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Let $U$ be an open and simply-connected subset of $ \mathbb{R}^3$. Then for every curl-free vector field $v \: \colon U \to \mathbb{R}^3$ there is a potential $\phi \in C^{\infty}(U; \mathbb{R})$ such that $v = \nabla \phi$ in $U$. This well-known result is a consequence of Poincare's lemma stating that $H_{\textrm{dR}}^1(U) = 0$. Here, $H_{\textrm{dR}}^1(U)$ denotes the first de Rham cohomology group of $U$.

Let us now consider the 3-torus $T^3$ and a smooth curl-free vector field $v \: \colon T^3 \to \mathbb{R}^3$. Now, $T^3$ is neither simply connected nor is its first de Rham cohomology group $H_{\textrm{dR}}^1(T^3)$ trivial. In fact, $H_{\textrm{dR}}^1(T^3) = \mathbb{R}^3$. As I am not a geometer by training, excuse my (probably simple) question:

Is it true that there exist a potential $\phi \in C^{\infty}(T^3, \mathbb{R})$ and curl-free functions $u_1, u_2, u_3\: \colon T^3 \to \mathbb{R^3}$ such that $v = \nabla \phi + u_1 + u_2 + u_3$?

Moreover, if we additionally require that $\int_{T^3} v = 0$ does this imply that $v = \nabla \phi$ for some suitable function $\phi$?

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Mike
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2 Answers2

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Your first question is trivial as written: just take $\phi = 0$, $u_1 = v$, $u_2=u_3=0$. Perhaps you meant for the $u_i$ to be independent of $v$? In this case it's almost true, but you need to allow arbitrary linear combinations of the $u_i$:

There are three curl-free, divergence-free vector fields $u_i$ on $\mathbb T^3$ such that for every curl-free vector field $v$ there is some potential $\phi$ and some coefficients $c_i \in \mathbb R$ such that $$v = \nabla \phi + c_1 u_1 + c_2 u_2 + c_3 u_3.$$

This is a special case of the Hodge decomposition for differential forms: the (Hodge dual of the) $u_i$ are a basis for the space of harmonic one-forms on $\mathbb T^3$, which in particular represent every de Rham cohomology class. If you're considering the usual flat torus $\mathbb T^3 = \mathbb R^3 / \mathbb Z^3$ then the standard unit vector fields work for $u_i$.

Your suggestion for eliminating the $u_i$ is not quite right - we instead require $\int_{\gamma_i} v = 0$ for $\gamma_i$ a basis of the first homology group. The intuition here is that each $u_i$ represents circulation around a "hole", so if we don't want any of these components we need to check that $v$ has zero circulation around each hole.

Anthony Carapetis
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  • Is there an easy way to prove that $u_1, u_2$, and $u_3$ are the standard unit vector fields? – Mike Jun 28 '16 at 19:21
  • @Mike: if you know all this theory then it's just because they are harmonic and cover all the cohomology classes. If you want to prove this whole thing in an elementary way, start by considering the torus as a cube with periodic boundary conditions and apply the Poincaré lemma. You should be able to fix up the failure of the resulting potential to be periodic by adding the right multiples of the $u_i$ to $v$ first. – Anthony Carapetis Jun 28 '16 at 22:33
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I think you had the following in mind: The three cycles $$\sigma_i:\quad t\mapsto t e_i\qquad(0\leq t\leq1)$$ form a basis of $H^1(T^3)$. Compute the periods of $v$ along the $\sigma_i$, i.e., the numbers $$\alpha_i:=\int_{\sigma_i}v\cdot dx\ .$$ Use these to define the constant field $$u:=(\alpha_1,\alpha_2,\alpha_3)\ .$$ Then there is a potential $\phi:\>T^3\to{\mathbb R}$ such that $$v=\nabla\phi + u\ .$$

Christian Blatter
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