Martin's Blog

The Faltings height and normed modules

Posted by martin on Saturday, 31 December 2011 at 15:31

In this post I shall give the definition of the Faltings height of an abelian variety over any number field. Last time we did this over $\mathbb{Q}$ only, and we used two properties of $\mathbb{Q}$: the integers are a PID and there is only one archimedean place. To do things more generally, we will introduce the technology of normed modules and their degrees.

Normed modules

Let $\mathcal{O}_K$ be the ring of integers of a number field. We define a normed $\mathcal{O}_K$-module to be a projective $\mathcal{O}_K$-module $M$ of finite rank together with a norm $\lVert \cdot \rVert_v : M \otimes_{\mathcal{O}_K} K_v \to \mathbb{R}$ for each archimedean place $v$ of $K$.

(A norm is a function $\lVert \cdot \rVert_v : M \otimes_{\mathcal{O}_K} K_v \to \mathbb{R}$ satisfying $\lVert x \rVert_v > 0$ if $x \neq 0$, $\lVert \lambda x \rVert_v = \lvert \lambda \rvert_v \lVert x \rVert_v$ if $\lambda \in K_v$, and the triangle inequality.)

Our primary example of a normed module is the canonical module of the Néron model of an abelian variety: let $A/K$ be an abelian variety of dimension $g$ and $\mathcal{A}/\mathcal{O}_K$ its Néron model. Then $M = \bigwedge^g T_e^\vee (\mathcal{A}/\mathcal{O}_K)$ is a projective $\mathcal{O}_K$-module of rank 1, and we get a norm on $M \otimes K_v$ as follows: if $\omega_0 \in M \otimes K_v$ then we can extend $\omega_0$ uniquely to a global canonical form $\omega$ on $A/K_v$. We define $ \lVert \omega_0 \rVert_v = \sqrt{\mathop{\mathrm{Vol}}(A(\bar{K_v}), \omega)}. $

We now define the degree of a normed module of rank 1, which gives us the Faltings height of an abelian variety. (It seems a bit odd to me to call this a degree, because I expect a degree to be an integer, but I believe that the justification comes from Arakelov intersection theory.)

If $\mathcal{O}_K = \mathbb{Z}$, then we simply choose a generator $m$ of $M$ (unique up to sign) and define $\deg M$ to be $-\log \lVert m \rVert_\infty$.

Over other number fields, there are two problems with this definition. Firstly there may be more than one archimedean place, so which one do we choose? We simply take a sum over all of them.

More seriously, the module $M$ might not be principal, so there is no generator. To get around this, we choose any $m \in M$ at all, and add on a term that measures how far $m$ is from being a generator. Specifically, $ \deg M = \log \#(M/\mathcal{O}_K.m) - \sum_{v \mid \infty} \epsilon_v \log \lVert m \rVert_v. $ Here $\epsilon_v = [K_v : \mathbb{R}]$ (which is 1 or 2).

Note that by the Chinese remainder theorem, we can write the first term of $\deg M$ as a sum over the finite places of $K$: $ \log \#(M/\mathcal{O}_K.m) = \sum \log \#((M \otimes_{\mathcal{O}_K} \mathcal{O}_v) / \mathcal{O}_v.m). $ The expression for $\deg M$ is independent of the choice of $m \in M$ by the product formula for absolute values.

Faltings heights

We define the (unstable) Faltings height of an abelian variety $A/K$ using the degree of the canonical module of its Néron model: $ h(A/K) = \frac{1}{[K:\mathbb{Q}]} \deg \bigwedge^g T_e^\vee (\mathcal{A}/\mathcal{O}_K). $

The factor $1/[K:\mathbb{Q}]$ reduces the dependence on the base field $K$, but does not eliminate it. Specifically, if $M$ is a normed $\mathcal{O}_K$-module, then we have $ \deg (M \otimes_{\mathcal{O}_K} \mathcal{O}_L) = [L:K] \deg M. $

However the Néron model of $A_L$ need not be the base change of the Néron model of $A$, so that $h(A_L) \neq h(A)$. For example an elliptic curve with additive reduction may have good reduction after a finite extension of the base field.

We say that an abelian variety $A$ has semistable reduction if every reduction of its Néron model is a semiabelian variety, that is an extension of an abelian variety by a torus. In the case of an elliptic curve, this is saying that it has good or multiplicative reduction everywhere.

We have the following two facts:

Fact 1. Any abelian variety $A/K$ has semistable reduction over some finite extension $L$ of $K$.

Fact 2. If $A$ has semistable reduction, then the identity component of its Néron model is unchanged by finite extensions of the base field.

So we define the stable Faltings height $h_F(A)$ to be $h(A_L)$ for some finite extension $L/K$ such that $A_L$ has semistable reduction. By fact 2 it does not matter which $L$ we pick, and the stable Faltings height is independent of the base field, so it is often more convenient than the unstable Faltings height.

Tags , , , ,

Trackbacks

Use the following link to trackback from your own site:
http://www.martinorr.name/blog/trackbacks?article_id=217

Comments

  1. Barinder Banwait said about 1 month later:

    Let $C/K$ be a curve over a number field, and $J$ its Jacobian variety. There are various notions of the height of $C$, and I’ll let you choose your favourite one. Is there any link between the height of $C$, and the Faltings height of $J$?

  2. Barinder Banwait said about 1 month later:

    Allow me a moment of shameless self-advertising to the readership of Martin’s blog.

    Barinder Banwait, frequent commentator on Martin’s blog, has setup his own blog at bbanwait.wordpress.com. He will be discussing Mazur’s Isogeny Theorem, with a view towards the Uniform Boundedness Theorem. But he may not get there. Why not check out what the fuss is about?

Comments are disabled

Archives

Syndicate