Understanding the conservative property of a vector field

Question

I am taking high school calculus and I noticed a connection between something I know and this new concept.

I read that a force field is conservative, if the work done from $A$ to $B$ is independent of the path taken, which reminded me of something I saw in my own study and it seems that this is saying:

For each homotopy class $[I,\partial I; X,\{a,b\}]$ of paths with $f(0)=a,f(1)=b$, for $X$ a simply connected surface - each of these classes has a specific work done associated to it.

But it doesn't say anything about what this is. Is a force field a section of a vector bundle? Then how does this conservative property work with this homotopy notion? I am confused.

Answer 1

A conservative force can be construed as a closed 1-form, where a 1-form is a section of the cotangent bundle. In $\mathbb{R}^3$ this means $\nabla\times F=0$ (which implies line integrals are path-independent by Stokes' theorem, so that this really is a conservative force). In $\mathbb{R}^3$ closed forms are exact, which means $F=\nabla U$ for a scalar function $U$. Generally De Rham cohomology is used to compare closed forms to exact forms. Apparently de Rham cohomology is a homotopy invariant (I don't know much algebraic topology so I don't know precisely how this relates to homotopy groups).

To summarize, conservative forces are like closed 1-forms. Closed forms, depending on topology of the space, may not be exact (exact corresponds to the force having a potential). For $\mathbb{R}^n$, closed forms are exact. You can come up with a simple counterexample on the 1d torus by considering $d\theta$. See https://en.wikipedia.org/wiki/Closed_and_exact_differential_forms.