Sign up ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

This question comes from an unanswered question on Math Stack Exchange.

A two-component link in $\mathbb{RP}^3$ is any embedding $S^1\uplus S^1\to \mathbb{RP}^3$. Two such links are homotopic if there exists a homotopy between the maps such that the images of the circles remain disjoint at each intermediate stage. (Note that each circle may pass through itself during the homotopy.)

What is the set of homotopy classes of such links?

If either of the circles is nontrivial in $\pi_1(\mathbb{RP}^3)$, then it seems to me that the possibilities are completely determined by the linking numbers of the various components of the preimages in $S^3$. However, if both of the circles are nullhomotopic then the situation in the cover is much more complicated.

share|cite|improve this question
Did you look at Milnor's original paper ? You can try to follow his arguments (section 5) in the $RP^3$ case and see if they imply that in the remaining case the $RP^3$-linking number is the only invariant (as is the case for 2-component links in the 3-space)? – Misha Jun 27 '13 at 10:25
The issue, as I see it, is to deform the link via homotopy so that one component of the new link is contained in an affine patch of $RP^3$ (I think, it is not too hard, just requires some work). If you could do this, then you can undo all the crossings in that component, convert it to a trivial knot and then follow Milnor. – Misha Jun 27 '13 at 11:13

1 Answer 1

If I understand your question correctly, the homotopy classes of links such that one of the circles is nullhomotopic in $\mathbb{RP}^3$ are completely classified by the first homotopy group of ($\mathbb{RP}^3$ minus a little unknotted curve), which can be given to you by Van Kampen's theorem.

share|cite|improve this answer
This could be true, but you should explain why. – Misha Jun 27 '13 at 10:38
I agree that this sounds likely. I think the fundamental group you mention is isomorphic $\mathbb{Z}*\mathbb{Z}_2$, so there ought to be one homotopy class for each conjugacy class in this group. – Jim Belk Jun 27 '13 at 13:49
Assume that one component $K_1$ of a link $L=K_1 \cup K_2$ is nullhomotopic, and fix a little unknotted circle $C$. From a generic knot homotopy from $K_1$ to $C$ one can deduce a link homotopy from $L$ to a link $C \cup K_2^\prime$. Now leaving $C$ fixed, link homotopy amounts to homotopy of $K_2^\prime$ in the complementary of $C$. What remains to see is what happens when you change the role of $K_1$ and $K_2$. But then you have to assume that both are nullhomotopic, and since the linking number is symmetric you do not have to mod out $\mathbb{Z}*\mathbb{Z}_2$ by anything. – Arnaud Mortier Jun 27 '13 at 14:51
@Misha: I don't agree, what I claim here is that you can always "unknot" $one$ component (as soon as it is nullhomotopic), possibly getting the other components in a very messy shape. Think how it works to exchange the components of a Whitehead link in the spherer to get an idea. – Arnaud Mortier Jun 27 '13 at 19:15
Well I can detail a little bit that argument. It works in any 3-manifold. A knot homotopy is an alternating sequence of diffeomorphisms and "crossing changes" - which you can see as Dehn twists, but this point of view is not necessary - the point is that none of these two processes pays any attention to the possible existence of other link components. Crossing changes happen within little balls, and if other link components pass through these balls, you can always push them out. – Arnaud Mortier Jun 27 '13 at 22:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.