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Zeroth question - am I right that in the "ordinary" sense an abelian variety does not possess any representations at all?

More precisely, a representation of an algebraic group $G$ (over an algebraically closed field $K$, say, over complex numbers) is a homomorphism $G\to\operatorname{GL}(V)$ for some $K$-vector space $V$, which is a morphism of algebraic varieties. I have vague feeling that if $G$ is a projective variety, hence an abelian variety, then there are no nontrivial such homomorphisms.

If the answer to this zeroth question is negative, then my actual ("number one") question would be whether there is a classification of such representations.

(As @Wojowu explains in a comment below, this is indeed true)

If it is positive, then the question is whether there exists a modification of the notion of representation that would give some meaningful result - mostly, would allow studying an abelian variety $G$ through such representations.

Possible approaches would, maybe, include allowing "representations with singularities", say, instead of polynomial homomorphisms to allow rational homomorphisms $G\to\operatorname{GL}(V)$. Or, say, one might consider $G$-equivariant vector bundles over $G$ (whatever this means). Or, one might look at algebraic homomorphisms $G\to\operatorname{Aut}(A)$ where $A$ is some commutative "thing" in $K$-varieties such that the algebraic group $\operatorname{Aut}(A)$ admits nontrivial algebraic homomorphisms from abelian varieties to it. Subquestion: are there such $A$? Can it be, say, another abelian variety?

Maybe one more hopefully simpler subquestion. Let $\operatorname{Aut}(|G|)$ be the algebraic group of all algebraic automorphisms of the underlying algebraic variety $|G|$ of $G$. Then (presumably) the map assigning to $x\in G$ the multiplication-by-$x$ operator $G\to G$ is an injective algebraic homomorphism $G\to\operatorname{Aut}(|G|)$, so $\operatorname{Aut}(|G|)$ contains a copy of $G$ as a subgroup. What are, if any, subgroups in between? How does $\operatorname{Aut}(G)$ sit inside $\operatorname{Aut}(|G|)$? Is this $\operatorname{Aut}(|G|)$ studied somewhere?

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    $\begingroup$ Your feeling in the zeroth question is correct. Given a morphism $G\to GL(V)$, we can compose it with $GL(V)\hookrightarrow\mathbb A^{\dim V}$ and then project onto a coordinate. If $G$ is projective, the resulting image is a point. It follows $G\to GL(V)$ is constant. There there are indeed no (nontrivial) reps in the usual sense. $\endgroup$
    – Wojowu
    Commented Sep 16, 2020 at 10:27
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    $\begingroup$ @LSpice Yes, I had in mind morphisms of algebraic groups there, as written. $\endgroup$ Commented Sep 16, 2020 at 10:39
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    $\begingroup$ @LSpice Well I would hope that $\operatorname{Aut}(|G|)$ is more manageable than the group of all permutations of $G$. In fact the former defines naturally an algebraic group (i. e. a functor from $K$-varieties to groups) while the latter - I don't think so. $\endgroup$ Commented Sep 16, 2020 at 10:45
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    $\begingroup$ Sorry; misread "algebraic gp." somehow as "Abelian var." (!). Also right, certainly for pos.-dim. var. $G$, but I meant $G$ finite (and smooth)—where $\operatorname{Aut}(\lvert G\rvert)$ is just the symmetric group. But I guess I'm misunderstanding something in the definition and a non-trivial finite group, or at least a non-Abelian one, isn't an Abelian variety. (Connected is part of def., maybe?) $\endgroup$
    – LSpice
    Commented Sep 16, 2020 at 10:46
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    $\begingroup$ To answer one of your questions, an abelian variety $G$ will act on itself by translation. If you try to linearize this by looking at the action on $\mathcal{O}(G)=K$, it would trivial. But the you could act things which are less linear, such as the action of $G$ on sheaves (in various senses) on $G$. $\endgroup$ Commented Sep 16, 2020 at 13:13

2 Answers 2

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I am also very far from an expert here, but I think there's a case to be made that the "correct notion" involves actions on categories of sheaves, as Donu says in the comments.

