$\DeclareMathOperator{\Q}{\mathbb{Q}}$Call "L-rig" any class $\mathcal{L}$ of L-functions of automorphic representations of $\operatorname{GL}_{n}(\mathbb{A}_{\Q})$ for some $n$ belonging to the Selberg class that be closed under both the usual product (which we'll denote by $\times$) and the Rankin-Selberg convolution (which we'll denote by $\otimes$), containing the respective neutral elements $s\mapsto 1$ and $\zeta$, and fulfilling the abstract algebraic properties making $(\mathcal{L},\times,\otimes,s\mapsto 1,\zeta)$ a rig (ring without negatives).

Does the main result in automorphy of $m$-fold tensor products of GL(2), Dieulefait 2020 imply the existence of infinitely many non trivial L-rigs?

Moreover, denoting by $\mathcal{M}$ the maximal L-rig under inclusion, can we see it as the analogue for L-rigs of the separable closure of a field? If yes, would it make $\operatorname{Aut}(\mathcal{M})$ isomorphic to some absolute Galois group like, say, $\operatorname{Gal}(\bar{\Q}/\Q)$?

Edit October 25th, 2020: there are at least 3 different L-rigs, namely the trivial one $\mathcal{L}_{0}$ generated by $s\mapsto 1$ and the Riemann Zeta function, $\mathcal{M}$ and its sub-L-rig $\mathcal{D}$ consisting of all self-dual L-functions. Assuming $\operatorname{Aut}(\mathcal{M})$ is isomorphic to some absolute Galois group and the analogue for L-rigs of the fundamental theorem of Galois theory, this absolute Galois group can't be finite (as all such Galois groups are of order at most $2$). It may then be possible to prove that $\operatorname{Aut}(\mathcal{M})$ is profinite.

Edit October 30th, 2020: perhaps a way to show we face a profinite group would be to prove that $\displaystyle{\mathcal{M}}$ is defined by a filtration $(\mathcal{L}_{i}):={(\mathcal{L}(F_{i}))}_{i\in I}$ so that $\mathcal{M}=\varinjlim_{i\in I}\mathcal{L}_{i}$ and $\displaystyle{\operatorname{Aut}(\mathcal{M})\cong\varprojlim_{i\in I}\operatorname{Gal}\left(\frac{\mathcal{L}_{i}}{\mathcal{L}_{0}}\right)}$, where $\mathcal{L}(F)$ is the L-rig generated by $F$, the sequence of intermediate L-rigs $\mathcal{L}_{i}$ being analogues of Galois extensions of $\mathcal{L}_{0}$ defined above.

More exactly the considered Galois group should be $\operatorname{Gal}(\mathcal{K}_{\mathcal{L_{i}}}/\mathcal{K}_{\mathcal{L}_{0}})$ with $\mathcal{K_{L}}$ the field generated by the L-ring $\mathcal{L}$, that we can call an "L-field". Proving $\mathcal{K}_{\mathcal{L}_{0}}\cong\mathbb{Q}$ may imply that $\operatorname {Aut}(\mathcal{M})\cong\operatorname{Gal}(\bar{\mathbb{Q}}/\mathbb{Q})$.

Edit November 1st, 2020: following the answer by nguyen quang do in
https://math.stackexchange.com/questions/2782069/abstract-properties-of-the-absolute-galois-group-over-mathbbq and assuming $\operatorname{Aut}(\mathcal{M})$ has the structure of an absolute Galois group, then it is a profinite group.

Edit November 11th 2020: as $\mathcal{L}_{0}$ is the L-ring generated by the neutral elements, $\mathcal{K}_{\mathcal{L}_{0}}$ is the L-field generated by those neutral elements, and as such is isomorphic to $\mathbb{Q}$. Now, the maximality of $\mathcal{M}$ implies that if the extension of $\mathbb{Q}$ isomorphic to $\mathcal{K}_{\mathcal{M}}$ is algebraic, then it is "its" algebraic closure $\bar{\mathbb{Q}}$.

