This question is a follow-up to Galois classes of L-functions. My goal here is to make things clearer.

Definition 1

Let $A$ be a subclass of the Selberg class containing $s\mapsto 1$, closed under products, and such that every element of $A$ can be factored in a unique fashion in a product of primitive elements of $\mathcal{S}$, these primitive elements belonging to $A$. Such a subclass of $\mathcal{S}$ will be called a Galois class of L-functions.

Definition 2

Let $A$ be a Galois class of L-functions. $T$ is an automorphism of $A$ iff the following properties simultaneously hold true:

1) $T$ is a bijective map from A to itself
2) $T$ maps a primitive element of $A$ to a primitive element of $A$
3) for all $F$ in $A$, the degree of $F$ and the degree of $T(F)$ are the same
4) for all $F$, $G$ in $A$, $T(F.G)=T(F).T(G)$

Let $M$ be the maximal Galois class of L functions, $F$ an element of $M$. Let's denote $G_{S}(F)$ the group of complex isometries preserving globally the multiset of non trivial zeros of $F$, $G_{F}$ the group of automorphisms of $M$ preserving $F$.

Hadamard's factorization theorem says that knowing $F$ is equivalent to knowing its multiset of non trivial zeros. Therefore, preserving this multiset is equivalent to preserving $F$.
My question is: does this mean that among the automorphisms of $M$ preserving $F$, there exist "good" automorphisms of $M$ preserving $F$ such that if $G_{F}^{good}$ is the group of such "good" automorphisms, then there exists a group homomorphism $\rho: G_{F}^{good}\to G_{S}(F)$ that is actually an isomorphism?

Thanks in advance.

EDIT September 18th 2013: maybe one could establish the existence of the desired isomorphism asking the question in the framework of category theory, for example showing that the two categories $\mathcal{C}$ and $\mathcal{D}$ consisting of elements of $M$ or their multisets of zeros with the desired morphisms between these objects are equivalent, hence showing that the corresponding groups of automorphisms are isomorphic. In other words, I expect the groups $Aut_{\mathcal{C}}(F)$ and $Aut_{\mathcal{D}}(Z(F))$ for $F$ an object of $\mathcal{C}$ and $Z(F)$ the corresponding object of $\mathcal{D}$ to be, respectively, equal to $G_{F}^{good}$ and $G_{S}(F)$. Is there any hope to prove this rigorously?

  • $\begingroup$ Is $G_{S}(F)$ nonempty for any nontrivial Selberg L function? $\endgroup$ – Will Sawin Sep 15 '13 at 16:10
  • $\begingroup$ Of course. At least it contains the identity. It also contains the complex conjugation if $F=\bar{F}$. $\endgroup$ – Sylvain JULIEN Sep 15 '13 at 16:21
  • $\begingroup$ Are there any other examples? $\endgroup$ – Will Sawin Sep 15 '13 at 20:20
  • $\begingroup$ I think a positive answer to my question would establish that the order of $G_{S}(F)$ can't be greater than $2$. $\endgroup$ – Sylvain JULIEN Sep 15 '13 at 20:33

The size of the automorphism group of a nontrivial Selberg $L$-function is at most $2$.

The reason is the distribution of zeroes of elements of the Selberg class. Such functions have no zeroes for $Re s>1$. For $Re s<0$, they have only the trivial zeroes. These live in a horizontal strip on linear progressions. In the critical strip, we have the non-trivial zeroes. There are infinitely many of these, and they are denser than any finite set of critical progressions.

Thus we can "see" the location of the critical strip and the region of trivial zeroes by looking at the function, so any automorphism must fix these locations. This implies that the automorphism cannot be a rotation, because otherwise it would not fix the direction of the axis running through the trivial zeros. It cannot be a a translation, because otherwise it would not fix the intersection of the trivial zeroes axis with the critical strip. It cannot be a reflection through any line other than a horizontal line through the trivial zeroes, because again it would not fix the direction of the trivial zeroes.

Hence it must be of the form $z \to \bar{z}+it$ for $t$ real. I am not sure whether we can rule out the case $t neq 0$. We can if it has a pole, since the pole must be unique and at $s=1$, but do all Selberg L-functions have poles? It sounds plausible but in my brief research I haven't seen a proof.

I'm not sure whether this means that there is an automorphism of the Galois class of L-functions. I don't see any reason why one couldn't have a conjugate-asymmetric class containing a conjugate-symmetric function.

  • $\begingroup$ Thanks for your answer. Nevertheless I'm not interested in the trivial zeroes, to me the big deal is to determine the automorphism group of the multiset of non-trivial zeroes of a given element of the Selberg class. Maybe the fact that such an L function is determined by both its full multiset of zeroes and its multiset of non trivial zeroes implies that this automorphism group is of order at most 2 too, but I'm not quite sure of it yet, that's why I can't accept this answer now. I will if I get a proof that "your" automorphism group and "mine" are isomorphic. $\endgroup$ – Sylvain JULIEN Sep 20 '13 at 19:11
  • $\begingroup$ So the multiset of nontrivial zeroes lives in a strip. Hence its automorphism group can consist of translations along that strip, reflections through that strip, and reflections through an axis perpendicular to the strip. The first type cannot exist, since it would imply the strip is periodic, hence the number of zeroes in a large interval would be approximately linear in the length, but in fact it is approximately nonlinear. The second kind of reflection must exist, due to the functional equation. The third kind might or might not. So the group is of order $2$ or $4$. $\endgroup$ – Will Sawin Sep 20 '13 at 20:23
  • $\begingroup$ Or of order $1$ or $2$ if the maps $s\mapsto 1-\bar{s}$ and $s\mapsto s$ have the same action on every non trivial zero and thus are identified. $\endgroup$ – Sylvain JULIEN Sep 20 '13 at 20:32
  • $\begingroup$ You defined $G_s(F)$ to be the group of complex isometries, not the group of permutations of the zeroes coming from complex isometries. $\endgroup$ – Will Sawin Sep 20 '13 at 21:25
  • $\begingroup$ I defined it as the group of complex isometries that preserve globally the multiset of non trivial zeroes, hence, as you say, the group of permutations of the (non trivial) zeroes coming from complex isometries. $\endgroup$ – Sylvain JULIEN Sep 20 '13 at 21:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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