You're completely right, of course -- what I said above is FALSE and only true "virtually". Thank you for the correction!

That should be $Q_p$-algebraic groups in the first line, and $GL_m(Q_p)$ in the third, sorry. Can't edit comments.

Any such is induced from a map of p-algebraic groups $SL_n \rightarrow SL_m$ . You can see this by passage to the Lie algebra. Given such a map, the image of $SL_n(Z_p)$ is conjugate by $GL_m(Z_p)$ to a subgroup of $SL_m(Z_p)$ . However, there can be multiple conjugacy classes of maps inducing the same algebraic morphism.

I agree with you, but couldn't see it either; the "vanishing" was the best I could do. Here's an interesting case which might be helpful to think about: Consider $SL_2$ over a real quadratic field $F$. It gives us some complex two-dimensional (Hilbert modular) variety. In this case, an elliptic curve over $F$ would contribute four dimensions to the $H^2$, one dimension in Hodge numbers $(0,2),(2,0)$ and two dimensions in Hodge number $(1,1)$. A CM elliptic curve only contributes half of this. What's going on in direct terms? I couldn't see this.

OK, if you really want to use MAGMA, you can use some trick like trying to replace M,N by p-adically close but definite forms M', N'; then use Magma's implemented functions to check if these are equivalent at p. (You do this at all primes p dividing the discriminant.) I'm not sure if Magma's implemented functions cover the "spinor" version of this, however.

Kevin -- In fact, I also don't know how to do it in a "concrete" way in general. In the case of curves, though, you can get by in classical language because you can express the de Rham cohomology in terms of meromorphic forms. In concrete terms: You can write down an meromorphic $f$ on the curve $E_{\lambda}$, depending on $\lambda$, so that, if $\omega = dx/\sqrt{\dots}$ is the pertinent differential above, then a linear combination of $\omega, \omega', \omega''$ equals $df$. So this linear combination is zero in cohomology.

Two small points. If you "extend $\gamma$ by parallel transport" the local derivatives are zero. Secondly, as you say, the section will "obviously" satisfy a DE; however, the key point, which does not follow from what you observe, is that DE is of algebraic origin. The point is that you can construct the linear dependence in the (algebraic) de Rham cohomology.

Alternately, for any function f holomorphic at 0, f(2^{-s}) is a Dirichlet series of the type you describe.

Thanks! I was able to get what I wanted by following the links. A variant with different signs of this formula is in Corollary 7.14 of Deligne-Lusztig, which one combines with Proposition 7.3 to get the result.

I'm confused, why can't you argue like this: For any two smooth points, we can join them by a sequence of curves. For these curves, the intersection with non-smooth set is Zariski closed, so finite, and removing them does not disconnect.