Edit December 12th 2020: define the "symmetry group" $\operatorname{Sym}(F_{\pi})$ of an element $F_{\pi}:s\mapsto L(\pi,s)=\prod_{v}L_{v}(\pi,s)$ of $\mathcal{M}$ as the stabilizer thereof under the action of $\operatorname{Aut}(\mathcal{M})$ on $\mathcal{M}$. Then any permutation $\sigma$ of the places $v$ leaves $F_{\pi}$ invariant, so that if it induces an automorphism of $\mathcal{M}$, that we'll denote by $\phi_{\sigma}$, the latter induces an isomorphism between $\mathbb{Q}_{v}$ and $\mathbb{Q}_{\sigma(v)}$ as a morphism between fields. But $v\neq v'\Longrightarrow\mathbb{Q}_{v}\not\cong\mathbb{Q}_{v'}$ and in particular, $\sigma$ induces an automorphism of $\mathbb{R}$, (when $v$ is the archimedean place), hence either the identity or the complex conjugation. Hence $\operatorname{Sym}(F_{\pi})\cong\operatorname{Gal}(\mathbb{C}/\mathbb{R})$ if $\pi$ is self-contragredient, and is trivial otherwise.

  • $\begingroup$ What if I start with 1, zeta, and the Dirichlet L-function of some quadratic character. Is the rig generated by those three objects just all finite products of those functions (because the Rankin-Selberg of that Dirichlet L-function with itself is the Riemann zeta function)? $\endgroup$ Oct 31, 2020 at 21:42
  • $\begingroup$ Logically, yes. $\endgroup$ Oct 31, 2020 at 22:03
  • 2
    $\begingroup$ So there are infinitely many. $\endgroup$ Oct 31, 2020 at 22:42
  • $\begingroup$ Feel free to post this remark as an answer. $\endgroup$ Nov 1, 2020 at 10:38
  • $\begingroup$ I guess an analogous reasoning can be made with Dirichlet characters of any order? In that case there would be infinitely many non isomorphic L-rigs. $\endgroup$ Nov 1, 2020 at 10:40

1 Answer 1


The Rankin-Selberg convolution of a quadratic Dirichlet L-function with itself is the Riemann zeta function. Therefore the rig generated by $\{1, \zeta(s), L(s, \chi_d)\}$ consists of all finite products (and powers) of $\zeta(s)$ and $L(s, \chi_d)$. In particular, there are infinitely many L-rigs.

If you start with $\{1, \zeta(s), L(s, \chi)\}$ where $\chi$ is a primitive Dirichlet character, then Rankin-Selberg convolution gives you $L(s, \chi^j)$ for any positive integer $j$. That L-rig is generated by a finite set, depending on the order of $\chi$. So you get infinitely many non-isomorphic L-rigs.

Note that if $\chi^j$ is not primitive, then $L(s, \chi^j)$ should be interpreted as the Dirichlet L-function of the inducing primitive character.

  • 1
    $\begingroup$ Thank you very much. Would this imply that $\operatorname{Aut}(\mathcal{M})\cong\operatorname{Gal}\left(\bar{\mathbb{Q}}/\mathbb{Q}\right)$? $\endgroup$ Nov 1, 2020 at 13:32
  • 9
    $\begingroup$ I think the real issue is whether or not there is anything to be gained by thinking about the collection of all L-rigs. Does it reveal anything beyond what we already believe (namely, that we know the set of L-functions is closed under products and we believe it is closed under R-S convolution)? I don't see it pointing in a useful direction. $\endgroup$ Nov 1, 2020 at 14:16
  • $\begingroup$ Well, I see one. If we consider the torsion subgroup of $\operatorname{Aut}(\mathcal{M})$ that preserve a given L-function $F$, it may be possible to prove this "symmetry group" of $F$ is isomorphic to the group of isometries of the complex plane preserving the multiset of its non trivial zeros as, loosely speaking, preserving an L-function is equivalent to preserving its (non trivial) zeros. And RH can be seen as the "minimality" of this isometry group (the latter being isomorphic to $C_{2}$ rather than to the Klein group as far as zeta is concerned). $\endgroup$ Nov 1, 2020 at 14:55
  • 3
    $\begingroup$ Those L-rigs may match up with products of cyclic groups, but I don't see there is any deep meaning to that. $\endgroup$ Nov 12, 2020 at 18:30
  • 9
    $\begingroup$ I don't see any useful mathematical content in that apparent connection, and I stand by my previous comment that these apparent connections are not pointing in a useful direction. $\endgroup$ Nov 12, 2020 at 19:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.