Amplifications, Diminutions, Subscorings for Categories, Allegories

26 Pages·2016·0.29 MB·English

Amplifications, Diminutions, Subscorings
for
Categories, Allegories
PETER FREYD
[email protected]
April 20, 2020
•ppppp33333 DDDDDiiiiimmmmmiiiiinnnnnuuuuutttttiiiiiooooonnnnn fffffooooorrrrr [[[[[11111.....11111]]]]] FFFFFooooouuuuurrrrr ooooofffff ttttthhhhheeeee eeeeeiiiiiggggghhhhhttttt aaaaaxxxxxiiiiiooooommmmmsssss aaaaarrrrreeeee rrrrreeeeeddddduuuuunnnnndddddaaaaannnnnttttt!!!!!
(Thanks to Martin Knopman):
The axioms:
xy is defined iff x(cid:50) = (cid:50)y,
((cid:50)x)x = x and x(x(cid:50)) = x,
x(yz) = (xy)z
•ppppp33333 [[[[[11111.....1111133333]]]]] ssssshhhhhooooouuuuulllllddddd nnnnnooooowwwww ssssstttttaaaaarrrrrttttt wwwwwiiiiittttthhhhh:::::
((cid:50)x)(cid:50) = (cid:50)x because ((cid:50)x)x is defined, similarly (cid:50)(x(cid:50)) = x(cid:50). If xy is defined then
(xy)(y(cid:50)) is defined (because (xy)(y(cid:50)) = x(y(y(cid:50)))) hence (xy)(cid:50) = (cid:50)(y(cid:50)) = y(cid:50). Similarly,
if xy is defined then (cid:50)(xy) = (cid:50)x. Using the convention for [1] we have (xy)(cid:50) y(cid:50) and
(cid:50)(xy) (cid:50)x.
•ppppp55555 [[[[[11111.....1111177777]]]]] AAAAA fffffooooooooootttttnnnnnooooottttteeeee fffffooooorrrrr ttttthhhhheeeee wwwwwooooorrrrrddddd “““““gggggrrrrrooooouuuuupppppoooooiiiiiddddd”””””:::::
The roots of category theory lie in algebraic topology and the word “groupoid” has been
used in this way since at least Paul Smith’s 1949 “Homotopy groups of certain algebraic
systems.” (Proc. Nat. Acad. Sci.) Smith cites Reidemeister’s 1932 Einfu¨hrung in die kombi-
natorische topologie (Braunschweig : F. Vieweg & So W.) for his usage The notion is usually
¨
credited to H Brandt, “Uber eine Verallgemeinerung des Gruppengriffes” Math. Ann. 96,
360-366, 1926. Indeed, there was a time when categories were most quickly described as
“Brandt semigroupoids.” But, be warned: in some communities “groupoid” means a set with
any binary operation without restriction.
•ppppp2222211111 AAAAA ooooonnnnneeeee-----tttttooooo-----ooooonnnnneeeee cccccooooorrrrrrrrrreeeeessssspppppooooonnnnndddddeeeeennnnnccccceeeee fffffrrrrrooooommmmm ttttthhhhheeeee lllllaaaaasssssttttt pppppaaaaarrrrraaaaagggggrrrrraaaaappppphhhhh ooooofffff [[[[[11111.....333336666644444]]]]] mmmmmaaaaadddddeeeee eeeeexxxxxpppppllllliiiiiccccciiiiittttt:::::
Given 0 ∈ S ⊂ A the Cantor-Bernstein-Schroeder theorem delivers a one-to-one corre-
spondence from S×A(cid:63) to A(cid:63). It can be described as follows, in which n is a natural number,
w ∈ A(cid:63), a ∈ A\S, s ∈ S \{0}:
(cid:104)0,0n(cid:105) → 0n
(cid:104)0,0naw(cid:105) → 0naw
(cid:104)0,0nsw(cid:105) → 0n+1sw
(cid:104)s,w(cid:105) → sw
[1] \begin{picture}(17,7)\qbezier(5,6.5)(6,5)(12,5)\qbezier(5,1.5)(6,3)(12,3)\end{picture}
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
•ppppp5555500000 AAAAAdddddddddd sssssmmmmmaaaaallllllllll-----ppppprrrrriiiiinnnnnttttt cccccooooommmmmmmmmmeeeeennnnnttttt tttttooooo eeeeennnnnddddd ooooofffff [[[[[11111.....444446666611111]]]]]::::: (Thanks to Peter Selinger!)
The category composed of local homeomorphisms has equalizers and, rather surprisingly,
it actually does have products, indeed all non-empty diagrams have limits. The construction
of pullbacks and equalizers, however, is much easier than the construction of even binary
products, Unlike pullbacks, products are not preserved by forgetful functors. The product of
the space of rational numbers with itself, for example, is uncountably large. (And for every
infinite cardinal there’s a space whose product with itself has the next power-cardinal as its
size.)
