Advanced Examples

Example: Local and Global Epsilon Factors for Dirichlet Characters

Example: Local and Global Epsilon Factors (ex-938a3a)

Run in calculator (opens in a new tab)Copy input

We illustrate the relation between local and global epsilon-factors for Dirichlet characters (or Artin representations). Pick any Dirichlet character \(\chi\); in this example we take the unique odd one of conductor \(5\cdot 19\), order 6, and \(\Im\,\chi(2)>0\).

> D:=FullDirichletGroup(5*19);
> L:=[chi: chi in Elements(D) | (Order(chi) eq 6) and
>   (Conductor(chi) eq 5*19) and IsOdd(chi) and (Imaginary(chi(2)) ge 0)];
> assert #L eq 1;
> chi:=L[1];

We compute its local Galois representations at all the bad places: 5, 19, and \(\infty\). (See §Hodge Structure for Hodge structures at \(\infty\).)

> G5:=GaloisRepresentation(chi,5); G5;
1-dim Galois representation (1,-1,-zeta(3)_3-1,zeta(3)_3,zeta(3)_3+1,-zeta(3)_3)
   with G=C6, I=C2, conductor 5^1 over Q5[40]
> G19:=GaloisRepresentation(chi,19); G19;
1-dim Galois representation (1,-1,zeta(3)_3,-zeta(3)_3-1,zeta(3)_3+1,-zeta(3)_3)
   with G=C6, I=C6, conductor 19^1 over Q19[40]
> Ginfty:=HodgeStructure(ArtinRepresentation(chi)); Ginfty;
Hodge structure of weight 0 given by <0,0,1>

Now compute the corresponding local root numbers - the way the conventions are set up for Artin representations, at \(\infty\) it is the root number of the Hodge structure that enters the functional equation.

> localrootno:=[ComplexField()| RootNumber(G5),RootNumber(G19),
>               RootNumber(Ginfty)];
> globalrootno:=&*localrootno;

The global root number is the sign of the L-series of \(\chi\), and we check that it agrees with the one determined numerically from the functional equation:

> L:=LSeries(chi);
> ok:=CheckFunctionalEquation(L);
> Sign(L);
0.910747215816471738723996800097 + 0.412964294924567358770699943664*I
> globalrootno;
0.910747215816471738723996800108 + 0.412964294924567358770699943647*I

Run in calculator (opens in a new tab)Copy input

Example: Reconstructing a Galois Representation from its Euler Factors

Example: Reconstruction from Euler Factors (ex-6f8047)

Run in calculator (opens in a new tab)Copy input

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

\[E/K: y^2 = x^3-26x, \qquad K={\mathbb{Q}}_{13}.\]

> 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 \({\mathbb{Q}}_{13}(\root d\of{13})\) 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 \(L={\mathbb{Q}}_{13}(\root 4\of{13})\), a cyclic extension of degree 4, but not over its subfields. So the inertia group must be \(C_4\),

\[{\rm InertiaGroup}(A) \,\,\cong\,\, I_{L/K} \,\,=\,\, \mathop{\rm Gal}\nolimits(L/K) \,\,\cong\,\, C_4.\]

Over \(L\) the representation \(U=\mathop{\rm Res}\nolimits_{L} A\) is unramified, and there it is determined by its Euler factor. It is a sum of two unramified characters, \(\mathop{\rm Frob}\nolimits_{L}^{-1}\mapsto -2\pm 3i\).

> 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 \(\sigma\oplus\sigma^{-1}\) where \(\sigma\) 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 \(\mathop{\rm Frob}\nolimits_K=\mathop{\rm Frob}\nolimits_L\) commutes with inertia in the Galois group

\[\mathop{\rm Gal}\nolimits(K^{nr}(\root 4 \of{13})/K) \cong\mathop{\rm Gal}\nolimits(K^{nr}/K) \times \mathop{\rm Gal}\nolimits(L/K) \cong\hat{\mathbb{Z}}\times C_4\]

through which \(A\) factors, \(A(\mathop{\rm Frob}\nolimits_K)\) and \(A(\sigma)\) are simultaneously diagonalizable in \(\mathop{\rm GL}\nolimits_2({\mathbb{C}})\), so \(A\) must be one of the following two representations \(A_1\) and \(A_2\):

> 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

Run in calculator (opens in a new tab)Copy input