We illustrate the relation between local and global epsilon-factors for
Dirichlet characters (or Artin representations).
Pick any Dirichlet character χ; in this example we take the unique odd one
of conductor 5.19, order 6 and Im χ(2)>0.
We compute its local Galois representations at all the bad places: 5, 19
and ∞. (See Para
Now compute the corresponding local root numbers - the way the conventions
are set up for Artin representations,
at ∞ it is the root
number of the Hodge structure that enters the functional equation.
The global root number is the sign of the L-series of χ,
and we check that it agrees with the one determined numerically
from the functional equation:
Every semisimple Galois representation A over a p-adic field K can be
uniquely recovered from its Euler factors over the extensions of K
(
[DD15, Thm. 1]). We illustrate this with a Galois representation
attached to an elliptic curve
E/K: y2 = x3 - 26x, qquad K=Q13.
> K:=pAdicField(13,20);
> E:=EllipticCurve([K|-26,0]);
> A:=GaloisRepresentation(E);
> Degree(A),IsSemisimple(A);
2 true
Without looking at A, let us reconstruct it from its Euler factors
over extensions of K. First we determine the inertia group
InertiaGroup(A) by looking for a field over which A is unramified
(i.e. E has good reduction). As the residue characteristic is >3,
one of the fields Q
13(root d of(13)) will do, for d=1, 2, 3, 4 or 6.
> R<x>:=PolynomialRing(K);
> [EulerFactor(BaseChange(A,ext<K|x^d-13>)): d in [1,2,3,4,6]];
[ 1, 1, 1, 13*x^2 + 4*x + 1, 1 ]
We see that the representation becomes unramified over L=Q
13(root 4of(13)),
a cyclic extension of degree 4, but not over its subfields.
So the inertia group must be C
4,
(InertiaGroup)(A) isomorphic to IL/K
= Gal(L/K) isomorphic to C4.
Over L the representation U=ResL A is unramified,
and there it is determined by its Euler factor. It is a sum of two unramified
characters, FrobL - 1 |-> - 2∓ 3i.
> L:=ext<K|x^4-13>;
> f1:=EulerFactor(BaseChange(A,L));
> U:=UnramifiedRepresentation(L,f1); U;
2-dim unramified Galois representation Unr(1+4*x+13*x^2) over
ext<Q13[20]|x^4-13>
> Decomposition(U);
[
1-dim unramified Galois representation Unr(-2+3*i) over ext<Q13[20]|x^4-13>,
1-dim unramified Galois representation Unr(-2-3*i) over ext<Q13[20]|x^4-13>
]
We are ready to reconstruct A. Its restriction to inertia is a
faithful 2-dimensional representation of C
4, of determinant 1
(as it comes from an elliptic curve), so it must be σ direct-sum σ
- 1
where σ is one of the faithful 1-dimensional characters of C
4.
> GroupName(InertiaGroup(A)),IsUnramified(Determinant(A)); // just checking
C4 true
> list:=GaloisRepresentations(x^4-13);
> sigma:=[g: g in list | Order(Character(g)) eq 4][1];
> sigma;
1-dim Galois representation (1,-1,zeta(4)_4,-zeta(4)_4) with G=C4, I=C4,
conductor 13^1 over Q13[20]
Since Frob
K=Frob
L commutes with inertia in the Galois group
Gal(Knr(root 4 of(13))/K) isomorphic to Gal(Knr/K) x Gal(L/K) isomorphic to hatZ x C4
through which A factors, A(FrobK) and A(σ) are simultaneously
diagonalizable in GL2(C), so A must be one of the
following two representations A1 and A2:
> Qi<i>:=CyclotomicField(4);
> A1 := sigma * UnramifiedCharacter(K,-2-3*i) +
> sigma^(-1)*UnramifiedCharacter(K,-2+3*i);
> A2 := sigma * UnramifiedCharacter(K,-2+3*i) +
> sigma^(-1)*UnramifiedCharacter(K,-2-3*i);
Finally, to determine which one it is, we pick
another extension of K
where A becomes unramified and compare the Euler factors.
Then we see that A must be A
2.
> L2:=ext<K|x^4-26>;
> EulerFactor(BaseChange(A,L2));
13*x^2 - 6*x + 1
> EulerFactor(BaseChange(A1,L2));
13*x^2 + 6*x + 1
> EulerFactor(BaseChange(A2,L2));
13*x^2 - 6*x + 1
> A eq A2;
true
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