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Let $G$ be a commutative connected algebraic group over a field $k$, with group operation $m:G\times G\to G$. If $k=\mathbb{F}_q$, we may use a character $\varphi:G(k)\to\overline{\mathbb{Q}}_\ell^\times$ to construct the "Lang sheaf" $\mathscr{L}_\varphi$ [$\S$1.4, SGA 4 $\frac{1}{2}$, Sommes trig.]. This is a rank 1 local system (= lisse sheaf) on $G$ satisfying "the character condition" $m^* \mathscr{L}_\varphi=\mathscr{L}_\varphi\boxtimes \mathscr{L}_\varphi$.

Question 1: is every rank 1 local system on $G$ satisfying the character condition a Lang sheaf?

If $k=\mathbb{C}$ and $G=\mathbb{G}_a$ (resp. $G=\mathbb{G}_m$), there's a rank 1 vector bundle with connection over $G$ satisfying the character condition. It's the exponential (resp. Kummer) D-module $e^{\alpha x}:=\mathcal{D}/\mathcal{D}(\partial -\alpha)$ (resp. $x^\alpha:=\mathcal{D}/\mathcal{D}(x\partial -\alpha)$).

Question 2: are those the only rank 1 vector bundles with connection satisfying the character condition on $G$?

Question 3: can we construct rank 1 vector bundles with connection satisfying the character condition on other groups $G$?

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Question 1: Yes. See Lemma 2.14 of my paper On the Ramanujan conjecture for automorphic forms over function fields with Nicolas Templier, although this simple argument is surely not original to us.

Question 2: Any rank one connection has the form $\partial-f$ for a function $f$ on $\mathbb G_a$ or $x \partial -f$ for a function $f$ on $\mathbb G_a$. (This uses crucially the fact that every rank one vector bundle on $G$ is trivial.) In both cases, the first term is the unique-up-to-scaling invariant vector field. Two different functions give the same connection if and only if there difference is the logarithmic derivative of a nowhere vanishing function. On $\mathbb G_a$ this occurs only if their difference vanishes and on $\mathbb G_m$ only if their difference is an integer multiple of $1$. In either case, this is a discrete set. Thus, a function gives a translation-invariant connection if and only if the function itself is translation-invariant, i.e. is a constant $\alpha$.

Question 3: Sure, just pick any basis $\nabla_1,\dots, \nabla_n$ for the $G$-invariant vector fields on $G$ and then mod out by $\nabla_i-\alpha_i$ for any $\alpha_1,\dots, \alpha_n \in \mathbb C$ (in other words, take the connection associated to a translation-invariant 1-form).

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  • $\begingroup$ Q1 and Q2: Wonderful! About Q3: do translation-invariant 1-forms always exist? In any case, do all character sheaves (in the language of your paper) appear in this way? $\endgroup$
    – Gabriel
    Commented Apr 29, 2022 at 14:15
  • $\begingroup$ @Gabriel Yes, the same construction that works in the Lie theory situation can be done algebraically (pick a linear form on the tangent bundle at the identity, and then define the 1-form at an arbitrary point by translating to the identity). For the second question, not if $G$ is an abelian variety, as then you can take any degree $0$ line bundle and any 1-form gives a connection. But I think one can show any character sheaf arises from this more general construction by basically the same argument. $\endgroup$
    – Will Sawin
    Commented Apr 29, 2022 at 14:24
  • $\begingroup$ Which "more general construction"? You mean the connection associated to a translation-invariant 1-form? $\endgroup$
    – Gabriel
    Commented Apr 29, 2022 at 14:26
  • $\begingroup$ @Gabriel Hmm, I guess what I said in the abelian variety case is not quite right, because for a nontrivial degree 0 line bundle, there's no algebraic way of relating connections to 1-forms (you can do it analytically using Hodge theory). So my claim is that, for general G, every such $D$-module arises from a translation-invariant connection on a degree $0$ line bundle, and for the trivial bundle, translation-invariant connections arise from translation-invariant $1$-forms, so therefore given any one translation-invariant connection on a nontrivial line bundle, the others arise from $1$-forms. $\endgroup$
    – Will Sawin
    Commented Apr 30, 2022 at 20:20

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