B-N-R Part 1: Twisted Endomorphisms

Alright, I’m back, and newly married (thus the long hiatus from posting).  And now it’s time to get back to math.  I’m currently attempting to read a paper by Beauville, Narasimhan and Ramanan titled “Spectral Curves and the Generalized Theta Divisor.”  This paper is one of the early uses of the Hitchin System to prove results about the moduli space of vector bundles over a curve.  The main result can be roughly stated as “a generic vector bundle is the pushforward of a line bundle.” This is a very nice result, and I’m going to work towards it in small steps, largely because I’m still trying to understand it myself.

First up comes some background material.  The background material is going to cover a couple of posts, and will likely include a detour through some basics of Abelian Varieties.  But for now, we fix a curve X, a vector bundle E of rank n, a line bundle L and a homomorphism \phi:E\to L\otimes E.  This is called a twisted homomorphism.  We can use it to define the trace of \phi as an element of \Gamma(X,L), by viewing the map \phi:E\to L\otimes E as a map \phi:E\otimes E^\vee\to L and setting \mathrm{tr}(\phi)=\phi(\mathrm{id}_E).  We can do this by taking a section s_E of E and t_{E^\vee} of E^\vee and setting \phi(s_E\otimes t_{E^\vee})=(t_{E^\vee}\otimes 1)(\phi(s_E)), which is a section of L.  Now, more generally, we define a_i\in\Gamma(X,L^i) to be (-1)^i\mathrm{tr}(\wedge^i \phi) to be the characteristic coefficient, and Cayley-Hamilton says that \phi satisfies \sum a_i\phi^{n-i}=0 as a homomorphism E\to L^n\otimes E.  Using this, we can define nilpotent twisted endomorphisms, unipotent ones, and anything else that we normally can express using the characteristic polynomial.

So, as a quick and stupid example, let X=\mathbb{P}^1.  The big result is only for hyperbolic curves, but this definition still makes sense, and we have an easy description of vector bundles that we can use.  We’ll be so specific, in fact, that we take E=\mathscr{O}(1)\oplus\mathscr{O}(1), and L=\mathscr{O}(2).  Set \phi:E\to E\otimes L by (x_0^2,x_1^2).  Really, we just need to pick a 2\times 2 matrix of quadratic forms, and that’s it, because our map is from \mathscr{O}(1)\oplus \mathscr{O}(1)\to \mathscr{O}(3)\oplus\mathscr{O}(3), because tensor product distributes over direct sum, and we’ve chosen a diagonal matrix.

So next up we need to turn that into a map \phi:E\otimes E^\vee\to L.  Now s_E=(\ell_1,\ell_2) is a pair of linear forms, so then \phi(s_E)=(x_0^2\ell_1,x_1^2\ell_2), a pair of cubic forms.  But then we apply t_{E^\vee}, which is a pair of functionals on linear forms, (t_1,t_2) and get x_0^2t_1(\ell_1)+x_1^2t_2(\ell_2), which is a single quadratic form, because t_i(\ell_i) is just a homogeneous function of degree 0.  So the trace, a_1=-(x_0^2+x_1^2), because we need the image of the identity.  For trivial reasons, a_0=\mathrm{id}=1.  So now, we still need to computer a_2=\wedge^2\phi(\mathrm{id}).  Now, \wedge^2\mathscr{O}(1)^{\oplus 2}\cong \mathscr{O}(2), and we need to map this into \wedge^2\mathscr{O}(3)^{\oplus 2}\cong\mathscr{O}(6).  We do this by taking \ell_1\wedge \ell_2, which is quadratic, to x_0^2\ell_1\wedge x_1^2\ell_2, which is degree 6.  And then the image of the identity is just x_0^2x_1^2, which is a_2.

Thus, in our example, we get that x_0^2x_1^2-(x_0^2+x_1^2)\phi+\phi^2:E\to E\otimes L^2 should be zero.  For our purposes, we note that \phi^2 is defined by E\stackrel{\phi}{\to}L\otimes E\stackrel{1\otimes\phi}{\to}L\otimes L\otimes E, and so is given by multiplication by (x_0^4,x_1^4).  Plugging \phi and \phi^2 into the characteristic polynomial, we find that x_0^2x_1^2(1,1)-(x_0^2+x_1^2)(x_0^2,x_1^2)+(x_0^4,x_1^4)=0, as desired.  So at least in simple situations, we can do computations with this formalism.  Naturally, things become much more difficult in general, when line and vector bundles are harder to describe explicitly, as we made regular use of the fact that \mathrm{Pic}(\mathbb{P}^2)\cong \mathbb{Z}, whereas it is \mathbb{Z} cross a dimension g abelian variety in general.


About Charles Siegel

This entry was posted in Abelian Varieties, Big Theorems, Curves, Vector Bundles. Bookmark the permalink.

5 Responses to B-N-R Part 1: Twisted Endomorphisms

  1. Pani says:

    Woah congratulations on getting married!

  2. Pingback: B-N-R Part 2: Moduli of Vector Bundles « Rigorous Trivialities

  3. Pingback: B-N-R Part 5: Spectral Curves « Rigorous Trivialities

  4. Math says:


    Sorry to bother but, in paragraph 3, do you mean hyperelliptic curves?

  5. No, the big results in the paper are for hyperbolic curves…curves of genus g\geq 2. That is, curves whose universal cover is the hyperbolic plane.

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