•ppppp5555522222 [[[[[11111.....444447777755555]]]]] rrrrreeeeemmmmmooooovvvvveeeee lllllaaaaasssssttttt ssssseeeeennnnnttttteeeeennnnnccccceeeee (((((ssssseeeeeeeeee CCCCCooooorrrrrrrrrreeeeeccccctttttiiiiiooooonnnnnsssss))))) aaaaannnnnddddd aaaaadddddddddd:::::
Foranexampleofaone-valuednon-specialcartesiancategorylettheobjectsbetopological
unital rings in which the topology is either discrete or the other extreme, indiscrete (the only
open sets are empty or entire). The maps are continuous ring homomorphisms. The one-
element ring is a strict terminator, that is, the only maps therefrom are isomorphisms. The
only constant maps in this category are those targeted at the terminator, hence for any map
either its source must be discrete or its target must be indiscrete. Equalizers can be obtained
using the standard construction. For binary products topologize the standard ring product
with the indiscrete topology when both factors are indiscrete (otherwise with the discrete
topology). Then A×− is faithful iff A is indiscrete. (Instead of rings one could use semi-
lattices or, for that matter, any equational theory [2] with constants 0 and 1 and a binary
operation such that 0x = 0 and 1x = x. For a less algebraic example use posets with top and
bottom.) By adding a strict coterminator to any one-valued non-special cartesian category
one may obtain a two-valued example.
•ppppp5555555555 FFFFFooooooooootttttnnnnnooooottttteeeee fffffooooorrrrr ttttthhhhheeeee fififififirrrrrsssssttttt pppppaaaaarrrrraaaaagggggrrrrraaaaappppphhhhh ooooofffff [[[[[11111.....444449999933333 sssssmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt]]]]]:::::
A more conceptual description: first lexicographically order the product of the f-targets.
Then (cid:104)T;f ,...,f (cid:105) ∈ τ iff the induced map from T to the product preserves order.
1 n
•ppppp6666666666 FFFFFooooooooootttttnnnnnooooottttteeeee fffffooooorrrrr ttttthhhhheeeee mmmmmiiiiiddddddddddllllleeeee pppppaaaaarrrrraaaaagggggrrrrraaaaappppphhhhh ooooofffff [[[[[11111.....44444(((((1111122222))))) sssssmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt]]]]]:::::
A more conceptual description: (cid:104)a ,...,a (cid:105) < (cid:104)b ,...,b (cid:105) iff for all sufficiently large x,
1 n 1 m
a xn +a xn−1 +···+a x < b xm +b xm−1 +···+b x.
1 2 n 1 2 m
•ppppp6666688888–––––6666699999 CCCCCooooorrrrrrrrrreeeeeccccctttttiiiiiooooonnnnn fffffooooorrrrr [[[[[11111.....555551111122222]]]]] rrrrreeeeemmmmmooooovvvvveeeee tttttoooooppppp ssssseeeeennnnnttttteeeeennnnnccccceeeee ooooonnnnn ppppp6666699999 aaaaannnnnddddd aaaaadddddddddd tttttwwwwwooooo nnnnneeeeewwwww eeeeennnnntttttrrrrriiiiieeeeesssss:::::
•ppppp6666699999 NNNNNeeeeewwwww eeeeennnnntttttrrrrryyyyy [[[[[11111.....555551111155555]]]]]:::::The class of covers is closed under left cancellation (if A→ B → C
is a cover, then so must be B → C). In a category with pullbacks the class of covers is closed
under composition.
because: Left cancellation is immediate For composition, suppose that A → B and
B → C are covers and that a monic D → C allows their composition. Let
E → B
↓ ↓
D → C
be a pullback. Since E → B is monic and allows A → B it must be an isomorphism. We thus
obtain a map from B to D which shows that D → C allows B → C. Hence D → C is an
isomorphism.
[2] Sometimes“algebraictheory.”Theproblemwiththatnameisthetheoryoffields—thequintessentialtheoryofalgebrafor
mostmathematicians—wouldnotbeanalgebraictheory.
2
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
•ppppp6666699999 NNNNNeeeeewwwww eeeeennnnntttttrrrrryyyyy [[[[[11111.....555551111166666 sssssmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt]]]]]:::::
In a category without pullbacks, covers need not be closed under composition. Consider
the subcategory of the category of sets with three objects named 1,2,3 each with the indicated
number of elements. Allow no maps from 1 or 2 into 3 but allow all other maps. (The only
maps targeted at 3 are endomorphisms.) Any endomorphism on 3 is, by default, a cover. To
obtain a counterexample, compose a constant endomorphism on 3 with an onto map from 3
to 2. Each is a cover but the composition is not.
•ppppp7777744444 NNNNNeeeeewwwww eeeeennnnntttttrrrrryyyyy [[[[[11111.....555553333366666]]]]]:::::
In the previous small-print exploration of τ-categories we defined a universal property
for slice categories [1.4(11)6]. It’s worth describing the material without the full uniqueness
allowed by the τ-structure:
For an object in a cartesian category B ∈ A define the generic point in A/B to be the
map g : ∆1 → ∆B carried by Σ to the diagonal map B → B×B.
