Bogomolov's paper "Sur l'algébricité des représentations l-adiques" proves that the image of the $\ell$-adic Galois representation associated to an abelian variety over a number field $K$ is open in the $\mathbb{Q}_ {\ell}$ points of its $\mathbb{Q}_{\ell}$-Zariski closure.

The Comptes-Rendus announcement has only sketched the proof. My French is limited, but I have tried to capture the key step below.

In the abelian case, it proceeds as follows. Let $H_{\ell}$ be the $\mathbb{Q}_ {\ell}$ Zariski-clsure of the image of the absolute Galois group. Replacing $K$ by a finite extension, we assume that the is connected. If $H_{\ell}$ is commutative, then $\rho$ is abelian. We apply properties of Hodge-Tate representations to deduce that $H_{\ell}$ is a torus. For otherwise, $H_{\ell}$ contains a factor of $\mathbb{G}_a$. Composing with $\rho$ would give an infinite unramified abelian extension of $K$. [** Why? **]

I do not understand the last step used to obtain an infinite unramified abelian extension. Perhaps somebody else can see it?

The Russian version is at http://www.mathnet.ru/links/c33c60182f2bcfa61bc2cb3a0b684c62/im1843.pdf, but I cannot read a word of Russian.

  • $\begingroup$ You have a nontrivial homomorphism from the Galois group to $G_a(\mathbb Z_l)$. Since the Galois group is combact, the image must be of the form $\mathbb G_a(\mathbb Z_l)$. This gives an abelian infinite quotient of the Galois group. it remains to show it is infinite unramified. The ramification groups of primes not lying over $l$ will be finite, by class field theory. So it remains to check that the extension is unramifed at $l$. This step requires $p$-adic Hodge theory, and more than my slim understanding of it. $\endgroup$
    – Will Sawin
    Feb 25 '13 at 6:00
  • $\begingroup$ which I guess in this case is then $\ell$-adic Hodge theory :) $\endgroup$ Feb 25 '13 at 6:38
  • $\begingroup$ Will -- one has to be careful because the corresponding purely local statement is false! The Tate module of an elliptic curve over $\mathbf{Q}_p$ with ordinary reduction but no CM (over an alg closure) will always in practice be a non-split extension of one character by another; however the representation would be Hodge-Tate -- even crystalline. $\endgroup$
    – user30035
    Feb 26 '13 at 0:21

What Bogomolov proved is the algebraicity of the corresponding $\ell$-adic representation. What you are trying to do is to prove its semisimplicity, which was done by Faltings 4 years later. The presence of additive factors $G_a$ in the Zariski closure of the Galois image is not an issue as far as you are interested only in the algebraicity. In fact, since every linear nilpotent Lie algebra (in char 0) is algebraic (coincides with the Lie algebra of a certain linear algebraic group), one may replace the original representation by its semisimplification and deal with a certain semisimple $\ell$-adic representation $\rho$ of the absolute Galois group $Gal(K)$ of a number field $K$. Replacing $K$ by its suitable finite algebraic extension, one may assume that the inertia groups of places not lying over $\ell$ act as unipotent operators and $H_{\ell}$ is connected. So, if $\rho$ is semisimple abelian then it is unramified outside divisors of $\ell$. Since it is of Hodge-Tate type, the images (under $\rho$) of the inertia groups of divisors of $\ell$ are open in their Zariski closures $U_{\lambda}$ (see Serre's Abelian $\ell$-adic representations and elliptic curves), i.e., the corresponding local $\ell$-adic representations are algebraic. Now, if $U$ is is the algebraic subgroup of $H_{\ell}$ that is generated by all $U_{\lambda}$'s, then the images of all the inertia groups generate (after taking the closure) an open subgroup in $U(\mathbf Q_{\ell})$. Now it suffices to check that $H_{\ell}=U$.

If $U$ is a proper subgroup of $H_{\ell}$ then the composition $$Gal(K)\to H_{\ell}(\mathbf Q_{\ell}) \twoheadrightarrow H_{\ell}(\mathbf Q_{\ell})/U(\mathbf Q_{\ell})$$ gives rise to a continuous surjective homomorphism from $Gal(K)$ to $\mathbf Z_{\ell}$ that corresponds to an infinite everywhere unramified abelian extension of $K$. Since such extensions do not exist, we get the desired contradiction.

By the way, Bogomolov's Izvestiya paper is available in English http://mr.crossref.org/iPage/?doi=10.1070%2FIM1981v017n01ABEH001329 .


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