The principal ideal problem in quaternion algebras IMB, Université de Bordeaux March 11, 2014 CIRM, Luminy
Elliptic curves Theorem (Wiles, Taylor, Diamond, Conrad, Breuil) Let E be an elliptic curve over Q with conductor N. Then there exists a weight 2, level N newform f such that for all p N we have a p (f) = p + 1 E(F p ). Easy to compute spaces of classical modular forms Lists of curves of given conductor tables (Cremona). What if we replace Q with a number field F? automorphic forms for GL(2)/F.
Elliptic curves Theorem (Wiles, Taylor, Diamond, Conrad, Breuil) Let E be an elliptic curve over Q with conductor N. Then there exists a weight 2, level N newform f such that for all p N we have a p (f) = p + 1 E(F p ). Easy to compute spaces of classical modular forms Lists of curves of given conductor tables (Cremona). What if we replace Q with a number field F? automorphic forms for GL(2)/F.
Elliptic curves Theorem (Wiles, Taylor, Diamond, Conrad, Breuil) Let E be an elliptic curve over Q with conductor N. Then there exists a weight 2, level N newform f such that for all p N we have a p (f) = p + 1 E(F p ). Easy to compute spaces of classical modular forms Lists of curves of given conductor tables (Cremona). What if we replace Q with a number field F? automorphic forms for GL(2)/F.
Elliptic curves Theorem (Wiles, Taylor, Diamond, Conrad, Breuil) Let E be an elliptic curve over Q with conductor N. Then there exists a weight 2, level N newform f such that for all p N we have a p (f) = p + 1 E(F p ). Easy to compute spaces of classical modular forms Lists of curves of given conductor tables (Cremona). What if we replace Q with a number field F? automorphic forms for GL(2)/F.
Goal: computing automorphic forms for GL(2)/F Computing = computing a finite dimensional vector space with an action of Hecke operators, that is isomorphic to a space of automorphic forms. Previous work: F = Q: classical modular forms. F totally real: Donnelly,, Greenberg, Voight. F imaginary quadratic: Bygott, Cremona, Figueiredo, Lingham, Şengün, Torrey, Whitley, Yasaki. some other field F of small degree: Gunnells, Yasaki. Our algorithm: any F with at most one complex place.
Goal: computing automorphic forms for GL(2)/F Computing = computing a finite dimensional vector space with an action of Hecke operators, that is isomorphic to a space of automorphic forms. Previous work: F = Q: classical modular forms. F totally real: Donnelly,, Greenberg, Voight. F imaginary quadratic: Bygott, Cremona, Figueiredo, Lingham, Şengün, Torrey, Whitley, Yasaki. some other field F of small degree: Gunnells, Yasaki. Our algorithm: any F with at most one complex place.
Goal: computing automorphic forms for GL(2)/F Computing = computing a finite dimensional vector space with an action of Hecke operators, that is isomorphic to a space of automorphic forms. Previous work: F = Q: classical modular forms. F totally real: Donnelly,, Greenberg, Voight. F imaginary quadratic: Bygott, Cremona, Figueiredo, Lingham, Şengün, Torrey, Whitley, Yasaki. some other field F of small degree: Gunnells, Yasaki. Our algorithm: any F with at most one complex place.
Computing automorphic forms for GL(2)/F : sketch Sketch of the method: Jacquet-Langlands: compute a suitable quaternion algebra over F and a maximal order O. Compute a fundamental domain for the unit group Γ = O. Compute the cohomology spaces H (Γ, M). Formulas for T p : compute generator of an O-ideal of norm p.
Computing automorphic forms for GL(2)/F : sketch Sketch of the method: Jacquet-Langlands: compute a suitable quaternion algebra over F and a maximal order O. Compute a fundamental domain for the unit group Γ = O. Compute the cohomology spaces H (Γ, M). Formulas for T p : compute generator of an O-ideal of norm p.
Computing automorphic forms for GL(2)/F : sketch Sketch of the method: Jacquet-Langlands: compute a suitable quaternion algebra over F and a maximal order O. Compute a fundamental domain for the unit group Γ = O. Compute the cohomology spaces H (Γ, M). Formulas for T p : compute generator of an O-ideal of norm p.
Computing automorphic forms for GL(2)/F : sketch Sketch of the method: Jacquet-Langlands: compute a suitable quaternion algebra over F and a maximal order O. Compute a fundamental domain for the unit group Γ = O. Compute the cohomology spaces H (Γ, M). Formulas for T p : compute generator of an O-ideal of norm p.
