Author: Eiko

Tags: differential, calculus, de rham cohomology

Time: 2024-09-12 16:18:11 - 2024-09-12 16:48:26 (UTC)

This article includes my understanting and thoughts on differentials and related curious questions.

The intuitive explanation of differential

A differential $\mathrm{d}f$ is just a linear part of this function $f$ (removing all constant parts and higher order parts). Talking about the space of differentials is the same as talking about the space of (continuous or even algebraic) sections of the bundle of linear forms on $X$,

$$s\in {\mathrm{\Omega }}_X^1(U)\iff \mathrm{\forall }x\in U,s_x:T_{x}X\to k.$$

And $\mathrm{d}:{\mathcal{O}}_X\to {\mathrm{\Omega }}_X^1$ is just a function that maps a function to the section that gives its linear part as the linear form at every point of $X$. Sometimes this $\mathrm{d}$ is also denoted as $d^0$ in ${\mathcal{O}}_X\stackrel{d^0}{\to }{\mathrm{\Omega }}_X^1$.

For this reason, we also call a differential a $1$-form.

Integration Inverts this process

Integration is the reverse of the above process of taking every linear part of a function, it aims to reconstruct the function $f$ when we are just getting the linear parts at every point of it at hand.

One question that come immediately is, can every differential be integrated? i.e. is every differential (section of linear form) locally given by functions? One might think this is obvious since a linear form is already a function, this should be possible.

But this is not necessarily true on dimension greater than $1$, because the process of taking a function to its differential, say in dimension $2$,

$$f\mapsto \mathrm{d}f=\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y$$

does not give every linear form section. It indeed can give every linear form at any particular point, but not the space of sections of linear forms. This is because the two functions $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$ are not independent as functions, they cannot be chosen as arbitrary functions and they satisfy the following constraint:

$$\frac{\partial }{\partial y}\left(\frac{\partial f}{\partial x}\right)=\frac{\partial }{\partial x}\left(\frac{\partial f}{\partial y}\right).$$

This means, if a differential $a\mathrm{d}x+b\mathrm{d}y$ is locally given by a function, it must satisfy $\frac{\partial a}{\partial y}=\frac{\partial b}{\partial x}$. This fact is captured in the de-Rham cohomology ${\mathrm{\Omega }}_X^1\stackrel{d^1}{\to }{\mathrm{\Omega }}_X^2$, given by

$$a\mathrm{d}x+b\mathrm{d}y\mapsto (\frac{\partial a}{\partial y}-\frac{\partial b}{\partial x})\mathrm{d}x\wedge \mathrm{d}y.$$

We see that a necessary condition for integrability of $\omega \in {\mathrm{\Omega }}_X^1$ is $d^1\omega =0$. The space of vanishing one-forms under $d^1$ is called $1$-cycles.

Boundaries: obvious cycles

Clearly all one-forms given directly by functions are already $1$-cycles. For this reason we can form the first de-Rham cohomolgoy group

$$H^1(X)=\frac{Z^1({\mathrm{\Omega }}^{\bullet }(X))}{B^1({\mathrm{\Omega }}^{\bullet }(X))}=\frac{\ker (d^1)}{\mathrm{Im}(d^0)}.$$

The de-Rham theorem tells us that these cohomology groups locally vanishes, which means that as long as the obvious constraints are satisfied, there is no more obstructions for you to locally construct their integration function $F$ out of $\omega$, so that $\mathrm{d}F=\omega$ locally.

Remark on Commutativity of derivatives

The commutativity of derivatives reflects something on flatness, no curvature. And this commutativity constraint is encoded in $D_X$, which is also why $D_X$-module structures correspond only to flat connections (or integrable connections) and not all connections.