EXAMPLES:
sage: Cusps
Set P^1(QQ) of all cusps
sage: Cusp(oo)
Infinity
Bases: sage.structure.element.Element
A cusp.
A cusp is either a rational number or infinity, i.e., an element of the projective line over Q. A Cusp is stored as a pair (a,b), where gcd(a,b)=1 and a,b are of type Integer.
EXAMPLES:
sage: a = Cusp(2/3); b = Cusp(oo)
sage: a.parent()
Set P^1(QQ) of all cusps
sage: a.parent() is b.parent()
True
Return g(self), where g=[a,b,c,d] is a list of length 4, which we view as a linear fractional transformation.
EXAMPLES: Apply the identity matrix:
sage: Cusp(0).apply([1,0,0,1])
0
sage: Cusp(0).apply([0,-1,1,0])
Infinity
sage: Cusp(0).apply([1,-3,0,1])
-3
Return the denominator of the cusp a/b.
EXAMPLES:
sage: x=Cusp(6,9); x
2/3
sage: x.denominator()
3
sage: Cusp(oo).denominator()
0
sage: Cusp(-5/10).denominator()
2
Suppose this cusp is \(\alpha\), \(G\) a congruence subgroup of level \(N\) and \(\sigma\) is the automorphism in the Galois group of \(\QQ(\zeta_N)/\QQ\) that sends \(\zeta_N\) to \(\zeta_N^t\). Then this function computes a cusp \(\beta\) such that \(\sigma([\alpha]) = [\beta]\), where \([\alpha]\) is the equivalence class of \(\alpha\) modulo \(G\).
This code only needs as input the level and not the group since the action of Galois for a congruence group \(G\) of level \(N\) is compatible with the action of the full congruence group \(\Gamma(N)\).
INPUT:
- \(t\) – integer that is coprime to N
- \(N\) – positive integer (level)
OUTPUT:
- a cusp
Warning
In some cases \(N\) must fit in a long long, i.e., there are cases where this algorithm isn’t fully implemented.
Note
Modular curves can have multiple non-isomorphic models over \(\QQ\). The action of Galois depends on such a model. The model over \(\QQ\) of \(X(G)\) used here is the model where the function field \(\QQ(X(G))\) is given by the functions whose fourier expansion at \(\infty\) have their coefficients in \(\QQ\). For \(X(N):=X(\Gamma(N))\) the corresponding moduli interpretation over \(\ZZ[1/N]\) is that \(X(N)\) parametrizes pairs \((E,a)\) where \(E\) is a (generalized) elliptic curve and \(a: \ZZ / N\ZZ \times \mu_N \to E\) is a closed immersion such that the Weil pairing of \(a(1,1)\) and \(a(0,\zeta_N)\) is \(\zeta_N\). In this parameterisation the point \(z \in H\) corresponds to the pair \((E_z,a_z)\) with \(E_z=\CC/(z \ZZ+\ZZ)\) and \(a_z: \ZZ / N\ZZ \times \mu_N \to E\) given by \(a_z(1,1) = z/N\) and \(a_z(0,\zeta_N) = 1/N\). Similarly \(X_1(N):=X(\Gamma_1(N))\) parametrizes pairs \((E,a)\) where \(a: \mu_N \to E\) is a closed immersion.
EXAMPLES:
sage: Cusp(1/10).galois_action(3, 50)
1/170
sage: Cusp(oo).galois_action(3, 50)
Infinity
sage: c=Cusp(0).galois_action(3, 50); c
50/67
sage: Gamma0(50).reduce_cusp(c)
0
Here we compute the permutations of the action for t=3 on cusps for Gamma0(50).
sage: N = 50; t=3; G = Gamma0(N); C = G.cusps()
sage: cl = lambda z: exists(C, lambda y:y.is_gamma0_equiv(z, N))[1]
sage: for i in range(5): print i, t^i, [cl(alpha.galois_action(t^i,N)) for alpha in C]
0 1 [0, 1/25, 1/10, 1/5, 3/10, 2/5, 1/2, 3/5, 7/10, 4/5, 9/10, Infinity]
1 3 [0, 1/25, 7/10, 2/5, 1/10, 4/5, 1/2, 1/5, 9/10, 3/5, 3/10, Infinity]
2 9 [0, 1/25, 9/10, 4/5, 7/10, 3/5, 1/2, 2/5, 3/10, 1/5, 1/10, Infinity]
3 27 [0, 1/25, 3/10, 3/5, 9/10, 1/5, 1/2, 4/5, 1/10, 2/5, 7/10, Infinity]
4 81 [0, 1/25, 1/10, 1/5, 3/10, 2/5, 1/2, 3/5, 7/10, 4/5, 9/10, Infinity]
TESTS:
Here we check that the Galois action is indeed a permutation on the cusps of Gamma1(48) and check that trac ticket #13253 is fixed.
sage: G=Gamma1(48)
sage: C=G.cusps()
sage: for i in Integers(48).unit_gens():
... C_permuted = [G.reduce_cusp(c.galois_action(i,48)) for c in C]
... assert len(set(C_permuted))==len(C)
We test that Gamma1(19) has 9 rational cusps and check that trac ticket #8998 is fixed.
sage: G = Gamma1(19)
sage: [c for c in G.cusps() if c.galois_action(2,19).is_gamma1_equiv(c,19)[0]]
[2/19, 3/19, 4/19, 5/19, 6/19, 7/19, 8/19, 9/19, Infinity]
REFERENCES:
- Section 1.3 of Glenn Stevens, “Arithmetic on Modular Curves”
- There is a long comment about our algorithm in the source code for this function.
