Page 29 - Leech, Jonathan E. 2020. Noncommutative Lattices: Skew Lattices, Skew Boolean Algebras and Beyond. Koper: University of Primorska Press
P. 29
I: Preliminaries
aaʹb (and bbʹa), abaʹ (and babʹ) and baaʹ (and abbʹ) are unambiguous. We consider only the non-
parenthesized cases, of which the amb, mab and baaʹ cases are trivial.
abm: Since bm is in the A-cell in M containing all images of b in M, (ab)m = ab = a(bm).
aaʹb: Since ab and aʹb lie in a common A-cell in M, (aaʹ)b = ab and a(aʹb) are images of a in
this cell and so are equal.
abaʹ: Note that baʹ is an image of b in M lying in the unique A-cell of M containing all images
of B in M. Thus (ab)aʹ = ab = a(baʹ). £
We consider next a class of necessarily regular bands for which architectural issues are
simplified. A band S satisfies the class covering condition (CCC) if for every comparable pair of
D-classes A ≥ B in S, and every b ∈ B an a ∈ A exists such that a ≥ b. Every b ∈ B is thus
“covered” by some a in A. The free left regular band on {a, b, c} does not satisfy the CCC.
Theorem 1.2.16. In a band S satisfying the class covering condition, the following hold:
i) S is regular.
ii) An element x lies in the center of S if and only if Dx = {x}
iii) Given x, y ∈S, Dxy is DxDy = {uv⎪u ∈Dx & v ∈Dy}.
iv) In particular, when Dx and Dy are finite, so is Dxy.
Proof. (i) Given x, y, z ∈ S, by the CCC, u, v ∈ Dx exists such that u ≥ xy and v ≥ zx. Since
uvw = uw holds in Dx we obtain: xyzx = (xyu)(vzx) = xy(uxv)zx = (xyu)x(vzx) = xyxzx. Thus S is
regular. (ii) If x commutes with all elements in S, then clearly Dx = {x}. Conversely, let
Dx = {x} and let y ∈ S be given. By the CCC, x ≥ xy and hence xy = xyx. Similarly, x ≥ yx so
that yx = xyx. Thus xy = yx and x lies in the center of S. (iii) Clearly DxDy ⊆ Dxy. Let z in Dxy
be given. Since Dx ≥ Dxy and Dy ≥ Dxy, by the CCC, u ∈Dx and v ∈Dy exist such that u ≥ z and
v ≥ z. Thus z = uvzuv = uv since both z and uv lie in Dxy. (iv) is now clear. £
The situation regarding cell-maps in the CCC case is much cleaner.
Theorem 1.2.17. In a band S satisfying the class covering condition, the following hold:
i) Given D-classes A > B, the A-cells in B are precisely the cosets, AbA.
ii) B is partitioned by the cosets of A in B and the cell-maps from A to B are
collectively surjective in that B = ∪b∈BAbA.
iii) Given a D-class chain A > B > C, all compositions of cell-maps from A to B
with cell-maps from B to C are cell-maps from A to C.
iv) Conversely, all cell-maps from A to C are obtained in this fashion.
Proof. (i) Given b ∈ B, by the CCC, a ∈ A exists such that a ≥ b. Hence b = aba ∈ AbA and
the cell containing b collapses to AbA. (ii) is now clear. (iii) Next, given A > B > C as stated,
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aaʹb (and bbʹa), abaʹ (and babʹ) and baaʹ (and abbʹ) are unambiguous. We consider only the non-
parenthesized cases, of which the amb, mab and baaʹ cases are trivial.
abm: Since bm is in the A-cell in M containing all images of b in M, (ab)m = ab = a(bm).
aaʹb: Since ab and aʹb lie in a common A-cell in M, (aaʹ)b = ab and a(aʹb) are images of a in
this cell and so are equal.
abaʹ: Note that baʹ is an image of b in M lying in the unique A-cell of M containing all images
of B in M. Thus (ab)aʹ = ab = a(baʹ). £
We consider next a class of necessarily regular bands for which architectural issues are
simplified. A band S satisfies the class covering condition (CCC) if for every comparable pair of
D-classes A ≥ B in S, and every b ∈ B an a ∈ A exists such that a ≥ b. Every b ∈ B is thus
“covered” by some a in A. The free left regular band on {a, b, c} does not satisfy the CCC.
Theorem 1.2.16. In a band S satisfying the class covering condition, the following hold:
i) S is regular.
ii) An element x lies in the center of S if and only if Dx = {x}
iii) Given x, y ∈S, Dxy is DxDy = {uv⎪u ∈Dx & v ∈Dy}.
iv) In particular, when Dx and Dy are finite, so is Dxy.
Proof. (i) Given x, y, z ∈ S, by the CCC, u, v ∈ Dx exists such that u ≥ xy and v ≥ zx. Since
uvw = uw holds in Dx we obtain: xyzx = (xyu)(vzx) = xy(uxv)zx = (xyu)x(vzx) = xyxzx. Thus S is
regular. (ii) If x commutes with all elements in S, then clearly Dx = {x}. Conversely, let
Dx = {x} and let y ∈ S be given. By the CCC, x ≥ xy and hence xy = xyx. Similarly, x ≥ yx so
that yx = xyx. Thus xy = yx and x lies in the center of S. (iii) Clearly DxDy ⊆ Dxy. Let z in Dxy
be given. Since Dx ≥ Dxy and Dy ≥ Dxy, by the CCC, u ∈Dx and v ∈Dy exist such that u ≥ z and
v ≥ z. Thus z = uvzuv = uv since both z and uv lie in Dxy. (iv) is now clear. £
The situation regarding cell-maps in the CCC case is much cleaner.
Theorem 1.2.17. In a band S satisfying the class covering condition, the following hold:
i) Given D-classes A > B, the A-cells in B are precisely the cosets, AbA.
ii) B is partitioned by the cosets of A in B and the cell-maps from A to B are
collectively surjective in that B = ∪b∈BAbA.
iii) Given a D-class chain A > B > C, all compositions of cell-maps from A to B
with cell-maps from B to C are cell-maps from A to C.
iv) Conversely, all cell-maps from A to C are obtained in this fashion.
Proof. (i) Given b ∈ B, by the CCC, a ∈ A exists such that a ≥ b. Hence b = aba ∈ AbA and
the cell containing b collapses to AbA. (ii) is now clear. (iii) Next, given A > B > C as stated,
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