Computing automorphic forms for GL(2)/F : sketch Sketch of the method: Jacquet-Langlands: compute a suitable quaternion algebra over F and a maximal order O. Compute a fundamental domain for the unit group Γ = O. Compute the cohomology spaces H (Γ, M). Formulas for T p : compute generator of an O-ideal of norm p.
Quaternion algebras A quaternion algebra over F is a central simple algebra A of dimension 4. Equivalently (char F 2), A is of the form ( ) a, b A = = F + Fi + Fj + Fij F with the relations where a, b F. The reduced norm is i 2 = a, j 2 = b and ij = ji nrd(x + yi + zj + tij) = x 2 ay 2 bz 2 + abt 2.
Quaternion algebras A quaternion algebra over F is a central simple algebra A of dimension 4. Equivalently (char F 2), A is of the form ( ) a, b A = = F + Fi + Fj + Fij F with the relations where a, b F. The reduced norm is i 2 = a, j 2 = b and ij = ji nrd(x + yi + zj + tij) = x 2 ay 2 bz 2 + abt 2.
Quaternion algebras A quaternion algebra over F is a central simple algebra A of dimension 4. Equivalently (char F 2), A is of the form ( ) a, b A = = F + Fi + Fj + Fij F with the relations where a, b F. The reduced norm is i 2 = a, j 2 = b and ij = ji nrd(x + yi + zj + tij) = x 2 ay 2 bz 2 + abt 2.
Ramification Only quaternion algebra over C: M 2 (C). Only quaternion algebras over R: M 2 (R) and H = ( ) 1, 1 R. F a p-adic field. Only two quaternion algebras over F: M 2 (F) and a division algebra. F a number field and A/F a quaternion algebra. v place of F: v splits if A F F v = M2 (F v ), ramifies otherwise. Reduced discriminant δ A = product of the ramified primes. Discriminant A = F 4 N(δ A ) 2.
Ramification Only quaternion algebra over C: M 2 (C). Only quaternion algebras over R: M 2 (R) and H = ( ) 1, 1 R. F a p-adic field. Only two quaternion algebras over F: M 2 (F) and a division algebra. F a number field and A/F a quaternion algebra. v place of F: v splits if A F F v = M2 (F v ), ramifies otherwise. Reduced discriminant δ A = product of the ramified primes. Discriminant A = F 4 N(δ A ) 2.
Ramification Only quaternion algebra over C: M 2 (C). Only quaternion algebras over R: M 2 (R) and H = ( ) 1, 1 R. F a p-adic field. Only two quaternion algebras over F: M 2 (F) and a division algebra. F a number field and A/F a quaternion algebra. v place of F: v splits if A F F v = M2 (F v ), ramifies otherwise. Reduced discriminant δ A = product of the ramified primes. Discriminant A = F 4 N(δ A ) 2.
Ramification Only quaternion algebra over C: M 2 (C). Only quaternion algebras over R: M 2 (R) and H = ( ) 1, 1 R. F a p-adic field. Only two quaternion algebras over F: M 2 (F) and a division algebra. F a number field and A/F a quaternion algebra. v place of F: v splits if A F F v = M2 (F v ), ramifies otherwise. Reduced discriminant δ A = product of the ramified primes. Discriminant A = F 4 N(δ A ) 2.
Ramification Only quaternion algebra over C: M 2 (C). Only quaternion algebras over R: M 2 (R) and H = ( ) 1, 1 R. F a p-adic field. Only two quaternion algebras over F: M 2 (F) and a division algebra. F a number field and A/F a quaternion algebra. v place of F: v splits if A F F v = M2 (F v ), ramifies otherwise. Reduced discriminant δ A = product of the ramified primes. Discriminant A = F 4 N(δ A ) 2.
Ramification Only quaternion algebra over C: M 2 (C). Only quaternion algebras over R: M 2 (R) and H = ( ) 1, 1 R. F a p-adic field. Only two quaternion algebras over F: M 2 (F) and a division algebra. F a number field and A/F a quaternion algebra. v place of F: v splits if A F F v = M2 (F v ), ramifies otherwise. Reduced discriminant δ A = product of the ramified primes. Discriminant A = F 4 N(δ A ) 2.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I)., integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Orders and ideals F number field with integers Z F, A/F quaternion algebra. A lattice in A is a finitely generatedz F module L such that FL = A. An order in A is a lattice O that is also a subring with unit. For a lattice I, the right order is O r (I) = {x A Ix x}, similarly left order O l (I). A right O-ideal is a lattice I such that O r (I) = O. Two-sided ideal if O r (I) = O l (I), integral if I O. nrd(i) is the ideal generated by {nrd(x) : x I}. I, J lattices such that O r (I) = O l (J): product IJ generated by {xy : x I, y J}.
