Semigroup Forum Vol. 43 (1991) 63-76 �9 1991 Springer-Verlag New York Inc.

R E S E A R C H A R T I C L E

A M A X I M A L C H A I N O F P R I N C I P A L I D E A L S IN T H E S E M I G R O U P O F B I N A R Y R E L A T I O N S O N A F I N I T E S E T

M i c h a e l B r e e n

Communicated by Boris M. Schein

A B S T R A C T :

A maximal chain of F ,+z - 1 principal ideals in the semigroup of the b inary relat ions on an n-element set X is constructed by represent ing a b inary relat ion as a Boolean matr ix . Here F , s tands for the n-th Fibonacci number.

S e c t i o n 1: I n t r o d u c t i o n

We denote the semigroup of b inary relations on a set X by Bx , where the operat ion is the usual opera t ion of composit ion. Since this opera t ion is unwieldy for all but the simplest b inary relat ions, we seek a semigroup isomorphic to Bx whose operat ion is computa t iona l ly easier. Such a semigroup is the semigroup of all n • n (0,1) Boolean matr ices with the usual ma t r ix mult ipl icat ion, where ]XI = n (recall tha t 1 + 1 = 1 in a Boolean algebra) , denoted Bn. When the elements of X have been enumera ted , the image of a relat ion p is the mat r ix A where Ai3 = 1 if and only if (i,j) E p �9

We are most interested in Green 's J - r e l a t i o n defined by A1 J A2 if (A~) - (Az ) where A1, A2 C Bn and (A) denotes the principal ideal of Bn generated by the matr ix A. We write Bn/J as J ( B n ) and denote an element A's J - c l a s s by JA. Note that in

n n , ~ = J [2]. Permut ing the rows and columns of a mat r ix results in a mat r ix which is J -

equivalent to the original mat r ix [8]. At times we also mention the s and Tr relation. For Aa and A2 in B , , Aa s A2 (A1 7~ A 2 ) i f the left ( r i gh t ) idea l s of B~ generated by A1 and A~ are equal.

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Occasionally we speak of Vn (V, t) the set of all (0,1) n-tuples (column vectors). This is the space that the rows (columns) of a Boolean matrix come from. A significant notion is the row (column) space of a matrix A. This is the 0-vector, 0, (0 t ) along with all finite sums of the rows (columns) of A.

For example, if

A = 1 1 , 0 1

then R(A)={000 ,100 ,110 ,011 ,111} , and

C(A) = {000 t, 110 t , 011 t, 001 t, 1 l l t }.

More details on some of the following results can be found in Kim[3]. For any matrix A, IR(A)I = IC(A)I �9 I call this number the row span or column span, respectively, of A. If we w r i t e r C w to mean vl = 1 implies that wl = 1 for each i , w e h a v e t h a t for any matr ix A, R(A) and C(A) are anti-isomorphic lattices. Note that R(XA) C R(A), and C(AY) C C(A) for any A, X, Y of B~. It is often helpful to speak of an element's height in the row (column) space. I adopt the following usage. An element 's level is the length of the longest maximal chain containing the element as its greatest element. So, 0 is on level 0. In the above example, in R(A), 100 and 011 are on level one, 110 is on level two, and 111 is on level three. The height of a row (column) space is the number of the highest level.

We write Ai# (A# i ) for the ith row (column) of A. In a notat ion similar to real vectors, ei means the vector whose j th coordinate is 1 if and only i f j -- i.

We call a set of vectors independent if none of the vectors is the 0-vector and if none is a sum of the others. A matr ix is called row-(eolumn-)reduced if its nonzero row (column) vectors form an independent set. The nonzero row (column) vectors in a row- (eolumn-)redueed matr ix are called the row (column) basis, writ ten Br (A) ( Bc ( A ) ) . A matr ix is called reduced if it is both row-redueed and column-reduced. Reducing a matrix does not affect its /f-class.

