Definition 30: Unipotent Subgroup for LPS Graphs #
For the LPS graph X_{p,q} = Cay(PGL(2,q), S_{p,q}), we define the unipotent subgroup
H = {[[1,x],[0,1]] | x โ ๐ฝ_q} โ PGL(2,q), which is isomorphic to (๐ฝ_q, +) โ
โค_q.
Main Definitions #
unipotentGL: The mapx โฆ [[1,x],[0,1]]intoGL(2, ๐ฝ_q).unipotentPGL: The composition with the projection toPGL(2,q).unipotentSubgroup: The subgroupH โ PGL(2,q).
Main Results #
unipotentPGL_injective: The mapx โฆ [[1,x],[0,1]]is injective into PGL.unipotentSubgroup_card:|H| = q.lpsGenSet_disjoint_unipotentSubgroup:S_{p,q} โฉ H = โ.
Unipotent GL elements #
The unipotent map sends 0 to 1.
The unipotent map sends a + b to unipotentGL a * unipotentGL b.
The unipotent element projected to PGL(2,q).
Equations
- LPS.unipotentPGL q x = (LPS.projPGL q) (LPS.unipotentGL q x)
Instances For
The map x โฆ [[1,x],[0,1]] as a group homomorphism Multiplicative(ZMod q) โ* PGL(2,q).
Equations
- LPS.unipotentPGLHom q = { toFun := fun (x : Multiplicative (ZMod q)) => LPS.unipotentPGL q (Multiplicative.toAdd x), map_one' := โฏ, map_mul' := โฏ }
Instances For
Injectivity of the unipotent map #
Two unipotent GL elements that map to the same PGL coset must be equal.
If [[1,a],[0,1]] and [[1,b],[0,1]] differ by a scalar in GL, then a = b.
The unipotent subgroup #
The unipotent subgroup H = {[[1,x],[0,1]] | x โ ๐ฝ_q} โ PGL(2,q).
Isomorphic to (๐ฝ_q, +) โ
โค_q.
Equations
Instances For
The cardinality of the unipotent subgroup is q.
The order of H is odd (since q is an odd prime).
Disjointness: S_{p,q} โฉ H = โ #
The determinant of a unipotent GL element is 1.
The LPS matrix M(a,b,c,d) has determinant aยฒ + bยฒ + cยฒ + dยฒ = p in ZMod q.
For (a,b,c,d) โ Sฬ_p, det M = p.
Key lemma: if g โ H (unipotent subgroup), then any GL lift of g has determinant
that is a perfect square in ๐ฝ_q^ร. Specifically, g = projPGL(unipGL(x)) where
det(unipGL(x)) = 1 = 1ยฒ.
p is not a square in ZMod q when legendreSym q p = -1. Uses
ZMod.nonsquare_of_jacobiSym_eq_neg_one via the Legendre-to-Jacobi conversion.
The generating set S_{p,q} is disjoint from the unipotent subgroup H.
This follows because elements of S_{p,q} have determinant p (not a QR mod q),
while elements of H have determinant 1 (a QR). In PGL, two GL elements project
to the same class iff they differ by a scalar, so their determinants differ by a
perfect square. Since (p/q) = -1, p is not a QR โ contradiction.
The stronger disjointness: S_{p,q} โฉ gHgโปยน = โ
for all g โ PGL(2,q).
This follows from the determinant argument: conjugation preserves determinant class,
so elements of gHgโปยน still have determinant class 1, while elements of S_{p,q}
have determinant class p โ 1 mod squares.