# Fundamental group of the complement of $3$ pairwise linked circles in $\mathbb R^{3}$

I'm reading through the Hatcher book for a course on algebraic topology. Here it is explained how to find the fundamental group for the complement of two linked circles. I'm trying to proof the case with three pairwise linked circles, but I can't seem to reproduce any results. I tried moving to $S^3$, and moving the point on infinity. This way you would get two linked circles and a line passing through them. This could be (I think) deformation retracted to a torus with two linked circles cut out, the circles going around the middle 'hole'. At that point I'm stuck on how to either:

a) Moving on from here to another deformation which gives me a space or wedge sum of spaces I know the fundamental group of.

b) Finding suitable open sets for using Van Kampen's theorem.

I'd be very interested in hearing anyone's thoughts on this, and also in seeing if there is a way to extend the answer to the complement of $n \in \mathbb N$ pairwise linked circles. (I expect the answer to be $\mathbb Z^n$, based on the case of 1 and 2 linked circles)

• Are the circles linked as in this question? math.stackexchange.com/questions/251364/… (NB there are no answers there but there is some useful-looking comments). Oct 26, 2015 at 15:58
• Yes they are, will edit the question
– Six
Oct 26, 2015 at 16:04
• There is actually no link in $S^3$ whose complement has fundamental group $\Bbb Z^n$, $n>2$.
– user98602
Oct 26, 2015 at 16:30
• On further inspection of the question: no not linked like borromean rings. Just every circle linked in the other two circles, removing one circle would leave two linked circles, not two unlinked ones. Sorry for the confusion!
– Six
Oct 26, 2015 at 16:39
• The other question's comments are still relevant. Look at the Hatcher question about the "Wirtinger presentation" for inspiration.
– user98602
Oct 26, 2015 at 16:41

Hatcher shows two unlinked circles have a fundamental group $\mathbb{F}_2$ which is humungous - it's the free group of all possible wors in two letters. The fundamental group of two linked circles is simply $\mathbb{Z} \oplus \mathbb{Z}$ - we can exploit the linking of the rings in order to express any entanglement as a special type. In order to find the homotopy group of the Borromean rings you may need to draw as many as 4 rings.

3 pairwise linked circles will have three generators, looping ones around each of the possible three circles. Do these generators commute? Sure we can imagine one the circles disappearing and getting two circles. We have shown:

$$\pi_1(3\text{ linked circles}) \subseteq \mathbb{Z}\oplus \mathbb{Z} \oplus \mathbb{Z}$$

• You can't just imagine one of the circles disappearing to argue for commutativity. The commutator homotopy will intersect the deleted circle, so won't be valid in the complement of all three. Oct 31, 2019 at 3:21