Also, let fi = { 1 , 2 , . . . , n } where n C N.

S e c t i o n 2: A n O r d e r R e l a t i o n o n B~ a n d J ( Bn )

D e f i n i t i o n 2.1. Let A,B C Bn. The statement B _< A means that there exist X, Y Bn such that XAY = B. Thus, < is what is usually called the f l - (or 79 -) quasi-order on Bn. The s ta tement B < A m e a n s B _< A a n d B ~ J a .

Indeed, the above can be interpreted as saying that B is divisible by A. Some facts are immediate and we do not prove them.

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R e m a r k 2.2. If A ,B E Ba, then i. B < I (where I i j = 1 if and only if i= j ) ,

ii. O < A ( O q = 0 for every i and j) , iii. B < A if and only if (B) C (A),

iv. (A) = U { J B : B < A}, v. B < A if and only if B t < A t

vi. i f B < A , B = B 1 , A = A l , t h e n B1 < A1.

Using iii and iv, we can consider the par t ia l order relat ion on J (Bn) or on { ( A ) : A E Bn}. There was some idea tha t for all n, the set of principal ideals (or J-classes) with this par t ia l order relat ion formed a la t t ice or perhaps even a chain. It was Plemmons and West [5] who first s ta ted tha t J (B= ) with the divisibil i ty relat ion was not a la t t ice for n > 2.

Two not as obvious results are tha t 1) A 3" B if and only if R(A) ~ R(B) (as lat t ices) [8] and 2) A s (TE) B if and only if the matr ices have the same row (column) space [51 .

The fundamenta l result on divisibil i ty is due to Zaretskii[8].

T h e o r e m 2.3. For A, B E B . , B < A if and only if there is a map f from R(B) into R(A) such that for every bl , b2 E R(B), f(bl } C f (bz) if and only if bl C_ bz.

I call such a map a Zaretskii map. Al though it is t empt ing to regard f as a la t t ice or semilat t ice homomorphism it

need not be either. Some immedia te results from Zaretski i ' s theorem follow. They are s ta ted without proof.

R e m a r k 2.4. For A, B E Bn,

i. if R(B) c R ( A ) t h e n B < A,

ii. if C(B) C C ( A ) t h e n B _< A, iii. if B < A then IR(B)I < IR(A)I, iv. if B < A then the height of R(B) is less than or equal to the height of R(A).

None of the above s ta tements ' converses is true.

Algor i thms for deciding left and right divisibil i ty are es tabl ished by Vagner [7] and Luce [4], yet there is no a lgor i thm for deciding (two-sided) divisibility. Using Zaretskii 's theorem to decide divisibi l i ty in Bn for n larger than 4 is imprac t ica l (for example, there are O0 J-classes in B4 ).

S e c t i o n 3. T h e C h a i n

We build the chain in J (B= ) along the following lines. Assume the chain has been constructed in J ( B n _ l ), so tha t we have a chain of classes with each class represented by

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a part icular element, A. The matr ix A is equivalent to a matr ix in B, formed from A by leaving rows and columns 1 through n - 1 intact and by writing O's in row and column n. Thus, each class in the chain in J( Bn-1 ) corresponds in a natura l and unique way to a class in the chain in J (Bn ).

If JA2 and JA1 are classes in the chain (in J ( B n - 1 )) and the former class covers the latter one, and the row span of A1 differs from the row span of A2 by more than one, then I say that there is a gap between the two classes. If there is a gap between two adjacent classes in the chain, then we will form matrices in B, whose classes will begin to bridge this gap by replacing row n of A1 (which is 0) with nonzero row vectors from V, . These rows will come from R( A2 ) (in Vn ) and the classes of the resulting matrices will lie between the classes of A1 and A2. If there is no gap between two adjacent classes in the chain, no construction of the chain need be done in this area of the chain.

