Van Kampen's theorem and fundamental groups

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I didn't see a topology forum, so I thought I'd post this question here. Can anyone give any pointers on using van Kampen's theorem? I understand the basic way it works, decompose a space X into open, path-connected sets, say U and V.

Then pi1(U) * pi1(V) = pi1(X)/N, where N is a normal subgroup. The N is what gives me trouble. How do calculate what the normal subgroup is?

Any help is appreciated!
 

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  • #2
matt grime
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It depends on what the decomposition is. It is more complicated to describe than do. Let me recap a bit:

Effectively you decompose T into two smaller sets U and V such that you know the fund. grps of U and V and UnV. Obviously, any loop in UnV defines a loop in U and a loop in V.

Then the fund grp of T is the (free) product of the fundamental groups of U and V modulo the relations imposed by the fact that loops in the overlap are identified.

Those relations are the "modulo N" part.

I guess this isn't a direct answer, but then I wouldn't calculate N, i'd describe explicitly the relations on the overlap, that implicitly tell you what N is.

Let's do an example where we already know the answer: the torus.

Let U be the torus less one point - the punctured torus.

Let V be a disc around that puncture point.

U can then be deformation retracted to the bouquet of two circles, thus its fundamental group is the free group on two generators F_2. Let the gens be g and h, we'll explain why in a second.

V has trivial fundamental group

UnV is homotopic to a circle. It's fundamental group is Z; let t be some (homotopy class of) loop generating this group. In V this is sent to the identity, as V is contractible. In U this loop is sent to the path ghg^{-1}h^{-1}. To see this, imagine the torus as the square with opposite sides identified. U is then this square with, the centre missing, thus the punctred torus retracts to the boundary of the square, which identifies to give two circles joined at a point. A loop around the hole retracts to be a path that goes along the top edge of the square, down the side along the bottom and up the other side again, right? So it is a path around one circle, then around the other, back along the first circle in the opposite direction, and then along the second circle, again in the opp. direction, ie ghg^{-1}h^{-1} in the fundamental group.


Phew, this is complicated to write, but draw a diagram to see what's going on.

Anyway, by the van Kampen Theorem, the fundamental group is

(F_2)*(e) = F_2 modulo the relation that ghg^{-1}h^{-1}=e; we identify those loops we described.

This means that we make gh=hg, ie abelianize F_2=ZxZ, so N in this case would be the commutator subgroup of F_2.


It sounds more complicated than it is, honest.

N is the subgroup generated by the relations we impose because of the overlap.
 
  • #3
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Sorry to resurrect an old thead. Thanks for the above explanation, it does help some. I tried copied your method for the double torus.

Let U=double torus - {point}
let V=Disk around the point.

Everything is open and connected like it should be and U u V = X = double torus.

U deformation retracts to a wedge of 4 circles--it's fundamental group is a free group with 4 generators, a1, b1, a2, b2, say. V is contractible.

U n V is homotopic to a circle. So tracing along the circle, we get a_1b_1a_1(^-1)b_1(^-1)a_2b_2a_2(^-1)b_2^(-1).

Okay, here's where its gets iffy for me. Since V is contractible, we equate the above string to e?

Then I get a_1b_1a_1(^-1)b_1(^-1)a_2b_2a_2(^-1)b_2^(-1)=e

a_1b_1a_1(^-1)b_1(^-1)=b_2a_2b_2^(-1)a_2^(-1)

So what is this?

From Hatcher's book I see that the presentation should be <a1,b1,a2,b2|[a1,b1][a2,b2]>

I guess my problem is figuring out what the relations should be.

Also, If V was not contractible, how would I proceed?


ETA:
[a1b1][a2b2] in the presentation just means multiplying the commutators together? If so, then I think I'm starting to get it. Still not sure what to do if fundamental group of V isn't the identity, though.
 
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  • #4
mathwonk
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all compact connect oriented surfaces of genus g are most easily computed as riemann did, by cutting open the surface into a 4g sided polygon.

for a geuns one torus we get generators a,b, and one relation going around the outside of the rectangel, namely the commutator a b a^-1 b^-1, as matt said

in general, the edges of the polygon seem to give the one relation
a1 b1 a1^(-1) b1^(-1).......ag bg ag^-1 bg^-1. not what you have above. ah yes you misread hatcher page 51.

as often happens this problem is easier done without the machinery of van kampens theorem, although topologists love this theorem and their are no doubt many situations where it is essential.

actually hatcher is essentially using riemanns presentation of the manifold via a 2 cell attached to a wedge of 2g circles.
 
  • #5
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mathwonk said:
in general, the edges of the polygon seem to give the one relation
a1 b1 a1^(-1) b1^(-1).......ag bg ag^-1 bg^-1. not what you have above. ah yes you misread hatcher page 51.
Unless I'm have a very stupid moment, isn't that what I wrote?
 
  • #6
mathwonk
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my mistake, that is what you wrote. then what is your question?

i.e.hatcher's presentation is what a_1b_1a_1(^-1)b_1(^-1)a_2b_2a_2(^-1)b_2^(-1)=e
means.
 
  • #7
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If the intersection of the decomposed spaces is a simple space, like S^1, and I know how to construct the original space from a polygon, like a genus 1 or 2 torus, then I think I can apply van kampen. That takes care of all orientable and nonorientable surfaces.

But what if it isn't so easy? One problem in Hatcher that I've been working on is show the fundamental group of R^2 - Q^2 is uncountable.
 
  • #8
matt grime
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That's a straight forward result, and doesn't need Van Kampen.

Pick irrational numbers r<s<t, and another irrational x consider the square loops round the points

(r,x) to (s,x) to (s,x+1) to (r,x+1) to (r,x)

and the same with s replaced by t. These loops are not homotopic, and thus we can easily find an uncountable number of non-homotopic loops.

As mathwonk said, Van Kampen is all well and good, but is almost always useless except in the nice cases.
 

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