2
$\begingroup$

I am trying to compute the fundamental group of a genus-$2$ surface using van Kampen. Let $U_1$ and $U_2$ be the component tori with $U_1 \cap U_2 = U_0$ homotopically equivalent to a circle. I am convinced that $U_1$ is homotopically equivalent to a torus missing a point, which has fundamental group $\mathbb{Z} * \mathbb{Z}$.

So the fundamental group of the genus-2 surface is $\pi_1(U_1) \cdot \pi_1(U_2) / N = \left \langle a, b, c, d \right \rangle / N$, where: $$ N = \{ \left \langle i_{1}(w) i_{2}(w^{-1}) \right \rangle : w \in \pi_1(U_0) \cong \mathbb{Z} \}$$

Here, $i_{1}:\pi_1(U_0) \rightarrow \pi_1(U_1)$ and $i_{2}:\pi_1(U_0) \rightarrow \pi_1(U_2)$ are induced by the inclusions $U_0 \hookrightarrow U_1$ and $U_0 \hookrightarrow U_2$. I have seen the solution so I know what $N$ should be, but I am confused why this is the case. I think my confusion lies in the descriptions of $i_1$ and $i_2$. Does anyone have an intuitive explanation for what is going on here? Any help / advice is appreciated. Thanks

$\endgroup$
0
$\begingroup$

In the identification of $U_1$ as a punctured torus, the boundary circle is a simple loop around the puncture. Now to identify the class of this loop as an element of $\mathbb{Z} * \mathbb{Z}$, you need to follow this loop in the identification $\pi_1(\text{punctured torus}) = \mathbb{Z} * \mathbb{Z}$. For instance, if you do this by taking a deformation retract to a wedge of two circles, then applying the deformation retract to this specific loop should produce a loop that traverses each circle twice in opposite orientations, and equivalent to the commutator of the generators.

This is relatively easy to see in the picture of the punctured torus as a square/rectangle with identifications and the puncture in the center. As the circle in the middle is stretched to the boundary, it goes around the boundary of the square in say counterclockwise order, but this matches and mismatches with each pair of opposite sides.

$\endgroup$
  • $\begingroup$ This is not quite true: a simple loop around the puncture will deformation retract to a loop that traverses each circle twice, in opposing orientations. $\endgroup$ – Rylee Lyman Jan 12 at 22:36
  • 1
    $\begingroup$ I appreciate the response. You explained the part I did not initially understand: that I should find the equivalence class of the loop around the puncture in terms of the generators of $\mathbb{Z} * \mathbb{Z}$. However, I agree with @RyleeLyman. I believe a nice way to see this is to start with a rectangle (as in the usual construction of a torus) with a puncture in the middle. Then the loop around the puncture is simply a loop around the edges of the rectangle: $aba^{-1} b^{-1}$. $\endgroup$ – jonan Jan 12 at 23:14
  • $\begingroup$ You're right, I fixed it and added some detail. $\endgroup$ – ronno Jan 13 at 17:07

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.