In this chain, if JA, < JA2 , where A1 and A2 are our representative elements, then R(A1 ) C R(A2 ). As an i l lustration, in J( B4 ), the following matrices represent two classes of the chain (100 ) (1000)

A1 = 0 1 0 ,A2 = 0 1 0 0 0 0 1 0 0 1 0 " 0 0 1 0 0 0 1

Here JA, < JA~, ]R(A1)] = 12 and IR(A2)g = 16. In J ( B s ) , each of the following matrices ' classes will lie between the classes containing the elements of Bs which correspond to the above elements.

B 1

( oooi) 1 0 0 0 1 0 ,B2 = 0 1 1 1 0 1

1000 )0 0 1 0 0 0 0 0 1 0 0 0 1 1 0 1 0 1

Note that when A1 and A2 have been expanded C R(B1) C R(B2) C R(A2). I use the term "lead 1" largest group of zeros at the beginning of a row its "lead chain.

so as to lie in Bs, we have R( A1 ) in the usual sense and I call the zeros." Let us now construct the

D e f i n i t i o n 3.1. Suppose n > m > kl > k2 > . . . > kt > 1, where m _< m + t < n. The element I[m;kl ,k2 , . . . , k t ] in Bn has the following form:

i. For eachi , 1 < i < m, r o w i i s e i ;

ii. When t > 1, the first k l - 1 coordinates of row m + 1 are zero, coordinates kl through m + 1 are one, and the remaining coordinates (if any) are zero;

iii. For t > 1, the remaining rows (from row m + 2 through row m + t) are formed based on the previous row. Row m + i, m + 2 <_ m + i < m + t, will be zero in the coordinate corresponding to the lead 1 of the previous row. It (row m + i) will be equal to the previous row to the right of this zero, zero in the first ki - 1 coordinates, and one elsewhere.

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iv. All o the r rows are 0.

R e m a r k 3 .2 .

1. T h e e lement I[m], some t imes wr i t t en Ira, is the m x m ident i ty m a t r i x (wi th zero rows and co lumns adjoined) .

2. We have m + t _< n, o therwise the e lement would not be in Bn.

3. In par t iii above, we know there is a lead 1 in the p rev ious row, o therwise it is the zero vector , and thus kl 1 - 0.

4. We may regard m as k0.

P r o p o s i t i o n 3 .3 . [R(I[rn; k l , . . . , kt])] = U '~ 4 2 k~-I _[_... + ~kt-1

P r o o f . By induc t ion on t. If t = 0, the result is obvious. For convenience , call I [m;k l , . . . , k , _ l ], A, and call I [ m ; k l , . . . , k , _ l ,kt] , B. Wri te w for row m + t of B. T h e n

IR(B)I = IR(A) U {v + w : v E R(A)}J = IR(A)I + J{v e Vn : v, : wl ,k t < i < n}l .

The new e lements b rough t to R(B) by w will be 0 in all coord ina tes g rea te r t han kt for which w is 0 (since vectors wi th l ' s in these coord ina tes are a l ready in R(A) and thus R(B)) and a rb i t r a ry in the first kt - 1 coord ina tes , since the first k t - 1 rows of A - and

B - span a subspace of Vn i somorphic to Vk,-1 �9 There fore , by the induc t ion hypothes is and by the above, the desired row span is 2 m + 2 kl 1 gv ' ' " -{- 2 k ' - ~ - I + 2 ~' 1.

P r o p o s i t i o n 3 .4 . For all t and s, I [ k o ; k l , . . . , k j , . . . , k t } < [ [ ~ 0 ; g l , . - . , g j , . - . , ~ , ] i f and only if there is a j; 0 < j < t , s ; such that kj < ~j and ki - ~i f o r i < j .

P r o o f .

Sufficiency.

This follows f rom the p reced ing propos i t ion and the fact t ha t ( separa te ly) the k 's and l 's form a s t r ic t ly decreas ing sequence. , f rom kl and 11 on.

Necessity.