The slice category A/B is the result of freely adjoining a point to the object B.
b
That is, given any representation of cartesian categories T : A → C and point 1 →T(B)
there is a representation, unique up to natural equivalence, T(cid:48) : A/B → C that carries the
generic point to b and each ∆f to Tf.
Given an object a : A → B construct T(cid:48)a by using the pullback:
T(cid:48)a →TA
↓ ↓ Ta
b
1 → TB
•ppppp111110000099999 AAAAAdddddddddd tttttooooo eeeeennnnnddddd ooooofffff [[[[[11111.....666664444411111]]]]]:::::
For the same reason, given any f : A → B in a boolean pre-logos, the map induced by
inverse-images, f# : Sub(B) → Sub(A), preserves complements.
•ppppp111111111177777 [[[[[11111.....77777]]]]] AAAAAdddddddddd tttttooooo vvvvveeeeerrrrryyyyy tttttoooooppppp:::::
Recall that in a regular category the inverse image and direct image operations satisfy:
f(A(cid:48)) ⊂ B(cid:48) iff A(cid:48) ⊂ f#(B(cid:48)) [1.51].
•ppppp111112222299999 SSSSSmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt aaaaaddddddddddiiiiitttttiiiiiooooonnnnn fffffooooorrrrr [[[[[11111.....7777744444(((((1111100000)))))]]]]]:::::
The Freyd curve may be described with a four-state automaton with states E(cid:48),E,N,N(cid:48), and
inputs “−,◦,+.” (we’re suppressing the 1s):
Next State Output
E(cid:48) E N N(cid:48) E(cid:48) E N N(cid:48)
+ E N E N + + + − −
◦ E E(cid:48) N(cid:48) N ◦
− E N E N − − − + +
For an open onto map any one of the states may be taken as initial. (To get the function
described above start at E.) Each next state is always an adjacent state as defined by the list
E(cid:48),E,N,N(cid:48). The ◦ input toggles E(cid:48),E and N,N(cid:48). The two non-◦ inputs have the same next-state
behavior: they both always target one of the two middle positions which fact together with
3
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
the fact that all next states are adjacent determines their action. In the diagram below the
gray (double) arrows show the next-state behavior for the ◦ input.
↔ ↔
E(cid:48) →E ↔N ←N(cid:48)
As for output: when ◦ is the input there is no output; in the first two states, E(cid:48),E the
output echoes the non-◦ input; in the last two, N,N(cid:48), it negates.
The first task is to show that the resulting function is defined not just on sequences but
on elements of [−1,+1], that is, the same output is engendered by sequences that name the
2 2
same interval element. We need to consider the output of two machines that have been fed
the same initial inputs but one machine will henceforth hear ◦++++··· and the other
+−−−−···. We’ll do better with a single machine but with two demons jumping from
state to state each according to the commands issued by its appointed sequence. We feed
them both the same initial sequence that brings them to the same state. For each of those
four possible states the jump then commanded by the input ◦+ produces the same output
as that commanded by + but the demons will land in different states Each will be at one of
the two center states, E,N, but not the same center state. Thereafter they will continue to
exchange position forever and that means that when one machine echoes the other negates.
Which is just what is needed for the output engendered by a constant sequence of +s on
one machine to be the same as that engendered by a constant sequence of −s on the other
machine. All of this easily dualizes for the pair ◦−−−−··· and −++++···.
Another four Freyd curves may also be described with a four-state automaton:
Next State Output
E E(cid:48) N N(cid:48) E E(cid:48) N N(cid:48)
+ E N N E + + + − −
◦ E(cid:48) E N(cid:48) N ◦
+ E N N E − − − + +
The ◦ input toggles E,E(cid:48) and N,N(cid:48). The two non-◦ inputs have the same next-state behavior:
E and N are stationary and the only targets. In the diagram below, the vertical gray (double)
arrows show the next-state behavior for the ◦ input:
(cid:108)
≺E←N(cid:48)
(cid:108) (cid:108)
E(cid:48)→N≺(cid:108)
As for output: when ◦ is the input there is no output; in the two left-hand states, E,E(cid:48), the
output echoes the non-◦ input; in the two right-hand states, N,N(cid:48), it negates.
Forthismachinewhenwestarttwodemonsinthesamestate,onelisteningto ◦++++···
and the other +−−−−··· the jump commanded by the input + again produces the same
output as that commanded by ◦+ but the demons will land in different states One will be
in E the other N and there they’ll stay forever. One of the demons will echo the input, the
other will negate it, which is—again—just what is needed for the output engendered by a
constant sequence of +s on one machine to be the same as that engendered by a constant
sequence of −s on the other machine. All this, again, easily dualizes.
