# A covering space in $\Bbb R^3$ exists iff there is an embedding $M_g$ of the 3-torus which induces a surjection

This is Exercise 1.3.19 from Hatcher's Algebraic Topology

I want to show the following statement:

For a closed orientable surface $$M_g$$ of genus $$g$$, there is a path-connected normal (regular) covering space $$\tilde X$$ of $$M_g$$ in $$\Bbb R^3$$ with deck transformation group isomorphic to $$\Bbb Z^3$$ iff there is an embedding of $$M_g$$ to the 3-torus $$T^3=S^1 \times S^1 \times S^1$$ such that the induced map is surjective.

First, it was easy to show that there is an embedding $$M_g \to T^3$$.

The map $$\tilde X \to \Bbb R^3 \to \Bbb R^3/ \Bbb Z^3=T^3$$ (the first map is inclusion, and the second is quotient map) induces the embedding of $$M_g$$ to $$T^3$$.

However, I am stuck showing the induced homomorphism is surjective.

On the other hand, I have no idea showing the converse. How do I have to proceed? Any hints?

• Have you done exercise 1.3.18? – Max Aug 18 '19 at 19:59
• @Max Yes I did. – user302934 Aug 18 '19 at 20:39

What the author asks in the book is the following:

Show that there is an embedding $$M_g\rightarrow T^3$$ inducing a surjection $$\pi_1(M_g)\rightarrow\pi_1(T^3)$$ iff there is a subspace $$X\subseteq \Bbb R^3$$ and a covering map $$X\xrightarrow p M_g$$ whose deck transformations are a group of translations of $$\Bbb R^3$$ isomorphic to $$\Bbb Z^3$$

$$(\Leftarrow)$$ By the hypothesis on the group of deck transformations it follows that this group is generated by translation of three linearly independent vectors. Let us denote by $$\Gamma$$ this group. Then as you observed we obtain an embedding $$M_g\cong X/\Gamma\rightarrow \Bbb R^3/\Gamma$$ from the inclusion $$X\subseteq \Bbb R^3$$; this is an embedding as $$M_g$$ is compact. We also have a covering map $$\pi:\Bbb R^3\rightarrow \Bbb R^3/\Gamma\cong T^3$$; as $$\Gamma$$ is generated by translations of three linearly independent vectors.

Fix $$x_0\in X$$. Then we have $$\pi^{-1}([x_0])\subseteq X$$; as both $$\pi$$ and $$\pi\upharpoonright X$$ are covering spaces of degree $$3$$, and also an induced map $$\pi_1(X/\Gamma,[x_0])\rightarrow\pi_1(\Bbb R^3/\Gamma,[x_0])$$, and this map is onto. This is because if $$[\alpha]\in\pi_1(\Bbb R^3/\Gamma,[x_0])$$, we can lift $$\alpha$$ to a map $$\tilde\alpha:[0,1]\rightarrow \Bbb R^3$$. Then as $$\tilde\alpha(1)\in X$$, consider a path $$\beta$$ in $$X$$ going from $$x_0$$ to $$\tilde\alpha(1)$$, then $$[\pi\circ\beta]=[\alpha]$$ in $$\pi_1(\Bbb R^3/\Gamma,[x_0])$$(why?) and $$[\pi\circ\beta]\in \pi_1(X/G,[x_0])$$.

$$(\Rightarrow)$$ In this case let $$\pi:\Bbb R^3\rightarrow T^3$$ be the covering map built using the exponential map. Suppose $$M_g\subseteq T^3$$. Fix $$y_0\in M_g$$. Let $$G$$ be the kernel of the induced map $$\pi_1(M_g,y_0)\rightarrow \pi_1(T^3,y_0)$$. Let $$p:X\rightarrow M_g$$ be a covering map such that $$p_\#(\pi_1(X,x_0))=G$$ for some $$x_0\in p^{-1}(y_0)$$. Then $$i_\#\circ p_\#(\pi_1(X,x_0))=0$$; where $$i:M_g\rightarrow T^3$$ is the inclusion, thus there is a map $$\tilde i: X\rightarrow \Bbb R^3$$ making the diagram $$\require{AMScd}$$

$$\begin{CD} X @>\tilde i>> \Bbb R^3\\@VVpV @VV\pi V\\ M_g @>i>> T^3. \end{CD}$$

Let us see $$\tilde i$$ must be inyective. Suppose, to get a contradiction, there are distinct $$x,y\in X$$ such that $$\tilde i(x)=\tilde i(y)$$. As $$i$$ is injective, we must have $$x$$ and $$y$$ must lie in the same fiber of $$p$$. Suppose this fiber is $$p^{-1}(y_0)$$. Let $$\beta$$ be a path in $$X$$ going ffrom $$x$$ to $$y$$. Then $$[p\circ\beta]\in \pi_1(M_g,y_0)\setminus G$$, so that $$[i\circ p\circ\beta]=i_\#([p\circ\beta])\neq e$$. However, $$[i\circ p\circ\beta]=[\pi\circ\tilde i\circ\beta]=\pi_\#([\tilde f\circ\beta])=e$$, as $$\Bbb R^3$$ is simply connected. To prove this for all fibers, show that for all $$y\in M_g$$ and all $$z\in p^{-1}(y)$$ we have

$$p_\#(\pi_1(X,z))=\ker(\pi_1(M_g,y)\xrightarrow{i_\#}\pi_1(T^3,y)).$$

As $$\pi, p$$ are local homeomorphisms and $$i$$ is an embedding, $$\tilde i$$ must also be an embedding. Thus we may assume $$X\subseteq \Bbb R^3$$, and from the diagram we can also assume $$p=\pi\upharpoonright X$$. In particular if $$\Delta_p$$ and $$\Delta_\pi$$ are the groups of deck transformations of $$p$$ and $$\pi$$ respectively, we have $$\{D\upharpoonright X:D\in \Delta_\pi\}\subseteq\Delta_p.$$

We claim this inclusion is actually an equality. Suppose not, pick $$D\in \Delta_p$$ which doesn't extend to an element of $$D_\pi$$. Then if we pick any $$x\in X$$ and set $$y:=D(x)$$, $$x$$ and $$y$$ must not be in the same fiber of $$\pi$$; for otherwise there'd be an element $$D'\in \Delta_\pi$$ with $$D(x)=y$$, and as $$D\upharpoonright X\in\Delta$$ this would imply $$D'\upharpoonright X=D$$. However, $$x$$ and $$y$$ are in the same fiber of $$p$$, which contradicts the commutativity of the above diagram.

• The reason $[\pi \circ \beta]=[\alpha]$ is because $\beta$ and $\tilde {\alpha}$ are path-homotopic in $\Bbb R^3$ ($\Bbb R^3$ is simply-connected). Am I right? – user302934 Aug 20 '19 at 3:57
• Thank you for your answer, but I have two questions in the $\Rightarrow$ direction. First, from $\tilde {i}(x)=\tilde {i}(y)$ I know that $p(x)=p(y)$ but how do I have to construct $\alpha$? Second, I know that $p_* (\pi_1(X, x_0))$ is a index 3 subgroup, but how this implies that the deck transformation group is $\Bbb Z^3$? – user302934 Aug 20 '19 at 4:22
• @Comol I have made yet another edit. I hope everything is clear enough. – Camilo Arosemena-Serrato Aug 23 '19 at 1:11