If j = 0, the resul t is obvious . For j > 0, the rows of the two mat r i ces (call t h e m A and B, where we are t ry ing to show tha t A < B) differ first in row m + j. There fore , rows 1 t h rough m + j - 1 of A are ob t a inab l e f rom B since the ma t r i ces are equal there. In row m + j, the ma t r i ces differ in coord ina tes kj t h rough lj - 1, where the lesser m a t r i x

~,1 i -] is 1 and the g rea te r m a t r i x is 0. Then , r o w m + j of A is row m + j of B plus z~kj el. Each of these s u m m a n d s is a row of B. Fu r the rmore , rows m + j + 1 t h rough row m + t of A are 0 in coo rd ina t e k j . T h e y m a y each have some l ' s in some coord ina tes to the left of k j , bu t this is not a concern. Since they are each zero in coord ina te k j , they conta in

row m + j of B. Thus , they can each be ob ta ined f rom row m + j of B (us ing this row and

some of rows 1 t h r o u g h m of B). Since R(A) C R(B) , A < B.

We encode each m a t r i x wi th a b inary (wha t else?) n u m e r a l which is the row span of the ma t r ix . We find the n u m b e r of classes in the chain in J ( Bn ) by f inding the n u m b e r of b inary numera l s possible. Obv ious ly each class has a un ique b ina ry numera l associa ted with it. It is also t rue tha t each n u m e r a l defines a m a t r i x of a class in the chain. For if the n u m b e r is 2 "~ + 2 kl + . . . + 2 k' , we define the m a t r i x so tha t for each i, each of rows

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1 through m is e i ; row m + 1 has kl O's followed by l ' s in coordinates kl + 1 through m + 1, etc.

I say a binary numeral has k digits if from the left-most 1 in the numeral to the right-most digit in the numeral , we have k digits ( that is, there are k significant digits).

P r o p o s i t i o n 8.5. In d (Bn ), the number of classes in the chain is the number of nonzero binary numerals such that the number of digits in the numeral plus the number of l 's in the numeral is at most n § ~.

P r o o f . Note that the numeral will have at most n + 1 digits. The number of digits tells us in what column (reading from the right and calling

the right-most column of the numeral , column 1) the left-most 1 is in, and thus what identi ty matrix, I[m], the matr ix is bui l t /based on. This 1 will be in column m + 1 of the numeral . Therefore, a matr ix I[m;kx , . . . , k t ] will be represented by an m + 1 digit binary numeral with t + 1 ones. Since m + t (the number of rows) is at most n,

( m + l ) + ( t + l ) = ( m + t ) + 2 < n + 2 .

Hence, every class and every matr ix will determine such a numeral , and conversely.

In the following, I write Fi to denote the ith term of the Fibonnacci sequence. Also, I call a class new to J (Bn ) if it does not contain elements which (upon deleting zero rows and /or columns) are in Bn-1 �9 (That is, for our chain, a class is called new if it isn ' t in effect already in the chain in J ( B n - 1 ).)

P r o p o s i t i o n 3.6. The number of new classes in the chain in J (Bn ) is F,~+I.

P r o o f . From the previous proposition, the number of new classes is the number of at most n + 1 digit binary numerals such that the number of digits plus the number of l ' s is

r - - ] n + 2. For each r and s, r > s, the number of r-digit binary numerals with s ones is ( , -1) (a 1 must go in c o l u m n r , then the other s - 1 ones are in columns 1 t h r o u g h r - 1).

So the number of new classes is

r W s = n + 2 t+u=n ~ <t<n

P r o p o s i t i o n 3.7. The height of J ( B k ) is at least F,~+z - e.

P r o o f . By convention, the height of a chain is one less than the number of elements in the chain. We induct on k, and count the number of classes in our chain. If k : 0, we have one class, hence a height of 0 = 2 - 2 = Fz - 2 = F0+z - 2.

