From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Periodic linear group Date: 2 Feb 1998 16:35:40 GMT In article , mjoao wrote: >It is possible to find a finite dimension vector space V over >an infinite field F, such that Aut(V) --the automorphism group-- >is periodic. No, you can't do it with F infinite, and even if F is finite, Aut(V) is only periodic if dim(V/F) is finite. For any V we have the automorphisms f(v) = a*v for all a in F-{0}. The statement that this f is periodic is equivalent to the statement that a is a root of unity. In particular, if Aut(V) is to be periodic, F must be algebraic over its prime field. Moreover, F cannot contain the rational numbers. So already we have the necessary condition that F is a subfield of the algebraic closure of some F_p. Actually, we can do better: if V = F e + V_0 is a decomposition into subspaces with codim(V_0)=1, define a Frobenius automorphism f(a e + v) = a^p e + v. Then f has order dividing N iff a^(p^N)=a for all a in F, which in particular implies F is finite. So F is some Galois field F_q. No matter what the field F is, if dim(V/F) is infinite, Aut(V) is not periodic. (I'm assuming you have no topological or other restrictions on which automorphisms you allow.) For example, partition the elements of (a countable subset of) a basis into {e_1, e_2}, {e_3, e_4, e_5}, ...; then let f be the automorphism which cyclically permutes the elements in each set. This f does not have finite order. So the vector spaces you seek must be finite-dimensional spaces over a finite field F_q. Conversely, the automorphism groups GL(n,q) of these vector spaces are finite (hence periodic) groups. dave ============================================================================== Newsgroups: sci.math From: Jeremy Rickard Subject: Re: Periodic linear group Date: Tue, 3 Feb 1998 16:36:18 GMT Dave Rusin wrote: > > In article , > mjoao wrote: > > >It is possible to find a finite dimension vector space V over > >an infinite field F, such that Aut(V) --the automorphism group-- > >is periodic. > > No, you can't do it with F infinite, You can, in fact. F has to be a subfield of the algebraic closure of some F_p, as Dave shows. For such an F, *all* automorphisms of an n-dimensional vector space V over F are periodic. To see this, choose a basis for V and describe an automorphism with respect to this basis by a matrix. Since this matrix contains only finitely many entries, all the entries lie in the finite field F_q for some power q of p. Hence the matrix lies in the *finite* group GL(n,F_q) and so has finite order. > For any V we have the automorphisms f(v) = a*v for all a in F-{0}. > The statement that this f is periodic is equivalent to the statement that > a is a root of unity. In particular, if Aut(V) is to be periodic, F > must be algebraic over its prime field. Moreover, F cannot contain the > rational numbers. So already we have the necessary condition that F > is a subfield of the algebraic closure of some F_p. > > Actually, we can do better: if V = F e + V_0 is a decomposition into > subspaces with codim(V_0)=1, define a Frobenius automorphism > f(a e + v) = a^p e + v. Then f has order dividing N iff a^(p^N)=a > for all a in F, which in particular implies F is finite. So F > is some Galois field F_q. But this f isn't linear. Jeremy. ============================================================================= Date: Tue, 3 Feb 1998 16:18:59 -0600 (CST) From: Dave Rusin To: j.rickard@bristol.ac.uk Subject: Re: Periodic linear group You're quite right. I had been thinking about the inverse limits but then got worried about automorphisms of infinite order whose image over every F_q was finite. Naturally this suggested the Frobenius but as you pointed out, it's only F_p-linear, not F_q-linear. Thanks for posting the correction. Must remember to think first then post! dave ============================================================================== From: mareg@csv.warwick.ac.uk (Dr D F Holt) Newsgroups: sci.math Subject: Re: Periodic linear group Date: 3 Feb 1998 18:35:23 -0000 In article <6b4sks\$otb\$1@gannett.math.niu.edu>, rusin@vesuvius.math.niu.edu (Dave Rusin) writes: >In article , >mjoao wrote: > >>It is possible to find a finite dimension vector space V over >>an infinite field F, such that Aut(V) --the automorphism group-- >>is periodic. > >No, you can't do it with F infinite, and even if F is finite, Aut(V) >is only periodic if dim(V/F) is finite. > > >For any V we have the automorphisms f(v) = a*v for all a in F-{0}. >The statement that this f is periodic is equivalent to the statement that >a is a root of unity. In particular, if Aut(V) is to be periodic, F >must be algebraic over its prime field. Moreover, F cannot contain the >rational numbers. So already we have the necessary condition that F >is a subfield of the algebraic closure of some F_p. > >Actually, we can do better: if V = F e + V_0 is a decomposition into >subspaces with codim(V_0)=1, define a Frobenius automorphism >f(a e + v) = a^p e + v. Then f has order dividing N iff a^(p^N)=a >for all a in F, which in particular implies F is finite. So F >is some Galois field F_q. Hmmm! I guess this depends on your interpretation of what Aut(V) means. My instinct was to take it to mean GL(V), the group of linear maps V->V. With that interpretation, the answer to the problem posed is yes. Take F to be the algebraic closure of F_p. For any element in GL(V), its matrix has entries lying in some finite field, and so it has finite order. Thus GL(V) is periodic. Of course, if you take Aut(V) to mean the group \Gamma L(V) of semilinear transformations V -> V, then you are correct. Derek Holt.