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.)


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.)


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


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 .


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.)

  • $\begingroup$ 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. $\endgroup$
    – jmc
    Nov 7 '12 at 8:30
  • $\begingroup$ That should be $\operatorname{End}(\kappa_{\text{s}}^{*})$, of course. $\endgroup$
    – jmc
    Nov 7 '12 at 9:14
  • 1
    $\begingroup$ 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. $\endgroup$ Nov 7 '12 at 10:30
  • $\begingroup$ True. I used the term in a bit more inclusive sense, I think. $\endgroup$
    – 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 $\pm 1$.

  • $\begingroup$ 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.) $\endgroup$
    – jmc
    Nov 7 '12 at 20:23
  • $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Nov 7 '12 at 21:45
  • $\begingroup$ 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 $\kappa$. $\endgroup$
    – jmc
    Nov 7 '12 at 23:39
  • $\begingroup$ 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! $\endgroup$
    – jmc
    Nov 8 '12 at 3:20
  • 1
    $\begingroup$ 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. $\endgroup$
    – Will Sawin
    Nov 8 '12 at 3:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.