# The Mordell and Bogomolov problems in linear groups

Many things in the arithmetic of abelian varieties have counterparts not only in linear tori, but also for semisimple linear groups. Two examples are the Tamagawa number and the conjectured finiteness of the Shafarevich-Tate group.

I wanted to ask about the Mordell conjecture (proved by Faltings and Vojta), which for a complex semiabelian variety $A$ and a finitely generated subgroup $\Gamma \subset A(\mathbb{C})$ states that the Zariski closure of any subset of $\Gamma$ is the union of finitely many cosets of algebraic subgroups of $A$.

For $\mathrm{SL}_2$ the obvious translation of this statement is false: the arithmetic subgroup $\mathrm{SL}_2(\mathbb{Z})$ is generated by two elements and contains a Zariski-dense subset from the algebraic set $\mathrm{tr}(A) = 0$. (One could still ask if there is any description of the possible Zariski closures $V$ of subsets of arbitrary finitely generated subgroups $\Gamma \subset G(\mathbb{C})$ of the complex points of a linear group $G$. Taking cue from this example I had asked whether, for instance, such a $V$ has to be defined over a number field. As Venkataramana explains in his answer below, this question is meaningless as it stands: at the very least, I must remove from $V$ a finite union of algebraic subgroup cosets, thereby focusing on the counterexamples to the literal Mordell-type statement.)

Perhaps the analogs of the Manin-Mumford and Bogomolov problems (which concern the Zariski closures of sets of points of finite order) could make more sense in linear groups. Instead of attempting to make any more hasty guesses of how the structure of the possible Zariski closures might look like, it would be more prudent to just record this as an open-ended problem:

Problem. 1. Describe the possible Zariski closures $V$ of sets of elements of finite order in $G(\mathbb{C})$. 2. Whatever these $V$ are, do they coincide with the set of subvarieties of $G$ which possess a Zariski-dense sequence of semisimple elements of $G(\bar{\mathbb{Q}})$ all of whose eigenvalues have canonical Weil heights approaching zero?

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Your statement that the Mordell conjecture says that "the Zariski closure of any subset of $\Gamma$ is an algebraic subgroup of $A$" is not correct. Indeed, take any single point $P\ne O$; the Zariski closure of the set $\{P\}$ is not a subgroup of $A$. The Mordell conjecture is really a statement about the intersection of $\Gamma$ with a subvariety $X \subset A$. Then the Faltings-Vojta theorem says that the Zariski closure of $\Gamma\cap X$ is a finite union of translates of algebraic subgroups of $A$. –  Joe Silverman Aug 15 '14 at 14:50
@JoeSilverman: Thanks! I should have said "a finite union of algebraic subgroup cosets." –  Vesselin Dimitrov Aug 15 '14 at 14:55

## 1 Answer

The answer to whether $V$ is defined over a number field is no. If $G$ is defined over the algebraic closure of $\mathbb Q$, then the number of possible choices for $V$ is countable, so, for measure theory reasons, we can choose an element $a$ in $G({\mathbb C})$ which does not lie in any $V$. So the Zariski closure of the group generated by the element $a$ is not contained in any $V$ defined over a number field.

There are too many other questions which I did not follow.

Borel density says that a lattice is Zariski dense; none of the statements in the question seem to contain the Borel density as a special case

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No, the subgroup generated by a single element is abelian. Unless you are considering the smallest algebraic group defined over the field of algebraic numbers, containing $a$. –  Venkataramana Aug 15 '14 at 15:59
Can I ask you one more question suggested by a statement about commutative algebraic groups: Do you know if the Zariski closure of a set of finite order elements in $G(\mathbb{C})$ can be anything other than a finite union of cosets of algebraic subgroups of $G$? –  Vesselin Dimitrov Aug 15 '14 at 16:10
I think not; you can take the conjugacy class of a finite order element, which does not lie in such a union –  Venkataramana Aug 15 '14 at 16:20
Indeed. I have to include not only the algebraic subgroup cosets, but also their conjugacy classes by algebraic subgroups of $G$ (which made no difference in the commutative case). Thank you! –  Vesselin Dimitrov Aug 15 '14 at 16:36