We describe a simple criterion for showing that a group has Serre’s property FA. By exhibiting a certain pattern of finite subgroups, we show that this criterion is satisfied by and when .
The Zieschang Gedenkschrift
\conferencestart5 September 2007
\conferenceend8 September 2007
\conferencenameConference in honour of Heiner Zieschang
Configurations of torsion and Property ]A condition that prevents groups
from acting nontrivially on trees Martin R Bridson]Martin R Bridson \givennameMartin R \surnameBridson \urladdrhttp://www.maths.ox.ac.uk/ bridson \dedicatoryDedicated to the memory of Heiner Zieschang \volumenumber14 \issuenumber \publicationyear2008 \papernumber08 \startpage129 \endpage133 \doi \MR \Zbl \arxivreference \subjectprimarymsc200020E08 \subjectprimarymsc200020F65 \published29 April 2008 \publishedonline29 April 2008 \proposed \seconded \corresponding \version \makeautorefnamethmaTheorem \makeautorefnamelemmaaLemma
We describe a simple criterion for showing that a group has Serre’s property FA. By exhibiting a certain pattern of finite subgroups, we show that this criterion is satisfied by Aut(F_n) and SL(n,Z) when n¿=3.
We describe a simple criterion for showing that a group has Serre’s property FA. By exhibiting a certain pattern of finite subgroups, we show that this criterion is satisfied by Aut(F¡sub¿n¡/sub¿) and SL(n,¡b¿Z¡/b¿) when n≥ 3.
An –tree is a geodesic metric space in which there is a unique arc connecting each pair of points. A group is said to have property if for every action of by isometries on an –tree, the fixed point set is nonempty. Serre’s property FA is similar except that one considers only actions on simplicial trees. A group has FA if and only if it cannot be expressed as a nontrivial amalgamated free product or HNN extension.
Let be a group that is generated by the union of the subsets . If has property for all , then also has property .
Let be connected subsets of an –tree. It is not difficult to see that if is nonempty for , then is nonempty (cf Serre [7, p 65]). Setting proves the lemma, since is assumed to be nonempty and . ∎
Every finite group has because the circumcentre of any –orbit in an –tree will be a fixed point.
Corollary 1 (The Triangle Criterion).
If is generated by and is finite for , then has property .
Let denote the automorphism group of the free group of rank .
If then and satisfy the Triangle Criterion and hence have property .
J-P Serre  was the first to prove that has FA if , and his argument shows that these groups actually have . Our argument is very similar to his except that he exploited the pattern of nilpotent subgroups rather than finite ones. In the light of a theorem of J Tits , Serre’s argument shows that all subgroups of finite index in have FA. In contrast, there is a subgroup of finite index that does not have property FA (see McCool ), while it is unknown if has such subgroups when . O Bogopolski  was the first to prove that has FA. M Culler and K Vogtmann  gave a short proof based on their idea of “minipotent” elements.
The obvious appeal of \fullrefmain lies in the final phrase, but the stronger fact that these groups satisfy the Triangle Criterion is useful in my work on fixed point theorems for actions of automorphism groups of free groups on higher-dimensional CAT spaces . One can extend the theorem in various ways (cf \fullrefvariation) but I shall not present the details here as to do so would obscure the simple and transparent proof that has property FA, which is the main point of this note. I hope that it is a proof that Zieschang would have enjoyed.
This research was supported by Fellowships from the EPSRC and by a Royal Society Wolfson Research Merit Award.
1 Generating and by finite subgroups
We assume that and fix a basis of . For , let be the automorphism of that sends to and fixes the other basis elements. J Nielsen proved that is generated by the right Nielsen transformations and the involutions .
Let be the group generated by permutations111We shall write to denote the transposition of and . of . Conjugation by a permutation sends to and to . Therefore is generated by and . In particular is generated by and the subgroup generated by and the . (The action of on the abelianisation of gives a epimorphism , and the image of under this map is the group of monomial matrices.)
We write and for the subgroups corresponding to the sub-basis . Let , let and , and note that each is an involution. Define
is generated by .
Conjugating by we get , which conjugates to and to . Thus and are in the subgroup generated by the ; hence and are too. We already noted that is generated by and . ∎
The groups are finite.
and commute with the involutions and , and has order , so . As , we have . And . ∎
These lemmas prove that satisfies the Triangle Criterion if , and by taking the images of the under the natural map we see that does too. When is odd, we obtain the corresponding result for by replacing the image of each by ; let denote the image of modified in this manner.
When is even we need to adjust the a little more. Let and note that is generated by the image of and the subgroup corresponding to the sub-basis . If then the groups and remain finite if we add to . Thus the sets , , demonstrate that satisfies the Triangle Criterion.
If then by modifying the sets slightly one can also show that , the inverse image in of , satisfies the Triangle Criterion.
1.1 The geometry of the
It would be unfair of me to leave the reader to guess the origin of the finite subgroups used in the above proof, so let me explain the geometry behind the construction.
Any finite subgroup of can be realised as a group of basepoint-preserving isometries of a graph of Euler characteristic . \fullreffigure222I am grateful to Karen Vogtmann for producing this figure. below gives such realisations for the groups . An important point is that if then cannot be realised as a group of symmetries of . I wanted to obtain the generating set that proved useful in my work with Karen Vogtmann . Thus, starting with the rose and the graph for , I looked for a third graph where could be realised together with a symmetry intertwining and .
2 Variations on the theme
I have concentrated on configurations of finite subgroups in this note but \fullreflem1 can also be applied to situations where the subgroups are infinite. For example, if lies in the commutator subgroup of its centralizer, then will be nonempty whenever acts by isometries on an –tree. (This is a special instance of a general fact about semisimple actions on CAT spaces .) By exploiting such facts in conjunction with \fullreflem1 one can prove, for example, that the mapping class group of a surface of genus at least 3 has property , a result first proved in .
One can also strengthen \fullreflem1 using an argument due to J-P Serre [7, p 64]: it suffices to require that the have and, in any action of on an –tree, that have a fixed point, for every and . To see this, one reduces to the case and argues that if the fixed point sets of and did not intersect then the point of closest to would be fixed by all with and , which is a contradiction.
A quite different strengthening begins with the observation that the behaviour of convex sets described in the proof of \fullreflem1 is a manifestation of the fact that trees are –dimensional objects. A suitable version of Helly’s Theorem provides constraints on the way in which convex sets can intersect in higher-dimensional CAT spaces, and by applying these constraints to the fixed point sets of finite subgroups one can prove far-reaching generalisations of \fullrefmain; this is the theme of .
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