Our other goal is to see how closely we needRandLto imitate Re andLe in order to get a
result like Theorem 8.6 whereLe ·eS e·Re = S. In essence, our construction is an attempt at
fusion between the Rees construction and Putcha’s fantastic result,Le·eS e·Re =S.
In a further effort to emulate Theorem 8.6 we will also assume that there exists a pair (B,A) ∈ R× Lso that φ(B,A) = 1. Our φmap acts as a stand in for, (r, `) 7→ r` ∈ eS e, the multiplication coming from L = Le, M = eS e, and R = Re. In the case we are generalising, L= Le, M =eS e, andR= Re, such a pair is already seen to exist, by lettingA= B=e.
Remark 9.8. If we know γ(P)δ(Q) = H then this is equivalent to the existence of a pair
(B,A)∈R×L so thatφ(B,A)∈H.
With our pair (B,A) ∈R×L observe that (A,1,B) is an idempotent in L× M×Rand by extension,e:= π(A,1,B) is an idempotent inS.
Lemma 9.9. θ : M → Sgiven byθ(m)= π(A,m,B)is a morphism of algebraic monoids and θ(M)=eSe withθ(H)= He.
Proof. Let m,m0 ∈ M. As we have defined θ(m) = π(A,m,B) we can quickly calculate,
θ(m)θ(m0) = π(A,m,B)π(A,m0,B) = π(A,mφ(B,A)m0,B) = π(A,mm0,B) = θ(mm0). So this is a monoid morphism. Now observe if,a ∈eSethen there exists a tuple, (`,m,r)∈ L×M×R
soa= π(A,1,B)π(`,m,r)π(A,1,B) = π(A, φ(B, `)mφ(r,A),B). We seeθ(φ(B, `)mφ(r,B)) = a. Thus,θ(M)= eSe.
9.2. ConstructingSemigroupsWithGreen’sRelations 131
As the group of units, we know H = M. So, by continuity and the fact thatθ(H) ≤ He, we
seeθ(H)⊆ He ⊆ θ(H)=θ(M)=eSe⊆ θ(H). But then it is clear thatθ(H) is a dense subgroup
ofHe. This impliesθ(H) = He as the image of an algebraic group is a closed subgroup of the
algebraic group codomain.
Theorem 9.10. eSe M
Proof. Consider the morphism,ψ:L×M×R→ Mgiven byψ(`,m,r)= φ(B, `)mφ(r,A). This morphism isP×Qinvariant. Indeed, for p∈P,q∈Q,
ψ(`p−1, γ(p)mδ(q−1),qr)=φ(B, `p−1)γ(p)mδ(q−1)φ(qr,A)
=φ(B, `)γ(p)−1γ(p)mδ(q)−1δ(q)φ(r,A)= φ(B, `)mφ(r,A).
Then there is a unique morphismeψ:S → M.
L×M×R M S ψ e ψ π
Sinceψis surjective (ψ(A,m,B)=m) it follows thateψis also surjective. Consider the map
θ: M → Sgiven byθ(m)=π(A,m,B). We claimθandψeare inverses.
Consider an element, m ∈ M. Then eψ(θ(m)) = eψ(π(A,m,B)) = ψ(A,m,B) = m. Addi- tionally, for any element s ∈ eSe we know that there existsm ∈ M so that π(A,m,B). Then
θ(eψ(s))=θ(eψ(π(A,m,B)))= θ(ψ(A,m,B))=θ(m)=π(A,m,B)= s.
It remains to check that these are monoid morphisms. θ is by our previous lemma, so consider s,s0 ∈ eSe. Then there exist m,m0 so that θ(m) = s andθ(m0) = s0. And so we see
that, e
ψ(s)eψ(s0)=ψ(A,m,B)ψ(A,m0,B)= φ(B,A)mφ(B,A)φ(B,A)m0φ(B,A)=φ(B,A)mm0φ(B,A)
sinceφ(B,A)= 1. So,φ(B,A)mm0φ(B,A)=ψ(A,mm0,B)=eψ(π(A,mm0,B))=eψ(ss0).
Corollary 9.11. He H.
