Timeline for Orientation of a smooth manifold using sheaves
Current License: CC BY-SA 2.5
8 events
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Mar 10, 2010 at 15:58 | comment | added | BCnrd | @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. | |
Mar 10, 2010 at 6:59 | comment | added | Dinakar Muthiah | @Brian: How about this construction of C: Let V be a 2-dim real vector space with an inner product. Pick a pair of orthogonal lines. Then there are exactly two operators on V that preserve the inner product, have positive determinant, and swap the two lines. The algebra generated by these two operators and the identity operator is an algebraic closure of R, but neither of the square roots is special. | |
Mar 10, 2010 at 6:08 | comment | added | BCnrd | @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). | |
Mar 10, 2010 at 6:05 | comment | added | BCnrd | @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.) | |
Mar 10, 2010 at 5:56 | comment | added | Dinakar Muthiah | I'm confused. How can you orient C without choosing a square root of -1? | |
Mar 10, 2010 at 5:43 | vote | accept | Harry Gindi | ||
Mar 10, 2010 at 5:42 | comment | added | S. Carnahan♦ | I had not realized this isomorphism worked without choosing a square root of -1. That is very neat. | |
Mar 10, 2010 at 5:32 | history | answered | BCnrd | CC BY-SA 2.5 |