The number of classes in the chain in J ( B n + l ) will be the number of classes in the chain in J( B, ) plus the number of new classes (in the chain in J( B~+I ) ). From the induct ion hypothesis and from the previous proposition, the number of classes desired is

F,~+z - 2 + F,~+2 = F,~+4 - 2 = F,~+1§ - 2.

So the height of J ( B k ) is at least this number .

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Sect ion 4: Maximal i ty o f the Chain

For convenience, I will refer to elements of adjacent classes as A1 and A2, where A1 < A2 and A2 has m + j nonzero rows. Some observations about A1 and A2 follow.

R e m a r k 4.1.

i. Each of the nonzero rows m + 1 through n are pairwise incomparable. ii. The number of l ' s in the nonzero rows of rows m + 1 through n is an increasing

sequence.

iii. The (m + 1)-st column of A1 is ~ , , ~ + l ( e i ) t , the (m + 1)-st column of A2 is ~ m + j i=m+l(e l ) t (Here I am assuming that we are trying to prove maximality, in which case A1 must have n nonzero rows; otherwise either we could create a class between the two mentioned classes, or there would be no gap.)

In the equation XAY = B, I say that i) X (Y) sums rows (columns) of A if there exist i,j,k such that Xij = Xik = 1 (Yj~ = Y ~ = 1); and ii) X (Y) ignores a row (column) of A i f t h e r e i s a n i such that Ai# # f a n d X # i = fit (A#i r f t and Yi# = 0) (the row (column) of A is not "used" in the product).

Now let us prove that the chain is maximal. We wish to show that if AI and A~ are elements of adjacent classes in the chain in J ( B n ) and A1 < B _< A2, then B _= A1 or B _= A2. Because of our assumption, we know that there exist X, X~, Y, Y1 in Bn such that 1) XA2Y = B and 2) X1BY1 = At .

Note that if A1 = I[m], then A2 = I[m;1] and the row spans of the two elements would differ by only one (hence, B would be as desired). We will assume that B is reduced (by construction, A~ and A2 are) and that J s covers JA1 in J ( B n ) .

We first show that the chain is maximal from the classes of I[n - 1] to I[n] in J ( B n ) . Since S > I[n - 1], B has the following form

B = w

where v t E ( V n - 1 ) t , w ~ Vn-1 , and i is 0 or 1. We know that I[n - 1] < A 1 =

I[n - 1;k] < I[n], therefore rows 1 through n - 1 of A1 are on level one of R(A1) each (separately) contained in chains with exactly n - 1 elements above them.

By Zaretskii 's theorem and by the preceding argument , R(B) must contain at least n - 1 incomparable row vectors each with at least n - 1 zeros (in order for the chain lengths from R(A1) to be preserved). Since R(B) C_ V n , w e can assume that v t = f t . If i = 0 , then B is not reduced. Hence, i = l and IR(B)I = 2 '~-~ + 2 ~ where s is the number of zeros of w. Thus, IR(B)I = IR(A1)[ or IR(B)t = IR(A2)I , as desired. This part of the chain is maximal.

The proof continues by establishing the following:

A. In equations 1 and 2 above, we can choose the X's and Y's so that Y and Y1 are permuta t ion matrices. That is, any summing done by these two matrices can be done by X and X 1. There can be no ignoring.

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B. Either X or X1 can sum but not both. Therefore, since permuting preserves the J-class, B _= A1 or B _= A2.

P r o o f of A.

We will accomplish this by demonstrat ing the following:

1. The matr ix XA2 has the following form: / i 0 ) S .

2. Neither X nor Y ignores.

3. In the matr ix XA~ (above), each of rows m + 1 through m + j of the submatr ix S is a sum involving the corresponding row of A2.

4. Any summing of the columns of X A~ or X ~ X A2 done by Y or Y 1 respectively, can be done by pre- mult iplying A2 by an appropriate matrix, hence we can assume that Y = Y~ = I[m + 1].

P r o o f o f 1.