Proof. This is just a combination of Lemma 9.9 and Theorem 9.10. With our ability to identify MandeSealong with H andHe, the natural question to ask is
Proposition 9.12. Define the morphisms, θL : L × M → S by θL(a,m) = π(a,m,B) and θR : M×R→ SbyθR(m,b)= π(A,m,b). Then,
(1)θL(L×M)=Se (2)θR(M×R)= eS
Proof. (1) Take anyπ(`,m,r) ∈ S. π(`,m,r)π(A,1,B) = π(`,mφ(r,A),B). SoθLis onto Se.
If ` ∈ L, m ∈ M, then θL(`,m) = π(`,m,B) = π(`,m,B)π(A,1,B) ∈ Se. ThusθL’s image is
contained inSe, henceθL(L×M)=Se. (2) is done similarly.
So this settles what Se andeS look like, but what about Le and Re? Sadly, Land R are
beginning to stray away from our original goal. That is, LandRare acting more likeS eand
eS than likeLe andRe.
Proposition 9.13. The following are equivalent for any`∈L (1)π(`,h,B)∈Le for all h∈H
(2)π(`,h,B)∈Le for some h∈H (3)φ(y, `)∈H for some y∈R.
The following are equivalent for any r ∈R (4)π(A,h,r)∈Re for all h∈H
(5)π(A,h,r)∈Re for some h∈H (6)φ(r,x)∈H for some x∈L.
Proof. (1)⇒(2) is clear. Suppose thatπ(`,h,B)∈Le. Then there exists an element,π(a,m,b),
soπ(a,m,b)π(`,h,B) = e. e = ee = π(A,1,B)π(a,m,b)π(`,h,B) = π(A, φ(B,a)mφ(b, `)h,B). By Corollary 9.11 it is clear thatφ(B,a)mφ(b, `)h=1 and it follows thatφ(b, `)∈ H.
Now lety∈Rbe such thatφ(y, `)∈ H. Consider π(`,h,B) for arbitrary fixedh∈ H. Then we can see the following multiplication is correct, π(A,h−1φ(y, `)−1,y)π(`,h,B) = eIt is clear that this suffices to showπ(`,h,B)∈Le.
(4)⇒(5)⇒(6)⇒(4) is proven similarly.
Definition 9.14. We define thequasiclasseswith respect toφto be the sets,
9.2. ConstructingSemigroupsWithGreen’sRelations 133
The following proposition gives a number of facts about these quasiclasses, showing how closely they approximate theL-classes andR-classes we are trying to imitate.
Proposition 9.15.
(1)`∈ L0if and only if there exists r ∈R0withφ(r, `)∈H. (2) r ∈R0if and only if there exists` ∈L0withφ(r, `)∈H. (3) L0×H =θL−1(Le)whereθLis defined as in Proposition 9.12. (4) H×R0 = θR−1(Re)whereθRis defined as in Proposition 9.12. (5) L0 is open in L, hence a quasiaffine variety.
(6) R0is open in R, hence a quasiaffine variety. (7) The action of P restricts to L0.
(8) The action of Q restricts to R0. (9) Ifγis surjective,π(L0,1,B)= Le. (10) Ifδis surjective,π(A,1,R0)= Re.
Proof. We will only prove the odd numbered results.
(1) This comes from considering the definition of bothL0andR0.
(3) Suppose that π(`,m,B) ∈ Le for ` ∈ L and m ∈ M. Then there exists π(a,n,b) so π(a,n,b)π(`,m,B)= e. e = ee= π(A,1,B)π(a,n,b)π(`,m,B) = π(A, φ(B,a)nφ(b, `)m,B). By Corollary 9.11 it is clear thatφ(B,a)nφ(b, `)m = 1 and it follows thatφ(b, `) ∈ H andm ∈ H. Thus` ∈L0 andm∈Has desired.
Conversely, suppose that ` ∈ L0 andh ∈ H. So there existsy ∈ R such that φ(y, `) ∈ H. Thenπ(A,h−1φ(y, `)−1,y)π(`,h,B)=eshows thatθ
L(L0×H)=Le.
