Conservative condition for (static) electric field $\mathcal{E}$ is usually defined as $\mathcal{E}$ being closed (curl-free). Now this clearly holds when for the given manifold $X$ we have $H_\text{dR}^{1}(X)\simeq 0$, but consider the case when it is not so.

For general $k$-manifolds:

Consider smooth $k$-manifold $X$ s.t. $H_\text{dR}^1(X)\not\simeq 0$. Denote by $\Phi$ the set of all closed 1-forms which are not exact. Let $\mathcal{E}=\mathcal{E_0}\varphi$ for some $\mathcal{E_0}\in\mathbf{R}$ and $\varphi\in \Phi$. This gives charge density for this setting (let units be s.t. $\epsilon_0 = 1$): $$ \rho = \star d\star \mathcal{E} $$ And, by definition of $\varphi$, we have $d\mathcal{E} = 0$. Then (by non-exactness) there exists a closed 1-path $\gamma\colon [0,1]\rightarrow X$ s.t. $$ \int_\gamma \mathcal{E} \neq 0 $$ Which evidently violates the condition of static electric field being conservative.

An example on $S^1$:

Suppose one is given $S^1$, and the electric field $\mathcal{E}\in\Gamma(TS^1)$ is a smooth section, defined by: $$ \mathcal{E}((x,y)) ~\stackrel{\text{proj.2.}}{=}~ \mathcal{E}_0 \frac{x~dy-y~dx}{x^2+y^2} $$ Evidently the value of $\mathcal{E}$ is a closed 1-form, furthermore $\mathcal{E}$ is divergence-free.

Assuming that there are no external fields, and all fields are time-independent, we have $\mathcal{E}$ as a solution to Maxwell's equations: $$ d\mathcal{E} = 0\qquad \star d\star\mathcal{E} = 0 = \rho $$ Thus $\rho$ is well defined, bounded and constant. However $\mathcal{E}((x,y)) = \mathcal{E}_0\mathrm{vol}_{S^1}$ is not exact, moreover we have: $$ \int_{S^1}\mathcal{E} = 2\pi \mathcal{E_0} \neq 0\qquad \text{by choosing}~\mathcal{E}_0\neq 0 $$ hence $\mathcal{E}$ admits no potential and is thus non-conservative.

Are Maxwell's equations restricted on (smooth) manifolds which have trivial de Rham cohomology, or is there some other condition I'm missing?


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