3
$\begingroup$

When studying Galois representations, we always assume that our representations are continuous. I'm new to studying these objects and am a bit struck by this assumption. What is the reasoning behind this? Do the interesting/meaningful representations all end up being continuous? Why is this the case?

$\endgroup$
5
  • 1
    $\begingroup$ Related. Unfortunately I lack the expertise to judge how much that might help you. $\endgroup$ Jun 8, 2022 at 6:01
  • 1
    $\begingroup$ Or this? $\endgroup$ Jun 8, 2022 at 6:20
  • 2
    $\begingroup$ Well, the Galois group naturally comes with a topology, so why not use it? Infinite groups have too many representations, so we like to restrict our attention to ones with good properties. $\endgroup$
    – Kenta S
    Jun 8, 2022 at 6:59
  • 3
    $\begingroup$ For example, general normal subgroups of Galois groups are not nice, but closed normal subgroups are (they correspond to sub-Galois extensions). Including non-continuous representations would mean representations coming from these not-as-nice quotients would also be considered. $\endgroup$
    – Kenta S
    Jun 8, 2022 at 7:01
  • $\begingroup$ @JyrkiLahtonen these are both helpful, thank you! $\endgroup$ Jun 8, 2022 at 22:53

1 Answer 1

4
$\begingroup$

This started as a comment, but it's grown a bit too long!

For starters there are plenty of examples of discontinuous representations. For example, you'll often see number theorists fixing an isomorphism $\overline{\mathbb Q}_\ell\cong \mathbb C$. This gives you a discontinuous ring homomorphism $\mathbb Z_{\ell}\hookrightarrow \mathbb C$, and hence a discontinuous group homomorphism $\mathbb Z_\ell^\times\hookrightarrow\mathrm{GL}_1(\mathbb C)$. Now $G_{\mathbb Q}$ surjects onto $\mathbb Z_\ell^\times\cong \mathrm{Gal}(\mathbb Q(\zeta_{l^{\infty}})/\mathbb Q)$. We obtain a homomorphism $$G_{\mathbb Q}\to \mathrm{GL}_1(\mathbb C).$$ But it's pretty clear that this homomorphism really sees nothing about the number theoretic properties of $G_{\mathbb Q}$! All the things that we care about when we talk about $\mathbb Z_{\ell}^\times$ are completely lost! There's no valuations, reduction mod $\ell$, etc.

As algebraic number theorists, what we really care about is the arithmetic of $\overline{\mathbb Q}$. In particular, we care about Galois groups of extensions of $\mathbb Q$.

The absolute Galois group is a convenient tool to package together the information from all Galois groups of all algebraic extensions of $\mathbb Q$. Indeed, the fundamental theorem of Galois theory says that these Galois groups are exactly the quotients of $G_{\mathbb Q}$ by closed normal subgroups. $G_{\mathbb Q}$ is, somehow, more than the sum of its parts. It also carries information about all the local Galois groups $G_{\mathbb Q_\ell}$, and (at least conjecturally), information about Dirichlet characters, modular forms, automorphic representations, etc.

On the other hand, $G_{\mathbb Q}$ is an extremely complex object, and is extremely hard to study. So instead of trying to understand the whole thing, we study it piecewise using is representations. The image of a representation is some quotient of $G_{\mathbb Q}$ that, conveniently, has the structure of linear maps on a vector space.

But, and here is the point, what we really care about is arithmetic. So we only care about those representations whose images can tell us something about arithmetic. In other words, we only care about those representations whose images are also Galois groups, and whose images behave like Galois groups.

What does it mean for the image of a representation $\rho\colon G_{\mathbb Q}\to \mathrm{GL}(V)$ to be a Galois group? By the fundamental theorem of Galois theory, it exactly means that the kernel is a closed normal subgroup of $G_{\mathbb Q}$. Moreover, we want the image to have Galois theoretic properties - so it should be profinite, and have a fundamental theorem. In short, we want continuity.

$\endgroup$
1
  • $\begingroup$ This is crystal clear! $\endgroup$ Jun 8, 2022 at 22:52

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .