# Is there a 3-sheeted covering map of torus S^1 × S^1? [closed]

Basically, I'm having trouble solving this question. Does anyone know how to approach this?

• I mean, there is a trivial one, but I'm assuming you are looking for a connected cover, not just any cover. – Don Thousand May 28 at 14:58
• Connected covers correspond bijectively to (conjugacy classes of) subgroups of the fundamental group. Under this bijection, the number of sheets is the same as the index of the subgroup. So if one is interested purely in existence, it suffices to come up with an index 3 subgroup of $\mathbb{Z} \times \mathbb{Z}$ which is easy to do. – Connor Malin May 28 at 15:02
• Here is a good exercise for you. Note that $\mathbb{Z}\times \mathbb{Z}$ is an additive subgroup of $\mathbb{R}\times \mathbb{R}$, so it has a natural action by translation. If you take any subgroup of the form $n\mathbb{Z} \times m\mathbb{Z}$, show that the quotient of $\mathbb{R}\times \mathbb{R}$ by this subgroup (with the above action) is a space with a natural covering map to $S^1 \times S^1$ with $nm$ sheets. – William May 28 at 15:12
• (I should specify that in my above comment $n$ and $m$ should both be positive) – William May 28 at 15:34

The torus is the quotient of $$\mathbb{R}\times \mathbb{R}$$ by $$f(x,y)=(x+1,y), g(x,y)=(x,y+1)$$ consider
the quotient of $$\mathbb{R}^2$$ by $$u(x,y)=(x+{1\over 3},y), v(x,y)=(x,y+1)$$ is a $$3$$-cover of the torus, in fact this cover is isomorphic to the torus itself.
Yes. The map $$m_3 : S^1 \to S^1, m_3(z) = z^3$$, is a $$3$$-sheeted covering. Hence $$m_3 \times id$$ is also a $$3$$-sheeted covering. Note that products of two coverings maps are coverings maps, and the number of sheets of such product maps is the product of the number of sheets of the two factors.
• I should have specified that I'm supposed to find a 3-sheeted covering $p: T^2 \to T^2$ – Andrea May 28 at 15:53
• @Andrea $p = m_3 \times id : S^1 \times S^1 \to S^1 \times S^1$ is a $3$-sheeted covering – Paul Frost May 28 at 21:16