4
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
•ppppp111111111188888 NNNNNeeeeewwwww eeeeennnnntttttrrrrryyyyy [[[[[11111.....777771111155555]]]]]:::::
A Boolean pre-logos is a logos.
because: f##(A(cid:48)) = ¬f(¬A(cid:48)). That is, B(cid:48) ⊂ ¬f(¬A(cid:48)) iff f(¬A(cid:48)) ⊂ ¬B(cid:48) iff
¬A(cid:48) ⊂ f(¬B(cid:48)) iff ¬A(cid:48) ⊂ ¬f(B(cid:48)) iff f#(B(cid:48)) ⊂ A(cid:48), using the definition of direct
image, [1.641] and the fact that complementation is order-reversing. Thus a Boolean pre-logos
is automatically a boolean logos, and representations of Boolean pre-logoi are automati-
cally representations of Boolean logoi.
•ppppp111113333333333 IIIIInnnnnssssseeeeerrrrrttttt aaaaafffffttttteeeeerrrrr fififififirrrrrsssssttttt tttttwwwwwooooo ssssseeeeennnnnttttteeeeennnnnccccceeeeesssss ooooofffff [[[[[11111.....7777777777]]]]]:::::
Note that (R◦)+ = (R+)◦. (If T is transitive then so is T,◦ and since R◦ ⊂ (R+)◦ we
have (R◦)+ ⊂ (R+)◦and since R = (R◦)◦ ⊂ ((R◦)+)◦we have R+ ⊂ ((R◦)+)◦therefore
(R+)◦ ⊂ (R◦)+. ) Similarly (R◦)(cid:63) = (R(cid:63)).◦
•ppppp111119999966666 AAAAAdddddddddd tttttooooo eeeeennnnnddddd ooooofffff [[[[[22222.....11111]]]]]:::::
See 2.113.
•ppppp111119999977777 IIIIInnnnnssssseeeeerrrrrttttt iiiiinnnnn [[[[[22222.....111111111133333 sssssmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt]]]]] aaaaafffffttttteeeeerrrrr “““““MMMMMOOOOODDDDDUUUUULLLLLAAAAARRRRR LLLLLAAAAATTTTTTTTTTIIIIICCCCCEEEEE”””””:::::
The Horn sentence is equivalent to the equation (A∪X)∩(A∪B) = A∪(X ∩(A∪B))
which is equivalent to the containment (A∪X)∩(A∪B) ⊂ A∪(X ∩(A∪B))
since in any lattice [3] A∪(X ∩(A∪B)) ⊂ (A∪X)∩(A∪B)
When catenation is to be interpreted as union the universally quantified contain-
[3]
ment in the middle line is easily equivalent with the universally quantified containment
RS ∩T ⊂ (R∩TS)S. [4]
•ppppp111119999988888 AAAAAdddddddddd tttttooooo eeeeennnnnddddd ooooofffff [[[[[22222.....1111133333]]]]]:::::
Theimportantfamiliesofendo-relations,reflexive/symmetric/transitive/coreflexive/equivalence,
are each closed under intersection:
Reflexive: 1 ⊂ 1∩1 ⊂ R∩S.
◦ ◦ ◦ ◦
Symmetric: (R∩S) ⊂ R and (R∩S) ⊂ S.
Transitive: (R∩S)2 ⊂ R2 ⊂ R and (R∩S)2 ⊂ S2 ⊂ S.
•ppppp111119999999999–––––222220000000000 AAAAA bbbbbeeeeetttttttttteeeeerrrrr vvvvveeeeerrrrrsssssiiiiiooooonnnnn ooooofffff [[[[[22222.....111113333355555]]]]]:::::
The isomorphisms in A and Map(A) coincide. That is:
If R is an isomorphism then R is a map and R-1 = R◦.
Or, using the convention for :
R-1 R◦hence all isomorphisms are maps.
This is an immediate consequence of a much better lemma [indeed, the most conspicuous
omission in the published book]:
◦
1 ⊂ RS and SR ⊂ 1 iff R is a map and S is its reciprocal (S = R ).
because: R is entire since RS is. Hence S ⊂ S1 ⊂ SRR◦ ⊂ 1R◦ ⊂ R◦. But we also
◦ ◦ ◦ ◦ ◦
have 1 ⊂ S R and R S ⊂ 1 and the same argument yields R ⊂ S.
[3] A ⊂ (A∪X), A ⊂ (A∪B), (X∩(A∪B)) ⊂ X ⊂ (A∪X) and (X∩(A∪B)) ⊂ (A∪B).
[4] Assuming the middle line specialize X,A,B to R,S,T for RS ∩T = (X ∪A)∩B ⊂ (A∪X)∩(A∪B) ⊂
A ∪ (X ∩ (A ∪ B)) = S(R ∩ ST) = (R ∩ TS)S. For the converse specialize R,S,T to X,A,(A ∪ B) for
(A∪X)∩(A∪B) = SR∩T = RS∩T ⊂ (R∩TS)S = (X∩((A∪B)∪A))∪A = A∪(X∩(A∪B)).
5
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
(If an allegory is viewed as a 2-category, then the hypothesis, 1 ⊂ RS and SR ⊂ 1,
says that S and R form an adjoint pair.)
