Today we’re back to some material from the first post in this series, and going to prove an actual theorem about vector bundles. Next time, we’ll be getting into the heart of the paper, and that may be my last post on B-N-R.

For this post, we’ll fix a smooth, irreducible curve , we want , and we want for . We start with some constructions. Let , and set the natural projection. Then we have a natural relatively ample line bundle on , which has on . Because the pushforward has a natural section , we get a section of , which we call .

Similarly, we have that , which has a natural section , which gives us a section of .

Now, we define a polynomial which is a section of . We denote the zero set of this by , and it is a curve in . Restricting to , we get a finite morphism of curves . In fact, it is an -sheeted ramified cover.

Let’s say we want to be integral. Can we do that? Well, we have . We can show that the set of these where is integral is open by noting that not being reduced and not being irreducible will both be closed conditions: we can write down equations that check them. It remains to check that the set of these is nonempty. To do that, we look at for , and choose so that isn’t equivalent to for any . If we can do this, then we have integral, and so get a nonempty open set such.

From now on, we assume that is integral. We’re going to give a second construction of , by working backwards from what we have. It’s a fact that and $V(y)$ are disjoint, and so is nowhere vanishing. Now, a line bundle with a nowhere vanishing section is trivial, so is the trivial line bundle. Similarly, we can check that is a section of . Now, if we push forward the structure sheaf of , we get . This is isomorphic to , where is the ideal given by the image of the map by taking the sum of the maps . If we do this, then we get that .

As an aside, we can calculate the genus of . We start with the fact that . But that will just be , then apply Riemann-Roch, and note that what you just calculated was , and you get that teh genus is .

That was a tangent, now we prove the first real result we have about vector bundles:

**Theorem**: Let be a smooth curve, a set of sections such that is integral. Then there exists a canonical bijection between torsion free sheaves of rank 1 on and pairs of a rank vector bundle on and a twisted endomorphism of such that the characteristic coefficients are the .

In fact, if is smooth, we can say line bundles.

*Proof*: Let be a torsion free sheaf of rank 1 on . Then is a vector bundle of rank . Additionally, it comes with the structure of a -module. Conversely, a vector bundle of rank on , which has -module structure, defines a sheaf on because is an affine morphism, and we’ll have . The integrality of will guarantee that is torsion free of rank 1.

Now, we have an -algebra, , and so being a -module is the same as a homomorphism , which is the same as a map such that the polynomial defined by the , called , has .

Now, as is integral, we have that is irreducible over , which means it must be the characteristic polynomial of . So given torsion free of rank 1, we’ve constructed the pair. All that remains is that for each pair we get a torsion free sheaf, and we’ve got it for vector bundles with -module structure. We just showed that that structure is the same as have with has , and so if we assume that the characteristic coefficients of are the , we get by Cayley-Hamilton, finishing the proof. .

September 22, 2009 at 11:34 am

[...] restrict even further to an open set where the spectral curve is smooth and irreducible. By the theorem from last time, we have that consists of pairs where is a stable bundle coming from the direct [...]

December 9, 2009 at 5:44 am

I have a question about this. If X and Y are algebraic curves and f : Y -> X is a finite proper morphism (e.g. X and Y are compact Riemann surfaces and Y is an m:1 cover of X), then why is the pushforward f_*L of any line bundle L->Y a locally-free sheaf on X? Essentially, this comes down to the behavior at branch points, but I can’t see why f_*L is necessarily free in the vicinity of each and every branch point. Do you have any thoughts on this?

March 22, 2010 at 1:26 pm

Why does the integrality of X_s guarantee that M is torsion-free?

Why does this integrality and the irreducibility of P_s over k(X) guarantee that P_s is the characteristic polynomial of \phi?

March 24, 2010 at 8:17 pm

For comment 2: Locally, a branch point is on . Try computing the pushforward of along the th power map to see what happens.

For comment 3: We know that is a polynomial that satisfies. Irreducibility then implies it is the minimal polynomial, and degree implies characteristic polynomial. As for your first point, I don’t have a proof off the top of my head, nor a quick reference.