Consider the following toy model: if $A$ is, say, a finite abelian group then its Pontryagin dual $\widehat{A}$ can be defined as the group $\text{Hom}(A, \mathbb{G}_m)$ of homomorphisms from $A$ into the multiplicative group $\mathbb{G}_m$ (say over the complex numbers, so $\mathbb{G}_m(\mathbb{C}) \cong \mathbb{C}^{\times}$, but any algebraically closed field of characteristic $0$ would do, or we could think in terms of Cartier duality). There is then a canonical pairing

$$A \times \widehat{A} \to \mathbb{G}_m$$

which is used, for example, to define the Fourier transform $L^2(A) \cong L^2(\widehat{A})$. In representation-theoretic terms these homomorphisms correspond to $1$-dimensional representations and give exactly the irreducible representations of $A$.

Abelian varieties $X$ also have duals $X^{\vee}$, but they're defined not in terms of maps into the multiplicative group $\mathbb{G}_m$ but in terms of line bundles, or equivalently in terms of maps into the classifying stack $B\mathbb{G}_m$ of line bundles (although we need to restrict to degree $0$ line bundles). There is again a canonical "pairing"

$$X \times X^{\vee} \to B \mathbb{G}_m,$$

namely the Poincaré bundle over $X \times X^{\vee}$, and it can be used to define the Fourier-Mukai transform $D(X) \cong D(X^{\vee})$ between derived categories of coherent sheaves. Among other things, what this analogy suggests is that the analogue of the "regular representation" for an abelian variety is its action on its derived category $D(X)$ by translation.

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This is not an answer, more an extended comment. Though I am far from being an expert, I would have rather thought of an abelian variety (say $X$) to be analoguous to a vector (or projective) space (say $V$) and $\textrm{End}(X)$, the ring of isogenies of $X$, to be analogous to $\mathrm{GL}(V)$. Note that there is a notion of dual abelian variety which is quite close to the classical duality for projective space.

Furthermore assume that $X$ is simple (this is a very common hypothesis), that is $X$ contains no non-trivial abelian sub-variety. Then, any proper non-trivial homomorphism $f : X \longrightarrow X$ must be surjective with finite kernel. By a classical Theorem on abelian varieties, this implies that $f$ is an isogeny. So basically any interesting homomorphism from $X$ to $X$ is in $\mathrm{End}(X)$.

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    $\begingroup$ We can also consider, say, an elementary abelian $p$-group to be a vector space and study representations on it, but it’s a different and interesting question to study representations of it. $\endgroup$ Commented Sep 16, 2020 at 21:39
  • $\begingroup$ Well, unlike vector spaces, projective spaces do not have a group structure. And, unlike projective spaces, vector spaces are not complete varieties. But your mention of the dual I find very interesting. Dual of a vector space, as well as (the Pontryagin) dual of an abelian group can be viewed as the set of its 1-dimensional representations, with appropriate structure. To follow this analogy then, one should view line bundles on an abelian variety as some sort of its 1-dimensional representations. $\endgroup$ Commented Sep 16, 2020 at 21:43
  • $\begingroup$ @Yuan if you want. Though I am not an expert, I believe that for abelian varieties it goes the other way round. The study of $\textrm{End}(X)$ (for instance complex multiplication, Rosati involution etc..) is currently much richer than the non-existing theory of vector representations for $X$. $\endgroup$
    – Libli
    Commented Sep 16, 2020 at 21:59
  • $\begingroup$ @მამუკაჯიბლაძე : sure! Otherwise I would not have made that a comment, but a full answer. If you read carefully, I used the word "analogous", not generalization... $\endgroup$
    – Libli
    Commented Sep 16, 2020 at 22:01

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