Hi,i want to understand how fundamental group of a closed orientedby seydunas Tags: fundamental, oriented 

#1
May1711, 04:45 PM

P: 39

Hi,
i want to understand how fundamental group of a closed oriented 3mfd determines all its homology and cohomology gorups. Please can you help me. 



#2
May1711, 06:48 PM

Sci Advisor
P: 1,716

 The first homology over Z is the abelianization of the fundamental group 



#3
May1811, 02:10 PM

P: 39

Yes, i understood why we need to Poincare duality. But my central question is that abelianization of the fundamental group is the first homology group, but others? second homology group and the third? i know the second homology group is the torsion part of the fundamental group but how? And the third guy is what and why?




#4
May1811, 03:03 PM

Sci Advisor
P: 1,716

Hi,i want to understand how fundamental group of a closed oriented
Every orientable manifold has top homology group equal to Z. There is a general proof that there is a fundamental cycle
that generates the homology. Poincare duality just restates this fact for the top and bottom dimensions. For the rest, compute the first cohomology then use Poincare duality  I think 



#5
May1811, 03:39 PM

P: 39

i just know the poicare duality and i use it when i need to get cohomology group of any manifold. Bu you said that compute the first cohomology group to get second homology group of oriented three manifold. i am not sucssesful doing such thing, how can you compute it? Or is there any simple way to figure out the torsion part of the fundamental group.




#6
May1911, 09:01 AM

Sci Advisor
P: 1,716





#7
May1911, 01:15 PM

P: 39

Ok, i know the universal coefficient theorem says that if g: H^1 (M)> Hom( H_1 (M)>Z) then g is surjective. It means that H^1(M) / ker(g) isomorphic to Hom( H_1 (M)>Z) and ker(g) is torsion part of H^1(M). So what is H^1 (M)?




#8
May1911, 05:33 PM

Sci Advisor
P: 1,716

In that case H^1 (M) is isomorphic to the free part of H_1 (M). This is not what you said so I am worried that this is wrong. 



#9
May2011, 05:02 AM

P: 39

In fact you are right, because the H^1 (M) will be the free part of the H_1(M). I was on the wrong way. Ok now the picture is more visible:) But torsion part of H^1(M)=ker(g) is Ext(H_0(M), Z)? i dont know the homological algebra. If i am convinced it , question will be comlpleted for me.



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