Ngày tải lên :
22/03/2014, 15:21
... )L(2s + 1, π, 2 ) L(1 − s, π) L(1 − 2s, π, 2 ) L(1 + s, π) L(1 + 2s, π, 2 ) If G = SO(2n), m(s) = L(1 − 2s, π, 2 ) L(2s, π, 2 ) = ε(2s, π, 2 )L(2s + 1, π, 2 ) L(1 + 2s, π, 2 ) 899 ON THE ... Proof The Langlands quotient is obtained as the image of the longest intertwining operator, which is the composition of the following intertwining 907 ON THE NONNEGATIVITY OF L( ,π) FOR SO2n+1 ... SO(2n + 1) then m(s) = L(2s, π, sym2 ) L(1 − 2s, π, sym2 ) = ε(2s, π, sym2 )L(2s + 1, π, sym2 ) L(1 + 2s, π, sym2 ) If G = Spn then m(s) = = L(2s, π, 2 ) L(s, π) ε(s, π)L(s + 1, π) ε(2s, π, ∧2...