Write down (without proof) a presentation for \(\pi_1(\Sigma_2,p)\), where \(\Sigma_2\) is a closed, connected, orientable genus \(2\) surface and \(p\) is any point in \(\Sigma_2\).
Show that \(\pi_1(\Sigma_2,p)\) is not abelian by showing that it surjects onto a free group of rank \(2\).
Show that there is no covering space map from \(\Sigma_2\) to \(S^1\times S^1\). You may use the fact that \(\pi_1(S^1\times S^1)\cong\ZZ^2\) together with the result in part (b).
Solution
1. A standard presentation is\[ \pi_1(\Sigma_2,p) \cong \left\langle a_1,b_1,a_2,b_2 \mathrel{\Big|} [a_1,b_1][a_2,b_2]=1 \right\rangle. \]
This is the requested presentation in part (a); no proof is required there.
2. There is a surjective homomorphism\[ \pi_1(\Sigma_2,p)\twoheadrightarrow F(x,y), \]where \(F(x,y)\) is the free group of rank \(2\).
Define a map on the generators in 1 by \[ a_1\longmapsto x, \qquad b_1\longmapsto y, \qquad a_2\longmapsto y, \qquad b_2\longmapsto x. \] The defining relator maps to \[ [x,y][y,x] =[x,y][x,y]^{-1} =1, \] so the assignment descends to a homomorphism \[ \varphi:\pi_1(\Sigma_2,p)\longrightarrow F(x,y). \] Its image contains both \(x=\varphi(a_1)\) and \(y=\varphi(b_1)\), so \(\varphi\) is surjective.
3. The group \(\pi_1(\Sigma_2,p)\) is not abelian.
Every quotient of an abelian group is abelian. But \(F(x,y)\) is nonabelian, since the reduced words \[ xy \qquad\text{and}\qquad yx \] are distinct. By 2, \(F(x,y)\) is a quotient of \(\pi_1(\Sigma_2,p)\). Therefore \(\pi_1(\Sigma_2,p)\) cannot be abelian.
4. There is no covering map\[ q:\Sigma_2\longrightarrow S^1\times S^1. \]
3. Hence no such covering map exists.