# Classification of fundamental groups of non-orientable surfaces

I want to compute the presentation of the fundamental group of the non orientable surfaces $N_h$, thus $\pi_1(N_h)$.

I notated with $N_h$ the sphere with $h$ crosscaps. Herefore I first have to compute the fundamental group of the dunce hat in topological sense.

But how to do this on a good mathematical way ? I know that we can use van Kampen Theorem. Then I want to compute the fundamental group of the projective plane and then try to generalize this to $N_h$ for arbitrary $h\in\Bbb{N}$.

Can someone explain how to use van Kampen Theorem (I know what the meaning is but I thought of applying) and how to solve this exercise ?!

Thank you.

I just consider the case $h=3$, but the argument is completely general.

From the classification of closed surfaces, we know that $N_3$ is the connected sum of three projective spaces, so that $N_3$ be the quotient of an hexagon $P$ by identifying its sides as indicated by the following figure:

Now let $U, V \subset N_3$ be subspaces illustrated by the figure below:

Applying van Kampen theorem, we deduce that $\pi_1(N_3)$ is the quotient of the free product $\pi_1(U) \ast \pi_1(V)$ identifying the images of $\pi_1(U \cap V)$ into $\pi_1(U)$ and $\pi_1(V)$. In fact, $\pi_1(V)$ is clearly trivial so that $\pi_1(N_3)$ turns out to be the quotient of $\pi_1(U)$ by its subgroup corresponding to $\pi_1(U \cap V)$.

Now, there is naturally a deformation retract $\phi$ from $U$ to the quotient of $\partial P$: it is just a graph $\Gamma$, more precisely a bouquet of three circles labelled by $a_1$, $a_2$ and $a_3$. Therefore, $\pi_1(U)$ is just the free group $\langle a_1,a_2,a_3 \mid \ \rangle$.

Then, there is a deformation retract from $U \cap V$ to the circle $\partial V$. Therefore, $\pi_1(U \cap V)$ is infinite cyclic and its image into $\pi_1(U)$ is the homotopy class of the circle $\phi(\partial V)$ in the quotient $\Gamma$ of $\partial P$. Such a loop in $\Gamma$ is labelled by $a_1^2a_2^2a_3^2$.

Finally, we conclude that $$\pi_1(N_3)= \langle a_1,a_2,a_3 \mid a_1^2a_2^2a_3^2 \rangle.$$

• for the record, $N_3$ is dubbed the Dyck's surface – janmarqz Nov 30 '14 at 20:13
• How did you get that diagram as a valid CW complex for $N_3$? I agree it holds in this case, but for example for $N_2$ the similar-looking square won't give us $N_2$, so I was wondering how you came up with it. The only way I can think of is to manually cut a disk out of $\mathbb{R}P^2$ and do the connected sum manually, but that gives a polygon with way too many sides :P – Juan Carlos Ortiz Mar 30 at 19:08

You might like to read Marc Lackenby's notes for the Oxford course 'Topology and Groups'. They include some examples of using the van Kampen Theorem to give a presentation for a fundamental group.