•ppppp222220000000000 NNNNNeeeeewwwww eeeeennnnntttttrrrrryyyyy [[[[[22222.....111113333377777 sssssmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt]]]]]:::::
A von Neumann allegory is an allegory that obeys the equation:
RR◦R = R [5]
Since one containment is a consequence of the standard axioms it is enough to require:
◦
RR R ⊂ R
A von Neumann regular category is a regular category whose associated allegory is
von Neumann.
An equational theory is a mal(cid:48)cev theory if a mal(cid:48)cev operator can be constructed,
to wit, a ternary operator, traditionally denoted as t and whose values are traditionally
denoted as txyz, that satisfies the equations:
txxz = z and txzz = x
Given any equational theory, the category of its models is regular and the forgetful functor to
sets is a faithful representation. As we sill see, the category of models of an equational theory
is von Neumann regular iff the theory is Malc(cid:48)ev.
First, a few examples of Mal(cid:48)cev theories. In any theory that contains the theory of
groups we can construct txyz as xy-1z. For Heyting algebras, txyz may be taken as
((x → y) → z) ∧ ((z → y) → x).[6][7] For a less symmetric example take the theory of
colonies: a single binary operation, with values denoted x:y, subject to just one equation
x:(y:x) = y (equivalently (x:y):x = y [8]) take txyz = (x:x):(z:(x:y)) [9] or any of the
other 23 terms of the same size that do the trick.[10] (Examples of colonial operations include
x:y = (xy)-1 in any group.)[11]
Colonies are a special case of the theory of quasigroups, that is, the theory of binary
operations—associative or not—with unique left and right division. The equational formu-
[5] AvonNeumannregularringisonewithananti-involutionsuchthat xx∗x=x all x.
[6] Toseethat txxz=z notethatthefirsthalf, ((x→x)→z), isequalto 1→z=z anditsufficestoshowthesecond
halfisatleast z, but z ≤ (z→x)→x isequivalentto z∧(z→x) ≤ x which,ofcourse,isequivalentto z→x ≤ z→x.
[7] An orthomodular lattice, beloved by physicists, is a lattice with 0, 1 and an anti-involution x(cid:48) such that x ∨ x(cid:48)
= 1, plus one consequence of modularity, to wit, a∨(a(cid:48) ∧(a∨b)) = a∨b (e.g., the lattice of closed subspaces of a
Hilbertspace).Take txyz=(x∨(y(cid:48)∧(y∨z)))∧(z∨(y(cid:48)∧(y∨x))). Then txzz = (x∨(z(cid:48)∧(z∨z)))∧(z∨(z(cid:48)∧(z∨x))) =
(x∨(z(cid:48)∧z))∧(z∨x) = (x∨0)∧(z∨x) = x.
[8] If (x:y):x = y all x,y then x:(y:x) = ((y:x):y):(y:x) = y.
[9] txxz = (x:x):(z:(x:x)) = z and txzz = (x:x):(z:(x:z)) = (x:x):x = (x:x):(x:(x:x)) = x.
[10] This is the simplest possible consistent Mal(cid:48)cev theory (inconsistent means all equations are provable). The proof is an
exerciseinequationaltheories.There’snowayofobtainingaternaryoperatorwithouthavingan n-aryoperationwith n≥2
andifwehaveanysuchthenwehaveonefor n=2. Weneed,ofcourse,atleastoneequation.There’snowayofdeducinga
singular equation—that is, one in the form s=y where y is a “naked” variable and s is not—from nonsingular equations,
henceifwehavejustoneequationithastobesingular.If y doesnotappearin s thenwededuce y=y(cid:48) (henceallequations)
from s = y and s = y(cid:48). If y appears in the extreme right position of s then a model of the theory is one in which all
termsareequaltotheirfarrightvariable,thatis,therearemodelsthatcannothaveaMal(cid:48)cevoperator.Dittoforthefarleft
side. In particular s can not have just one instance of the binary operation. Assuming just two instances, there are two ways
ofputtingintheparenthesesbutsymmetrysays—forourpresentpurpose—thatitdoesn’tmatterwhich.Iftheoneequationis
(x:y):z=y thenwecandeduce x:z=y (since x:z=((x:x):y):z=y)thusreturningtothecase s=y where y isabsent
from s.Allofwhichsaysthattheonlyequationalaxiomsleftare (x:y):x = y and x:(y:x) = y.
[11] Themotivatingexampleofcoloniesforthiswriterarosefrom“orientablesurfaceswithoutdiagonals,”thatis,triangulated
orientablesurfacesinwhichthe1-skeletonisthecompletegraphonthevertices.Define x:y fordistinctvertices x,y tobethe
vertexonereacheswhenonemovesfromxtoyandthentakesthe“firstrightturn.”Define x:x=x. Thetwoknownexamples
yield colonies arising from the Galois fields of order 4 and 7 with x:y = −cx−c2y where c is a primitive cube root of 1
(i.e., c2+c+1=0).
