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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.

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    $\begingroup$ 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. $\endgroup$ – Danny Ruberman Dec 5 '18 at 14:42
  • $\begingroup$ 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?). $\endgroup$ – Klaas Dec 6 '18 at 17:44
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The following is my own attempt at an answer. I would appreciate critical proofreading.

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.

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