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I don't see the point in using homology and cohomology with coefficients in the field $\mathbb{Z}/2\mathbb{Z}$.

Can you provide some examples for why this is useful?

fosco
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    One quick reason: a manifold is always Z/2Z orientable. This allows results like Poincare duality to work. –  Jun 23 '11 at 08:21
  • Thanks! Where can I find a reference for this? I tried in Bott-Tu (Differential Forms etc.) and Irapidly noticed that the third book of Fomenko "Modern Geometry" spends lots of words about Z/2 cohomology, but I'm lacking a precise place. – fosco Jun 23 '11 at 08:29
  • Have you checked in Massey's Algebraic Topology book, or Bredon's Geometry and Topology book? –  Jun 23 '11 at 08:36
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    Another (more technical) thing is that it's easier to work in char 2, because you can ignore all signs — which makes many computations much, much easier (say, try to compute homology — or just Euler char — of a real Grassmannian...). – Grigory M Jun 23 '11 at 08:48
  • I asked a similar question - why do we use different coefficient's at all - and there are some interesting responses here:

    http://math.stackexchange.com/questions/37148/applications-of-universal-coefficient-theorem

    Have a look at Section 3.3 of Hatcher's Book for a nice discussion of Poincare Duality and $\mathbb{Z}/2\mathbb{Z}$ coefficients.

     – Juan S Jun 23 '11 at 12:45
  • Thanks a lot to everybody, I found a couple of good examples in your answers: is there a way to "accept" a comment? – fosco Jun 25 '11 at 21:12

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  1. A lot of times it's easier to work with $\mathbb Z_2$ coefficients. You don't have to worry about sign and calculations are easier.

  2. Field coefficients are nice because there is a precise duality between homology and cohomology, without worrying about the universal coefficient theorem.

  3. Knowledge of $\mathbb Z_2$ coefficients can help you say things about integral homology. For example, if $H_i(X;\mathbb Z_p)=0$ for all primes $p$, then $H_i(X;\mathbb Z)=0$.

  4. Some spaces have a nicer presentation with $\mathbb Z_2$ coefficients. The cohomology ring $H^*(\mathbb{RP}^n;\mathbb Z_2)$ is isomorphic to $\mathbb Z_2[x]/x^{n+1}$, whereas the integral cohomology ring is messier to write down.

Cheerful Parsnip
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Here is another one to add to the list: I am currently reading about cohomology operations, in particular the Steenrod squares.

These are functions

$$\operatorname{Sq^i}:H^k(X,\mathbb{Z}/2\mathbb{Z}) \to H^{k+i}(X,\mathbb{Z}/2\mathbb{Z})$$

which satisfies a whole bunch of nice properties (or axioms depending on how one looks at things)

Adams used these cohomology operations to solve the vector fields on spheres problem.

See the very excellent book by Mosher & Tangora's "Cohomology Operations and Applications in Hommotopy Theory" for the details

Juan S
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