6
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
lation requires three binary operations and four equations, to wit, any two of the following
three rows:
(x/y)y = x = (xy)/y
y(y\x) = x = y\(yx)
(y/x)\y = x = y/(x\y) [12]
[F12o]r quasigroups one may take txyz = (x/x)\((x/y)z). [13]
We will see yet another Mal(cid:48)cev theory in [2.34] (below, in these amplifications). For the
(cid:92).
∨
theory of scales,[14] take txyz = (x|z)|(y|⊥).
For the connection between the von Neumann equation on the allegory of relations and
the existence of a Mal(cid:48)cev operator, first observe that given a relation R:A →B between
Mal(cid:48)cev algebras, suppose that aRb, a(cid:48)Rb, and a(cid:48)Rb(cid:48). If we view R as a subalgebra of
A×B then t(cid:104)a,b(cid:105)(cid:104)a(cid:48),b(cid:105)(cid:104)a(cid:48),b(cid:48)(cid:105) = (cid:104)taa(cid:48)a(cid:48),tbbb(cid:48)(cid:105) = (cid:104)a,b(cid:48)(cid:105) must be in R, in other words, aRR◦Rb(cid:48)
implies aRb(cid:48). A corollary, using [1.593], is that
Abelian categories are von Neumann regular categories.
For the reverse connection, suppose that the von Neumann equation holds in the allegory
of relations between algebras of a given equational theory. Let F be the free model on two
genterators, a and b. Let R ⊂ F×F be the submodel generated by pairs (cid:104)a,a(cid:105),(cid:104)a,b(cid:105),(cid:104)b,b(cid:105).
When we view R as a relation the von Newmann equation says that RR◦R ⊂ R and since
bRbR◦aRa we obtain bRa hence (cid:104)b,a(cid:105) ∈ R. But for any equational theory we obtain the
subalgebra generated by three elements as the set of values of all ternary operators (primitive
and derived) applied to those three elements. That is, there must be a ternary operator t
suchthatt(cid:104)a.a(cid:105)(cid:104)a,b(cid:105)(cid:104)b,b(cid:105) = (cid:104)b,a(cid:105)whichsays,ofcourse,thatt(cid:104)a.a(cid:105)(cid:104)a,b(cid:105)(cid:104)b,b(cid:105) = (cid:104)taab,tbba(cid:105) =
(cid:104)b,a(cid:105), hence that t is a Mal(cid:48)cev operator.
In a von Neumann allegory in which symmetric idempotents split we may define (up to
◦ ◦
isomorphism) an image for a relation R; because RR and R R are symmetric idempotents—
◦
one on the domain of R, one on its codomain—with R and R yielding an inverse pair of
isomorphisms between their splittings.
The von Neumann equation implies two conditions on endomorphisms: reflexive implies
symmetric (rrrrriiiiisssss), that is, 1 ⊂ R implies R◦ = R; and reflexive implies transitive (rrrrriiiiittttt),
◦ ◦ ◦
that is, 1 ⊂ R implies RR ⊂ R (if 1 ⊂ R then R ⊂ 1R 1 ⊂ RR R ⊂ R and
RR ⊂ R1R ⊂ RR◦R ⊂ R.) [15]
Both ris and rit imply equivalence relations commute (eeeeerrrrrccccc): if E and F are equivalence
relations then since 1 ⊂ EF ris implies EF = (EF)◦= F◦E◦= FE and rit implies FE ⊂
1FE1 ⊂ EFEF = (EF)2 ⊂ EF, similarly EF ⊂ FE. When EF = FE we know that
[12] Coloniesarequasigroupsinwhich xy=y\x=y/x=x:y.
[13] Firstusethe4thequationtoobtain txxz=(x/x)\((x/x)z)=z, thenusethe1s,t 4t,h andthe1stequationagaintoobtain
txzz=(x/x)\((x/z)z)=(x/x)\x=(x/x)\((x/x)x)=x. Itiscuriousthatonlytwoofthedefiningequationsareneeded.There
is a transitive group of symmetries on the six-element set of operations and their “twists” that preserve the equations and we
obtain six pairs of equations each of which suffices for the construction of a Mal(cid:48)cev operator. There are no other such pairs.
There are usually an infinity of Mal(cid:48)cev terms if there’s one, e.g., tx(tx(txyz)z)z. (There’s only one txyz in the free abelian
group generated by x,y,z. When the word “abelian” is dropped there are infinitely many, e.g., xz-1xy-1zx-1z.) In the case
ofquasigroupsifwesticktotermsofthesamesizethereare72versions.Heavenly.
[14] http://www.math.upenn.edu/~pjf/analysis.pdf
[15] Both ris and rit can be strengthened: it is enough to assume that a morphism in a von Neumann allegory con-
tains both its domain and codomain to imply symmetry and it is enough to assume that it contains the intersection of its
domain and codomain to imply transitivity: if Dom(R), Dom(R◦) ⊂ R then Dom(R◦)R◦Dom(R) ⊂ RR◦R ⊂ R;
if Dom(R◦)∩Dom(R) ⊂ R then(using2.121) RR ⊂ RDom(R◦)Dom(R)R = R`Dom(R◦)∩Dom(R)´R ⊂ RR◦R.
