# Borromean Lines of three $\mathbb{R}^1$ in $\mathbb{R}^3$ and analogous Milnor link invariants

It is know that Borromean rings can be detected by Milnor invariant
$$\bar{\mu}(\gamma_1,\gamma_2,\gamma_3)= \# (\Sigma_1 \cap \Sigma_2 \cap \Sigma_3)-\frac{1}{2}\sum_{I,J,K}\epsilon_{IJK} \sum_{\scriptsize\begin{array}{c}a\in \gamma_{I}\cap \Sigma_K \\ b\in \gamma_{I}\cap \Sigma_J\\ x_a>x_b \end{array}}(-1)^{\epsilon(a)+\epsilon(b)}$$ or Milnor's triple linking number. See Refs:

• B. Mellor and P. Melvin, A geometric interpretation of milnor’s triple linking numbers, Algebraic & Geometric Topology 3 (2003) 557–568, [math/0110001].

• Braiding Statistics and Link Invariants of Bosonic/Fermionic Topological Quantum Matter in 2+1 and 3+1 dimensions Putrov, Wang, Yau; 2016 Annals Phys. 384 (2017) 254-287 (2017-09) DOI: 10.1016/j.aop.2017.06.019 e-Print: https://arxiv.org/abs/1612.09298

From 1612.09298: It is easy to evaluate for the Borromean rings link. Consider realization of Borromean rings shown in Figure with natural choice of Seifert surfaces $$\Sigma_I$$ lying in three pairwise orthogonal planes. It is easy to see that the first term in is $$1$$ while all other terms vanish. Terms are defined in 1612.09298.

My question: The above well-known Milnor invariant is evaluated in $$S^3$$, and there are three $$S^1$$ circles. Can we define a link invariant, such that we have a set of Borromean lines (three of $$R^1$$) link in a infinite larger space $$R^3$$? (I think I can define the 3 lines along the $$T^3$$ via a surgery.) But I wish to have the following conditions satisfy:

• triple linking number defined in $$R^3$$.

• The three $$R^1$$ lines are extended in infinity in the space $$R^3$$.

• the three $$R^1$$ are linked in the similar manner as Borromean rings (or lines).

Comments added: For instance, two $$S^1$$ circles in $$R^3$$ can form a Hopf link. One can stretch one $$S^1$$ into two parallel $$R^1$$ lines in $$x$$ direction. And we further stretch the other $$S^1$$ into two parallel $$R^1$$ lines in $$y$$ direction, where one of the $$y$$ line lies in between the two $$x$$ lines. These four $$R^1$$ lines form nontrivial linking, which we interpret as an infinite version of the Hopf link. Our question is whether there exists an infinite version of the Borromean ring.

Any Refs/comments are welcome!

If the answer is negative, can we give a proof for impossibility?

• I'm maybe missing something, but there are no obstructions that would limit three mutually nonintersecting, mutually non-parallel $\mathbb{R}^1$ lines from winding up in any configuration whatsoever, so there's no way of distinguishing 'linked' from 'unlinked' lines in a way that would be invariant even under linear transformations. – Steven Stadnicki Sep 11 at 20:47
• @StevenStadnicki Thanks! A comment is added in the end, which explains what we mean by $R^1$ lines forming a Hopf link. The question is then about an analogous version for Borromean ring. – user34104 Sep 11 at 21:19
• Thank you - the comment is very helpful! Note that just like in the Hopf case, you'll likely need to use two copies of $\mathbb{R}^1$ for each circle, so you may want to talk about six copies of $\mathbb{R}^1$, not three... – Steven Stadnicki Sep 12 at 3:02