[Math] the number field analogue of the Narasimhan-Seshadri theorem

nt.number-theory

In his famous 1940 letter from prison in Rouen to his sister Simone, André Weil talks about the analogy between number fields and functions fields (in one variable) over finite fields, and the analogy between these functions fields and functions fields over $\mathbf{C}$ (or equivalently compact connected curves over $\mathbf{C}$). This letter is reproduced in his Scientific papers and has been recently translated into English (Notices of the AMS 52(3) 2005).

Question What is the number field analogue of the Narasimhan-Seshadri theorem (Stable and unitary vector bundles on a compact Riemann surface. Ann. of Math. (2) 82 1965 540–567) ?

Addendum (in response to Felipe's comment) The original paper of Narasimhan and Seshadri is available on JSTOR. An excerpt from their introduction : D. Mumford has defined the notion of a stable vector bundle on a compact
Riemann surface $X$ and proved that the set of equivalence classes of stable
bundles (of fixed rank and degree) has a natural structure of a non-singular,
quasi-projective, algebraic variety [13]. We prove in this paper that, if $X$ has
genus $\ge2$, the stable vector bundles are precisely the holomorphic vector bundles
on $X$ which arise from certain irreducible unitary representations of suitably defined fuchsian groups acting on the unit disc and having $X$ as quotient
(Theorem 2, $\S12$).
[…]
A particular case of our result is that a holomorphic vector bundle of
degree zero on $X$ is stable if and only if it arises from an irreducible unitary
representation of the fundamental group of $X$. As a consequence one sees that
a holomorphic vector bundle on $X$ arises from a unitary representation of the
fundamental group of $X$ if and only if each of its indecomposable components
is of degree zero and stable.

Their main result is summarised by Atiyah (MR0170350) and Le Potier (Séminaire Bourbaki Exposé 737) as follows :

Atiyah: Let $X$ be a compact Riemann surface. If $W$ is a (holomorphic) vector bundle of rank $n$ over $X$ we define $d(W)$ to be the degree of the associated line bundle $\bigwedge^n W$. A bundle $W$ is stable, in the sense of Mumford, if $(\mathrm{rank}W)d(V)<(\mathrm{rank}V)d(W)$ for all proper sub-bundles $V$ of $W$. According to Mumford [Proc. Internat. Congr. Mathematicians (Stockholm, 1962), pp. 526–530, Inst. Mittag-Leffler, Djursholm, 1963], the set of isomorphism classes of stable bundles of rank $n$ and degree $q$ over $X$ has a natural structure of an algebraic variety. In this paper the authors give a complete characterization of stable bundles in terms of unitary representations of a certain discrete group (provided genus $X$ $≥2$).

Their main theorem runs as follows. Given integers n and q, with $-n< q \le0$, we can choose (i) a discrete group $\pi$ acting on a simply connected Riemann surface $Y$ with $Y/\pi=X$ and with the map $p:Y\to X$ being ramified over only one point $x_0\in X$; (ii) a representation $\tau:\pi_{y_0}→\mathrm{GL}(n,\mathbf{C})$ of the isotropy group of $\pi$ at a point $y_0\in p^{−1}(x_0)$ by scalars such that the following holds. A vector bundle over $X$ of rank $n$ and degree $q$ is stable if and only if the corresponding sheaf is isomorphic to a sheaf of the form $p_∗^\pi(\mathbf{V})$, where $\mathbf{V}$ denotes the $\pi$-sheaf of holomorphic mappings $Y\to V$, $V$ is an irreducible unitary representation of $\pi$ coinciding with $\tau$ when restricted to $\pi_{y_0}$, $p_∗$ is the direct image functor and $p_∗^\pi$ denotes the subsheaf invariant under $\pi$. Moreover, two such stable bundles are isomorphic on $X$ if and only if the corresponding unitary representations of $\pi$ are equivalent.

It should be observed that the inequality $-n< q \le0$ presents no essential restriction since it can always be realized by tensoring with a line bundle $L$ and, on the other hand, the definition of stable bundle shows that $W$ is stable if and only if $W\otimes L$ is stable.

