Footnotes on the Actual Talk:
0. Geometric monodromy is over the algebraic closure. Inner equivalence is about arithmetic monodromy of a cover, Galois closure group over a given definition field for a cover.
1. Action of braid q, gives Aq on M(v2,v3) in SL2(Z/p). Check for the two generators q2 and the shift (sh). Example: up to inner equivalence, the shift maps: M(v2,v3) → M(v3 - v2,v3 - 2v2). Represented by the 2x2 matrix with rows (-1 1) and (-2 1) of Det 1.
<Ash, Aq2> = SL2(Z/p), but M(v2,v3) and M(-v2,-v3) are inner equivalent by conjugation by M(-1 0): SL2(Z/p)/<±1> is the geometric monodromy of the j-line cover by the modular curve denoted X(p) (or X(Z/p)).
In GL2 case we come upon components, but they are conjugate over Q(ζp), ζp, a primitive p th root of 1. This is one of the possibilities explained traditionally through the Weil Pairing on pk+1 division points.
2. The last half of this talk features the "lift invariant." In the 2nd Geometry talk we will find that Hurwitz spaces include the Weil pairing on Abelian varieties directly. This is part of finding the definition field of components of Hurwitz spaces.
3. Recall the permutation representation attached to a degree n cover. The spaces are moduli spaces. In this alternating group case they have fine moduli properties. For Inner spaces that is equivalence to G has no center. For absolute spaces that means group stabilizing an integer in the representation self-normalizes.
4. A (G,G) realization (over K) is also called a regular realization (of G over K). This is how simple groups have been realized as Galois groups. For G=Dp, NSp(Dp) = Z/p xs (Z/p)*. Even dihedral groups – the easiest nonabelian groups – pose a challenge with involution realizations, even though regular realizations of dihedral groups appear in a 1st year graduate algebra course.
5. Indecomposable over Q, is equivalent to <M,Gp> is primitive. To get solutions to Prob2, for example, apply Hilbert's irreducibility theorem: a general matrix M will have the property it is irreducible on (Zp)2.
6. For CM pairs (f,K), K s must be extensions of a complex multiplication field, but they can't give a Dp regular realization.
6a. Going to the Galois closure must extend the constant field to get an exceptional cover: Gar must be different from G.
6b. A cyclotomic (torsion) point on an abelian variety over K is an n-division point v for which GK acts on the elements of {uv | u in (Z/n)* as if these were the primitive powers of the nth roots of 1.
7. Solution in GL2 case: Assuming such an M exists, then f mod p is exceptional is equivalent to f indecomposable mod p.
A rational function over finite field Fq is exceptional if and only if it gives a one-one map over ∞-ly many extensions of Fq. Guralnick-Müller-Saxl showed that other than the list in the talk, only finitely degrees for an f give pairs for which (f,K) is exceptional over infinitely many residue fields of K.
8. We took elliptic curves of Ogg, and used these as motivation for specific cases of Langland's conjecture to develop L-series whose Euler factors reveal the exceptional primes.
9. The lift invariant is more general. When (|C|, |SMG|)=1 will suffice here and in Geometry Talk 2.
10. In ker(Spinn → An) = Z/2 we use the multiplicative notation writing the result as <±1>.
11. Serre asked me in 1989 about the 3-cycle case for the (genus 0) lift invariant formula. He proved the general case. [Fr10, Cor. 2.3] shows the 3-cycle case gives the general case. All (allowable) values of r appear in these results.
12. More about the two types of 3-cycle components separated by their lift invariants.
12a. From [FrV91]: H(An,C3)inn → H(An,C3)abs is 2-1. If the outer automorphism of An is not braidable, then there could be two components of the inner space mapping to a given component of of the absolute. That is not the case: Braiding that outer automorphism is one of the main issues in figuring the components. In these cases: Lift invariant separates the components.
12b. HM representatives of Nielsen classes are defined by elements that have the form (g1, (g1)-1,…, gs, (gs)-1), where 2s = r. In even the most general case HM representatives have trivial lift invariant.