A combinatorially motivated component weight function

We continue to assumePsatisfies EC. If one does not care about the relationship between combinatorial properties and eigenvalues, Theorem 3.2.8 said that a representation ofn₊carried byF I(P) can be extended tob0₊without requiring coloring properties for Pbeyond EC, NA, and I3ND: Create a component weight function by choosing for each component ofF I(P) a Γ-set of complex numbers and a split. Then use the corresponding component diagonal operators{Ha}a∈Γto extend the action ofn+. Here we construct a

particular weight function{µa}a∈ΓonF I(P) whose values are determined by the local structure ofP. WhenP

has two new additional properties beyond EC, in Proposition 4.1.2 we use Corollary 3.2.4 to show that it is a component weight function. As we work toward obtaining the attractive (upper)P-minuscule representations ofg0(andb0₊), in the next two sections we will obtain relationships between further coloring properties and thish0-weight.

We prepare to define our new Z-valued weight function {µa}a∈Γ. Fix a color b ∈ Γ. To construct µb:F I(P)→_Z, we introduce auxiliary functionsυb:F I(P)→_Nandψb:F I(P)→_N:

We first defineυb :F I(P)→_Nin stages. Let (F,I) be a split. IfPb∩Idoes not have a maximal element,

then setυb(F,I) :=1. Now suppose thatPb∩Ihas a maximal elementy. By EC the elementyis unique. We

three requirements:

(i) The elementzis greater thany, (ii) Its colorc:=κ(z) is adjacent tob, and

(iii) The number of elements greater thanythat are inPc∩Iis finite.

If there is some colora ∼bsuch that there are infinitely many elements greater thanyinPa∩I, then set υb(F,I) :=|Υb(F,I)|+1. Otherwise setυb(F,I) :=|Υb(F,I)|.

Figure 4.1 illustrates the three possible scenarios for computing{υi}i∈Γon aΓ-colored posetPat a split

(F,I). Since there is no maximal element of colord inI, one hasυd(F,I) = 1. Becauseg ∼ d, we get

υg(F,I)=|Υg(F,I)|+1=2. Lastly, we haveυa(F,I)=|Υa(F,I)|=0.

We also defineψb:F I(P)→_Nin dually analogous stages. Let (F,I) be a split. IfPb∩Fdoes not have

a minimal element, then setψb(F,I) :=1. Now suppose thatPb∩Fhas a minimal elementy. By EC the

elementyis unique. We build up another setΨb(F,I)⊆ Ffrom the empty set∅. Letz∈F. We placezinto

Ψb(F,I) if it meets the following three requirements:

(i) The elementzis less thany,

(ii) Its colorc:=κ(z) is adjacent tob, and

(iii) The number of elements less thanythat are inPc∩Fis finite.

If there is some colora ∼ b such that there are infinitely many elements less thany inPa ∩F, then set ψb(F,I) :=|Ψb(F,I)|+1. Otherwise setψb(F,I) :=|Ψb(F,I)|.

We can now define the weight function{µa}a∈Γ. Let (F,I) be a split and letb∈Γ. IfPb∩I ,∅, then

• • • • • • • • • • • · · · · · · • • • • • • • g a b c d e f g a b c a b d e f d e F ↑ I ↓

defineµb(F,I) :=1−υb(F,I). IfPb∩I =∅, then defineµb(F,I) :=−1+ψb(F,I). Finally, define theΓ-set of

operators{Ma}a∈Γto be the diagonal operators withh0-weight{µa}a∈Γ. These are theµ-diagonal operators.

Foreshadowing Proposition 4.2.1, it is already easy to see that Mb.hF,Ii= +hF,IiifI has a maximal element of colorb. Here are some technical comments: When definingυb(F,I), we added+1 to|Υb(F,I)|

when there was a colora∼bsuch that there were infinitely many elements inPa∩Igreater than the maximal

element of colorb. For Sections 4.1–4.2 we could have instead added any complex number to|Υb(F,I)|. But

this number must be a positive integer for the proof of Theorem 5.1.1 to work. The dual comment applies to

ψb(F,I). The definition of{µa}a∈Γis not symmetric with respect toFandI; an alternate construction can be

made for theµ-diagonal operators that emphasizes filters instead of ideals. Letb∈Γand (F,I)∈ F I(P). If Pb∩F=∅, then setµ0_b(F,I) :=1−υb(F,I). IfPb∩F ,∅, then setµ0b(F,I) :=−1+ψb(F,I).

Proposition 4.1.1. Suppose P satisfies EC and the following additional properties: (AC): Elements with adjacent colors are comparable, and

(I2A): For every a∈Γ: The open interval between any two consecutive elements of color a contains exactly two elements whose colors are adjacent to a.

Thenµ0_b=µbfor all b∈Γ.

It is easy to see that this new property I2A implies both ND and I3ND.

