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I ran into the following concept in passing here. Let $G$ be a group and let $\phi$ be an automorphism of $G$. Let $P$ be a presentation for $G$ with $X$ the set of generators in $P$. Form a new group $P^\phi$ by appending to $P$ relations of the form $x = \phi(x)$ for all $x \in X$.

Is the isomorphism type of $P^\phi$ independent of the choice of $P$? Maybe I should assume $G$ is finitely presented and we are just working with finite presentations.

We say that $\phi$ "kills" $G$ (I know a bit dramatic) if $P^\phi = 1$ for some (all?) choices of $P$. Is there an equivalent way of saying that $\phi$ kills $G$ that doesn't mention presentations?

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    $\begingroup$ As to your first question, yes, $P^{\phi}$ is independent of the choice of $P$, as $x = \phi(x)$ for all $x \in X$ is equivalent to $g = \phi(g)$ for all $g \in G$. $\endgroup$ – Andreas Caranti Apr 24 at 7:09
  • $\begingroup$ This is usually called a semidirect product of $G$ with $Z$. $\endgroup$ – Moishe Kohan Apr 24 at 15:31
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Since you like topology, let $M$ be a space with $\pi_1(M)=G$ and let $\phi$ be induced by a map $f:M\to M$. If you construct the mapping torus $$M_f=M\times [0,1]/(x,1)\sim(f(x),0),$$ its fundamental group is given by an HNN extension $$\pi_1(M_f) = \langle G,t\mid tgt^{-1}=\phi(g)\text{ for all $g\in G$}\rangle.$$ The $t$ corresponds to a loop that closes up the path ${*}\times [0,1]$ with some arc in $M$ (or: choose $f$ so that the basepoint of $M$ is a fixed point). If you then glue in a disk $D$ along this loop, you get $$\pi_1(M_f\cup D)=\langle G \mid g=\phi(g)\text{ for all $g\in G$}\rangle.$$ That this is independent of $M$, $f$, and $D$ is that the group presentation on the right does not mention them. Also, notice that if $G$ is finitely presented, so is $\pi_1(M_f\cup D)$ since using the presentation complex for $M$ gives $M_f\cup D$ a finite CW structure. (Equivalently, one can take $\pi_1(M_f/\partial D)$ to get the group in question.)

Underlying this is the normal closure of the subgroup $\langle g^{-1}\phi(g):g\in G\rangle$. The map $\phi$ "kills" $G$ iff the normal closure is all of $G$.

A weird equivalent statement is that $\phi$ "kills" $G$ iff, for every normal covering space of $M_f$ that $\partial D$ lifts to, that covering space has only one sheet.

Something I couldn't find a use for is that if $M$ is a $K(G,1)$, then $M_f$ is a $K(\pi_1(M_f),1)$. The problem is that collapsing the $*\times [0,1]$ loop seems like it might create nontrivial higher homotopy groups even if $\phi$ "kills" $G$. It would be interesting if $\phi$ "killing" $G$ means the collapsed $K(\pi_1(M_f),1)$ is contractible.

As an example, for a fibered knot $K\subset S^3$, the complement $S^3-\nu(K)$ is a mapping torus, with the monodromy inducing an automorphism $\phi:\pi_1(S)\to \pi_1(S)$ where $S$ is some leaf, a once-punctured compact oriented surface. The monodromy is isotopic to one with a fixed point on the boundary of $S$, so there is a meridian that serves the role of $t$ for $\pi_1(S^3-\nu(K))$ as an HNN extension. Gluing in a thickened disk along this meridian results in a space that is homeomorphic to $S^3-B^3$, so not only does $\phi$ "kill" $\pi_1(S)$, but the resulting collapsed mapping torus is contractible.

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