# A “prequestion” about meromorphic representations of algebraic groups

In a comment exchange around an answer to Is a group scheme determined by its category of representations? there arose the issue of Tannakian reconstruction for non-affine algebraic groups (e. g. abelian varieties, in particular, elliptic curves). Some very naïve approach came to my mind, and then @AntonFetisov mentioned one proper way to do it, via representations of the category of quasicoherent sheaves, in view of Lurie's Tannakian reconstruction theorem. He then suggested to post a separate question, which I promised to do, but formulating it in a comprehensive way takes much more time and effort than I thought. So I decided in the meanwhile to make a preliminary version of the most primitive part of the question. Here it is.

Let $G$ be an algebraic group. Call a meromorphic representation of $G$ a collection of rational functions $(\varrho_{ij})_{1\leqslant i,j\leqslant n}$ on $G$ such that $$\varrho_{ij}(xy)=\sum_{1\leqslant k\leqslant n}\varrho_{ik}(x)\varrho_{kj}(y)$$ for all $x,y\in G$ (and all $1\leqslant i,j\leqslant n$).

Have these been studied anywhere? Are there any nontrivial ones (that is, not equivalent, in one way or another, to "ordinary" representations)? Is it possible to classify such representations for $G$, say, an elliptic curve?

• If I properly understand your notation, you are specifying a dense Zariski open subset $U$ and a regular morphism $\rho:U\to \textbf{GL}_n$ such that for the open subset $V=(U\times U)\cap m^{-1}(U)$, the restriction $\rho\circ m|_V$ equals $m\circ (\rho\times \rho)|_V$. In that case, for every $x\in U$, $\rho$ extends uniquely and regularly to the open subset $x^{-1}U$ via $y\mapsto \rho(x)^{-1}\rho(xy)$. Thus, the maximal open domain of definition $U$ is stable for all translates $x^{-1}U$. Assuming that $G$ is a finite type group scheme, it follows that $U$ equals $G$. – Jason Starr Feb 4 '17 at 9:47
• @JasonStarr Mmm is it obvious that translates of $U$ by $x^{-1}$ with $x\in U$ cover everything? If yes, this settles it of course, so could you make this an answer? – მამუკა ჯიბლაძე Feb 4 '17 at 9:54
• For every $y\in G$, since both $U$ and $Uy^{-1}$ are dense open subsets of $G$, $U\cap (Uy^{-1})$ is also a dense open. In particular, $U\cap (Uy^{-1})$ is nonempty. For every $x\in U\cap (Uy^{-1})$, $x$ is an element of $U$ such that $y$ is in $x^{-1}U$. Thus, the translates $x^{-1}U$ cover $G$. – Jason Starr Feb 4 '17 at 10:04

Edit. Thanks to nfdc23 for pointing out a couple of corrections. As nfdc23 points out, these kinds of things are important in passing from a birational group law to a (regular) group law, so they may go back to Weil. Also, I vaguely remember something about some of this in SGA 3, so that might also be a reference.

As requested, I am posting my comments as an answer. I cannot quite remember who first proved these things, but it might have been Rosenlicht (I think that is where I learned them, anyway).

Let $G$ be a group scheme over $k$ (edit. and assume that $G$ is of finite type, and also assume that $k$ is perfect to avoid working over Artinian rings). Denote by $m:G\times_{\text{Spec}\ k} G \to G$ the multiplication morphism. Denote by $i:G\to G$ the group inverse morphism. Denote by $m':G\times_{\text{Spec}\ k}G \to G$ the composition $m\circ (i\circ\text{pr}_1,\text{pr}_2)$. For every dense open subscheme $U$ of $G$ (edit. and assume that $U$ is schematically dense), the morphism $$m'|_{U\times U}: U\times U \to G,$$ is surjective. This can be checked on geometric points. Thus, let $y\in G(\text{Spec}\ k)$ be an element. Since $U$ and $Uy^{-1}$ are both dense open subschemes of $G$, also $U\cap (Uy^{-1})$ is a dense open. For every geometric point $x$ in $U\cap (Uy^{-1})$, $x$ is in $U$ and $z=xy$ is also in $U$. Thus $(x,z)$ is in $U\times U$, and $m'(x,z)$ equals $y$. Therefore $m'|_{U\times U}$ is surjective.

Now let $H$ be a separated group scheme over $k$ with multiplication morphism $n:H\times_{\text{Spec}\ k}H\to H$. Let $U\subset G$ be a dense open subscheme. Let $\rho_U:U\to H$ be a $k$-morphism. On the dense open subscheme of $G\times_{\text{Spec}\ k} G$, $V=(U\times_{\text{Spec}\ k} U)\cap m^{-1}(U)$, assume that the composition $\rho_U\circ m|_V$ equals the composition $n\circ (\rho_U\times \rho_U)|_V$. Then there exists a unique morphism $\rho:G\to H$ that extends $\rho_U$, and this is a morphism of $k$-group schemes.

Since $H$ is separated and since $\rho_U$ is defined on a dense open, all extensions are unique if they exist. For the same reason, for the unique extension $\rho$, since $\rho_U\circ m|_V$ equals $n\circ (\rho_U\times \rho_U)|_V$, also $\rho\circ m$ equals $n\circ (\rho\times \rho)$. Thus, the unique extension is a morphism of $k$-group schemes. Thus, it suffices to prove that local extensions exist. Again, using uniqueness to confirm the cocycle condition for fpqc descent, it suffices to construct the extension after an fpqc base change.

Probably it is best to use directly the faithfully flat morphism $m'|_{U\times U}$. However, conceptually it is simpler to use the base change from $k$ to its algebraic closure. Then $U(k)$ is dense. Thus, there exist elements $x\in U(k)$ such that the translates $x^{-1}U$ cover $G$. On each open $x^{-1}U$, the unique extension of $\rho_U|_{U\cap (x^{-1}U)}$ is defined by $$\rho(y) = \rho_U(x)^{-1}\cdot \rho_U(xy)$$ (I am supressing $m$ and $n$ in this formula).

• You want $G$ to be (at least locally) of finite type over $k$, and "dense open" should be "schematically dense open" (if allowing non-smooth group schemes, as you are doing). Group schemes over a field are always separated, since the identity section is a closed immersion (as rational points over a field are always closed). Lastly, the appeal to fpqc descent (with finite extensions of $k$, maybe not separable, to avoid the non-noetherian $\overline{k}\otimes_k\overline{k}$) entails working over artin rings, so a non-reduced base, and hence one has to be careful about "density" considerations. – nfdc23 Feb 4 '17 at 15:03
• The fact that (in older language, with smooth groups) "rational representations" extend uniquely to actual representations might pre-date Rosenlicht since I think Weil worked earlier with "birational group laws" (which leads one into many of these same issues). However, I'm not entirely sure of the chronology. It may be hard to attribute this to one specific person first. – nfdc23 Feb 4 '17 at 15:05
• Does the first comment by @nfdc23 mean that there might still exist some kind of nontrivial "infinitesimal" rational representations? – მამუკა ჯიბლაძე Feb 5 '17 at 21:19
• It seems very unlikely. If over an artin local ring instead of a field then one has to assume the group scheme is flat (anything less would be unlikely to be of any interest, let alone be workable) and be more precise about a useful meaning for "rational function", sure to entail $U$ being relatively schematically dense. Thus, by the Zariski-density of the Cohen-Macaulay locus of the special fiber there would exist a plentiful supply of "finite flat quasi-sections" (i.e., points valued in finite flat algebras) over the artin local base, enough to do translation arguments. – nfdc23 Feb 6 '17 at 0:18