Quaternion algebras: examples Let A = ( ) 3, 1 Q. It is ramified at 2 and 3. O = Z+Zi +Zj +Zij is an order in A. O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2 is a maximal order in A. I 1 = 19O is an integral two-sided O-ideal and nrd(i 1 ) = 19 2 Z. Let x = 3 4i + j A. I 2 = xo is an integral right O-ideal but O l (I 2 ) = xox 1 O. nrd(x) = 38 so nrd(i 2 ) = 38Z. I = xo + 19O is an integral right O-ideal, nrd(i) = 19Z.
Quaternion algebras: examples Let A = ( ) 3, 1 Q. It is ramified at 2 and 3. O = Z+Zi +Zj +Zij is an order in A. O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2 is a maximal order in A. I 1 = 19O is an integral two-sided O-ideal and nrd(i 1 ) = 19 2 Z. Let x = 3 4i + j A. I 2 = xo is an integral right O-ideal but O l (I 2 ) = xox 1 O. nrd(x) = 38 so nrd(i 2 ) = 38Z. I = xo + 19O is an integral right O-ideal, nrd(i) = 19Z.
Quaternion algebras: examples Let A = ( ) 3, 1 Q. It is ramified at 2 and 3. O = Z+Zi +Zj +Zij is an order in A. O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2 is a maximal order in A. I 1 = 19O is an integral two-sided O-ideal and nrd(i 1 ) = 19 2 Z. Let x = 3 4i + j A. I 2 = xo is an integral right O-ideal but O l (I 2 ) = xox 1 O. nrd(x) = 38 so nrd(i 2 ) = 38Z. I = xo + 19O is an integral right O-ideal, nrd(i) = 19Z.
Quaternion algebras: examples Let A = ( ) 3, 1 Q. It is ramified at 2 and 3. O = Z+Zi +Zj +Zij is an order in A. O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2 is a maximal order in A. I 1 = 19O is an integral two-sided O-ideal and nrd(i 1 ) = 19 2 Z. Let x = 3 4i + j A. I 2 = xo is an integral right O-ideal but O l (I 2 ) = xox 1 O. nrd(x) = 38 so nrd(i 2 ) = 38Z. I = xo + 19O is an integral right O-ideal, nrd(i) = 19Z.
Quaternion algebras: examples Let A = ( ) 3, 1 Q. It is ramified at 2 and 3. O = Z+Zi +Zj +Zij is an order in A. O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2 is a maximal order in A. I 1 = 19O is an integral two-sided O-ideal and nrd(i 1 ) = 19 2 Z. Let x = 3 4i + j A. I 2 = xo is an integral right O-ideal but O l (I 2 ) = xox 1 O. nrd(x) = 38 so nrd(i 2 ) = 38Z. I = xo + 19O is an integral right O-ideal, nrd(i) = 19Z.
Quaternion algebras: examples Let A = ( ) 3, 1 Q. It is ramified at 2 and 3. O = Z+Zi +Zj +Zij is an order in A. O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2 is a maximal order in A. I 1 = 19O is an integral two-sided O-ideal and nrd(i 1 ) = 19 2 Z. Let x = 3 4i + j A. I 2 = xo is an integral right O-ideal but O l (I 2 ) = xox 1 O. nrd(x) = 38 so nrd(i 2 ) = 38Z. I = xo + 19O is an integral right O-ideal, nrd(i) = 19Z.
Principal ideal problem Two right O-ideals I, J are equivalent if we have I = xj for some x A. An ideal I is principal if I = xo for some x A. Set of equivalence classes of right O-ideals: Cl(O). The principal ideal problem is: Problem Given a right O-ideal I, decide whether I is principal. In that case, compute a generator. We will restrict to maximal orders.
Principal ideal problem Two right O-ideals I, J are equivalent if we have I = xj for some x A. An ideal I is principal if I = xo for some x A. Set of equivalence classes of right O-ideals: Cl(O). The principal ideal problem is: Problem Given a right O-ideal I, decide whether I is principal. In that case, compute a generator. We will restrict to maximal orders.
Principal ideal problem Two right O-ideals I, J are equivalent if we have I = xj for some x A. An ideal I is principal if I = xo for some x A. Set of equivalence classes of right O-ideals: Cl(O). The principal ideal problem is: Problem Given a right O-ideal I, decide whether I is principal. In that case, compute a generator. We will restrict to maximal orders.
