Page 25 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 25
I: Preliminaries
Theorem 1.2.7. A band is left regular [right regular] if and only if xyx = xy [xyx = yx].
In general, a band is regular, if and only if it satisfies the identity xyxzx = xyzx.
Proof. Given x, y in a band S, xyx R xy. Conversely, given x R y in S, x = xyx and y = xy. Thus
S is left regular (R = Δ and D = L) precisely when xyx = xy holds on S. Dually S is right regular
(L = Δ and D = R) precisely when xyx = yx holds on S. If S is regular, then consider the
canonical epimorphisms S → S/L and S → S/R. From the rectangular structure of D-classes of
S, it is follows that S/L is right regular and S/R is left regular. Thus both bands satisfy either
xyx = xy or xyx = yx and hence the identity xyxzx = xyzx. Since the two epimorphisms induce an
embedding of S into S/L × S/R which satisfies xyxzx = xyzx, so does S. Conversely, let S satisfy
xyxzx = xyzx. Suppose u L v in S, so that uv = v and vu = v. Then for all w,
(uw)(vw) = uvwuv = uvw = uw and likewise (vw)(uw) = vw. The assumed identity then gives us
(wu)(wv) = wuwvu = wuwuvu = wuwu = wu and likewise (wv)(wu) = wv showing that L is a
indeed congruence. In similar fashion R is seen to be a congruence. £
Corollary 1.2.8. A band S being regular is equivalent to either of the following:
(i) Given e ≻ a, b in S, aeb = ab.
(ii) S satisfies xyxʹzxʺ = xyzxʺ, given xʹ, xʺ D x.
Proof. If (i) holds, then xyxzx = (xy)x(zx) = xyzx follows and S is regular. Conversely, if S is
regular and e ≻ a, b in S, xyxzx = xyzx gives us (i): aeb = aebaeb = aeababeb = aababb = ab.
Clearly S is regular if (ii) holds. Conversely, (i) implies xyxʹzxʺ = (xy)xʹ(zxʺ) = xyzxʺ. £
The function ζ: S → S/L × S/R defined by ζ(x) = (Lx, Rx) is always 1-1 for any band.
The product S/L × S/R is naturally a band and ζ is a homomorphism (and thus a monomorphism)
precisely when S is regular. In this case, the image ζ[S] is the fibered product S/L ×S/D S/R of
S/L with S/R over the common maximal semilattice image, S/D. Thus the following commuting
diagram of natural epimorphisms is a pullback. The isomorphism S ≅ S/R ×S/D S/L is called the
Kimura factorization, after its discoverer, Naoki Kimura.
S ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯⎯ ⎯ → S/L
⎜⎜
↓↓
S/R ⎯ ⎯ ⎯ ⎯⎯ ⎯ ⎯ ⎯ ⎯→ S/D
Corollary 1.2.9. Every normal band is regular. Normal bands are also characterized by
the identity xyzx = xzyx. In particular, normal left [right] regular bands are characterized by the
identity uxy = uyx [xyu = yxu].
Proof. Normal bands clearly satisfy this identity, and any band S satisfying this identity is
regular: xyxzx = xxzyx = xzyx = xyzx. But a left regular band satisfies this identity it and only if
it satisfies xyz = xzy, thus making it normal. Likewise a right regular band satisfies this identity if
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Theorem 1.2.7. A band is left regular [right regular] if and only if xyx = xy [xyx = yx].
In general, a band is regular, if and only if it satisfies the identity xyxzx = xyzx.
Proof. Given x, y in a band S, xyx R xy. Conversely, given x R y in S, x = xyx and y = xy. Thus
S is left regular (R = Δ and D = L) precisely when xyx = xy holds on S. Dually S is right regular
(L = Δ and D = R) precisely when xyx = yx holds on S. If S is regular, then consider the
canonical epimorphisms S → S/L and S → S/R. From the rectangular structure of D-classes of
S, it is follows that S/L is right regular and S/R is left regular. Thus both bands satisfy either
xyx = xy or xyx = yx and hence the identity xyxzx = xyzx. Since the two epimorphisms induce an
embedding of S into S/L × S/R which satisfies xyxzx = xyzx, so does S. Conversely, let S satisfy
xyxzx = xyzx. Suppose u L v in S, so that uv = v and vu = v. Then for all w,
(uw)(vw) = uvwuv = uvw = uw and likewise (vw)(uw) = vw. The assumed identity then gives us
(wu)(wv) = wuwvu = wuwuvu = wuwu = wu and likewise (wv)(wu) = wv showing that L is a
indeed congruence. In similar fashion R is seen to be a congruence. £
Corollary 1.2.8. A band S being regular is equivalent to either of the following:
(i) Given e ≻ a, b in S, aeb = ab.
(ii) S satisfies xyxʹzxʺ = xyzxʺ, given xʹ, xʺ D x.
Proof. If (i) holds, then xyxzx = (xy)x(zx) = xyzx follows and S is regular. Conversely, if S is
regular and e ≻ a, b in S, xyxzx = xyzx gives us (i): aeb = aebaeb = aeababeb = aababb = ab.
Clearly S is regular if (ii) holds. Conversely, (i) implies xyxʹzxʺ = (xy)xʹ(zxʺ) = xyzxʺ. £
The function ζ: S → S/L × S/R defined by ζ(x) = (Lx, Rx) is always 1-1 for any band.
The product S/L × S/R is naturally a band and ζ is a homomorphism (and thus a monomorphism)
precisely when S is regular. In this case, the image ζ[S] is the fibered product S/L ×S/D S/R of
S/L with S/R over the common maximal semilattice image, S/D. Thus the following commuting
diagram of natural epimorphisms is a pullback. The isomorphism S ≅ S/R ×S/D S/L is called the
Kimura factorization, after its discoverer, Naoki Kimura.
S ⎯ ⎯ ⎯ ⎯ ⎯ ⎯ ⎯⎯ ⎯ → S/L
⎜⎜
↓↓
S/R ⎯ ⎯ ⎯ ⎯⎯ ⎯ ⎯ ⎯ ⎯→ S/D
Corollary 1.2.9. Every normal band is regular. Normal bands are also characterized by
the identity xyzx = xzyx. In particular, normal left [right] regular bands are characterized by the
identity uxy = uyx [xyu = yxu].
Proof. Normal bands clearly satisfy this identity, and any band S satisfying this identity is
regular: xyxzx = xxzyx = xzyx = xyzx. But a left regular band satisfies this identity it and only if
it satisfies xyz = xzy, thus making it normal. Likewise a right regular band satisfies this identity if
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