Le Potier: En 1965 Narasimhan et Seshadri établissaient une correspondence bijective entre l’ensemble des classes d’équivalence de représentations unitaires irréductibles du groupe fondamentale $\pi$ d’une surface de Riemann compacte $X$, et l’ensemble des classes d’isomorphisme de fibrés vectoriels stables de degré $0$ sur $X$ : ils associent à une representation $\rho:\pi\to\mathbf{U}(r)$ le fibré vectoriel holomorphe $E_\rho$ défini par
$$
E_\rho=\tilde X\times_\pi\mathbf{C}^r
$$
où $\tilde X$ est le revêtement universel de $X$, et où le produit ci-dessus est le quotient de $\tilde X\times\mathbf{C}^r$ par l’action de $\pi$ définie par
$(\gamma,(x,v))\mapsto(x\gamma^{-1},\gamma v)$ pour $\gamma\in\pi$ et $(x,v)\in \tilde X\times\mathbf{C}^r$.

Best Answer

Theorem of Narasimhan and Seshadri is a special case of what Carlos Simpson calls nonabelian Hodge theory developed by Hitchin and Simpson. This theory was generalized to the characteristic $p$ case in the paper of Ogus and Vologodsky, Nonabelian Hodge Theory in Characteristic p. Hope this helps.

Update: Below, on Alexander's request, is a brief explanation of relation between nonabelian Hodge theory and NS theorem. Consider vector bundles with vanishing 1st and 2nd Chern classes (I will call this condition ($\star$)), then the story in higher dimensions is exactly the same as for complex curves. The detailed explanation is in the pages 12-19 of Simpson's 1992 paper [S1992]. From there, it follows that flat unitary connections correspond exactly to vanishing Higgs fields (subject to ($\star$)). Briefly, every semistable Higgs bundle $E=(V, \bar\partial, \theta)$ has a hermitian YM metric $K$. Define the connection $D_K$ (as in [S1992], page 13), then then $D_K$ is flat (subject to ($\star$), page 17 of [S1992]).

If Higgs field $\theta$ vanishes then $D_K=\partial_K+ \bar\partial$ and, hence, by definition of $\partial_K$, connection $D_K$ preserves the metric $K$. Thus, our bundle reduces to a flat unitary bundle. Conversely, if bundle is flat unitary (with unitary metric denoted $K$) then the associated (multivalued) map $\Phi_K$ defined on page 16 of [S1992], is constant, so it has zero derivative. But its derivative is $0=d\Phi_K=\theta+ \bar\theta$ (here $\theta$ is the Higgs field determined by $K$). Since $\theta, \bar\theta$ have different types, the only way we can have $\theta+ \bar\theta=0$ is that $\theta=0$.

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Free Subquotient of Galois Representations from Hida Theory

$\newcommand\T{\mathbf{T}_{\mathfrak{m}}}$ $\newcommand\Q{\mathbf{Q}}$ $\newcommand\m{\mathfrak{m}}$ $\newcommand\F{\mathbf{F}}$ $\newcommand\Frob{\mathrm{Frob}}$ $\newcommand\rhobar{\overline{\rho}}$ $\newcommand\eps{\epsilon}$

First, as Professor Emerton mentions, the construction of $L^{+}$ you gave is not necessarily free over $\T$. Thus, I will interpret your question as asking the following: does there exist an exact sequence:

$$0 \rightarrow L^{+} \rightarrow (\T)^2 \rightarrow L^{-} \rightarrow 0$$ of $\T[G_{\Q_p}]$-modules where $L^{+}$ and $L^{-}$ are free $\T$-modules of rank one.

( Edit Perhaps this extra remark might be useful. Suppose that $L = (\T)^2$ admits a free rank one quotient $L^{-}$. Since $L^{-}$ is free, it admits a section $L^{-} \rightarrow L$, and hence the kernel $L^{+}$ of $L \rightarrow L^{-}$ is also free. Thus the existence of a free rank one quotient asked for in the question is equivalent to the existence of the exact sequence above.)

The answer to this question, in general, is no. The following argument is implicitly contained in papers of Wiese on the failure of multiplicity one and weight one forms.

The action of $G_{\Q_p}$ on $L^{+}$ is unramified and so acts via $G_{\mathbf{F}_p}$. Thus $\Frob_p$ acts on a basis vector as multiplication by some element of $\T$. Since $\T$ is determined by its action on classical eigenforms, one may identify this element with the Hecke operator $U$. In particular, $U \in \T$ (it wasn't clear whether your $\T$ included $U$ or not).

The exact sequence remains exact after tensoring with $\T/\m$, for dimension reasons. It follows that the sequence is split as a sequence of $\T$-modules. Hence it remains exact after quotienting out by any ideal of $\T$.

