B-N-R Part 5: Spectral Curves

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 X, we want L\in\mathrm{Pic}(X), and we want s_k\in \Gamma(X,L^k) for k=1,\ldots,n.  We start with some constructions.  Let P=\mathbb{P}(\mathscr{O}\oplus L^{-1}), and set p:P\to X the natural projection.  Then we have a natural relatively ample line bundle \mathscr{O}(1) on P, which has p_*\mathscr{O}(1)\cong \mathscr{O}\oplus L^{-1} on X.  Because the pushforward has a natural section (1,0), we get a section of \mathscr{O}(1), which we call y.

Similarly, we have that p_*(p^*L\otimes \mathscr{O}(1))\cong L\oplus \mathscr{O}, which has a natural section (0,1), which gives us a section x of p^*L\otimes \mathscr{O}(1).

Now, we define a polynomial x^n+(p^*s_1)yx^{n-1}+\ldots+(p^*s_n)y^n which is a section of p^*L^n\otimes\mathscr{O}(n).  We denote the zero set of this by X_s, and it is a curve in P.  Restricting p to X_s, we get a finite morphism of curves \pi:X_s\to X.  In fact, it is an n-sheeted ramified cover.

Let’s say we want X_s to be integral.  Can we do that? Well, we have s\in \Gamma(L)\oplus\ldots\oplus\Gamma(L^n).  We can show that the set of these where X_s 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 s_i=0 for i\neq n, and choose s_n\in \Gamma(L^n) so that s_n isn’t equivalent to mD for any m|n.  If we can do this, then we have X_s integral, and so get a nonempty open set such.

From now on, we assume that X_s is integral.  We’re going to give a second construction of X_s, by working backwards from what we have.  It’s a fact that V(x) and $V(y)$ are disjoint, and so y|_{X_s} is nowhere vanishing.  Now, a line bundle with a nowhere vanishing section is trivial, so \mathscr{O}(1)|_{X_s} is the trivial line bundle.  Similarly, we can check that x|_{X_s} is a section of \pi^*L.  Now, if we push forward the structure sheaf of X_s, we get \mathscr{O}\oplus L^{-1}\oplus\ldots\oplus L^{-(n-1)}.  This is isomorphic to \mathrm{Sym}(L^{-1})/\mathscr{I}, where \mathscr{I} is the ideal given by the image of the map L^{-n}\to \mathrm{Sym}(L^{-1}) by taking the sum of the maps s_i:L^{-n}\to L^{-(n-i)}.  If we do this, then we get that X_s=\underline{\mathrm{Spec}}(\mathrm{Sym}(L^{-1})/\mathscr{I}).

As an aside, we can calculate the genus of X_s.  We start with the fact that \chi(X_s,\mathscr{O}_{X_s})=\chi(X,\pi_*\mathscr{O}_{X_s}).  But that will just be \sum \chi(X,L^i), then apply Riemann-Roch, and note that what you just calculated was 1-g_{X_s}, and you get that teh genus is (\deg L)n(n-1)/2+n(g-1)+1.

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

Theorem: Let X be a smooth curve, s=(s_i) a set of sections such that X_s is integral.  Then there exists a canonical bijection between torsion free sheaves of rank 1 on X_s and pairs (E,\phi) of a rank n vector bundle E on X and a twisted endomorphism \phi of E such that the characteristic coefficients are the s_i.

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

Proof: Let M be a torsion free sheaf of rank 1 on X_s.  Then \pi_*M is a vector bundle of rank n.  Additionally, it comes with the structure of a \pi_*(\mathscr{O}_{X_s})-module.  Conversely, a vector bundle of rank n on X, E which has \pi_*(\mathscr{O}_{X_s})-module structure, defines a sheaf M on X_s because \pi is an affine morphism, and we’ll have \pi_*M\cong E.  The integrality of X_s will guarantee that M is torsion free of rank 1.

Now, we have an \mathscr{O}_X-algebra, \pi_*(\mathscr{O}_{X_s})\cong \mathrm{Sym}(L^{-1})/\mathscr{I}, and so being a \pi_*(\mathscr{O}_{X_s})-module is the same as a homomorphism \mathrm{Sym}(L^{-1})/\mathscr{I}\to\mathscr{E}nd(E), which is the same as a map \phi:E\to L\otimes E such that the polynomial defined by the s_i, called P_s, has P_s(\phi)=0.

Now, as X_s is integral, we have that P_s is irreducible over k(X), which means it must be the characteristic polynomial of \phi.  So given M 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 \pi_*(\mathscr{O}_{X_s})-module structure.  We just showed that that structure is the same as have \phi:E\to L\otimes E with has P_s(\phi)=0, and so if we assume that the characteristic coefficients of \phi are the s_i, we get P_s(\phi)=0 by Cayley-Hamilton, finishing the proof.  \Box.

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About Charles Siegel

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
This entry was posted in Abelian Varieties, Big Theorems, Curves, Vector Bundles. Bookmark the permalink.

4 Responses to B-N-R Part 5: Spectral Curves

  1. Pingback: B-N-R Part 6: Theorem 1 « Rigorous Trivialities

  2. Math says:

    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?

  3. Anonymous says:

    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?

  4. For comment 2: Locally, a branch point is z\mapsto z^n on \mathbb{C}. Try computing the pushforward of \mathscr{O}_{\mathbb{P}^1} along the nth power map to see what happens.

    For comment 3: We know that P_s is a polynomial that \phi 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.

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