## 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.

Charles Siegel is currently a postdoc at Kavli IPMU in Japan. He works on the geometry of the moduli space of curves.
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### 5 Responses to B-N-R Part 1: Twisted Endomorphisms

1. Pani says:

Woah congratulations on getting married!

2. Math says:

Hi,

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

3. 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.