Subject: Re: question and answer
From: asikora@buffalo.edu
Date: Tue, 01 Feb 2005 13:55:41 -0500

Here is an answer to Dan Grubb's question:

n(X)<=d(X) 

In fact we have a sharper inequality:
We say that a set {v_1,...,v_n} of elements of H1(X;Z) is primitive 
if 
(v_1,...,v_n): H_1(X;Z) -> Z^n is an epimorphism.
Let c(X) be the largest cardinality of a primitive set of elements 
of H1(X) such that all cup products between elements of
this set are 0.

Then for any simplicial complex,
n(X)<=c(X) 

One might generalize this result as follows:
Consider the largest number of primitive elements v_1,...,v_k in H1
(X;Z)
such that all Messay products of n-tuples of these elements 
are uniquely defined and equal 0. Denote k by d_n(X).

Let F_{n,k} denote the k-th group in the lower central series of the
free group F_n, and let f_k(G) be the largest n for which there is an
epi G->F_n/F_{n,k}.

Then f_n(\pi_1(X))<=d_n(X) 

Further details and references can be found in my paper
http://www.arxiv.org/abs/math.GT/0112106


Best Regards, Adam Sikora
_________________________________________________________

Subject: Re: question and answer
From: Allen Hatcher <hatcher@math.cornell.edu>
Date: Tue, 1 Feb 2005 14:12:12 -0500

Here's a comment on NIck Kuhn's solution to today's homework problem. 
Spectral sequences aren't really needed here, of course. 
One could start by taking a circle with a 2-cell attached by a map of degree 4. 
Then with Z/4 coefficients the square of a generator of H1 is twice a 
generator of H2 (see Example 3.9 of my book, pp. 208-209, for this elementary 
calculation) so with Z/2 coefficients the square of a generator is zero. 
This gives an example with (d,n) = (1,0). For the general case, take the 
wedge of k copies of this complex together with n circles to get a 
complex X with (d(X),n(X)) = (k+n,n). (To compute n(X), abelianize pi_1.)

Sorry about the shameless self-promotion!

Allen Hatcher

> Subject: Re: question abt sbgps of H1
> From: "Nicholas J. Kuhn" <njk4x@cms.mail.virginia.edu>
> Date: Tue, 01 Feb 2005 11:18:56 -0500
> To:
> Don Davis <dmd1@Lehigh.EDU>
>
> Here is an answer to the question sent in by Dave Rusin:
>
> With d(X) and n(X) as in the message (copied below),
>
> (1) d(X) is at least n(X), and
> (2) (d(X), n(x)) can be any pair subject to (1).
>
> Proof of (1):  (This would make a good exercise for a graduate course.) 
A surjection pi_1(X) --> F_n can be realized by a map f: X --> W(n)=wedge of
> n circles.  This maps splits, so f^* is monic, and the image of H1(W(n)) --> H1(X)
> is an isotropic submodule of dimension n.
>
> Proof of (2):  Let V be a finite dimensional Z/2 vector space (and let homology
> and cohomology have Z/2 coefficients).  There is a `universal' central group
> extension  H_2(V) --> G(V) --> V whose d_2 differential in the Serre spectral
> sequence for H^*(G(V)) is an isomorphism from H2(V)=E_2^{0,1} to E_2^{2,0}=H2(V).
>
> By construction, H1(V) --> H1(G(V)) is an isomorphism, and all products in H2(G(V))
> are zero.  Let BG(V) be the classifying space for the finite group G(V). 
Then n(BG(V))=0 (pi_1 is finite), but d(G(V)) = dimension of V.
>
> The first example is BZ/4 with d=1, n=0.  The next example, with V=Z/2xZ/2
> so BG(V) has d=2 and n=0, is the classifying space of a group of order 32
> (#18 on Jon Carlson's website).
>
> These examples can now be wedged with W(n)'s to create examples realizing any
> pair (d,n) with d at least n.
>
> Remark:  G(V) is the universal 2-group with quotient V having all elements of
> order p central in G. (a `pC' group in Alex Adem's terminology).
>
> Nick Kuhn
> University of Virginia
>
> --On Tuesday, February 01, 2005 7:00 AM -0500 Don Davis <dmd1@Lehigh.EDU> wrote:
>
> > Subject: Isotropic subgroups of H1
> > From: Dave Rusin <rusin@math.niu.edu>
> > Date: Mon, 31 Jan 2005 11:22:03 -0600 (CST)
> > One posting.........DMD
> > __________________________________________________
> >
> > My colleague Dan Grubb sent me this question to post to topologists:
> >
> > Let X be a 'sufficiently nice' space, say at least locally
> > path connected.
> >
> > An isotropic submodule M of H1 (X) is one where x,y \in M
> > imply that x\cup y=0 in H2(X).  Let d(X) denote the maximal
> > rank of an isotropic submodule of H1(X) (coefficients in the
> > integers mod 2 although integers would be good for oriented
> > manifolds).
> >
> > Now let n(X) denote the largest integar so that there is a
> > surjection from pi_1 (X) to the free group on n(X) letters.
> >
> > How do d(X) and n(X) relate?
> >
> > I know that (with Z/(2) coefficients), they are equal for
> > 2-dimensional closed manifolds. Also, if U and V are connected
> > open sets which cover X, then the number of components
> > of U\cap V is at most 1+max{d(X),n(X)}. Equality is preserved
> > by joins, but I am not sure if it is by products.
> >
> > --Dan
> >
>
>
>
>
>