From: rusin@math.niu.edu (Dave Rusin)
Newsgroups: sci.math.research
Subject: Re: Seifert/Van Kampen Theorem
Date: 26 Mar 1998 04:50:51 GMT
In article ,
MS wrote:
>Let /\ mean intersection, and U be union.
>
>I have a question about the Seifert\Van Kampen theorem.
>Let A and B be 3-manifolds (with boundary). Let's say x is
>a generator of A /\ B. I cannot figure out what the image of
>x is under inclusion into A, but I can tell that x^n goes to
>a word a in Pi1(A). If inclusion into B gives x = b, can I just
>list b^n = a as one of the relators of Pi1(A U B), along with
>the others?
Algebraically, you're asking the following. Given two groups G and H
and maps from a cyclic group g: -> G and h: -> H, you wish
to compute the free product with amalgamation X = G *_{} H. You already
know that if G = < gens | rels > and H = are
presentations of these groups, then
X = < gens U gens' | rels U rels' U {g(x)=h(x)} >
You're hoping to replace this with the group presentation
Y = < gens U gens' | rels U rels' U {g(x)^n=h(x)^n} >
for some n. Unfortunately X and Y are not in general isomorphic.
There is a natural surjection Y -> X, but there's no reason to expect it
to be an isomorphism, and indeed it's easy to write down examples in
which X and Y are non-isomorphic, e.g.
G = Z/2, H = {1}, g=surjection, n=2
gives |X| = 1, |Y| = 2.
This example can be realized geometrically as well, even just with
2-manifolds with empty boundary (cross with the interval if you want
3-manifolds): let M be a Moebius strip; identify opposite points of
its (circular) boundary to form A = RP^2. Also let B be the unit sphere
in R^3. Identify the center line of M (in A) with the equator of B.
The resulting union of A and B is simply connected, but you'd be
presenting its fundamental group as = pi1(A) = Z/2Z since a = b = identity.
for n=2.
Perhaps I've misunderstood the geometric constraints under which
you're operating?
dave