Principal ideal problem: definite case Two natural cases: 1 A is definite if every infinite place of F is real and ramified. Tr(nrd) positive definite quadratic form lattice enumeration method. Theorem (Donnelly, Dembélé, Kirschmer, Voight) There exists an explicit algorithm for solving the principal ideal problem in definite algebras that runs in polynomial time in the size of the input when the base field is fixed.
Principal ideal problem: definite case Two natural cases: 1 A is definite if every infinite place of F is real and ramified. Tr(nrd) positive definite quadratic form lattice enumeration method. Theorem (Donnelly, Dembélé, Kirschmer, Voight) There exists an explicit algorithm for solving the principal ideal problem in definite algebras that runs in polynomial time in the size of the input when the base field is fixed.
Principal ideal problem: definite case Two natural cases: 1 A is definite if every infinite place of F is real and ramified. Tr(nrd) positive definite quadratic form lattice enumeration method. Theorem (Donnelly, Dembélé, Kirschmer, Voight) There exists an explicit algorithm for solving the principal ideal problem in definite algebras that runs in polynomial time in the size of the input when the base field is fixed.
Principal ideal problem: indefinite case 2 A is indefinite otherwise. Cl A (F): ray class field with modulus the product of the ramified real places. Theorem (Eichler) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Then the reduced norm induces a bijection Cl(O) Cl A (F). Testing equivalence same problem over the base field. Computing a generator: algorithm by Kirschmer and Voight based on lattice enumeration, complexity is unknown.
Principal ideal problem: indefinite case 2 A is indefinite otherwise. Cl A (F): ray class field with modulus the product of the ramified real places. Theorem (Eichler) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Then the reduced norm induces a bijection Cl(O) Cl A (F). Testing equivalence same problem over the base field. Computing a generator: algorithm by Kirschmer and Voight based on lattice enumeration, complexity is unknown.
Principal ideal problem: indefinite case 2 A is indefinite otherwise. Cl A (F): ray class field with modulus the product of the ramified real places. Theorem (Eichler) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Then the reduced norm induces a bijection Cl(O) Cl A (F). Testing equivalence same problem over the base field. Computing a generator: algorithm by Kirschmer and Voight based on lattice enumeration, complexity is unknown.
Principal ideal problem: indefinite case 2 A is indefinite otherwise. Cl A (F): ray class field with modulus the product of the ramified real places. Theorem (Eichler) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Then the reduced norm induces a bijection Cl(O) Cl A (F). Testing equivalence same problem over the base field. Computing a generator: algorithm by Kirschmer and Voight based on lattice enumeration, complexity is unknown.
Principal ideal problem: indefinite case 2 A is indefinite otherwise. Cl A (F): ray class field with modulus the product of the ramified real places. Theorem (Eichler) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Then the reduced norm induces a bijection Cl(O) Cl A (F). Testing equivalence same problem over the base field. Computing a generator: algorithm by Kirschmer and Voight based on lattice enumeration, complexity is unknown.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Buchmann s algorithm, sketch Assuming the generalized Riemann hypothesis: Choose a set of primes in F that generates Cl(F): the factor base B. Look for random smooth elements in Z F : the relations R. Stop when B / R = Cl(F). Given a fractional ideal I: Look for a random element x I 1 such that xi is smooth. Do linear algebra.
Smooth numbers Theorem (Canfield, Erdös, Pomerance) Let ψ(x, y) = {n x, n is y-smooth}. If we set L(x) = exp( ln x ln ln x), then ψ(x, L(x) a ) = xl(x) 1/(2a)+o(1).
Adapting Buchmann s algorithm New algorithm for indefinite quaternion algebras: subexponential complexity under GRH and some heuristics. Set of ideals is not a group! Would work if every ideal was two-sided: group generated by maximal two-sided ideals. Eichler, indefinite case: up to equivalence, every right ideal is almost two-sided. nrd(i) = p n = I Jp n/2 with nrd(j) = p n mod 2. Make this effective? Need to understand the local case.
Adapting Buchmann s algorithm New algorithm for indefinite quaternion algebras: subexponential complexity under GRH and some heuristics. Set of ideals is not a group! Would work if every ideal was two-sided: group generated by maximal two-sided ideals. Eichler, indefinite case: up to equivalence, every right ideal is almost two-sided. nrd(i) = p n = I Jp n/2 with nrd(j) = p n mod 2. Make this effective? Need to understand the local case.
Adapting Buchmann s algorithm New algorithm for indefinite quaternion algebras: subexponential complexity under GRH and some heuristics. Set of ideals is not a group! Would work if every ideal was two-sided: group generated by maximal two-sided ideals. Eichler, indefinite case: up to equivalence, every right ideal is almost two-sided. nrd(i) = p n = I Jp n/2 with nrd(j) = p n mod 2. Make this effective? Need to understand the local case.
