From: "Underground Liberation" Subject: Re: some fundamental group of a topological space Date: Tue, 20 Jun 2000 12:28:05 +0200 Newsgroups: sci.math Summary: [missing] Hi Dave, I want to simplify a few things if you allowe me. Dave Rusin schreef in berichtnieuws 8ik9e6$9u8$1@gannett.math.niu.edu... > In article <8if4sc$na2$1@spectra.a2000.nl>, > Underground Liberation wrote: > >If (Y,p) is a universal coveringspace of a topologocal space X, (where > >p:Y-->X is the natural transformation; continues, open & surjective) we have > >an isomorphism between Aut(p^(-1)(x)) and the fundamental group of X with > >basepoint x. > > > >I wonder if there is a topological space X according to the above written > >properties where the fundamental group is isomorphic with SL(F_q, n), a > >simple finite Lie group? (q is a prime number and n is any natural number) > > Yes. To any finite group G (indeed, any discrete group) we may associate a > space BG which plays the role of your X. (The space is uniquely > determined, up to homotopy equivalence, by the additional condition that Y > be not only simply connected but contractible,) > > There is a construction of this space which is fairly natural, although I > don't think I've seen it anywhere with quite this level of concreteness. > Let me outline the steps for you. Note that what we really do is construct > Y, a space on which G acts, and then define X to be the quotient space > X = Y/G. > It is a good start to consider a topological group Y and a discrete normal subgroup N of Y. Then we can define a natural projection p: Y ---> Y/N. This is well defined. Now the thing is that if (Y,p) is a coveringspace of Y/N, then N = ker(p). (you can compare this situation with how you build up the arguments for the first isomorphism theorem) So we have a coveringspace (Y,p) of Y/(ker(p)). To make the covering universal, Y must be simply connected. (I'm not sure if you need the extra condition of Y to be contractible. It depents on how your construction looks like) If we take the unit element of Y/(ker(p)), wich is ker(p), to be the base point of Y/(ker(p)), then p^(-1)(ker(p)) = ker(p). OK. So we have an isomorphism Aut(ker(p)) --> PI_1(Y/(ker(p)), ker(p)). This looks nice. We must have an isomorphism Aut(ker(p)) --> SL(F_q,n). If we know the order of SL(F_q,n), then we know the order of ker(p). And then we know how many times Y/(ker(p)) is covered. (you know that there is a 1-1 correspondence with the fiber and the monodromy group of the covering) > Here is Y: First view G as the set of standard basis vectors in R^n > where n is the order of G. Thus G is a (discrete) topological space. > Now let L be the set of piece-wise continuous functions f:[0,1] -> G > which have only a finite number of discontinuities. Well, that's a > little high-falutin', since G is discrete: f is really a > piece-wise _constant_ function! Give L a topology by viewing it as > a subspace of the function space L^1( [0,1], R^n ). As you probably know, > the elements of this space are not functions, really, but rather > a.e.-equivalence classes of functions; so it isn't really L which > is a subspace of this function space; it's a quotient space of L, which we > will call Y. (Actually it's usually called EG .) So that's Y: > the set of (equivalence classes of) piecewise-constant functions > f: [0,1] -> G having only a finite number of discontinuities; the > topology is given by the metric d(f1, f2) = \int_0^1 || f1(t) - f2(t) || dt. > > The group G acts on Y pointwise, that is, by (g(f))(t) = g*(f(t)), > where * is the group action on G. (Recall the values of f are > restricted to just the n elements of R^n which we identify with the > elements of G.) We let X be the quotient space X = Y/G. > > That's really all you need to do to construct X, but there are many > things which can be checked, first to verify that Y is the universal > covering space of X with covering group G, and then to derive other > interesting properties. Here are some facts you can check as exercises. > > (1) In Y, d(f1, f2) = sqrt(2) * (measure of set where f1(t) \not= f2(t) ) > > (2) Elements of G act as isometries, and in particular the action > of G in Y is continuous. > > (3) g(f) = f iff g = 1. Indeed, if B is the ball of radius sqrt(2)/2 > around any fixed f in Y, then g(B) \intersect B = \emptyset unless > g = 1. > > (4) Y is contractible (hint: use H(f,u)(t) = {f(t), if tu}) > > (5) For any sequence g0, g1, ..., g_k in G, such that g_i is > different from g_{i+1} for every i, we define the "cell" > [g0 | g1 | ... | g_k] to be the set of elements of Y which take > on precisely the values g0, g1, ..., g_k as we traverse the domain > from 