R is a θ-acute rank 2 residue

If R is θ-acute, then the rank of Q = proj_R(θ(R)) is 1, i.e., Q is a panel. Since we are in a Moufang polygon,Qis endowed with the structure of a Moufang set. Recall that θ induces an automorphism of the Moufang set Q, namely θ⁰ := proj_Q◦θ.

DefineT :=Q\Q_θ, the complement of the induced flip-flop system inQ. Equiva-lently, T is the complement of the set of elements moved maximally byθ⁰. Hence if θ⁰ = id|Q, then T is the empty set; otherwise it is the set of chambers inQ fixed by θ⁰. In either case, by Lemma 3.3.4, T is a proper generalized Moufang subset of Q.

Proposition 4.6.4. If R is θ-acute, then direct descent into R^θ is possible.

Proof. Proposition 4.6.1 implies that the chambers inR^θare those which are opposite the residueQ and which are projected onto a chamber in Qθ =Q\T.

Direct descent into R^θ is possible by a similar argument as in Proposition 4.6.3:

For any chamber c in R, if c is not opposite T, then we find an adjacent chamber c⁰ which is farther away from Q, and hence has lower numerical θ-codistance. If c is already opposite Q, then it is contained in a panel P opposite Q. Then P must contain a chamberc⁰⁰ which is also opposite Qand projects to a chamber inQ_θ, and which hence is in R^θ.

We now turn to the question whether R^θ is connected as a chamber system. For digons this is trivially true: The incidence graph of the point-line geometry of a digon is a complete bipartite graph. The flip-flop system is a subset of the points and lines (containing elements of both types, as it consists of chambers). Therefore the induced incidence graph for the flip-flop system is once again a complete bipartite graph, thus connected. So we next consider triangles.

Proposition 4.6.5. IfRis aθ-acute Moufang projective plane, thenR^θ is connected.

Proof. By invoking duality, we may assume that Q corresponds to a line K, and henceT corresponds to a proper subset of the point row ofK. By Proposition 4.6.1, the elements ofR^θ are all chambers (p, L) where p is not on K, andL meetsK in a point q outside of T.

p₁ L₁

L₂

L₃

p₃ p₂

K r T r’

Figure 4.3.: Connecting good chambers inside a θ-acute Moufang projective plane.

Let (p₁, L₁) and (p₂, L₂) be two such chambers. Assume L₁ and L₂ meet in some point r. If r is a point outside K, we have a connection and are done. So assume r∈K. By Lemma 3.3.6, there must be a second point r⁰ inQ\T, different fromr.

Then there exists a chamber (p₃, L₃) with p₃ not on K and L₃ meeting K inr (see Figure 4.3). By what we said previously, this new chamber is connected insideR^θ to our two original chambers.

Proposition 4.6.6. Let R be a θ-acute Moufang quadrangle of order (s, t). Then R^θ is connected unless the order (s, t) of R is (2,2), (2,4), (4,2), (3,3) or (4,4) (i.e., associated to one of the groupsC₂(Fq)∼= Sp₄(Fq) forq ∈ {2,3,4}or ²A₃(F2)∼= PGU₄(F2)).

Proof. By invoking duality, we may assume thatQcorresponds to a lineK, and hence T corresponds to a (proper) subset of the point row of K. By Proposition 4.6.1, the elements in R^θ are all chambers (p, L) where L is opposite K and p projects to a point on K outside of T. We call points and lines satisfying these properties good points and lines. Observe that all but one of the lines in the pencil of a good point are good lines. Moreover, by Lemma 3.3.6, we have |K \T| ≥ 2, therefore every good line contains at least two good points.

