I need to prove by induction $\pi_1(\Sigma_g)= \left\langle a_1,b_1,\dots ,a_g,b_g\mid \prod_i [a_i,b_i] \right\rangle$. For genus 1 this holds since $\pi_1(T^2)\cong \mathbb Z\times \mathbb Z$. For the general case I thought of using the connected sum formula $$\pi_1(\Sigma_g)\cong \pi_1(\Sigma_{g-1} \# T^2)\cong \pi_1 (\Sigma_{g-1}\setminus D)\amalg_\mathbb{Z}\pi_1(T^2\setminus D)$$ but I'm kind of lost now. I thin $\pi_1(T^2\setminus D)\cong \mathbb Z\ast \mathbb Z$ since its a bouquet of 2 spheres, but I don't know how to proceed...

  • 1
    $\begingroup$ A way to proceed is the following: As a first step, proof by induction that for a surface $\Sigma_{g}^{1}$ of genus $g$ with one boundary circle, and a point $x$ on the boundary, the fundamental group is a free group $\pi_1(\Sigma_{g}^{1},x) = \langle a_1, b_1, \ldots, a_{g}, b_{g} \rangle$, where the boundary circle maps to the element $\prod_{i} [a_i,b_i]$. In this case, the induction step is a direct application of the Seifert-van-Kampen theorem. $\endgroup$ – Oles Wohnzimmer May 27 '16 at 15:04
  • $\begingroup$ Do you know the polygonal representation of $\sum_g$? $\endgroup$ – Anubhav Mukherjee May 27 '16 at 15:06
  • 1
    $\begingroup$ @Anubhav.K doesn't the polygonal representation circumvents induction? $\endgroup$ – popo May 27 '16 at 15:09
  • $\begingroup$ @OlesWohnzimmer can you post this as an answer, with some details? $\endgroup$ – popo May 27 '16 at 15:13
  • $\begingroup$ @OlesWohnzimmer how do you see the boundary circle maps to the product of the commutators? $\endgroup$ – user153312 May 28 '16 at 9:11

Let $\Sigma_{g}^{1}$ be denote a surface of genus $g$ with one boundary circle. Let $\iota_g\colon \textrm{S}^{1} \to \partial\Sigma_{g}^{1}$ be a homeomorphism on the boundary. For convenience later on, we write $\textrm{S}^{1} = [0,1]/\{0 \sim 1\}$, and we fix base points $x_g = \iota_g(0) \in \Sigma_{g}^{1}$.

I assume that the following statement is known:

$\pi_1(\Sigma_{1}^{1},x_1)$ is a free group, and if $a_1$ and $b_1$ denote the homotopy classes of the longitude and the latitude circle starting at $x_1$, one has $\pi_1(\Sigma_{1}^{1},x_1) = \langle a_1,b_1 \rangle$. In this notation, the boundary $\iota_1 \colon \textrm{S}^{1} \to \partial \Sigma_{1}^{1}$ represents the class $[a_1,b_1] = a_1 b_1 a_1^{-1} b_1^{-1}$.

One can prove this statement by constructing $\Sigma_{1}^{1}$ as a quotient of a square with a disc removed (e.g. $[-2,2]^{2} \setminus \{x \in \mathbb{R}^{2} \mid ||x||< 1\}$) with opposite edges identified.

Now one can inductively construct $\Sigma_{g}^{1}$:

Consider the intervals $\iota_1([0,\frac{1}{2}])$ , $\iota_{g}([0,\frac{1}{2}])$ in the boundaries of $\Sigma_{1}^{1}$ and $\Sigma_{g}^{1}$. Identifying these intervals yields a homeomorphism: $$ \phi_{g} \colon \Sigma_{1}^{1} \cup_{[0,\frac{1}{2}]} \Sigma_{g}^{1} \,\tilde{\to}\, \Sigma_{g+1}^{1}, \quad \phi_g(x_1) = \phi_g(x_g) = x_{g+1}$$

Now the intersection of the subspaces $\phi_{g}(\Sigma_{1}^{1})$ and $\phi_{g}(\Sigma_{g}^{1})$ in $\Sigma_{g+1}^{1}$ is an interval, hence contractible. The Seifert-van-Kampen theorem then computes $$ \pi_1(\Sigma_{g+1}^{1},x_{g+1}) \,\tilde{=}\ \pi_1(\Sigma_{1}^{1},x_1) \,\ast\, \pi_1(\Sigma_{g}^{1},x_g) \,\tilde{=} \, \langle a_1,b_1,\ldots,a_{g+1}, b_{g+1} \rangle,$$ where the indices of the generators of $\pi_1(\Sigma_{g}^{1},x_g)$ were shifted by $1$ to avoid conflicting notation. The new boundary circle is homotopic to the concatenation of the two old boundary circle, hence: $$[ \iota_{g+1} ] = [\iota_1] \cdot [\iota_{g}] = [a_1,b_1] \cdot \prod_{i=1}^{g} [ a_{i+1}, b_{i+1}] = \prod_{i=1}^{g+1} [ a_{i}, b_{i}] \in \pi_1(\Sigma_{g+1}^{1},x_{g+1})$$.

As a last step, one constructs $\Sigma_{g}$ by gluing a disk to $\Sigma_{g}^{1}$ at the boundary circle. Another application of the Seifert-van-Kampen theorem yields the desired result: $$ \pi_1(\Sigma_{g},x_g) = \langle a_1,b_1,\ldots, a_g,b_g\mid \prod_{i = }^{g} [a_i,b_i] \rangle$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.