0 to 1 . (So e.g. [ g ] contains only the constant function > f(t) = g for all t; [g0 | g1] is the set of all functions of the > form f(t) = { g0, if t < u; g1, if t > u } and there are as many of > these as there are elements u in [0,1].) Then [g0 | ... | gk] > is homeomorphic to an open k-simplex; its closure has boundary > consisting of the union of the cells [g0 | ... g_{i-1} | g_{i+1} | ... | g_k] > (i=0, 1, ..., k). This has to be treated carefully when g_{i-1} = g_{i+1}. > Thus Y is a CW complex. > > (6) G acts on Y cellularly; thus when passing to the quotient, X > also inherits the structure of a CW complex. (It has only one 0-cell.) > > (7) If G = {1, a} is a group of order 2, then Y has two k-cells, > conjugate under the action of G: [1|a|1|a|...] is homeomorphic to > the k-simplex, i.e. the k-ball. The k-skeleton is isomorphic to the > standard "hemispheric" construction of the k-sphere. Carried over to this > model, the action of G is the antipodal map. Thus X in this case > is the infinite projective space RP^\infty. > > (8) Y is a group under pointwise group operations. The set of constant > functions (the 0-skeleton of Y) forms a subgroup of Y isomorphic to G. > If G is Abelian, Y is also Abelian, so G is normal in Y and > X=BG = Y/G is also a group. > > (9) A homomorphism phi : G -> H induces a map phi_* : BG -> BH by > acting on functions pointwise. For example, if phi is the inclusion > of a subgroup, phi_* is the inclusion of a subspace. If phi is a > surjection, then phi_* is a fibration with fibre BN. > > Let me remark that (as this construction probably illustrates) I find it > much more fruitful to think of coverings by thinking of Y first and > then viewing X as the quotient of Y modulo some equivalence relation > determined by a group action. _You_ may think of a surface of genus 2 > as some sort of pretzel; _I_ think of it as the hyperbolic disk tiled > by octagons. (But of course, my kids have always maintained I'm a > little strange...) > > dave ============================================================================== From: "Charles H. Giffen" Subject: Re: some fundamental group of a topological space Date: Wed, 21 Jun 2000 14:34:07 -0400 Newsgroups: sci.math To: Ed Hook Summary: [missing] Ed Hook wrote: > > In article <8if4sc$na2$1@spectra.a2000.nl>, > "Underground Liberation" writes: > |> If (Y,p) is a universal coveringspace of a topologocal space X, (where > |> p:Y-->X is the natural transformation; continues, open & surjective) we have > |> an isomorphism between Aut(p^(-1)(x)) and the fundamental group of X with > |> basepoint x. > > |> I wonder if there is a topological space X according to the above written > |> properties where the fundamental group is isomorphic with SL(F_q, n), a > |> simple finite Lie group? (q is a prime number and n is any natural number) > |> You can also choose some q and n if there is no general result. > > Given any finitely-presented group (so, in particular, > any _finite_ group), there is a construction of a closed > 4-manifold having that group as its fundamental group. > (Here, '4' can be replaced by any integer greater than > 3.) If you take X above to be the manifold constructed > for SL(F_q,n) and Y to be X's universal cover, you get > an example that meets your requirements. > > There are similar (typically, easier) constructions that > build a CW complex or (the geometric realization of) a > simplicial complex, instead of a manifold. > > I have a feeling that you wanted something a bit more > concrete, but I don't know of a "naturally occurring" > example of the sort you're seeking ... > > -- > Ed Hook | Copula eam, se non posit > Computer Sciences Corporation | acceptera jocularum. > NAS, NASA Ames Research Center | All opinions herein expressed are > Internet: hook@nas.nasa.gov | mine alone Indeed, for any finite presentation of a group, there is a cell complex with one vertex plus one 1-cell for each generator in the presentation plus one 2-cell for each relator in the presentation --- and the fundamental group of this 2 dimensional complex is canonically isomorphic to the group being presented. Alternatively, for any group G, let X = G*G*G equal the three-fold join of copies of G with itself. Then X is simply connected and G operates freely on X in the evident manner (by translation), so the orbit space X/G has fundamental group G, and the projection X -> X/G is a universal covering. If you iterate the join, getting Y = G*G*G*... then in the same fashion you get a contractible Y on which G acts freely, so that Y -> Y/G is a universal principal G bundle, and Y/B is thus a model for BG, the "classifying space" of G. --Chuck Giffen