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I'm reading "Principles of algebraic geometry" by Griffiths and Harris.

While reading the first chapter, I keep running into the same problem, which I'll illustrate using some examples:

Proven: "On a real $C^\infty$ manifold $H_{DR}^*(M)\cong H_{sing}^*(M,\mathbb{R})$".

Used: "Using the de Rham isomorphism $H_{DR}^*(M)\cong H_{sing}^*(M,\mathbb{C})$"

Proven: "$H^*(K,\mathbb{Z})\cong \check H^*(M,\mathbb{Z})$"

Used: "$H^*(K,\mathbb{R})\cong \check H^*(M,\mathbb{R})$"

Proven: "$H_k(M,\mathbb{Z})\times H_{n-k}(M,\mathbb{Z})\to \mathbb{Z}$ is unimodular"

Used: $H_k(M\mathbb{Q})\cong H^{n-k}(M,\mathbb{Q})$

I don't understand this. I have two questions about this:

  • Why does this reasoning make sense? Why can be prove the statement for say $Z$, and use it with $\mathbb{Q}$
  • Why does this way of working make sense? Why prove something for $\mathbb{Z}$ in the first place, if you are only going to use it applied to $\mathbb{Q}$
user2520938
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    Do you know the universal coefficients theorem ? – Captain Lama Apr 22 '16 at 21:46
  • @CaptainLama Not really (just heard of it). Will this solve my problems? – user2520938 Apr 22 '16 at 21:47
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    Yes, it will. Basically if $A$ is a torsion-free group (and for instance if $A$ is actually a field) $H_(X,\mathbb{Z})\otimes A \simeq H_(X,A)$ (the theorem is more powerful than that). – Captain Lama Apr 22 '16 at 21:50
  • Griffiths and Harris is a bad place to learn about cohomology. Anyway, like Captain Lama said, what you want is the universal coefficient theorems. – Qiaochu Yuan Apr 23 '16 at 05:35
  • @QiaochuYuan To apply the UCT to the first statement (about the de Rham cohomologies), we will need http://math.stackexchange.com/a/1142827/95162 and the statement about how $\Gamma (E)\otimes \Gamma (B)=\Gamma(E\otimes B)$ for smooth manifolds, to make sure we have the appropriate relation between $H_{DR}^p(M,\mathbb{R})$ (real valued forms) and $H_{DR}^p(M,\mathbb{C})$ (complex valued forms), right? If we didn't have that we would be able to take the tensor out of the sections, so we wouldn't have something in a form we can apply UCT to. – user2520938 Apr 23 '16 at 08:27
  • @QiaochuYuan Also, do you have a suggestion for other sources that cover more of less the same material as the first chapter of Griffiths and Harris? I've been using Hatcher and R.O. Wells (differential analysis on complex manifolds), but more suggestions are always welcome. – user2520938 Apr 23 '16 at 08:28
  • @CaptainLama Could you please take a look at my follow up question http://math.stackexchange.com/questions/1755348/why-is-h-drpm-mathbbc-cong-h-drpm-mathbbr-otimes-mathbbr-mat?lq=1 – user2520938 Apr 23 '16 at 12:30
  • @QiaochuYuan I created a seperated question for this follow up, see http://math.stackexchange.com/questions/1755348/why-is-h-drpm-mathbbc-cong-h-drpm-mathbbr-otimes-mathbbr-mat?lq=1 – user2520938 Apr 23 '16 at 12:31

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In intuitive terms, results about $\mathbb{Z}$ are the most general. They are in a sense arithmetic with all the subtleties that implies. Results about $\mathbb{Q}$ or $\mathbb{R}$ or really any field are much cruder. The arithmetic subtleties are lost and all considerations of torsion are erased. Of course in some situations this is just what is needed. They are algebraic rather that arithmetic. If you look at all of the transfers in the question they are from a more to a less general situation. Formally of course you use the universal coefficients as the comments have pointed out.

Rene Schipperus
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