# Braid group and knot group

It is well known that braid groups and knot groups share many common properties. For example, they have the same $H_1$ and they are both residually finite and (hence) Hopfian. On the other hand, we know that $B_2$ and $B_3$ are isomorphic to the knot groups of unknot and trefoil knot respectively. My question is, for any $n\geq 4$, is there a knot with knot group $B_n$?

If not, how about high dimensional knot $S^n$ in $S^{n+2}$?

• The first question is answered (negatively) here: math.stackexchange.com/questions/2118701/… – Igor Rivin Aug 8 '17 at 2:50
• To summarize Igor's link, for $n\ge 4$, $B_n$ contains a copy of $\mathbf{Z}^3$ (use $\sigma_1$, $\sigma_3$, and a generator of the center), while an aspherical 3-manifold group can't contain $\mathbf{Z}^3$ unless it's virtually abelian. – YCor Aug 8 '17 at 15:07

## 2 Answers

This is a simple addendum to Danny's answer (because I can't use the citation thing in comments): Homology of braid groups has been computed by a number of people, and an extensive survey is:

Vershinin, Vladimir V., Homology of braid groups and their generalizations, Jones, Vaughan F. R. (ed.) et al., Knot theory. Proceedings of the mini-semester, Warsaw, Poland, July 13--August 17, 1995. Warszawa: Polish Academy of Sciences, Institute of Mathematics, Banach Cent. Publ. 42, 421-446 (1998). ZBL0905.20032.

Unless I am misinterpreting the results, there is nontrivial homology in all degrees, so the answer to the OP's second question is negative also.

This is an incomplete answer to the last question, but perhaps someone can supply the last piece of information to complete it. The question of whether a finitely presented group G is a higher-dimensional knot group was resolved by Kervaire. Necessary and sufficient conditions are (1) G is the normal closure of one element, (2) $H_1(G) \cong {\mathbb Z}$, and (3) $H_2(G) = 0$.

The first two of these hold for the braid groups. I'm sure that the calculation of $H_2(B_n)$ is somewhere in the literature, but I couldn't locate it.