Adapting Buchmann s algorithm New algorithm for indefinite quaternion algebras: subexponential complexity under GRH and some heuristics. Set of ideals is not a group! Would work if every ideal was two-sided: group generated by maximal two-sided ideals. Eichler, indefinite case: up to equivalence, every right ideal is almost two-sided. nrd(i) = p n = I Jp n/2 with nrd(j) = p n mod 2. Make this effective? Need to understand the local case.
Adapting Buchmann s algorithm New algorithm for indefinite quaternion algebras: subexponential complexity under GRH and some heuristics. Set of ideals is not a group! Would work if every ideal was two-sided: group generated by maximal two-sided ideals. Eichler, indefinite case: up to equivalence, every right ideal is almost two-sided. nrd(i) = p n = I Jp n/2 with nrd(j) = p n mod 2. Make this effective? Need to understand the local case.
Adapting Buchmann s algorithm New algorithm for indefinite quaternion algebras: subexponential complexity under GRH and some heuristics. Set of ideals is not a group! Would work if every ideal was two-sided: group generated by maximal two-sided ideals. Eichler, indefinite case: up to equivalence, every right ideal is almost two-sided. nrd(i) = p n = I Jp n/2 with nrd(j) = p n mod 2. Make this effective? Need to understand the local case.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
Local case F a p-adic field, R its ring of integers. The division quaternion algebra over F has a unique maximal order and every ideal is two-sided. In M 2 (F), maximal order O = M 2 (R). Every right O-ideal I is principal, generated by an element g GL 2 (F). I is two-sided if and only if g F GL 2 (R). need to understand GL 2 (F)/F GL 2 (R). Hermite normal form? Not sufficiently geometric.
The Bruhat-Tits tree T F p-adic field, R its ring of integers, π uniformizer. Lattice in V = F 2 : rank 2 R-submodule L such that FL = V. Vertices of T : homothety classes of lattices in F 2. Distance between vertices P, P represented by L, L : write L = Re 1 + Re 2 and L = Rπ a e 1 + Rπ b e 2. d(p, P ) = a b. Edge between vertices at distance 1. GL 2 (F) acts transitively on the vertices, the stabilizer of the class of L = R 2 is F GL 2 (R).
The Bruhat-Tits tree T F p-adic field, R its ring of integers, π uniformizer. Lattice in V = F 2 : rank 2 R-submodule L such that FL = V. Vertices of T : homothety classes of lattices in F 2. Distance between vertices P, P represented by L, L : write L = Re 1 + Re 2 and L = Rπ a e 1 + Rπ b e 2. d(p, P ) = a b. Edge between vertices at distance 1. GL 2 (F) acts transitively on the vertices, the stabilizer of the class of L = R 2 is F GL 2 (R).
The Bruhat-Tits tree T F p-adic field, R its ring of integers, π uniformizer. Lattice in V = F 2 : rank 2 R-submodule L such that FL = V. Vertices of T : homothety classes of lattices in F 2. Distance between vertices P, P represented by L, L : write L = Re 1 + Re 2 and L = Rπ a e 1 + Rπ b e 2. d(p, P ) = a b. Edge between vertices at distance 1. GL 2 (F) acts transitively on the vertices, the stabilizer of the class of L = R 2 is F GL 2 (R).
The Bruhat-Tits tree T F p-adic field, R its ring of integers, π uniformizer. Lattice in V = F 2 : rank 2 R-submodule L such that FL = V. Vertices of T : homothety classes of lattices in F 2. Distance between vertices P, P represented by L, L : write L = Re 1 + Re 2 and L = Rπ a e 1 + Rπ b e 2. d(p, P ) = a b. Edge between vertices at distance 1. GL 2 (F) acts transitively on the vertices, the stabilizer of the class of L = R 2 is F GL 2 (R).
The Bruhat-Tits tree T F p-adic field, R its ring of integers, π uniformizer. Lattice in V = F 2 : rank 2 R-submodule L such that FL = V. Vertices of T : homothety classes of lattices in F 2. Distance between vertices P, P represented by L, L : write L = Re 1 + Re 2 and L = Rπ a e 1 + Rπ b e 2. d(p, P ) = a b. Edge between vertices at distance 1. GL 2 (F) acts transitively on the vertices, the stabilizer of the class of L = R 2 is F GL 2 (R).
