Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with universal theorem of coefficients for homology

  1. Sep 9, 2007 #1
    *note before reading: pretend that the superscripts are really subscripts, for some reason , Latex is making the subscripts into superscripts*

    I'm not sure if I'm interpreting the universal coefficient theorem for homology correctly. Let's say I have a Homology group H(X;Z). This is a homology group using ring of integers. Now, let's say I want to compute Homology over the Field [tex]Z[/tex][tex]_{2}[/tex] (integers modulo 2), in order to simplify matters. By the way, I am trying to calculate the Betti numbers for a simplicial complex. Now, let's say I have a matrix M whose nullspace is isomorphic to the homology group. Now, I want to make all of the entries in the matrix modulo 2, if possible, in order to greatly simplify calculations.

    So far we have:
    kerM = H(X; Z)

    Now, the universal theorem gives us this: 0[tex]\rightarrow[/tex][tex]H[/tex][tex]_{k}[/tex]([tex]X[/tex]) [tex]\otimes[/tex]G[tex]\rightarrow[/tex][tex]H[/tex][tex]_{k}[/tex](G)[tex]\rightarrow[/tex]Tor(H[tex]_{k+1}[/tex]([tex]Z[/tex]),G)[tex]\rightarrow[/tex]0

    Basically, in this case, we can replace G with [tex]Z[/tex][tex]_{2}[/tex]. Also, the torsion should go to zero, since we have a free group. Now, since kerM is isomorphic to [tex]H[/tex][tex]_{k}[/tex]([tex]X[/tex]), can I say that ker(M)[tex]\otimes[/tex] [tex]Z[/tex][tex]_{2}[/tex] is isomorphic to [tex]H[/tex]([tex]X[/tex])[tex]\otimes[/tex] Z2= H(X;Z[tex]_{}2)[/tex][tex]Z[/tex][tex]_{2}[/tex] ? Can i also say that Ker(M) [tex]\otimes[/tex] Z[tex]_{}2[/tex] is the same as Ker(M[tex]\otimes[/tex]Z[tex]_{}2[/tex]) ?

    Back to my goal again, I want to make the entries in the matrix M, integers modulo 2. However, I realize that the nullspace of M modulo 2 and and the nullspace of M are different. By applying the universal coefficient theorem, can I say that these two nullspaces are isomorphic? If this universal coefficient theorem is correct, can I just convert the entries of the new matrix ([tex]M[/tex][tex]\otimes[/tex][tex]Z[/tex][tex]_{2}[/tex]) into 1's and 0's? It seems I would need to tensor the matrix [tex]M[/tex] with [tex]Z[/tex][tex]_{2}[/tex], but what would [tex]Z[/tex][tex]_{2}[/tex] in matrix form be? The identity matrix with modulo 2 entries? (basically just ones on the diagonal?). If I am totally wrong, can anyone suggest a method of making the nullspace of M and the nullspace of M modulo 2 isomorphic?
    Last edited: Sep 10, 2007
  2. jcsd
  3. Sep 10, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    You need to put consecutive symbols all within the same [ tex ] ... [ /tex ] tag to get the formatting right. To make it line up nicely with texts in paragraphs, you should use [ itex ] ... [ /itex ] instead of [ tex ] ... [ /tex ].
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Help with universal theorem of coefficients for homology
  1. Homology question (Replies: 2)

  2. Simplicial homology (Replies: 2)