7
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
EF is itself an equivalence relation, which makes it the smallest equivalence relation contain-
ing both E and F. In [2.113], particularly as amplified on page 5 of these amplifications, any
lattice of pairwise commuting equivalence relations was seen to be automatically modular (for
Mal(cid:48)cev theories this is usually stated as the modularity of the congruence lattices).[16]
Without further conditions ris and rit are independent. For a one-object allegory where
ris holds but not rit let R be a reflexive symmetric morphism in any allegory. Consider
the suballegory generated by R. All of its elements remain reflexive symmetric. They are all
transitive only if R was transitive. (If one starts with the allegory of relations on a three-
elementsetandtakesRtobeanyoneofthethreereflexivesymmetricnon-transitiverelations,
this construction yields a three-element allegory. If one starts with the natural numbers and
takes R to be the adjacency-relation-made-reflexive then this construction yields an allegory
with an infinite number of reflexive relations all of which are symmetric and only one is
transitive.)
Foraone-objectallegorywhereritholdsbutnotrisletRbeareflexivetransitiverelation
in any allegory and suppose further that R defines a total order (e.g., on a 2 element set).
Assuming that R (cid:54)= 1, the suballegory generated by R has exactly four elements all of which
are reflexive transitive but only two of which are symmetric.
But for allegories arising from regular categories (or equivalently, as explained below, for
tabular allegories) both ris and rit imply the von Neumann equation because each implies
erc and erc implies the von Neumann equation: given a relation R that is spanned by
a pair of maps, f, g, that is, such that R = f◦g, erc implies RR◦R = (f◦g)(f◦g)◦(f◦g) =
f◦(gg◦)(ff◦)g = f◦(ff◦)(gg◦)g = (f◦f)f◦g(g◦g) ⊂ 1f◦g1 = R.Tosummarizetheconnections,
let vvvvvnnnnneeeee mean the von Neumann equation Then for all allegories:
ris
(cid:37) (cid:38)
vne erc
(cid:38) (cid:37)
rit
The two counterexamples above show that neither vertical arrow may be inserted (and hence
none of the arrows may be reversed) in the context of all allegories. But—as just seen—for
those allegories in which there are enough maps:
ris
(cid:37) (cid:38)
vne ←− erc
(cid:38) (cid:37)
rit
[17]
[17]
[16] erc is equivalent to equivalence relations being closed under composition. Indeed, for equivalence relations EF = FE
iff EF is an equivalence relation: if EF = FE then EF is immediately symmetric and imminently transitive (EFEF =
EEFF = EF); if EF is an equivalence relation then it’s symmetric and EF = (EF)◦= F◦E◦= FE. Put another way:
erccouldjustaswellhavestoodforequivalence relations compose.
[17] What does it mean if there’s a Mal(cid:48)cev operator in a category, in particular suppose there’s a Mal(cid:48)cev operator in the
category of models of a Mal(cid:48)cev theory. Put another way suppose that t is a Mal(cid:48)cev operator that is a homomorphism with
respectto t.
In general, given an m’ary operator f and an n’ary operator g then f is a homomorphism with respect to g means that
for any m×n matrix of variables if we apply f to each of the n columns and then g to the resulting row, we obtain the
same as if we had applied g to each of the m rows and then f to the resulting column. (Yes, it’s a symmetric condition: f
isahomomorphismwithrespectto g iff g isahomomorphismwithrespectto f.)
8
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
•ppppp222220000055555 AAAAAmmmmmpppppllllliiiiififififificccccaaaaatttttiiiiiooooonnnnnsssss fffffooooorrrrr [[[[[22222.....111115555577777 sssssmmmmmaaaaallllllllll ppppprrrrriiiiinnnnnttttt]]]]]:::::
We say that a modular lattice whose only elements aside from the top and bottom are its
atoms and co-atoms satisfies the theorem of Desargues iff for any six atoms a ,b ,c ,a ,b ,c :
1 1 1 2 2 2
(a ∨a )∧(b ∨b ) ≤= c ∨c
1 2 1 2 1 2
implies
(cid:0) (cid:1) (cid:0) (cid:1)
(a ∨b )∧(a ∨b ) ≤= (a ∨c )∧(a ∨c ) ∨ (c ∨b )∧(c ∨b )
1 1 2 2 1 1 2 2 1 1 2 2
Note that every “−∨−” is a co-atom.[18]
[See diagram on next page.]
When f =g=t andm=n=3andthematrixofvariablesis:
a b c
d e f
g h i
weobtaintheequation
t(tabc)(tdef)(tghi)=t(tadg)(tbeh)(tcfi)
ThisequationplusthetwoMal(cid:48)cevequationsturnouttobedefiningwhatmaybedescribedastheresultofremovingzerofrom
thetheoryofabeliangroups.Inimitation(butpronounceably)ofJacobsonwe’llcallthisthetheoryofgrups.
