0
$\begingroup$

In $\mathbb{R}^3$, consider the space missing a circle and a finite number of distinct lines passing through the center of the circle. In the case of one line, one can show that it deform retracts to the torus. With several lines I was trying to use Van Kampen but I'm not entirely sure how to apply it.

$\endgroup$
  • $\begingroup$ Maybe Seifert-van Kampen + induction on the number of lines? $\endgroup$ – Matthew Leingang Jan 15 '15 at 1:41
  • $\begingroup$ I'm not sure how to go past n=1, can you give me the case of n=2? $\endgroup$ – Yon Kim Jan 15 '15 at 1:44
  • $\begingroup$ Just curious: In general if you have $g$ lines, is the complement homotopic equivalent to a surface of genus $g$? $\endgroup$ – user99914 Jan 15 '15 at 2:40
  • $\begingroup$ Are the lines intersecting at the center or just parallel? $\endgroup$ – WWK Jan 15 '15 at 2:43
  • $\begingroup$ Do you assume the lines to be disjoint from the circle? $\endgroup$ – Moishe Kohan Jan 15 '15 at 3:12
1
$\begingroup$

The whole space deformation retracts onto a solid sphere, with $n$ lines drilled through its center and a circle removed. Say this space is $X$

Now consider $X=U\cup V$, where $U$ is a portion of the sphere only containing a part of the missing circle and no lines, $V$ is the rest of the sphere and a bit more, so that $U\cap V$ is simply a solid cylinder with two lines drilled out.

Now, $U\cong D^3\setminus \text{\{a line\}}$, which deformation retracts onto $S^1$. Similarly $V\cong D^3 \setminus \{n+1\text{ lines}\}$, which is homotopy equivalent to wedge sum of $n+1$ circles. And $U\cap V$ is homotopy equivalent to $S^1\vee S^1$.

Say, $\pi_1(U)=\langle a\rangle, \pi_1(V)=\langle l_1,l_2,...,l_n,b \rangle, \pi_1(U\cap V)=\langle u,v\rangle$. Then by van-Kampen's theorem, $$\pi_1(X)=\langle a,b,l_1,...l_n|ab^{-1}\rangle=\langle a,l_1,...l_n\rangle$$ So $\pi_1(X)$ is the free group on $n+1$ generators.

Added: In my solution above, I assumed the lines to be disjoint from each other, which is not the case here. So here's is the corrected version.

$X$ is again the solid sphere as before. Cut a cylinder through $X$, so that it only contains the lines. Say the outside of this cylinder, which contains the circle, is $U$ and the cylinder itself is $V$. Adjust the spaces, so that $U\cap V$ is like a thick walled ordinary cylinder.

$U$ deformation retracts onto a torus. $U\cap V$ retracts onto an annulus and then to $S^1$. Now $V$ is homeomorphic to a solid sphere, with $n$ lines drilled through it and intersecting at the center, say. This can be deformed to $S^2$ with $2n$ points missing, by radially outward transformation starting from the center. Then $V$ is homotopy equivalent to wedge of $2n-1$ circles.

Say, $\pi_1(U)=\langle a,b\rangle, \pi_1(V)=\langle w_1, w_2,...,w_{2n-1} \rangle, \pi_1(U\cap V)=\langle c\rangle$. Then by van-Kampen's theorem, $$\pi_1(X)=\langle a,b,w_1,...,w_{2n-1}|b(w_1...w_n)^{-1}\rangle=\langle a,w_1,...w_{2n-1}\rangle$$

So $\pi_1(X)$ is free on $2n$ generators.

$\endgroup$
  • $\begingroup$ You are not accounting for the circle, which produces free factor isomorphic to $Z^2$. $\endgroup$ – Moishe Kohan Jan 16 '15 at 18:27
  • $\begingroup$ The correct answer is $Z^2 * F_{2(n-1)}$, where $n$ is the number of lines, assuming that the lines are disjoint from the circle. $\endgroup$ – Moishe Kohan Jan 17 '15 at 3:37
  • $\begingroup$ I see that now, $\pi_1(U)\cong \langle a,b | ab=ba\rangle$. But then, $\pi_1(X)=\langle a,w_1,...w_{2n-1}|aw_1...w_n=w_1...w_na\rangle$. How's this same as $\mathbb{Z}^2 * F_{2(n-1)}$? $\endgroup$ – ChesterX Jan 17 '15 at 4:59

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.