9
$\begingroup$

This summer I am in charge of hosting a prep course for my university's graduate qualifying examination in differential geometry/topology. I've done this before, so I'm fairly confident in my understanding of differential geometry and point-set topology. However I'm currently going through the most recent qualifying exam in the area, and, to my horror, there is an algebraic topology question on it. It is stated as follows:

Let $X$ be the connected sum of the torus with the Klein Bottle. Compute the fundamental group of $X$

Now I'm reasonably familiar with fundamental groups, and to my understanding the "connected sum" is created by deleting a ball from each space and gluing together the resulting boundary spheres. The problem does not specify precisely where we are gluing the two spaces together, so I can only assume this doesn't change the resulting fundamental group (as is intuitively true).

After doing some reading, I've come across the following version of the Seifert-van Kampen Theorem:

[Corollary 70.3 in Munkres] Let $X=U\cup V$, where $U$ and $V$ are open in $X$; assume $U$, $V$, and $U\cap V$ are path-connected. Fix $x_0\in U\cap V$. If $U\cap V$ is simply connected, then there is an isomorphism $$ k:\pi_1(U,x_0)*\pi_1(V,x_0)\to\pi_1(X,x_0). $$ [Here $*$ denotes the free product.]

Since the fundamental groups of the torus and Klein bottle are $\mathbb{Z}\times\mathbb{Z}$ and $<a,b:aba^{-1}b=1>$, respectively, it seems that I may apply the above theorem to say that $$ \pi_1(X,x_0)=(\mathbb{Z}\times\mathbb{Z})*<a,b:aba^{-1}b=1>. $$ Is this true? If so, can this representation be simplified more? I'm out of my element here, so I figure I should at least run my thoughts by more capable people than I before the prep course begins. Any comments or references to similar material is greatly appreciated. Thank you.

$\endgroup$
3
  • 3
    $\begingroup$ Are you trying to use Seifert-Van Kampen seeing the connected sum as the union of the parts of the torus and the projective plane?, in this case the intersection is homotopic to a circle, and this is not simply connected. $\endgroup$ Jun 27 '18 at 2:31
  • $\begingroup$ @CamiloArosemena-Serrato Ahh, I see the error in my thought. Thank you. I was imagining the gluing incorrectly. I take it this problem will require a different approach then? $\endgroup$
    – Blake
    Jun 27 '18 at 2:41
  • $\begingroup$ This answer should clarify what you have to do math.stackexchange.com/a/1470130 $\endgroup$ Jun 27 '18 at 3:01
5
$\begingroup$

After Camilo Arosemena-Serrato's helpful comments. I put together this "solution" that I believe is correct. If anyone could verify this with me I'll mark the question as answered. Thanks again.

"Answer"

Recall that the connected sum is constructed by deleting a ball from each space and gluing together the resulting boundary spheres. Let $T$ be the torus with some ball removed, and let $K$ be the Klein bottle with some ball removed. Let $\iota_1:T\to X$, $\iota_2:K\to X$ be the natural inclusions. Fix $x_0\in T\cap K$. According to the classical version of the Seifer-van Kampen theorem (Theorem 70.2 in Munkres), the homomorphism $$ j:\pi_1(T,x_0)*\pi_1(K,x_0)\to\pi_1(X,x_0) $$ is surjective, and its kernel is generated by all elements of the free product of the form $\iota_1(g)^{-1}\iota_2(g)$ and their conjugates, where $g\in \pi_1(T\cap K,x_0)$. Let $a_1,b_1$ be the generators of the fundamental group of the torus, and $a_2,b_2$ be the generators of the fundamental group of the Klein bottle. Now $T\cap K$ is homotopic to a circle, so $$ \iota_1(g)=a_1b_1a_1^{-1}b_1^{-1}\hspace{ .5 in}\text{and}\hspace{.5 in}\iota_2(g)=a_2b_2a_2b_2^{-1}. $$ We conclude by the first group isomorphism theorem that $$ \pi_1(X,x_0)=\big\langle a_1,a_1,b_1,b_2\,:\,a_1b_1a_1^{-1}b_1^{-1}a_{2}b_{2}a_{2}b_{2}^{-1}=1\big\rangle. $$ In other words the first fundamental group of $X$ is the free group generated by $a_1,a_2,b_1,b_2$ modulo the normal subgroup generated by $a_1b_1a_1^{-1}b_1^{-1}a_{2}b_{2}a_{2}b_{2}^{-1}$

$\endgroup$

This site is temporarily in read only mode and not accepting new answers.

Not the answer you're looking for? Browse other questions tagged .