$.

$\mathcal{C}^2$ functions are twice differentiable meaning they are "smooth" in some definitions of the word. You can read more about them here.

Answer 2

Comment/Remark:

For a continuous surface $z(x,y)$ cross derivative is related to torsional /shear curvature at a point. The rate of change of slope taken along parameter x changing with respect to $y$ and it does not change for surface continuity. Torsional curvature of a line is uniquely defined by either order in exchange for commutativity to hold good.

The commutativity can be looked at from a more general / functionally comprehensive complex variable viewpoint.

Taking a complex holomorphic differentiable function

$$ \phi(x,y) = f(x,y)+i g(x,y)$$

On the first of Cauchy- Riemann relations partial differentiate w.r.t $v$

$$ f_{u}= g_{v };\; f_{u,v}= g_{v,v}\tag1$$

and on the second C-R relation differentiate w.r.t $u$ $$ f_{v}=- g_{u }\; f_{v,u}= -g_{u,u}\tag2$$

Subtracting $$f_{u,v}-f_{v,u}=0\tag{3.1}$$

$$ g_{u,u}+g_{v,v}=0 \tag{3.2}$$

Since the second equation (3.2) is the Laplace equation that includes differentiability and smoothness, (3.1 ) must be also valid along with it, so the order of partial differentiation of $f$ with respect to $ {u,v} $ should not matter.

Similarly

$$ g_{u,v}= g_{v,u}; \; f_{u,u}+f_{v,v} =0 \tag4$$

Here also order of partial differentiation of $g$ also with respect to $ u,v $does not matter.