(5) Le is open inSeby Proposition 2.15. So then by (3) and Proposition 9.12 we see that L0×His open inL×M. Since projection is an open map we can then conclude thatL0is open inL. Quasiaffineness ofL0follows, as an open set of a quasiaffine variety is also quasiaffine.
(7) Suppose` ∈L0 and fix p∈P. Then there existsr∈R0 so thatφ(r, `)∈H. Observe that
φ(r, `p)=φ(r, `)γ(p)∈H.
(9) Consider π(`,m,r) ∈ Le. As we showed in (3) we may assume r = B, m ∈ H, and ` ∈ L0. Sinceγ is a surjective morphism we know that there exists p ∈ Psoγ(p) = m. Then
Our quasiclasses are now closer to Le andRe than LandRare, but they still are not quite
there. We would like something closer to Theorem 9.10, with a single isomorphism. However, this is still a noteworthy result and leads us to a discussion of whenSis regular.
Proposition 9.16. L0×H×R0 = π−1(Je)and Je is the unique maximalJ-class ofS.
Proof. Suppose π(`,m,r) = j ∈ Je. Then it follows that there exist elements s,s0 ∈ S1 sat-
isfying, s js0 = e. Now, since e = eee = es js0e we may assume that s = π(A,n,y) and
s0 = π(x,n0,B). Then e= π(A,1,B) = π(A,nφ(y, `)mφ(r,x)n0,B) = s js0. By Corollary 9.11 it follows thatφ(y, `),m, φ(r,x)∈Hand hence`∈L0,r ∈R0. Thus,π−1(Je)⊆L0×H×R0.
For the reverse, take any ` ∈ L0, h ∈ H, and r ∈ R0. There is x ∈ L0, y ∈ R0 so
φ(y, `), φ(r,x) ∈ H. e = π(A,h−1φ(y, `)−1,y)π(`,h,r)π(x, φ(r,x)−1,B) ∈ S1π(`,h,r)S1 and π(`,h,r)= π(`,h,B)π(A,1,B)π(A,1,r)∈ S1eS1. So it follows thatL0×
H×R0 =π−1(Je).
It remains to show that Je is the unique maximal J-class of S. Let J ⊆ S be any other J-class. It suffices to show that for any s∈ J, s ∈S1eS1. Let s = π(`,m,r). Then by now it
is quick to check that,π(`,m,B)π(A,1,B)π(A,1,r)=π(`,m,r).
This gives us a remarkable similarity to regular semigroups, as we know they have a unique maximalJ-class. In fact, as the next theorem indicates, if we were to perform our construc- tion withL0 andR0in place ofLandR, we would get a regular semigroup forS.
Theorem 9.17. IfS(L0,P,M,Q,R0)exists, then it is regular.
Proof. Consider (a,m,b) ∈ L0× M×R0. Let x ∈ L0andy ∈R0 be so thatφ(b,x), φ(y,a) ∈ H. Letn∈ Mbe such thatmnm=m(which exists asM is regular). Then,
(a,m,b)(x, φ(b,x)−1nφ(y,a)−1,y)(a,m,b)= (a,mφ(b,x)φ(b,x)−1nφ(y,a)−1φ(y,a)m,b),
which reduces to (a,m,b). So thenL0×M×R0 is regular. And since we are assuming thatπis
surjective thenS, as the image of a regular semigroup, is also regular.
With this theorem we now see that our idempotent, e= π(A,1,B) is shown to be an idem- potent of the maximal J-class of our regular semigroup, which is further imitation of the setup in [24]. This is what we have been shooting for since the start of our section. So when doesS(L0,P,M,Q,R0) exist?
9.3. Normality 135
The trivial answer is: whenO(L0 ×M×R0)P×Qis finitely generated. Obviously we would like a more substantial answer and one that perhaps relates to our original choice of varietiesL
andR. So when doesS(L,P,M,Q,R) exist?
Another naive answer would be when L and R are affine and P× Q is reductive (recall Theorem 8.10). Suppose thatLandRare affine andP×Qis reductive. When can we show that S(L0,P,M,Q,R0) exists? And let us not forget, we will also needπto be surjective, as many of our previous results have employed this assumption. The following section gives us a possible direction to pursue.