Steinberg homology, modular forms, and real quadratic fields

Abstract

We compare the homology of a congruence subgroup Gamma of GL2(Z) with coefficients in the Steinberg modules over Q and over E, where E is a real quadratic field. If R is any commutative base ring, the last connecting homomorphism psiGamma,E in the long exact sequence of homology stemming from this comparison has image in H0(Gamma, St(Q2;R)) generated by classes zβ indexed by beta in E \ Q. We investigate this image. When R=C, H0(Gamma, St(Q2;C)) is isomorphic to a space of classical modular forms of weight 2, and the image lies inside the cuspidal part. In this case, zbeta is closely related to periods of modular forms over the geodesic in the upper half plane from beta to its conjugate beta'. Assuming GRH we prove that the image of ,E equals the entire cuspidal part. When R=Z, we have an integral version of the situation. We define the cuspidal part of the Steinberg homology, H0cusp(Gamma, St(Q2;Z)). Assuming GRH we prove that for any congruence subgroup, psiGamma,E always has finite index in H0cusp(Gamma, St(Q2;Z)), and if Gamma=Gamma1(N)pm or 1(N), then the image is all of H0cusp(Gamma, St(Q2;Z)). If Gamma=Gamma0(N)pm or Gamma0(N), we prove (still assuming GRH) an upper bound for the size of H0cusp(Gamma, St(Q2;Z))/image(psiGamma,E). We conjecture that the results in this paragraph are true unconditionally. We also report on extensive computations of the image of psiGamma,E that we made for Gamma=Gamma0(N)pm and Gamma=Gamma0(N). Based on these computations, we believe that the image of psiGamma,E is not all of H0cusp(Gamma, St(Q2;Z)) for these groups, for general N.

0

Turn this paper into a lesson

ArcXiv compiles a structured reading guide from this paper's metadata: plain-English importance, contributions, prerequisite concepts, which sections to read first, flashcards, and a quiz. Grounded in the abstract, never invented.

Discussion (0)

Sign in to join the discussion.

Loading comments…