The Kervaire Conjecture

Think about your favourite non-trivial group. Now add a generator to it and then add any relation. Is the resulting group still non-trivial?

The above question is known as the Kervaire conjecture. Phrased more concretely, if \(G\) is any non-trivial group and \(r \in G \ast \mathbb{Z}\), is \((G \ast \mathbb{Z})/\langle\!\langle r\rangle\!\rangle\) again non-trivial?

This is an example of a problem which is extremely easy to state but which is still unsolved in complete generality. The best partial results so far (and that I could find by a quick search) are the following:

  • It is true for finite groups, proven by Gerstenhaber and Rothaus.
  • The above implies the conjecture for all residually finite groups and even all hyperlinear ones (the latter includes all sofic groups), noticed by Pestov.
  • For torsion-free groups, proven by Klyachko.
  • Instead of putting restrictions on the group, Klyachko and Lurye proved the conjecture under a restriction on the relation, namely that it is a proper power (i.e. \(r = w^k\) for some \(k \ge 2\)).

Since we don’t know any non-sofic group (which is one of the major open problems in group theory), a counter-example to the Kervaire conjecture currently seems to be completely out of reach.

The origin of Kervaire’s conjecture is his paper on the classification of high-dimensional knot groups. He proved the following: Given \(n \ge 3\). The group \(\pi\) is isomorphic to \(\pi_1(S^{n+2} – f(S^n))\) for some differential imbedding \(f\colon S^n \to S^{n+2}\) if and only if \(\pi/\pi^\prime \cong \mathbb{Z}\), the weight of \(\pi\) is \(1\), and \(H_2(\pi) = 0\).

In the above statement \(\pi^\prime\) denotes the commutator subgroup of \(\pi\), the weight of \(\pi\) is the smallest integer \(k\) with the property that there exists a set of \(k\) elements in \(\pi\) whose normal closure equals \(\pi\), and \(H_2(\pi)\) is the second homology group of \(\pi\) with integral coefficients and trivial action of \(\pi\) on \(\mathbb{Z}\).

At the end of his paper, Kervaire shows that the conditions \(\pi/\pi^\prime \cong \mathbb{Z}\) and \(H_2(\pi) = 0\) do not imply that the weight of \(\pi\) is \(1\), i.e. that the latter is not a redundant condition in the theorem. His example is of the form \(\pi = G \ast \mathbb{Z}\). Hence the question whether a non-trivial free product \(G \ast \mathbb{Z}\) can ever be of weight \(1\) arose.

Leave a Reply

Your email address will not be published.