This amounts to showing that we can assume that for all i in ~ , row i of XA2 is el. a. R(XA2 ) C R(A2 ) so each element of the former matr ix 's row space has at most

m + 1 ones.

b. By construction, Aa _< B < XA~. Thus, there exists a Zaretskii map f : R(A~) , R(XA2). Recall that for each i in ~ , r o w i of A1 i se l .

Claim: We can assume that for each such i, f(ei) = e i .

Since the submatr ix of A1 consisting of its first m rows is equal to Im, there exist vi~ , vlz ,. �9 Vim (with each bot tom-most subscript corresponding both to the number of summands and to the number of l ' s ) such that for each i, ei C vi~ C via C . . . C vi,,, �9 Also, v~,. + ( A1 ) k# has m + 1 ones whenever k > m + 1 (such a k exists because A1 # I[m]), these l ' s being in coordinates 1 through m + 1. Hence, for each i we have a chain in R(Ax), el C vi~ C . . . vi,, C v, whe rev has m + 1ones . There is only one such v because it is the greatest element of R(A~ ). By Zaretskii 's theorem, in R(XA2) we have 0 = f(0) C f(el) C f(vi~ )C . . . Q f(vi,, )C f(v) for each i. From a. (above), the greatest element of R(XA2 ) also has m + 1 ones. The number of l ' s in this chain must be strictly increasing so f(ei) must have exactly one 1 for each i.

Since { e l , . . . , e m } C_ R(A2) and since we can assume that em+l ~ R(A2), then for each i, f(ei) = e j where f is a permuta t ion on {ei : i e rh} . (V~re can assume that e,,~+~ 6 R(A2 ), for if it is then A2 = I[m + 1] and we can permute the rows and columns of A2 so that f(el) is as stated above).

If f(ei) # ei for some i, permute the rows of X so that it is. This will not affect the J-class of B. The claim is true.

P r o o f of 2 ( N e i t h e r X n o r Y c a n i g n o r e ) .

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a. Y can not ignore .

Let A \ v (v t ) deno te the m a t r i x ob ta ined f rom A by rep la r lng v (v t ) wi th O (6~).

If Y ignores a nonzero co lumn of A2, say co lumn i, then X ( A 2 \ c o l u m n i)Y = X A 2 Y = B. But , A 2 \ c o l u m n i is i somorphic to a subspace of Vm SO tha t IIm ] < Aa _< B _< IIm], a con t rad ic t ion .

b. X can not ignore .

i. Rows 1 t h rough m of A2 are not ignored since each is in R(X A2 ).

ii. Row m + j of A2 ( the last nonzero row of A2) is not ignored because A1 and A2 are equal in rows 1 t h r o u g h m + j - 1, in which case X A 1 Y = B and the m a t r i x B would then be equiva len t to Aa.

iii. We now show tha t no o the r row could be ignored by X. Assume row m + s of A2 is ignored by X. Wr i te

Zi = {coo rd ina t e s which conta in the lead zeros of row m + s + i } , zl = ]Zil, and

Wi = {coo rd ina t e s for which r o w m + s + i i s 1 and r o w m + s + i - l i s 0 } .

For example , if row m + s is 000011011101011, then Z0 = { 1 , 2 , 3 , 4 } , z0 = 4 and

w0 = {5, 6}. Recal l t ha t

[R(A2)] = 2 m + 2 k~-I + " - + 2 k ' - I + . - ' + 2 k' 1

= 2 m + 2 k1-1 + . . . + 2 z ' ' + . . . + 2 Z , -" .

(for convenience call j - s, r).

Cla im: ] R ( A ~ \ r o w m + s ) l = 2 m + . . . + 2 k"-~-a + 2 *~+1 + - . . + 2 *r+l .