Suppose that $\rhobar: G_{\Q} \rightarrow \mathrm{GL}_2(\F_p)$ is irreducible and modular (mod-$p$) of weight $1$. Suppose, moreover, that $\rhobar(\Frob_p)$ acts by a scalar $\lambda$. Associated to $\rhobar$ is a mod-$p$ weight $1$ form $f = \sum a_n q^n \in \F_p[[q]]$. If $A$ is the Hasse invariant, then then $Af$ and $f^p$ are both mod-$p$ modular forms of weight $p$. One can check that all elements of the $\F_p$-vector space $\{Af,f^p\}$ are eigenvalues for all the Hecke operators $T_l$ for $(l,p) = 1$, but the operator $T$ (and so $U$, which is the same as $T$ in weight $> 1$) satisfies $(U - \lambda)^2 = 0$ but does not act by a scalar. Since $U$ acts invertibly on this vector space, it gives rise to a surjective map: $$\T \rightarrow \F_p[\eps]/\eps^2,$$ where the image of $T_l$ lands in $\F_p$ for all $(l,p) = 1$, but $U$ does not act by a scalar. Let $I$ be the kernel.

The Galois representation on $(\T)^2/I \simeq (\F_p[\eps]/\eps^2)^2$ is equal to $\rhobar \oplus \rhobar$. This follows from a result of Boston-Lenstra-Ribet, since $T_l$ is acting by a scalar for each $(l,p) = 1$. It follows, by assumption, that the action of $G_{\Q_p}$ on $L^{+}/I L^{+} \simeq \T/I$ must also be trivial, because this is a sub-representation of $\rhobar \oplus \rhobar$. On the other hand, as we have seen, the action of Frobenius on $L^{+}$ and thus $L^{+}/I L^{+} = \T/I$ is given by $U$, which is acting non-trivially $\T/I$ by the construction of $I$. This is a contradiction.

Such representations $\rhobar$ exist (for example, with $p = 2$, and level $\Gamma_0(431)$) as mentioned in Professor Emerton's answer.

Computing First Few Non-Trivial Zeros of Zeta Function of a Number Field – Number Theory

Step I: Put the degree 24 polynomial into Magma, make it a number field, and call LSeries on it. This divides the $L$-function into a product of 7 distinct ones (Dokchitsers code, under an attribute called "prod" on the L-series object), given by Artin representations. So my plan was to compute zeros for each of these $L$-functions, with them being a subset of those of $\zeta_M$ naturally, and recombine. As representations for $SL(2,3)$ this is as $$1\oplus \omega\oplus\bar\omega\oplus 2\tau_2\oplus 2\tau_2\omega\oplus 2\tau_2\bar\omega\oplus 3\kappa_3.$$ Or as an $L$-function product $$\zeta_M=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\tau_2)^2\cdot L(\tau_2\omega)^2\cdot L(\tau_2\bar\omega)^2\cdot L(\kappa_3)^3.$$

EDIT: Oh I see now, you wanted it for $M$ w/o assuming analyticity of Artin constituent parts, but Magma automatically dissembles it, and assumes. But now at this point, I see you were trying to avoid this decomposition perhaps, but then I really don't see how you could work with $\zeta_M$ directly, as the conductor is too big, being $163^{16}$. From the standpoint of scientific evidence, it is probably enough that CheckFunctionalEquation in Magma for each of the above constituents gives an answer indistinguishable from zero, though the feel of zeros is also nice.

Another way to test individual analyticity is to use subfields. There is a unique (up to isomorphism) quartic subfield $K_4\subset M$, with $$\zeta_K=\zeta\cdot L(\kappa_3).$$ The conductor of $\zeta_K$ is small enough to compute with it directly (no decomposition), so the analyticity of $L(\kappa_3)=\zeta_K/\zeta$ can be checked (numerically) by seeing if Riemann zeros are a subset of those of $\zeta_K$. EDIT: Of course, $\kappa_3$ is induced from the a nontrivial linear character of the $Q_8$ subgroup of $SL(2,3)$, so analyticity already follows by induction.

There is also a unique (up to isomorphism) sextic subfield $L_6\subset M$ with $$\zeta_L=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\kappa_3),$$ so the linear parts can be isolated here by $\zeta_L/\zeta_K$, though maybe superfluous ("easy" theorem)? In that matter, already the cubic subfield $C_3\subset M$ has $\zeta_C=\zeta\cdot L(\omega)\cdot L(\bar\omega)$ if desired.

