Definition 4.2. ForT =J,L,R orH, and r ∈ R, we define,
Tr ={s∈ R | sT r} Tr0 ={s∈ M| sTr where˙ r is a preimage of r in M˙ }
This gives us two different notions of what theT -class of an element ofRlooks like. One may be concerned about the definition ofTr0, that it might vary based on the choice of preimage.
The following proposition shows that this is not the case.
Proposition 4.3. Tr0 is well-defined, regardless of our choice of preimage of r in the definition. Proof. Unfortunately, there is no general way to prove this, and the result is merely a coinci- dence based on the definitions of our various Green’s relations.
IfT = J, take two preimages ofr ∈ R, say p,q ∈ M. Then we know pT = qT, so we can findt∈T such that p= qt. Then, sinceJr0 =G pG=GqtG =GqG, ast ∈T ⊆G.
IfT =L, take two preimages ofr ∈ R, sayp,q∈ M. Then we knowT p =T q, so we can findt ∈T such that p=tq. Then, sinceL0r =G p=Gtq=Gq, ast∈T ⊆G.
IfT = R, take two preimages ofr∈ R, say p,q∈ M. Then we know pT =qT, so we can findt ∈T such that p=qt. Then, sinceR0r= pG =qtG =qG, ast ∈T ⊆G.
IfT =H , our previous work tells us thatL0
randR
0
rare well-defined. SinceH =L ∩R,
we see thatH0r= L
0
r∩R
0
r is well-defined.
With our new notion, we see that our fatT-class, BTrB= Fs∈Tr BsBis not only the given definition, but also makes sense as a symbol. An interesting question is, what would happen if we “fattened” theT -class ofrinM. That is, what doesBTr0Blook like? ForT = J,L, R andH we will see thatBTrB= BTr0B, and this is very useful for our analysis of fatT -classes.
First notice that for anyT,Tr⊆ Tr0by definition, so it is immediately clear thatBTrB⊆ BTr0B.
To approach the fatJ-classes, we will follow along with a suggestion coming from Sec- tion 6 in [27] by Renner. To show the independence of definition in this case, we will first need a result coming from that paper. This result presents an analogue to reductive monoids of Tits’ axiom.
Proposition 4.4. Let S be the set of simple reflections of the Weyl group, W. Then if r ∈ R, we have the following results,
4.1. EquivalentDefinitions 35
sBr ⊆ BrB∪BsrB rBs⊆ BrB∪BrsB
Proof. This is Proposition 5.3 and Remark 5.4 in [27].
Proposition 4.5. For r ∈ R, BJrB= BJr0B= J
0
r.
Proof. As has already been mentioned, it is clear that BJrB ⊆ BJr0B, so it suffices to show
the reverse inclusion. Observe that BJ0
rB = BGrGB = GrG = J
0
r (taking care of the second
equality) and r ∈ BJrB = Ft∈Jr BtB. So the result will be shown if we can demonstrate that
BJrBis closed under multiplication on the right byG and closed under multiplication on the
left byG. We will just show thatGBJrB⊆ BJrB.
The Bruhat decomposition for our reductive group, G, tells us that G = F
w∈W BwB. So it
suffices to show that (BwB)(BJrB) ⊆ BJrBfor allw ∈W. Write w = vsfor s ∈S andv ∈ W
such that`(w)= `(v)+1. Now,
(BwB)(BJrB)= BwB(Ft∈JrBtB)= BvBsB( F t∈Jr BtB) = BvB(F t∈Jr sBtB)⊆ BvB( F t∈Jr(BtB∪BstB)B) = BvB(F t∈Jr BtB∪BstB)⊆ BvB(BJrB) sincet∈Jr =⇒ st∈ Jr.
⊆ BJrB by induction on the length ofw.
Our proof is completed upon the statement of our base case,`(w) =0 =⇒ w= 1, and we
can clearly see that, B(BJrB)= BJrB.
Already we can see why the equivalence of BJrB = BJr0Bis useful. Just looking at BJrB
there does not seem to be a nice way to describe it, but BJr0B= GrG, which is an orbit of the
action ofG×Gon M. Orbits of algebraic group actions are well studied and we will make use of them as we go on.
Both fatL-classes and fatR-classes can be handled in a similar manner.
Proposition 4.6. For r ∈ R, BLrB= BLr0B and BRrB= BR0rB.
Proof. We will show this result for fat R-classes, as the proof is similar for fat L-classes. It is clear that BRrB ⊆ BR0rB, so we just need to show the reverse inclusion. Observe that BR0rB = BrGB = BrG and r ∈ BRrB = Ft∈RrBtB. So the result will be shown if we can demonstrate thatBRrBis closed under multiplication on the right byG.
As was remarked in Proposition 4.5, it suffices to show that (BRrB)(BwB) ⊆ BRrBfor all w∈W. Writew= svfors∈S andv∈W such that`(w)=`(v)+1. Now,
(BRrB)(BwB)= (Ft∈Rr BtB)BwB= ( F t∈Rr BtB)BsBvB =(F t∈Rr BtBs)BvB⊆( F t∈RrB(BtB∪BtsB))BvB =(F t∈Rr BtB∪BtsB)BvB ⊆(BRrB)BvB sincet ∈Rr =⇒ st∈Rr.
⊆ BRrB by induction on the length ofw.
Our proof is completed upon the statement of our base case,`(w) =0 =⇒ w= 1, and we
can clearly see that, (BRrB)B= BRrB.
The problem withH -classes is that they are not defined by a coset or double coset relation (ierLsif and only ifGr =Gs). So our previous work would not apply to dealing with the fat
H -classes. Instead we must take into account the definition ofH , namelyH = R∩L. It is from here, and with our previous results, that we can address fatH -classes.
Proposition 4.7. For r ∈ R, BL0rB∩BR
0
rB= BHrB= BHr0B.
Proof. We will achieve the result by proving the following containments,
BL0 rB∩BR 0 rB⊆ BHrB⊆ BHr0B⊆ BL 0 rB∩BR 0 rB
The last two containments are clear, asHr ⊆ Hr0,H
0 r ⊆ L 0 randH 0 r ⊆ R 0 r. By Proposition 4.6 we know that BL0rB∩BR 0
rB = BLrB∩BRrB. Suppose thatm ∈ BLrB∩BRrB. Then we can find s,t ∈ Rwith sLr andtRr so that m ∈ BsB and m ∈ BtB. But then BsB∩ BtB , ∅. Thus
BsB= BtBand sos= t. It follows that sHrandm∈ BsB⊆ BHrB.
Now that we have a well-defined notion of fat T-classes, we can begin to tackle the four problems listed above.