Help with universal theorem of coefficients for homology

[Coolphreak]

*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$$\rightarrow$$$$H$$$$_{k}$$($$X$$) $$\otimes$$G$$\rightarrow$$$$H$$$$_{k}$$(G)$$\rightarrow$$Tor(H$$_{k+1}$$($$Z$$),G)$$\rightarrow$$0

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$$\otimes$$Z$$_{}2$$) ?

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 ($$M$$$$\otimes$$$$Z$$$$_{2}$$) 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?

 

[Hurkyl]

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 ].

 

[tacman]

The universal coefficient theorem is definitely a useful tool for computing homology groups with different coefficient rings. In your case, you are correct in saying that the torsion term will go to zero since we have a free group. And yes, you can replace G with Z_{2} in the theorem to get the desired result.

As for your question about the nullspaces, the theorem states that the kernel of M is isomorphic to H_{k}(X) and the kernel of M tensor Z_{2} is isomorphic to H_{k}(X; Z_{2}). So, in a sense, they are "equivalent" and can be thought of as the same group. However, they are not exactly the same and cannot be directly compared or manipulated in the same way.

To make the entries in the matrix M modulo 2, you can indeed tensor M with Z_{2}. This would result in a matrix with entries in Z_{2} instead of Z. The identity matrix with modulo 2 entries would be a good representation of Z_{2} in matrix form. And yes, you can convert the entries of the new matrix (M tensor Z_{2}) into 1's and 0's to simplify calculations.

I hope this helps clarify things for you. If you have any further questions or need more assistance, don't hesitate to reach out for help. The universal coefficient theorem can be a bit tricky to grasp at first, but with practice and understanding of its applications, it can be a powerful tool in homology computations.