GivenanelementainagrupGwecanobtainanabelian-groupstructurebydefining
0a = a
x+ay = txay
−ax = taxa
u a v
The Mal(cid:48)cev equations immediately yield x+a0a =x=0a+ax. Using the variables a a a we obtain the equation
x a y
(u+av)+a(x+ay)=(u+ax)+a(v+ay) whichin[1.591]onp88wasshowntoimplythat+a iscommutativeandassociative.
a a x
Usethevariables x a a toobtain −ax+ax=0a. We’lldenotetheresultingabeliangroupas Ga.
a a a
x a b
Givenapairofelementsin G thecanonicalisomorphismfrom Ga to Gb isgivenby txab. Usethevariables a a b to
y a b
obtain t(x+ay)ab=(txab)+b(tyab). When a=b wegettheidentityfunction.Ifwefirstapplythefunctionnamedbytxab
x a b
thentheonenamedby txbc use a a b toprovethatweobtaintheonenamedbytxac.
c c c
Anabeliangrouphas,ofcourse,agrupstructure(asalwaysinthepresenceofanabeliangroupstructure txyz is x−y+z).
Note that translation by a, that is, the function that sends x to x+a is a grup-automorphism. Given a pair of abelian
groups,thegrup-homomorphismsfromthefirsttothesecondarepreciselythegroup-homomorphismsfollowedbyatranslation
by an element of the second. (When abelian groups are replaced by vector spaces such functions are, of course, called “affine
transformations.”) The resulting category just misses being equivalent to the category of grups. What we miss hitting is the
emptygrup.Henceaddanewstrictinitialobjecttothecategoryofabeliangroupsandgrup-homomorphismsandweobtaina
categoryequivalenttothecategoryofgrups.
Finally,whatifwehaveapairofMal(cid:48)cevoperators, t,t(cid:48) eachahomomorphismwithrespecttotheother?Justasin[1.591]
wecanprovet=t(cid:48).
[18] We’reusingthefactthatthevariablesnamesixatoms.
9
AMPLIFICAATIONS, DIMINUTIONS, SUBSCOORINGS
(a ∨a )∧(b ∨b ) = p p
1 2 1 2
(a ∨b )∧(a ∨b ) = w
1 1 2 2
(a ∨c )∧(a ∨c ) = u
1 1 2 2
(c ∨b )∧(c ∨b ) = v
1 1 2 2
c
1
a
1
u b
1
v w
a
2 b
2
c
2
In the allegory of sets (hence for relations in any regular category)
(A A ∩B B ) ⊂ C C
1 2 1 2 1 2
implies
◦ ◦ ◦ ◦ ◦ ◦
(A B ∩A B ) ⊂ (A C ∩A C )(C B ∩C B ).
1 1 2 2 1 1 2 2 1 1 2 2
Given the hypothesis and elements x,y such that x(A◦B ∩A B◦)y
1 1 2 2
we need to show x(A◦C ∩A C◦)(C◦B ∩C B◦)y.
1 1 2 2 1 1 2 2
Actually we’re given elements x,s ,s ,y satisfying:
1 2
xA◦s B y and xA s B◦y.
1 1 1 2 2 2
The hypothesis then says that since
s (A A ∩B B )s [because s A xA s and s B yB s ]
1 1 2 1 2 2 1 1 2 2 1 1 2 2
there must be z such that s C zC s . And that yields
1 1 2 2
x(A◦C ∩A C◦)z [because xA◦s C z and xA s C◦z] and
1 1 2 2 1 1 1 2 2 2
z(C◦B ∩C B◦)y [because zC◦s B y and zC s B◦y] hence,
1 1 2 2 1 1 1 2 2 2
finally, x(A◦C ∩A C◦)z(C◦B ∩C B◦)y.
1 1 2 2 1 1 2 2
•ppppp222221111111111 AAAAAdddddddddd tttttooooo eeeeennnnnddddd ooooofffff [[[[[22222.....111116666622222]]]]]:::::
Note that R is necessarily simple.
•ppppp222221111133333 AAAAA sssssiiiiimmmmmpppppllllliiiiififififificccccaaaaatttttiiiiiooooonnnnn fffffooooorrrrr [[[[[22222.....222221111166666(((((1111100000)))))]]]]] 1111144444ttttthhhhhllllliiiiinnnnneeeee uuuuuppppp:::::
Define G(cid:48) by G(cid:48) = G∩FR.
•ppppp222221111144444 AAAAAdddddddddd tttttooooo eeeeennnnnddddd ooooofffff [[[[[22222.....1111166666(((((1111111111)))))]]]]]:::::
(The word “neighbors” was chosen for its intransitivity. Let a,b,c be idempotent functions
on the set {0,1,2} defined by the array:
a b c
0 0 0 1
1 2 0 1
2 2 2 2
Then a and b are neighbors; b and c are neighbors; a and c are not.)
10