AUTHORS:
- William Stein, 2009-04-18
Return whether self and other are equivalent modulo the action of \(\Gamma_0(N)\) via linear fractional transformations.
INPUT:
OUTPUT:
EXAMPLES:
sage: x = Cusp(2,3)
sage: y = Cusp(4,5)
sage: x.is_gamma0_equiv(y, 2)
True
sage: _, ga = x.is_gamma0_equiv(y, 2, 'matrix'); ga
[-1 2]
[-2 3]
sage: x.is_gamma0_equiv(y, 3)
False
sage: x.is_gamma0_equiv(y, 3, 'matrix')
(False, None)
sage: Cusp(1/2).is_gamma0_equiv(1/3,11,'corner')
(True, 19)
sage: Cusp(1,0)
Infinity
sage: z = Cusp(1,0)
sage: x.is_gamma0_equiv(z, 3, 'matrix')
(
[-1 1]
True, [-3 2]
)
ALGORITHM: See Proposition 2.2.3 of Cremona’s book ‘Algorithms for Modular Elliptic Curves’, or Prop 2.27 of Stein’s Ph.D. thesis.
Return whether self and other are equivalent modulo the action of Gamma_1(N) via linear fractional transformations.
INPUT:
OUTPUT:
EXAMPLES:
sage: x = Cusp(2,3)
sage: y = Cusp(4,5)
sage: x.is_gamma1_equiv(y,2)
(True, 1)
sage: x.is_gamma1_equiv(y,3)
(False, 0)
sage: z = Cusp(QQ(x) + 10)
sage: x.is_gamma1_equiv(z,10)
(True, 1)
sage: z = Cusp(1,0)
sage: x.is_gamma1_equiv(z, 3)
(True, -1)
sage: Cusp(0).is_gamma1_equiv(oo, 1)
(True, 1)
sage: Cusp(0).is_gamma1_equiv(oo, 3)
(False, 0)
Return a pair (b, t), where b is True or False as self and other are equivalent under the action of G, and t is 1 or -1, as described below.
Two cusps \(u1/v1\) and \(u2/v2\) are equivalent modulo Gamma_H(N) if and only if \(v1 = h*v2 (\mathrm{mod} N)\) and \(u1 = h^{(-1)}*u2 (\mathrm{mod} gcd(v1,N))\) or \(v1 = -h*v2 (mod N)\) and \(u1 = -h^{(-1)}*u2 (\mathrm{mod} gcd(v1,N))\) for some \(h \in H\). Then t is 1 or -1 as c and c’ fall into the first or second case, respectively.
INPUT:
OUTPUT:
EXAMPLES:
sage: x = Cusp(2,3)
sage: y = Cusp(4,5)
sage: x.is_gamma_h_equiv(y,GammaH(13,[2]))
(True, 1)
sage: x.is_gamma_h_equiv(y,GammaH(13,[5]))
(False, 0)
sage: x.is_gamma_h_equiv(y,GammaH(5,[]))
(False, 0)
sage: x.is_gamma_h_equiv(y,GammaH(23,[4]))
(True, -1)
Enumerating the cusps for a space of modular symbols uses this function.
sage: G = GammaH(25,[6]) ; M = G.modular_symbols() ; M
Modular Symbols space of dimension 11 for Congruence Subgroup Gamma_H(25) with H generated by [6] of weight 2 with sign 0 and over Rational Field
sage: M.cusps()
[37/75, 1/2, 31/125, 1/4, -2/5, 2/5, -1/5, 1/10, -3/10, 1/15, 7/15, 9/20]
sage: len(M.cusps())
12
This is always one more than the associated space of weight 2 Eisenstein series.
sage: G.dimension_eis(2)
11
sage: M.cuspidal_subspace()
Modular Symbols subspace of dimension 0 of Modular Symbols space of dimension 11 for Congruence Subgroup Gamma_H(25) with H generated by [6] of weight 2 with sign 0 and over Rational Field
sage: G.dimension_cusp_forms(2)
0
Returns True if this is the cusp infinity.
EXAMPLES:
sage: Cusp(3/5).is_infinity()
False
sage: Cusp(1,0).is_infinity()
True
sage: Cusp(0,1).is_infinity()
False
Return the numerator of the cusp a/b.
EXAMPLES:
sage: x=Cusp(6,9); x
2/3
sage: x.numerator()
2
sage: Cusp(oo).numerator()
1
sage: Cusp(-5/10).numerator()
-1
Bases: sage.structure.parent_base.ParentWithBase
The set of cusps.
EXAMPLES:
sage: C = Cusps; C
Set P^1(QQ) of all cusps
sage: loads(C.dumps()) == C
True
Return the zero cusp.
NOTE:
The existence of this method is assumed by some parts of Sage’s coercion model.
EXAMPLE:
sage: Cusps.zero()
0
Deprecated: Use zero() instead. See trac ticket #17694 for details.