From: Severian Subject: Re: Mayer Vietoris sequence ; Calculation of Homology groups. Date: Mon, 22 May 2000 20:52:32 +0100 Newsgroups: sci.math To: Pete Bartlett Summary: [missing] Pete Bartlett wrote: > > During my revision for wretched exams I came across the following question: > > "Find the homology groups H_i(S^2 x S^2) for all integers i" > > The only tool I know about for doing this sort of thing is to use the > Mayer-Vietoris sequence for simplicial complexes. I know S^2 is triangulable > so that's Ok. There's also the Kunneth theorem. As every homology group in sight is torsion-free, the Tor part of the exact sequence will vanish, and we just get the tensor product. Obviously H_0 and H_4 are Z while H_2 is Z^2 and the rest are zero. > Writing S^2 x S^2 = K = S^2 union S^2 with 1 point P in the intersection I Do you really, really want S^2 x S^2 = K here? > believe we have: > > ........----> H_n+1(K) ----> H_n(P) ------> H_n(S^2) + H_n(S^2) ------> > H_n(K)---->.... > > Now I know H_i(P) is trivial unless i=0 in which case the group is > isomorphic to the integers. > I think H_i(S^2) is trivial unless i=0 or 2 when we again have the integers. Unless you *know* this then computing the homology of S^2 x S^2 will be a bit difficult! > I find that H_i(K) is trivial for i<0 and i>2. The i=2 case is OK too. > > However for H_1 and H_0 I get the following exact sequence: > > 0 ---> H_1(K) ----> Z -----> Z+Z -----> H_0(K) -----> 0 > > I have messed around this and can't get anywhere. In fact, using just this > sequence, I don't believe I can detemine H_1 and H_0 uniquely (i.e. upto > isomorphism). Does this mean that the there is more information to be had > from the Mayer-Vietoris sequence than just "the sequence exists". Do I know > anything more about the homomorphisms in the sequence? The proof of the MV > theorem in my notes appears to indicate that I should! Unfortunately I don't > understand it! Think about what the Z and Z+Z are in this sequence. Z+Z is H_0(S^2) + H_0(S^2) which are copies of Z. Of course H_0(K) is isomorphic to Z as K is path connected. With the natural identifications the final surjection in the exact sequence is (a,b) |-> a+b. Its kernel is {(a,-a)} and so this is the image of the middle map in the sequence which must be injective (also this can be seen by considering the definitionss of the maps in the MV sequence). Hence H_1(K) = 0. If you's used reduced homology instead the exact sequence would have become 0 ---> H_1(K) ----> 0 -----> 0 -----> 0 -----> 0 :-) So now to compute the homology for S^2 x S^2 ! -- Severian --------------------------------------------------------------- Victor Meldrew: "The police can use sperm now as a way of fingerprinting people." Mrs Warboys: "I don't see what was wrong with the old inkpads." --------------------------------------------------------------- David Renwick, _One Foot in the Grave_