Proof. Letb∈Γand (F,I)∈ F I(P). Sinceκis surjective, there are three cases to consider: (i)Pb∩I =∅

andPb∩F , ∅; (ii) Pb∩I , ∅and Pb∩F , ∅; and (iii) Pb∩I , ∅and Pb∩F = ∅. For (i) we have µb(F,I)= −1+ψb(F,I)= µ0_b(F,I). For (iii) we haveµb(F,I)= 1−υb(F,I)= µ0_b(F,I). For Case (ii) we haveµb(F,I)=1−υb(F,I) andµ0_b(F,I)=−1+ψb(F,I). By local finitenessI has a maximal elementyof colorbandFhas a minimal elementzof colorb. By EC we know thaty<z, and these must be consecutive elements of colorb. By I2A we know (y,z) has two elementsuandvwith colors adjacent tob. By AC, any element inΨb(F,I) orΥb(F,I) must be in (y,z). Thus we haveΨb(F,I)⊆ {u,v}andΥb(F,I)⊆ {u,v}. Note

thatuandvmust each be in exactly one of the setsΨb(F,I) orΥb(F,I). Thus we haveψb(F,I)+υb(F,I)=2

sinceψb(F,I)= |Ψb(F,I)| andυb(F,I)= |Υb(F,I)|. Henceµb(F,I) =1−υb(F,I)= 1−(2−ψb(F,I)) =

−1+ψb(F,I)=µ0_b(F,I).

Proposition 4.1.2. Suppose P satisfies EC, AC, and I2A. Then the weight function{µa}a∈Γis a component

weight function. If P further satisfies NA, then the corresponding operators{Ma}a∈Γcan be used to extend the

representations of Theorem 3.1.4 fromn₊andn−to X- and Y-square nilpotent representations ofb0₊andb0−.

Proof. We use Corollary 3.2.4 to show{µa}a∈Γis a component weight function. Letb∈Γ, let (F,I)∈ F I(P),

and letxbe minimal inF. Definea:=κ(x). We must showµb(F−x,I+x)−µb(F,I)=θab.

First suppose a = b. Start with the case Pa ∩I = ∅. Sincex is minimal in F, we haveψa(F,I) =

|Ψa(F,I)|=0 andµa(F,I)=−1. HerePa∩(I+x),∅. Sincexis maximal inI+x, we haveυa(F−x,I+x)=

|Υa(F−x,I+x)|=0 andµa(F−x,I+x)=1. We getµa(F−x,I+x)−µa(F,I)=2=θaa. Otherwise we

have the casePa∩I , ∅. Letz∈ Pa∩I and note thatPa∩[z,x] is finite by local finiteness for [z,x]. So

Pa∩Ihas a maximal elementy. Herey< xare consecutive occurrences of the colora. By I2A there are

exactly two elementsu,v∈(y,x) with colors adjacent toa. By AC all elements greater thanyinIwith colors adjacent toaare in (y,x). Henceuandvare the only such elements. This shows bothuandvare inΥa(F,I),

and no other elements can be inΥa(F,I). Thusυa(F,I)=|Υb(F,I)| =2 andµa(F,I)=−1. We still have µa(F−x,I+x)=1, and so againµa(F−x,I+x)−µa(F,I)=2=θaa.

Now supposea ∼ b. Again start with the casePb∩I = ∅. HerePb∩(I +x)= ∅as well. We know

Pb∩F ,∅sinceκis surjective. Letz∈Pb∩F. Notex<zby AC and that [x,z] andPb∩[x,z] are finite. So

Pb∩Fhas a minimal elementy. Then xsatisfies all three criteria to be inΨb(F,I). Butx<Ψb(F−x,I+x)

sincex<F−x. Observe thatyis also minimal inPb∩(F−x). So we haveΨb(F−x,I+x)= Ψb(F,I)− {x}.

Thus|Ψb(F−x,I+x)|=|Ψb(F,I)| −1. Note that there is some colorc∼bsuch that there are infinitely many

elements less thanyinPc∩Fif and only if the same statement is true forF−x. Thus whether or not such a

colorcexists we haveψb(F−x,I+x)=ψb(F,I)−1. Hereµb(F−x,I+x)=−1+ψb(F−x,I+x) and

µb(F,I)=−1+ψb(F,I). Thusµb(F−x,I+x)−µb(F,I)=−1=θab.

Otherwise fora∼bwe have the casePb∩I ,∅. HerePb∩(I+x),∅as well. Letz∈Pb∩I. Notez< x

by AC and that [z,x] andPb∩[z,x] are finite. SoPb∩Ihas a maximal elementy. Note thatxsatisfies all three

criteria to be inΥb(F−x,I+x). Butx<Υb(F,I) sincex<I. Observe thatyis also maximal inPb∩(I+x). So

we haveΥb(F−x,I+x)= Υb(F,I)∪ {x}. Thus|Υb(F−x,I+x)|=|Υb(F,I)|+1. Note that there is some color

c∼bsuch that there are infinitely many elements greater thanyinPc∩I if and only if the same statement

is true forI+x. Thus whether or not such a colorcexists we haveυb(F−x,I+x)=υb(F,I)+1. Here

Finally supposea;b. Again start with the casePb∩I =∅. HerePb∩(I+x)=∅as well. An elementy

is minimal inPb∩Fif and only if it is minimal in Pb∩(F−x). If no such minimal element exists, then ψb(F,I)=1=ψb(F−x,I+x). Otherwise there is an elementyminimal in bothPb∩FandPb∩(F−x).

Sincea ; b, there is a color c ∼ b such that there are infinitely many elements less thanyin Pc∩F if

and only if the same statement is true forPc∩(F−x). Thus whether or not such a colorcexists we have ψb(F−x,I+x)=ψb(F,I). Henceµb(F−x,I+x)−µb(F,I)=0=θab. Otherwise fora;bwe have the case

Pb∩I ,∅. Dualize the above argument withυbreplacingψbto again getµb(F−x,I+x)−µb(F,I)=0=θab.

So{µa}a∈Γis a component weight function. Since I2A implies I3ND, we can apply Corollary 3.2.6(b) to

get the last statement.