There is a unique (up to isomorphism) octic subfield $N_8\subset M$ with $$\zeta_N=\zeta\cdot L(\tau_2\omega)\cdot L(\tau_2\bar\omega)\cdot L(\kappa_3).$$ The integer ring has discriminant $163^4$ so $\zeta_N$ is likely still computable directly. By quotient $\zeta_N/\zeta_K$, this can tell about the product of conjugate 2-dimensional $L$-functions. EDIT: see below for another idea, using twists of $\zeta_N$.

Finally the unique duodecic subfield $R_{12}\subset M$ has $$\zeta_R=\zeta\cdot L(\omega)\cdot L(\bar\omega)\cdot L(\kappa_3)^3.$$ So this gives nothing new, and the discriminant is too large anyway.

Note that none of the parts has $L(\tau_2)$ directly, only $\zeta_M$ itself that is too hard to compute. EDIT: However, by Rankin-Selberg I think(?) it follows that the analyticity of $L(\tau_2)$ is equivalent to that after twisting by $\omega$ to get $L(\tau_2\omega)$.

Answer?: For that matter, recurring to the octic subfield $N_8$, instead of using just the Dedekind $\zeta$-function of $N_8$, twisting it by $\omega$ could be profitable, achieving $$L(\sigma_8\omega)=L(\omega)\cdot L(\tau_2\bar\omega)\cdot L(\tau_2)\cdot L(\kappa_3),$$ $$L(\sigma_8\bar\omega)=L(\bar\omega)\cdot L(\tau_2)\cdot L(\tau_2\omega)\cdot L(\kappa_3),$$ where $\sigma_8=1\oplus\tau_2\omega\oplus\tau_2\bar\omega\oplus\kappa_3$ is for the Dedekind of $N_8$. Also $\kappa_3\omega=\kappa_3$ nicely. Noting here the twisting gives $$L(\sigma_8)L(\sigma_8\omega)L(\sigma_8\bar\omega)=\zeta_M,$$ this could provide a way to compute $\zeta_M$, as each part of the left is known as analytic I presume. The conductor of $L(\sigma_8)$ is $163^4$, and that for the twists is $163^6$. Or one can avoid $\zeta_M$ alternatively, as my computation is $$L(\tau_2)^2={L(\sigma_8\omega)L(\sigma_8\bar\omega)\zeta^2\over\zeta_{N_8}\zeta_{L_6}},$$ where each factor on the right should be known (easily?) to be holomorphic away from the $\zeta$-pole. Note this exploits the linear characters of $SL(2,3)$, and you have none for your next case of $SL(2,5)$.

In all cases, zeros need to be computed, and the right tool is L-calc. But I don't know if it is really feasible to go too far for $L(\sigma_8\omega)$ of conductor $163^6$, without intensive effort.

Part II: Zeros of $L(\tau_2)$ computation (example): I compute its first few zeros for the 2-dimensional representation of real character. For this representation, with $10^5$ coeffs (taking 6sec in Magma), I exported these to Lcalc (a somewhat hackish tool of M. O. Rubinstein, included in Sage I suppose but I did it direct), which returns after 12 seconds, listing the imaginary parts of the first 10 zeros on the half-line (also proving RH up to that point by A Turing method):

 0
 0.99014365233
 1.38830360231
 2.35103235859
 3.45296629741
 4.32708276131
 4.73989005257
 5.42392092883
 5.99574967707
 6.70167394143

The first zero is at $s=1/2$ as the sign is $-1$. The second is about $s\approx 1/2+0.990i$

For reference, here are the L-calc settings I used, hackish as I say:

1
0
100000
0
2
.5
0.5 0.0
.5
0.5 0.0
51.884511447957879460656106859439682023
-1.0 0.0
0

As explained in their help, the first "1" says the coeffs are integers, the second "0" says they are not special, the 3rd "100000" says they are that many, the 4th "0" says not periodic, the 5th "2" says degree 2, the 6th ".5" and "0.5 0.0" say the form of the gamma factor, the "51.884" is $\sqrt{163^2/\pi^2}$ as the analytic conductor, the "-1.0 0.0" is the sign, and the final "0" says no poles. Then the 100000 coeffs are given as integers.

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Free Subquotient of Galois Representations from Hida Theory

Computing First Few Non-Trivial Zeros of Zeta Function of a Number Field – Number Theory

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