# Topological spaces with unknown fundamental group

Are there any well known topological spaces for which the fundamental group is not known yet?

• Since $\pi_n(X) \cong \pi_1 (\Omega^{n-1} X)$, this is the same as asking "are there any unknown homotopy groups"? And the answer is... yeah. – Zach L. Nov 7 '14 at 0:19
• @ZachL. what is $\Omega$? – Seth Nov 7 '14 at 0:20
• maybe I should rephrase my question and ask: are there any particular spaces among those, where one would like to know the fundamental group? – Loreno Heer Nov 7 '14 at 0:21
• $\Omega X$ is the space of based maps $S^1 \rightarrow X$ with a suitable topology (like compact open). Since the homotopy groups of spheres are pretty mysterious, calculating $\pi_1(\Omega^m S^n)$ is a fundamental group calculation lots of people would be interested in. – Zach L. Nov 7 '14 at 0:25
• @sanjab, sure. $\pi_1(\Omega^{m-1}S^n)$ is both important and unknown for most pairs with $m >n$. – user172541 Nov 7 '14 at 0:25

In general, it depends on what you mean by knowing a group. For example, the complement of a knot is a space with a fundamental group, the knot group of the knot, given by the Wirtinger presentation. Unfortunately, many properties of a group are undecidable given only a presentation of it: most importantly, it's undecidable whether two presentations are presentations of the same group. So it's unclear whether the Wirtinger presentation counts as an answer to the question "do we know the fundamental groups of all knot complements?"

As Zach L. alludes to in the comments, the iterated loop space construction $\Omega^n X$ (the space of pointed maps from an $n$-sphere $S^n$ into a pointed space $X$) has the property of shifting down the higher homotopy groups of a space:

$$\pi_k(\Omega^n X) \cong \pi_{k+n}(X).$$

In particular, asking questions about the fundamental groups $\pi_1(\Omega^n X)$ of iterated loop spaces of a space is equivalent to asking questions about the higher homotopy groups $\pi_{n+1}(X)$ of our space.

This is actually a good thing: the higher homotopy groups are abelian, and in reasonable cases (e.g. if $X$ is a simply connected finite CW complex) are finitely presented. Unlike the case of finitely presented groups, finitely presented abelian groups are much easier to work with algorithmically, and it's easy to give a complete list of invariants describing such a group via the structure theorem and easy to tell when two such groups are not isomorphic.

In particular, it's easy to describe what it means to not know such a group. There's a finite list of numbers (the number of times $\mathbb{Z}$ appears in the group, and the number of times $\mathbb{Z}_{p^n}$ appears in the group for all primes $p$ and for all positive integers $n$) that characterize a finitely presented abelian group, and we don't know the group if we don't know at least one of these numbers.

In that sense, we don't even know all of the higher homotopy groups of the $2$-sphere $S^2$! The study of the homotopy groups of spheres is a foundational and very hard question in homotopy theory, although I can't give precise statements as to exactly what is and is not known. I would guess that we don't know, say, $\pi_{70}(S^2)$.

• On p.62 of Rolfsen's Knots and Links, he states "The group of a knot is not a complete knot invariant (that is, $K\leadsto \pi_1(\mathbb{R}^3-K)$ is not one-to-one)." The specific example he gives is the square and granny knots. I have heard however that if you restrict to prime knots, then the knot group determines the knot, though I don't know a reference off the top of my head. – curious Nov 7 '14 at 1:08
• @curious: oops. Thanks for the correction! – Qiaochu Yuan Nov 7 '14 at 1:10
• Just a note: the isomorphism problem is solvable for closed 3-manifolds, and I think it should be fairly easy to deduce that it is solvable for knot complements. There's a good survey paper (with some new results) by Aschenbrenner, Friedl and Wilton. – Cheerful Parsnip Nov 7 '14 at 1:37
• @GrumpyParsnip: The isomorphism problem for prime 3-manifolds with incompressible boundary, such as complements of nontrivial knots, was solved by Waldhausen well before the general isomorphism problem for closed 3-manifolds. – Lee Mosher Nov 7 '14 at 22:13
• Thanks for the clarification, @LeeMosher! – Cheerful Parsnip Nov 7 '14 at 23:40