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There is a two-fold branched covering from 2-torus to the 2-sphere, $T^2 \rightarrow S^2$, whose covering transformation group is generated by the map $x \mapsto -x$ (Note that $T^2$ is an abelian group).

I heard that there is a three-fold branched covering from the 3-torus to the 3-sphere. Then what would be the covering transformation group of this case?

Probably it is trivial for topologists but could anyone can help me out?

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    $\begingroup$ In what you heard, was the covering Galois (= regular)? If so, I would start with trying to construct a representation of $\mathbb Z/3$ on $\mathbb Z^3$ whose tensor by $\mathbb Q$ or $\mathbb Z/p$. $p\ne3$, has no invariant vectors. $\endgroup$
    – Alex Degtyarev
    Commented Jul 8, 2015 at 19:40
  • $\begingroup$ Well, See arxiv.org/pdf/1212.6282.pdf for example. $\endgroup$
    – Creg
    Commented Jul 8, 2015 at 23:11
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    $\begingroup$ Well, See arxiv.org/pdf/1212.6282.pdf for example. In the introduction of the article, it says that the 3-torus is not a 2-fold branched cover of the 3-sphere. Am I mistaken? @Anton Petrunin, Could you explain or give some reference for the double cover? $\endgroup$
    – Creg
    Commented Jul 8, 2015 at 23:18
  • $\begingroup$ @AlexDegtyarev That's not possible. Every representation of $\mathbb{Z}/3$ over $\mathbb{Q}$ is a direct sum of the trivial rep and $\left( \begin{smallmatrix} 0 & 1 \\ -1 & -1 \end{smallmatrix} \right)$. So any odd dimensional representation of $\mathbb{Z}/3$ defined over $\mathbb{Q}$ has an invariant vector. $\endgroup$
    – David E Speyer
    Commented Jul 9, 2015 at 0:13
  • $\begingroup$ The 3-torus can't be a branched double cover because the triple cup product on $H^1$ would then be zero. $\endgroup$
    – Steven Sivek
    Commented Jul 9, 2015 at 5:44

2 Answers 2

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There is an algorithm, due to Montesinos, for converting a surgery diagram of a 3-manifold $M$ into a description of $M$ as a 3-fold (irregular; as remarked above, there is no regular branched cyclic covering $T^3 \to S^3$) cover of $S^3$. It is described nicely in Rolfsen, Knots and Links, Chapter 10.G. You need to start with a surgery description where all of the framings are $\pm 1$.

For the 3-torus, such a description is easily found. $T^3$ is surgery on the Borromean rings, with framings 0 on each component. Add a $+1$ framed meridianal circle to each component of the Borromean rings, changing the framing on that component to $1$. The picture below shows what I mean. If you follow the description in Rolfsen's book, you will have the branched covering. I haven't tried to draw it, though.

enter image description here

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First I think, the covering $T^2\to S^2$ has deck group $Z/2Z\oplus Z/2Z$ generated by $(x,y)\to (-x,y)$ and $(x,y)\to (x,-y)$.

Then for the 3-dimensional case, you consider the action of $$Z/2Z\oplus Z/2Z\oplus Z/2Z$$ on $T^3$, where the 3 generators send $$(x,y,z)$$ respectively to $(-x,y,z)$ or $(x,-y,z)$ or $(x,y,-z)$. The quotient by this action should be the 3-sphere, in analogy to the 2-dimensional case.

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  • $\begingroup$ your action is free, so you don't get a branched cover. And there's no nontrivial unbranched cover of $S^2$ since it is simply-connected. Instead, the quotient of $T^2$ by this group is another copy of $T^2$. All this holds likewise if you replace $2$ by $3$. $\endgroup$
    – Achim Krause
    Commented Jul 8, 2015 at 14:52
  • $\begingroup$ It seems that your deck group for $T^2$ induces a four-fold covering and that your deck group for $T^3$ induces a 8-fold covering. It is doubtful that the quotient is a sphere. $\endgroup$
    – Creg
    Commented Jul 8, 2015 at 14:57
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    $\begingroup$ @AchimKrause The action is not free, even though the quotient is certainly not a sphere: it is a square (cube etc.) $\endgroup$
    – Alex Degtyarev
    Commented Jul 8, 2015 at 15:15
  • $\begingroup$ @AlexDegtyarev I thought we're writing the Torus as $S^1\times S^1$ with $S^1$ unit complex numbers. I agree that the action of $-1$ on $\mathbb{R^2}/\Gamma$ is not free. $\endgroup$
    – Achim Krause
    Commented Jul 8, 2015 at 18:35

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