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First of all, my background in Homology and Algebraic Topology is very limited and doesn't include much except for the basic concepts; I'm trying to understand some Homology techniques as applications relevant to a course in combinatorics I'm taking.

Anyway, in class we've seen the example of computing the homology groups of $\mathbb{P}^2$ - The projective plane over $\mathbb{R}$. It is known that $\beta_0(\mathbb{P}^2 ; \mathbb{F})=1$ since the projective plane is connected. Now, it turns out that $H_2(\mathbb{P}^2 ; \mathbb{Q})=0$ while $H_2(\mathbb{P}^2 ; \mathbb{F}_2) \cong \mathbb{F}_2$, but the lecturer hasn't really explained how it is computed.

Any explanation on the computing method would be greatly appreciated, as will be any intuition on this whole concept, or references to further reading (I know Hatcher's book on Algebraic Topology, but didn't find it very helpful).

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3 Answers 3

up vote 6 down vote accepted

It turns out that the homology groups $H_{\bullet}(X,\mathbb{Z})$ of a space $X$ with $\mathbb{Z}$-coefficients determine the homology groups $H_{\bullet}(X,A)$ with coefficients in any abelian group $A$. The key result here is the (very well named!) Universal Coefficient Theorem.

The basic idea is that our first guess at $H_i(X,A)$ is simply $H_i(X,\mathbb{Z}) \otimes A$. This is a good first guess in that in all cases there is a natural injective map $H_i(X,\mathbb{Z}) \otimes A \rightarrow H_i(X,A)$. As you have seen, this map need not be an isomorphism. The universal coefficient theorem tells you that its cokernel is $\operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A)$ and also that the sequence is (non-canonically) split, i.e.,

$$H_i(X,A) \cong (H_i(X,\mathbb{Z}) \otimes A) \oplus \operatorname{Tor}(H_{i-1}(X,\mathbb{Z}),A).$$

Here $\operatorname{Tor}( \ , \ )$ is the first "Tor group" of homological algebra. It may well be that you don't know what this gadget is. (I didn't when I first learned algebraic topology.) So I found it helpful to write down a "cheatsheet" for $\operatorname{Tor}(X,Y)$ when $X$ and $Y$ are both finitely generated abelian groups. Indeed, since $\operatorname{Tor}$ is bi-additive and symmetric, it is enough to know that for all $m,n \in \mathbb{Z}^+$,

$\operatorname{Tor}(\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) = 0$ and
$\operatorname{Tor}(\mathbb{Z}/m\mathbb{Z},\mathbb{Z}/n\mathbb{Z}) \cong \mathbb{Z}/\operatorname{gcd}(m,n) \mathbb{Z}$.

As a first exercise, try to use all this information to confirm that the homology groups of $\mathbb{R} \mathbb{P}^2$ with $\mathbb{Z}/2\mathbb{Z}$-coefficients are as you said.

(There is also a Universal Coefficient Theorem for cohomology in which the correction term involves $\operatorname{Ext} = \operatorname{Ext}^1$ instead of $\operatorname{Tor}$...)

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I think that we can get along just fine without spectral sequences or the universal coefficient theorem :) (even though both of those are probably more useful in general- it's nice to actually do a real computation at least once in your life!).

For the case of $\mathbb{R}P^2$ we happen to have a nice description of this as a $\Delta$-complex (I'm assuming you saw this in Hatcher). So you can compute the homology groups just as Hatcher described for a $\Delta$-complex in general, except that when he uses $\mathbb{Z}$ (i.e. looking at the "free abelian group generated by") just replace it with $\mathbb{Z}/2$ or $\mathbb{Q}$ and look at the vector space.

More specifically: Given a $\Delta$-complex $X$, let $\Delta_n(X, F)$ be the vector space with basis the open $n$-simplices of $X$. Now look at Hatcher's Example 2.4 and see what happens with these new coefficients!

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My favorite way to compute the (co)homology is using the Serre spectral sequence, but I think that is a bit of over kill for this problem. I think the best way to do it would be to come up with a model for $\mathbb{R}P^2$ as a cell complex and then compute the cellular homology. To do this look at the model for $S^2$ that has a 2 cells in every dimension (less than or equal to 2). Now $\mathbb{R}P^2$ is a quotient of $S^2$ by a free $\mathbb{Z}/2$ action so we identify use as a model for $\mathbb{R}P^2$ the quotient of the model for $S^2$ under this action. Then we compute the cellular Homology. Tensor the complex with whatever coefficients you like and then take homology.

I will give you some time to think about the model for $S^2$ and come back later with more.

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@Sean - Thank you for your answer, but unfortunately I know nothing about cellular homology - only simplicial homology. –  Pandora Feb 11 '11 at 14:01
    
maybe I will add a bit of that as well. Cellular homology confused me a bit at first, but it is really helpful, it shows you how much control the homology has over the cell structure. –  Sean Tilson Feb 11 '11 at 14:04
    
@Sean: What do you mean when you say "Cellular homology [...] shows you how much control the homology has over the cell structure."? –  Rasmus Feb 11 '11 at 15:52
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I am always amused when someone opens an answer to a question like «how do I compute the homology of the projective plane» with a reference to spectral sequences :D –  Mariano Suárez-Alvarez Feb 11 '11 at 17:22
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@Sean, and computing the homology of projective planes by hand is, as dirty goes, pretty clean :) –  Mariano Suárez-Alvarez Feb 12 '11 at 16:47

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