Take two such chambers (p₁, L₁) and (p₂, L₂). Denote the projections ofp₁ andp₂ to K by q₁ and q₂, and the projection lines by K₁ and K₂. If q₁ and q₂ are equal,

4.6. Rank 2residues use the fact that L₂ contains a second good point p⁰₂. Then the chambers (p₂, L₂) and (p⁰₂, L₂) are connected in R^θ, and p⁰₂ projects to a point on K different from q₁. Hence it suffices to deal with the case where q₁ and q₂ differ to establish our claim in full generality.

p₁

K₁ K₂

L₂ p₂

q₁

K q₂

T T₁

L₁' p₁'

(a) Projecting fromK toL⁰₁toK₁ to obtainT₁

T₂ p₁

K₁ K₂

L₂ p₂

q₁

K q₂

T T₁

(b) Projecting betweenK₁ andK₂

Figure 4.4.: Connecting good chambers inside a θ-acute Moufang quadrangle.

Again by Lemma 3.3.6, there must be a second point p⁰₁ on L₁ which does not project toT. Take any line L⁰₁ through that point not meeting K. Then project T to L⁰₁ and from there to K₁, to obtain a set T₁ of “bad” points on K₁ (see Figure 4.4a). Note that all projection lines between L⁰₁ and K₁, except for the one through q₁, are opposite K. All points on K₁ which are not in T₁ and are not q₁ are good points, and by construction reachable from (p₁, L₁).

By symmetry, we can do the same with the second chamber to obtain a similar subset T₂ of K₂. Finally, we project from K₁ to K₂ (see Figure 4.4b) and apply Lemma 3.3.7 if the point order is at least 5: K₁ cannot be covered by T₁, the projection of T₂, and the single point q₁ (which also is the projection of q₂). This implies that there must be a projection line between K₁ and K₂ but different from K which meets those two lines in good points. This line clearly does not meet K, hence is good, and so we have constructed a suitable connection between our two starting chambers withinR^θ.

This leaves a finite number of potential exceptions for quadrangles, namely the Moufang quadrangles satisfyings+t ≤8. By computer calculations (see Appendix A.1), it turns out that the only counterexamples exist in the quadrangles of order (2,2), (2,4) (and its dual), (3,3) and (4,4) – these are the smallest existing Moufang quadrangles.

For Moufang hexagons (and possibly octagons), I believe that a similar statement holds, but no general proof is known to me at this time. However, the following counting argument proves connectedness for most finite Moufang hexagons.

Proposition 4.6.7. LetR be aθ-acute finite Moufang hexagon of order (s, t). Then R^θ is connected if s and t both are at least 19 and are not divisible by 2 or 3.

Sketch of proof. By invoking duality, we may assume thatQcorresponds to a lineK, and henceT corresponds to a (proper) subset of the point row ofK. By Proposition 4.6.1, the elements inR^θare all chambers (p, L) whereLis oppositeK andpprojects to a point on K outside of T. We call points and lines satisfying these properties good points and lines.

The number of good lines hence equals the number of lines opposite K, which is s²t³. Letx:=|T| −1. Then each good line containss+ 1− |T|=s−xgood points.

Hence R^θ consists of (s−x)s²t³ good chambers.

On the other hand, starting at a good line L, one can reach at least 1 + (s−x)(t−1) + (s−x)(t−1)(s−x−1)(t−1)

+ (t−1)(s−x−1)(t−1)(s−x−1)(t−1)

= t·1 + (s−x−1)(t−1) + (s−x−1)²(t−1)²

good lines. Therefore, the size of a connected component of R^θ is at least (s−x) times this number. Dividing the number of good chambers by this number, we obtain an upper bound on the number of connected components:

#connected components≤ s²t²

1 + (s−x−1)(t−1) + (s−x−1)²(t−1)². (4.1) If s and t are not divisible by 2 and 3, then by Lemma 3.3.6, we get x ≤ ₅^s (as 5 is the smallest prime number bigger than 2 and 3). Combining this with inequality (4.1) yields that for s and t at least 19, the number of connected components is less than 2.

Remark 4.6.8. A similar counting approach can be used for finite Moufang quad-rangles, but yields worse bounds than Proposition 4.6.6, and does not cover charac-teristic 2. For this reason a counting argument also fails for Moufang octagons as these only exist in characteristic 2.