# Understanding the determinant of the action of Frobenius on the character group of the toric part of the reduction of the Jacobian of a curve.

I am trying to understand a certain sentence in a paper that I am reading. Let me start with some notation/background. (For a motivation of why this should be interesting, see below, under the questions.)

## Notation/background

First the (pretty standard) setup:

• $k$ a field with a discrete valuation $v$;
• $\mathcal{O}_{v}$ the ring of integers;
• $\kappa$ the residue field (we assume it is finite);
• $\kappa_{\text{s}}$ a seperable closure of $\kappa$;
• $G_{\kappa}$ the absolute Galois group of $\kappa$;
• $F$ the Frobenius generator ($x \to x^{|\kappa|}$ on sections);
• $X/k$ a curve with semistable model $\mathcal{X} \to \operatorname{Spec} \mathcal{O}_{v}$.

We can look at the Jacobian of $X$, which is an abelian variety $\operatorname{Jac} X$ over $k$. By the theory of Ńeron models we can form a smooth model $\mathcal{J}$ over $\operatorname{Spec} \mathcal{O}_{v}$. Let $\tilde{J}$ denote the reduction at $v$, i.e., $\mathcal{J} \times_{\mathcal{O}_{v}} \kappa$. This is a commutative group scheme, and the component of the identity, $\tilde{J}^{0}$ fits is the extension of an abelian variety $A/\kappa$ by a linear group.

Minor question: If I am not mistaken, this linear group is a torus $T/\kappa$, because our curve $X$ has a semistable model. Is this correct?

Thus we have an exact sequence of commutative $\kappa$-group schemes: $1 \to T \to \tilde{J}^{0} \to A \to 0.$

The paper that I am reading now considers

[...] $\tau = \pm 1$, the determinant of the action of $F$ on the character group of $T$.

I searched the literature and the interwebs to get a hang of what is going on here, but I am not really confident of what I found. (Especially because I do not get $\tau = \pm 1$.)

## My guess

According to http://www.encyclopediaofmath.org/index.php/Character_group the character group of $T$ is $\operatorname{Hom}(T, \mathbb{G}_{\text{m}})$, i.e., $\mathbb{G}_{\text{m}}(T)$. However, I could not think of any Galois action on this. I proceeded by guessing that $X(T) = \operatorname{Hom}(T(\kappa_{\text{s}}), \mathbb{G}_{\text{m}}(\kappa_{\text{s}}))$ would be a good candidate for this character group, and moreover it carries a natural action of $G_{\kappa}$ given by $f \cdot \sigma = f \circ \sigma$.

Let $e$ denote the dimension of $T$. Then we have the identities $X(T) = \operatorname{Hom}((\kappa_{\text{s}}^{*})^{e}, \kappa_{\text{s}}^{*}) = \operatorname{Hom}(\kappa_{\text{s}}^{*}, \kappa_{\text{s}}^{*})^{e}.$ (By definition of algebraic torus and the universal property of direct sums.)

Now I wanted to understand the determinant of $F$ acting on $X(T)$. It seemed natural to me to view $X(T)$ as free module of rank $e$ over $R = \operatorname{End}(\kappa_{\text{s}}^{*})$. The action of $F$ would then be given by the scalar matrix $|\kappa| \cdot I$. Its determinant would then be $|\kappa|^{e}$. Unless $e = 0$ (in the case of good reduction) this is not equal to $\pm 1$.

Likely I am messing things up horribly. First of all my computation of $\tau$ is not equal to $\pm 1$, and secondly it seems to depend only on $e$. (I guess it should be more intricately connected to $T$ as $k$-scheme, instead of only $T_{\kappa_{\text{s}}}$, the base change to the seperable closure.)

# Question

1. Where did I take the wrong turn?
2. What is the right way of computing $\tau$.

## Motivation

Given the computation of $\tau$, we can 'easily' compute a certain local root number $\epsilon_{v}$. This local root number is a local factor in the sign $\epsilon$ of the (conjectured) functional equation of the $L$-function of a certain motive $M$ associated to $X$.