The Bruhat-Tits tree T F p-adic field, R its ring of integers, π uniformizer. Lattice in V = F 2 : rank 2 R-submodule L such that FL = V. Vertices of T : homothety classes of lattices in F 2. Distance between vertices P, P represented by L, L : write L = Re 1 + Re 2 and L = Rπ a e 1 + Rπ b e 2. d(p, P ) = a b. Edge between vertices at distance 1. GL 2 (F) acts transitively on the vertices, the stabilizer of the class of L = R 2 is F GL 2 (R).
Reduction theory of SL 2 Let P 0 be the vertex corresponding to R 2 and P 1 be a vertex at distance 1 from P 0. SL 2 (R) acts transitively on the set of vertices at fixed distance from P 0. Theorem The action of the group G = SL 2 (F) on the vertices of T has exactly two orbits G P 0 and G P 1.
Reduction theory of SL 2 Let P 0 be the vertex corresponding to R 2 and P 1 be a vertex at distance 1 from P 0. SL 2 (R) acts transitively on the set of vertices at fixed distance from P 0. Theorem The action of the group G = SL 2 (F) on the vertices of T has exactly two orbits G P 0 and G P 1.
Reduction theory of SL 2 Let P 0 be the vertex corresponding to R 2 and P 1 be a vertex at distance 1 from P 0. SL 2 (R) acts transitively on the set of vertices at fixed distance from P 0. Theorem The action of the group G = SL 2 (F) on the vertices of T has exactly two orbits G P 0 and G P 1.
Using the reduction theory of SL 2 Returning to ideals: Let I be an integral ideal with norm p n g local generator of I P = g P 0, d(p 0, P) n h A with nrd(h) = 1 such that h P = P n mod 2 then hi = Jp n/2.
Using the reduction theory of SL 2 Returning to ideals: Let I be an integral ideal with norm p n g local generator of I P = g P 0, d(p 0, P) n h A with nrd(h) = 1 such that h P = P n mod 2 then hi = Jp n/2.
Using the reduction theory of SL 2 Returning to ideals: Let I be an integral ideal with norm p n g local generator of I P = g P 0, d(p 0, P) n h A with nrd(h) = 1 such that h P = P n mod 2 then hi = Jp n/2.
Using the reduction theory of SL 2 Returning to ideals: Let I be an integral ideal with norm p n g local generator of I P = g P 0, d(p 0, P) n h A with nrd(h) = 1 such that h P = P n mod 2 then hi = Jp n/2.
Using the reduction theory of SL 2 Returning to ideals: Let I be an integral ideal with norm p n g local generator of I P = g P 0, d(p 0, P) n h A with nrd(h) = 1 such that h P = P n mod 2 then hi = Jp n/2.
Reduction theory, globally Use the reduction algorithm prime by prime. The element h should be in O l for all l p. Turns an ideal with norm having even valuation at every split prime into a two-sided ideal with the same norm. Let O = O l (J) where J is an integral right O-ideal of norm p. Look for units in O and O acting transitively on the projective line of the residue field. Are there enough units?
Reduction theory, globally Use the reduction algorithm prime by prime. The element h should be in O l for all l p. Turns an ideal with norm having even valuation at every split prime into a two-sided ideal with the same norm. Let O = O l (J) where J is an integral right O-ideal of norm p. Look for units in O and O acting transitively on the projective line of the residue field. Are there enough units?
Reduction theory, globally Use the reduction algorithm prime by prime. The element h should be in O l for all l p. Turns an ideal with norm having even valuation at every split prime into a two-sided ideal with the same norm. Let O = O l (J) where J is an integral right O-ideal of norm p. Look for units in O and O acting transitively on the projective line of the residue field. Are there enough units?
Reduction theory, globally Use the reduction algorithm prime by prime. The element h should be in O l for all l p. Turns an ideal with norm having even valuation at every split prime into a two-sided ideal with the same norm. Let O = O l (J) where J is an integral right O-ideal of norm p. Look for units in O and O acting transitively on the projective line of the residue field. Are there enough units?
Reduction theory, globally Use the reduction algorithm prime by prime. The element h should be in O l for all l p. Turns an ideal with norm having even valuation at every split prime into a two-sided ideal with the same norm. Let O = O l (J) where J is an integral right O-ideal of norm p. Look for units in O and O acting transitively on the projective line of the residue field. Are there enough units?
Reduction theory, globally Use the reduction algorithm prime by prime. The element h should be in O l for all l p. Turns an ideal with norm having even valuation at every split prime into a two-sided ideal with the same norm. Let O = O l (J) where J is an integral right O-ideal of norm p. Look for units in O and O acting transitively on the projective line of the residue field. Are there enough units?
