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Coolphreak
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*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 Z_{2} (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\rightarrowH_{k}(X) \otimesG\rightarrowH_{k}(G)\rightarrowTor(H_{k+1}(Z),G)\rightarrow0
Basically, in this case, we can replace G with Z_{2}. Also, the torsion should go to zero, since we have a free group. Now, since kerM is isomorphic to H_{k}(X), can I say that ker(M)\otimes Z_{2} is isomorphic to H(X)\otimes Z2= H(X;Z_{}2)Z_{2} ? Can i also say that Ker(M) \otimes Z_{}2 is the same as Ker(M\otimesZ_{}2) ?
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 (M\otimesZ_{2}) into 1's and 0's? It seems I would need to tensor the matrix M with Z_{2}, but what would Z_{2} 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?
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 Z_{2} (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\rightarrowH_{k}(X) \otimesG\rightarrowH_{k}(G)\rightarrowTor(H_{k+1}(Z),G)\rightarrow0
Basically, in this case, we can replace G with Z_{2}. Also, the torsion should go to zero, since we have a free group. Now, since kerM is isomorphic to H_{k}(X), can I say that ker(M)\otimes Z_{2} is isomorphic to H(X)\otimes Z2= H(X;Z_{}2)Z_{2} ? Can i also say that Ker(M) \otimes Z_{}2 is the same as Ker(M\otimesZ_{}2) ?
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 (M\otimesZ_{2}) into 1's and 0's? It seems I would need to tensor the matrix M with Z_{2}, but what would Z_{2} 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?
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