The Beilinson-Bloch conjecture link the order of vanishing (at a certain critical point) of this $L$-function to the rank of the Chow group of $M$. Under certain conditions on $X$, one can construct a non-trivial element $\Delta_{\xi}$ of $\operatorname{Ch}(M)$, hence proving that its rank is strictly positive. Assuming the truth of this conjecture, if $\epsilon = 1$, it follows that the rank is at least $2$.

And yep, that is why I think it is interesting to compute $\tau$.

For more information I refer to section 5 of Shou-Wu Zhang's paper “Gross–Schoen Cycles and Dualising Sheaves”, available at http://arxiv.org/abs/0812.0371 .

## Edits

As noted in my comment below. I stupidly overlooked the fact that $R$ is not a commutative ring.

Further I also found http://www.martinorr.name/blog/2010/01/24/character-groups-of-algebraic-tori which is really helpful. I have not fully figured out how to compute $\tau$. But at least it points in a different direction than my guess. (And I think the new direction is more promising.)

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Hmm, by now I think the answer to Q1 might have to do with the fact that $\operatorname{End}(\kappa_{\text{s}})$ is not a commutative ring. –  jmc Nov 7 '12 at 8:30
That should be $\operatorname{End}(\kappa_{\text{s}}^{*})$, of course. –  jmc Nov 7 '12 at 9:14
Minor nitpick: It's not the theory of Neron models that gives you a smooth model. The existence of a smooth model is elementary. You use the theory of Neron models to obtain a smooth model with the Neron mapping property. –  Jan Hendrik Nov 7 '12 at 10:30
True. I used the term in a bit more inclusive sense, I think. –  jmc Nov 7 '12 at 11:04

You need to work with just regular homomorphisms between those two groups, i.e. algebraic characters. That does two things.

First, note that $\operatorname{End}(\kappa_s^*)=\mathbb Z$.

Second, $f \circ \sigma$ is not in general a regular homomorphism. You need to take $\sigma^{-1} \circ f \circ \sigma$.

The key point is that this action on $X(T)$ is invertible, so lies in $GL_n(\mathbb Z)$, so has determinant $\pm 1$.

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Thank you very much for your reply! I figured out your first two lines using the blogpost of Martin Orr (mentioned in the edit of my question). I am still struggling with the action though. Since everything is affine, we can convert to Hopf algebra's. I would say that $\sigma$ acts on the coefficients of a polynomial. On the other hand, $f$ is determined by mapping $X \in \mathbb{Z}[X,X^{-1}]$ to a monomial in $\mathcal{O}_{T}(T)$, and is the identity on coefficients. It seems to me that $\sigma^{-1} \circ f \circ \sigma = f$. What am I doing wrong now? (Your last sentence is clear to me.) –  jmc Nov 7 '12 at 20:23
You want to look at maps from $\mathbb Z[X, X^{-1}]$ to $\mathcal O_T(T) \otimes_\kappa \kappa_s$. These are determined by $X \in \mathcal O_T(T) \otimes \kappa \kappa_s$, and the Galois action on these is indued by the tensor product. –  Will Sawin Nov 7 '12 at 21:45
Thanks for your reply. My confusion is more about why the action is non-trivial. It seems to me that $\sigma^{-1} \circ f \circ \sigma = f$ for all $f$; because the action is only on coefficients, while $f$ only changes monomials. (This wording is a bit sloppy.) If I understand the theory right, I am confused about how general $f$'s look. I seem to think that every map is defined over $\kappa$. –  jmc Nov 7 '12 at 23:39
Ok, if you can help me out why $\sigma^{-1} \circ f \circ \sigma$ is again a character, then I would be very grateful. It is now the only thing that I do not understand. You have been very helpful, thanks a lot! –  jmc Nov 8 '12 at 3:20
It's clear that it's a character, because it's a composition of 3 group homomorphisms. What's subtle is that it's a regular function. The obvious way to prove that is to just use the Hopf algebra perpective. In other words, $f$ is a function, say $f(x,y,z)$ is a polynomial in the coordinates $x,y,z$ of $T$. $\sigma^{-1} \circ f \circ \sigma$ is what you get when you apply $\sigma{-1}$ to the coefficients of $f$, so it's still a polynomial function. –  Will Sawin Nov 8 '12 at 3:23