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This was asked as part of an earlier question. But since this part did not attract many answers, I am asking it separately.

We consider the homology definition of an orientation for a manifold, as you define fundamental class., ie as some generator of some homology modules, satisfying some compatibility conditions. See for instance the book of Greenberg and Harper. What does it mean to say that a manifold is orientable, over rings other than $\mathbb Z$?

It is nice when the base ring is $\mathbb{Z}/2\mathbb{Z}$; every manifold is orientable here, and has a unique orientation. And thus you can do Poincare duality, etc.. But what on earth does it mean to have $4$ possible orientations for the circle or real line for instance, when you take the base ring for homology to be $\mathbb{Z}/5\mathbb{Z}$?

Maybe it is just a formalism; maybe we do not really have to bother about orientations except the ones given by $+1$ and $-1$ in a ring, and the rest are just matters of additional generators giving some extra vacuous information. But I keep wondering. I hope somebody can clarify.

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I just wanted to mention that while orientability for cohomology with arbitrary coefficients is governed solely by cohomology with coefficients in ℤ, there are other cohomology theories for which is is not true. For example, if you have an action of $\pi_1(X)$ on an abelian group M, then you can talk about (co)homology with twisted coefficients in M. For any vector bundle there is a coefficient module such that the bundle is orientable with respect to these twisted coefficients (or, to paraphrase Matthew Ando, "every bundle is orientable if you're twisted enough").

Also, one can ask whether a vector bundle is orientable with respect to topological K-theory, real or complex, or many other generalized cohomology theories, which capture interesting information about the manifold.

So while ℤ/m-coefficients may not be the most interesting coefficient systems to study orientability in, they're part of a larger systematic family of questions (and they don't take much extra work if you're already doing ℤ and ℤ/2-coefficients).

Finally, for something like real coefficients you might think of orientations that differ by a real scalar geometrically, e.g. according to some volume form.

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+1 for the groan-worthy pun/mnemonic in addition to the topology. No surprise that Matt's responsible for it. –  Eric Peterson Jan 20 '10 at 23:47

Orientability over $\mathbb Z_n$ for $n \geq 3$ is equivalent to orientability over $\mathbb Z$. This is because when you consider the local coherence condition for an orientation, if two orientations aren't compatible the map is given by negation. After all, this is what happens in $\mathbb Z$ coefficients, so use universal coefficients to deduce the same for $\mathbb Z_n$.

Having more than one orientation isn't such an exciting idea since they're all multiples of each other.

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Indeed, from the universal coefficients it would appear that homology of manifolds over rings other than $\mathbb Z$ contains no additional information. So are all the extra stuff just a formalism, maybe except for the nice thing that everything is orientable over $\mathbb{Z}/2\mathbb{Z}$? –  Anweshi Jan 20 '10 at 18:09
    
Non-integer coefficients play only a few roles. From the perspective of orientation and Poincare duality only $\mathbb Z$ and $\mathbb Z_2$ are relevant. But more generally coefficients are used for "localization" type arguments -- where you compute the (co)homology with integer coefficients by computing the (co)homology in various simpler coefficient systems and then assemble the data later. A good example would be the proof of the Thom isomorphism theorem in Milnor and Stasheff's "Characteristic classes" but you can also find many examples of this in Hatcher's Alg. Top. book. –  Ryan Budney Jan 20 '10 at 18:19
    
If $L=L(p,q)$ is a lens space, $H_2(L,\mathbb Z)$ is zero, yet $H_2(L,\mathbb Z_p)$ is not, so the answer to your question, Anweshi, depends on your definition of "additional information" :) –  Mariano Suárez-Alvarez Jan 20 '10 at 18:22
    
@Mariano. From the statement of univ. coeffts thm, it is clear that the homology with arbitrary coefficients can be computed by tensoring the ring with integral homology, the obstruction to this coming from the Tor of the ring with the integral homology of one lower dimension. This is what I meant. In your example I suppose the difference is coming from $H_1$ –  Anweshi Jan 20 '10 at 19:00
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Dear Anweshi, there are many reasons to consider (co)homology groups with coefficients other than $\mathbb Z$. They arise in obstruction theory, in studying cohomology operations, in de Rham theory and Hodge theory, in arithmetic geometry, in the theory of automorphic forms, ... . –  Emerton Jan 21 '10 at 5:42

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