Strong approximation Let O 1 O be the kernel of the reduced norm. Theorem (Strong approximation) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Let p be a prime of Z F that splits in A and k a positive integer. Then the map is surjective. O 1 SL 2 (Z F /p k ) Can construct units in O in subexponential time: choose a commutative suborder, use Buchmann s algorithm.
Strong approximation Let O 1 O be the kernel of the reduced norm. Theorem (Strong approximation) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Let p be a prime of Z F that splits in A and k a positive integer. Then the map is surjective. O 1 SL 2 (Z F /p k ) Can construct units in O in subexponential time: choose a commutative suborder, use Buchmann s algorithm.
Strong approximation Let O 1 O be the kernel of the reduced norm. Theorem (Strong approximation) Let A be an indefinite quaternion algebra over a number field F and O a maximal order in A. Let p be a prime of Z F that splits in A and k a positive integer. Then the map is surjective. O 1 SL 2 (Z F /p k ) Can construct units in O in subexponential time: choose a commutative suborder, use Buchmann s algorithm.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: structure Algorithm: 1 Choose a set of primes in F that generates Cl A (F): the factor base B. 2 Look for random smooth elements in O: the relations R. 3 Stop when B / nrd(r) = Cl A (F). 4 For each p B that splits: Let P 0 fixed by O and choose P 1 at distance 1 from P 0. Compute the maximal order O such that O fixes P 1. Compute units in O until they act transitively on the vertices at distance 1 from P 0. Compute units in O until they act transitively on the vertices at distance 1 from P 1.
Sketch of the algorithm: computing a generator Given a right O-ideal I: 1 Test whether nrd(i) is trivial in Cl A (F). 2 Look for a random element x I 1 such that xi is smooth. 3 Do linear algebra with the relations to find y such that nrd(yxi) = Z F. 4 For all p B, apply the reduction algorithm in the tree at p. Now hyxi = O, i.e. I = (hyx) 1 O.
Sketch of the algorithm: computing a generator Given a right O-ideal I: 1 Test whether nrd(i) is trivial in Cl A (F). 2 Look for a random element x I 1 such that xi is smooth. 3 Do linear algebra with the relations to find y such that nrd(yxi) = Z F. 4 For all p B, apply the reduction algorithm in the tree at p. Now hyxi = O, i.e. I = (hyx) 1 O.
Sketch of the algorithm: computing a generator Given a right O-ideal I: 1 Test whether nrd(i) is trivial in Cl A (F). 2 Look for a random element x I 1 such that xi is smooth. 3 Do linear algebra with the relations to find y such that nrd(yxi) = Z F. 4 For all p B, apply the reduction algorithm in the tree at p. Now hyxi = O, i.e. I = (hyx) 1 O.
Sketch of the algorithm: computing a generator Given a right O-ideal I: 1 Test whether nrd(i) is trivial in Cl A (F). 2 Look for a random element x I 1 such that xi is smooth. 3 Do linear algebra with the relations to find y such that nrd(yxi) = Z F. 4 For all p B, apply the reduction algorithm in the tree at p. Now hyxi = O, i.e. I = (hyx) 1 O.
Sketch of the algorithm: computing a generator Given a right O-ideal I: 1 Test whether nrd(i) is trivial in Cl A (F). 2 Look for a random element x I 1 such that xi is smooth. 3 Do linear algebra with the relations to find y such that nrd(yxi) = Z F. 4 For all p B, apply the reduction algorithm in the tree at p. Now hyxi = O, i.e. I = (hyx) 1 O.
Sketch of the algorithm: computing a generator Given a right O-ideal I: 1 Test whether nrd(i) is trivial in Cl A (F). 2 Look for a random element x I 1 such that xi is smooth. 3 Do linear algebra with the relations to find y such that nrd(yxi) = Z F. 4 For all p B, apply the reduction algorithm in the tree at p. Now hyxi = O, i.e. I = (hyx) 1 O.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Example ( ) A = 3, 1 Q, O = Z+Zi +Zj +Zω where ω = (1+i + j + ij)/2, I = xo + 19O where x = 3 4i + j A. Factor base B = {2, 3, 5, 7, 11, 13, 17}. 1 Cl A (Q) = 1, so I is principal. 2 Find x = (7+i 9j 3ω)/19 I 1 such that nrd(xi) = 7Z: xi is smooth. 3 Linear algebra: c = 1 2i j +ω, cxi/7 = J/7 where J = 49O+wO with w = 17 8i + j. 4 Local reduction at 7: h = ( 9 5i 7j 3ω)/7. Multiply out everything: 3+4i 3j 11ω has norm 19, generator of the ideal I.
