# Topology of connected subsets of the $3$-torus

Consider the $$3$$-torus $$Y=T^3$$, a subset $$\Sigma\subset Y$$, and $$\Sigma^*=Y\setminus\overline\Sigma$$. We assume both $$\Sigma$$ and $$\Sigma^*$$ to be open, connected, and smoothly bounded.

I am concerned with the subsets $$A = \ker (H^1(Y)\to H^1(\Sigma)), \quad B=\ker (H^1(Y)\to H^1(\Sigma^*))$$ of the first de Rham cohomology of $$Y$$. The maps are induced by the restriction maps.

My conjecture is that one of the following three statements must hold true:

• $$A=0$$ or
• $$B=0$$ or
• $$A=B$$ and $$\dim A=1$$.

Can you help me prove or disprove this conjecture?

Why I expect the conjecture to be true: Honestly, I have only very crude intuitive arguments for this. If $$a\in A\subset\mathbb R^3$$ is nonzero, then $$a$$ can be written as a gradient in $$\Sigma$$. This means that there cannot be a closed loop $$\gamma$$ in $$\Sigma$$ "going in the direction of $$a$$", meaning that the fundamental group class of that loop in $$Y$$ (considered as an element in $$\mathbb R^3$$) is non-orthogonal to $$a$$. But the existence of such a loop is obstructed only by $$\Sigma^*$$, which means that there must be some sort of two-dimensional plane inside $$\Sigma^*$$ which is "orthogonal" to $$a$$. But then, given any direction $$b\in\mathbb R^3$$ orthogonal to $$a$$, we can find a closed loop in that plane (thus in $$\Sigma^*$$) "going in the direction of $$b$$". This shows that $$b\notin A$$.

What I have already done: I have already looked at the Mayer Vietoris sequence, but it does not seem to yield enough information. But it helps me to draw conclusions in case I already know the conjecture to be true. Indeed, denoting by $$k$$ the number of connected components of $$\partial\Sigma$$, we then know that \begin{align*} 1 &\ge\dim A\cap B = \dim\ker(H^1(Y)\to H^1(\Sigma)\oplus H^1(\Sigma^*)) \\ &= \dim \operatorname{im} (H^0(\partial \Sigma)\to H^1(Y)) \\ &= k - \dim\ker(H^0(\partial\Sigma)\to H^1(Y)) \\ &= k - \dim\operatorname{im}(H^0(\Sigma)\oplus H^0(\Sigma^*)\to H^0(\partial\Sigma) = k-1, \end{align*} showing that $$\partial\Sigma$$ can have at most $$2$$ connected components. I have no independet proof of this result, so my second question would be if this statement is true.

• You're on the right track with the Mayer-Vietoris sequence; have a look at the "half-lives, half-dies" principle. It tells you about the image of the restriction maps on $H^1$. Eg Lemma 3.5 of Hatcher's 3-manifold notes pi.math.cornell.edu/~hatcher/3M/3M.pdf. – Danny Ruberman Dec 5 '18 at 14:42
• Thank you, this is interesting stuff! Can you expand a little on how this principle would help? In the Mayer Vietoris sequence there is only the difference of two such restriction maps. What makes me a bit skeptical is that my intuitive argument works in any dimension, that is, not just for the $3$-torus, whereas using the "half-lives, half-dies" principle would work only in dimension $3$ (or, any odd dimension?). – Klaas Dec 6 '18 at 17:44

The exact sequences of the pairs $$(Y,\Sigma)$$ and $$(Y,\Sigma^*)$$ show that $$A=\operatorname{im}(i^*:H^1(Y,\Sigma)\to H^1(Y)), \quad A=\operatorname{im}(j^*:H^1(Y,\Sigma^*)\to H^1(Y))$$
Take any $$[u]\in H^1(Y,\Sigma)$$ and $$[v]\in H^1(Y,\Sigma^*)$$. Here, $$u$$ and $$v$$ are closed $$1$$-forms on $$Y$$, and we can assume that $$u$$ vanishes of $$\Sigma$$ and $$v$$ vanishes on $$\Sigma^*$$. Consequently, $$0 = [u\wedge v]_{H^1(Y)} = [u]_{H^1(Y)} \wedge [v]_{H^1(Y)} = i^*[u]_{H^1(Y,\Sigma)} \wedge j^*[v]_{H^1(Y,\Sigma^*)}$$
We have thus shown that $$A\wedge B=0$$ which is but a concise formulation of the original claim.