Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Is there any way to define the orientation of an orientable smooth manifold using sheaves (when our smooth manifold is viewed as a locally ringed space) without our definition being overly contrived?

share|improve this question
    
To elaborate without bumping up the question, is there a definition that isn't contrived, and if not, is there a moral reason why not? –  Harry Gindi Mar 10 '10 at 4:15
3  
Is this relevant? mathoverflow.net/questions/10966/… –  Gjergji Zaimi Mar 10 '10 at 4:17
    
I don't think you should worry about bumping it up if you are adding important content. (I don't think you should worry anyway when you just asked the question.) –  Jonas Meyer Mar 10 '10 at 4:18
4  
Dear fpqc, You may want to look at my answer to the question mentioned in Gjergji Zaimi's comment, where I discuss the orientation sheaf of a manifold and related notions. Orientation is a notion defined for manifolds; it is related to the topology of Euclidean space (essentially, that punctured neighbourhoods are spheres, which admits one of two orientations; the orientation sheaf then measures how these vary as you move around the manifold). It is not a general locally ringd space concept, as far as I understand. –  Emerton Mar 10 '10 at 4:53
3  
Sheaf, singular, Cech, or simplicial (after you choose some triangulation). In the context of $X \setminus \{x\}$, these all measure the same thing and have the same meaning. (If you take real coefficients you could use de Rham as well.) The point is not the particular way you compute the cohomology, but the underlying geometric fact: a sphere around a point in Euclidean space has two sides. Cohomology is just a convenient way to formalize this notion, and to measure what happens to these two sides as one moves around the manifold. –  Emerton Mar 10 '10 at 5:29

2 Answers 2

up vote 7 down vote accepted

In the study of (finite-dimensional?) paracompact and locally compact (?) spaces there is Verdier's topological duality theorem, expressed in terms of a dualizing complex (which is built up from a sheafification process using duals of compactly-supported cohomologies of open subspaces, or something like that). It is pure topology, having nothing to do with ringed spaces (just like the orientation sheaf!). In the special case of smooth (paracompact) manifolds, this recovers the orientation sheaf up to a shift on each connected component. It is analogous to the fact that the super-abstract dualizing complexes in Grothendieck duality for (quasi-)coherent cohomology collapses to the old friend "top-degree relative differentials" (up to shifting) in the smooth case.

But that's all just fancy mumbo-jumbo which puts the orientation sheaf into a broader perspective (like many duality theories for cohomologies). This does not qualify as a good way to initially "define" the orientation sheaf, much as appealing to Grothendieck duality would be a strange (and even circular, from some viewpoints) way to "define" top-degree relative differentials in the smooth case. To get a real theorem out, we have to put some content in.

It seems more illuminating at a basic level to understand how the orientation sheaf is constructed using punctured neighborhoods along the lines of Emerton's comment or the oriented double cover as in David Roberts' answer, and how one can remove some orientability hypotheses in some classical results on "constant coefficient" cohomology by instead allowing coefficients in the locally constant sheaf given by the orientation sheaf. And likewise to understand why the constant sheaf associated to $\mathbf{Z}(1) = \ker(\exp)$ has $n$th tensor power that serves as an orientation sheaf on a complex manifold of dimension $n$ (and so the natural isomorphism $\mathbf{C}(1) = \mathbf{C}$ via multiplication explains the absence of needing to choose orientations for various cohomological calculations on complex manifolds (very relevant if one is to have a hope to translating things into algebraic geometry).

share|improve this answer
    
I had not realized this isomorphism worked without choosing a square root of -1. That is very neat. –  S. Carnahan Mar 10 '10 at 5:42
    
I'm confused. How can you orient C without choosing a square root of -1? –  Dinakar Muthiah Mar 10 '10 at 5:56
1  
@Scott: I don't know any way to actually "write down" $\mathbf{C}$ without essentially singling out a preferred square root of $-1$. (Abstractly one can say "choose an algebraic closure of $\mathbf{R}$", but if one makes that choice specific in any sense then a choice of $i$ drops out. Years ago I discussed this with Lurie, who thought he could possibly prove it couldn't be avoided. I found this very disturbing, since for $\overline{\mathbf{Q}}_ p$ nobody mentions $i$, say for $p$ = 3 mod 4.) –  BCnrd Mar 10 '10 at 6:05
    
@Dinakar: the point is that by using the orientation sheaf, we don't need to choose a global trivialization of it (i.e., a choice of $i$). Considering that Poincare duality in etale cohomology agrees with Poincare duality in topological cohomology (with finite or $\ell$-adic coefficients) over $\mathbf{C}$, and the Artin comparison morphism has nothing to do with orientations, in the guts of the proof of these compatibilities there has to be a way to proceed without any mention of $i$ (and there is, since these are theorems). –  BCnrd Mar 10 '10 at 6:08
1  
@Dinakar: Maybe if one is more explicit about the process of "writing down" a pair of orthogonal lines, it comes out that one has to order the choice of the two lines and thereby single out a preferred direction of rotation. But I don't really know anything about logic and such stuff (to define "writing down"), so I can't be more precise. Likewise, I can say "let C be a splitting field of x^2 + 1 over R" without seeming to order the set of 2 roots, but if one tries to "write down" such a field then an ordering among them seems to pop out. It's sort of a silly discussion anyway. –  BCnrd Mar 10 '10 at 15:58

A manifold supports a so-called orientation sheaf, which unfortunately I cannot recall how to define except to say it is the sheaf of sections of the orientable double cover of the manifold. I vaguely recall that this may be discussed in Dold's book on algebraic topology. An orientation is then a global section of the orientation sheaf. As the orientable double cover has fibres isomorphic to a two element set, there are at most two orientations.

share|improve this answer
1  
Right! The orientation sheaf is the sheaf of sections in the orientation bundle. The latter can be reconstructed from transition functions defined as the determinants of the transitions functions of an atlas of the manifold. –  Konrad Waldorf Mar 10 '10 at 4:26
    
Is such a sheaf definable on any locally ringed space? –  Harry Gindi Mar 10 '10 at 4:33
3  
Slightly more specifically, the sheaf of sections of the oriented double cover is a $\mu_2$-torsor, but $\mu_2 = {\rm{Aut}}(\mathbf{Z})$, so it corresponds to a unique $\mathbf{Z}$-torsor which "is" the usual orientation sheaf. –  BCnrd Mar 10 '10 at 5:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.