Jacquet-Langlands correspondence and cohomology Let F be imaginary quadratic, p,q primes in Z F. Let A be ramified at p,q and O A a maximal order. Let Γ 0 (pq) be the subgroup of PGL 2 (Z F ) of elements that are upper triangular modulo pq. Jacquet-Langlands: injection of Hecke-modules H 1 (O /Z F,C) H 1(Γ 0 (pq),c) What happens if we replace C with another ring, say F p?
Jacquet-Langlands correspondence and cohomology Let F be imaginary quadratic, p,q primes in Z F. Let A be ramified at p,q and O A a maximal order. Let Γ 0 (pq) be the subgroup of PGL 2 (Z F ) of elements that are upper triangular modulo pq. Jacquet-Langlands: injection of Hecke-modules H 1 (O /Z F,C) H 1(Γ 0 (pq),c) What happens if we replace C with another ring, say F p?
Jacquet-Langlands correspondence and cohomology Let F be imaginary quadratic, p,q primes in Z F. Let A be ramified at p,q and O A a maximal order. Let Γ 0 (pq) be the subgroup of PGL 2 (Z F ) of elements that are upper triangular modulo pq. Jacquet-Langlands: injection of Hecke-modules H 1 (O /Z F,C) H 1(Γ 0 (pq),c) What happens if we replace C with another ring, say F p?
Jacquet-Langlands correspondence and cohomology Let F be imaginary quadratic, p,q primes in Z F. Let A be ramified at p,q and O A a maximal order. Let Γ 0 (pq) be the subgroup of PGL 2 (Z F ) of elements that are upper triangular modulo pq. Jacquet-Langlands: injection of Hecke-modules H 1 (O /Z F,C) H 1(Γ 0 (pq),c) What happens if we replace C with another ring, say F p?
Jacquet-Langlands correspondence and cohomology Let F be imaginary quadratic, p,q primes in Z F. Let A be ramified at p,q and O A a maximal order. Let Γ 0 (pq) be the subgroup of PGL 2 (Z F ) of elements that are upper triangular modulo pq. Jacquet-Langlands: injection of Hecke-modules H 1 (O /Z F,C) H 1(Γ 0 (pq),c) What happens if we replace C with another ring, say F p?
Modulo p cohomology of arithmetic groups Work of Calegari and Venkatesh: Relate the size of H 1 (O /Z F,Z) tors and of H 1 (Γ 0 (pq),z) tors. Work of Scholze: Existence of Galois representations attached to Hecke eigenforms in H 1 (Γ 0 (N),F p ).
Modulo p cohomology of arithmetic groups Work of Calegari and Venkatesh: Relate the size of H 1 (O /Z F,Z) tors and of H 1 (Γ 0 (pq),z) tors. Work of Scholze: Existence of Galois representations attached to Hecke eigenforms in H 1 (Γ 0 (N),F p ).
A modulo p Jacquet-Langlands correspondence? Joint work with M. H. Şengün (in progress). Let F = Q(ζ 3 ), p = (7, 2+ζ 3 ), q = (31, 25+ζ 3 ). Let A be a quaternion algebra ramified exactly at p,q. Let O be a maximal order in A, and Γ = O /Z F. We have H 1 (Γ,C) = 0, and H 1 (Γ 0 (pq),c) = 0. Let p = 5. Then H 1 (Γ,F p ) = F p c 1, and H 1 (Γ 0 (pq),f p ) = F p c 2 +F p c 3.
A modulo p Jacquet-Langlands correspondence? Joint work with M. H. Şengün (in progress). Let F = Q(ζ 3 ), p = (7, 2+ζ 3 ), q = (31, 25+ζ 3 ). Let A be a quaternion algebra ramified exactly at p,q. Let O be a maximal order in A, and Γ = O /Z F. We have H 1 (Γ,C) = 0, and H 1 (Γ 0 (pq),c) = 0. Let p = 5. Then H 1 (Γ,F p ) = F p c 1, and H 1 (Γ 0 (pq),f p ) = F p c 2 +F p c 3.
A modulo p Jacquet-Langlands correspondence? Joint work with M. H. Şengün (in progress). Let F = Q(ζ 3 ), p = (7, 2+ζ 3 ), q = (31, 25+ζ 3 ). Let A be a quaternion algebra ramified exactly at p,q. Let O be a maximal order in A, and Γ = O /Z F. We have H 1 (Γ,C) = 0, and H 1 (Γ 0 (pq),c) = 0. Let p = 5. Then H 1 (Γ,F p ) = F p c 1, and H 1 (Γ 0 (pq),f p ) = F p c 2 +F p c 3.