T h e first s u m m a n d s are obvious. For the rest , row m + s + 1 in t roduces new vectors to the row space of the m a t r i x (which consists of rows 1 t h rough m + s - 1 and row m + s + 1 of A2) which are a rb i t r a ry in the coord ina tes which make up the set Za ; 1 in each of the e lements of W1 ; e i ther 0 or 1 in the least e lement of W 0 , 1 e lsewhere in W0 ; and equal to row In + s of A2 in all coord ina tes g rea te r t h a n the larges t coord ina te of W o . There fo re , this row adds 2 z~+l e lements to the row space. The a r g u m e n t for rows m + s + 2 t h r o u g h m + j is s imil iar . The c la im is es tabl ished.

F r o m the above, we have tha t

[R(A2)J - IR(A2 \ row m + s)l : 2 . . . . (2 ~ + . . - + 2 "r) > 2 ~ - 2 ~+1 + 2 ~r ~ 2 ~

(this is t rue since z0 > z l ) > 2 k' > 2 k ' - a .

This last n u m b e r is the difference in the row spans of A1 and A2. Thus , IR(B)I <_ IR(XA2)I _< I R ( A 2 \ r o w m + s)I < IR(A1)I . This is a con t rad ic t ion . T h e m a t r i x X can not ignore.

There fo re , ne i the r X nor Y can ignore. We show la te r tha t ne i ther X 1 nor Y l ignore, also.

P r o o f o f 3 ( t h e f o r m o f X A ~ ) .

F r o m 1, we know tha t XA2 is of the fol lowing form:

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Im I o

S

We have jus t shown that none of the rows of A2 is ignored by X. W h a t are the possibil i t ies for the rows in the submat r ix S? They are sums from either:

i. I,,~ only,

or ii. S only, or iii. I,r, and S.

Case i is impossible otherwise XA2 is not reduced. Fur thermore , we can choose X so that case ii does not arise. For any sum of this type can be accomplished by choosing one summand from S and adding to it appropr ia t e rows from Im (or more precisely, Im 12 O) since all rows in S are 1 in coordinate m + 1. Since no row of A2 is ignored, and since each row of S can be assumed to be a sum as in case iii, then we can assume that each row of S is the corresponding row (the row in the same posit ion) of A~ plus v where v is in the row space of I,~ t 3 0 .

P r o o f o f 4 (we c a n a s s u m e t h a t Y = Y1 = I [ m + 1]).

a. Y > I[m]

This is t rue since X A 2 Y = B, Y > B > A1 > I[m]. Therefore, we can assume tha t Y is equivalent to a ma t r ix of the following form:

(i �9 / u where T could be the zero matr ix .

b. We can arrive at the above mat r ix by pe rmut ing the columns of Y only.

This is a desirable conclusion, because permut ing the columns of Y will affect B only. Pe rmuta t ions of the rows of Y will have to be compensa ted for by permut ing the columns of A2, which we would like to avoid.

Note tha t the ma t r ix Y has exact ly m + l rows. Assume tha t the columns of Y have been pe rmuted (without loss of generality, call this mat r ix Y) so tha t only a row pe rmuta t ion is necessary to arrive at the required form. Let II be that pe rmuta t ion matr ix . Then there is a map P : m + 1 , m + 1 such tha t Yi# is (1]Y)p(1)# . Let a be such tha t P ( a ) = m + l . If a = m + l , we are done - the desired form can be achieved by permut ing columns only. If a # m + l , let P ( m + l ) = b and define Y' as follows:

{ Y#P(i) if i E ~ \ { a } }

Y~r = Y#b if i = a �9

Y#i otherwise

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Except for possibly row a, Y' contains Im as a submatrix. We can permute rows a and m + 1 of y t to achieve the desired submatrix, but we do not need to. Since we are missing only column a from our desired result, none of the columns of T can have a 1 in its a-th coordinate (otherwise we could permute that column with column a to achieve the result). Thus, any column of A2Y' which contains column a must be in rh (actually, rh \ {a}) , and the summing of column a can be accomplished using the rows of A2 as follows.

For i E ~ \ { a } , column i of A2Y' is column i of A2 plus vi where vi C Vn t

Define Si = {k : ( A 2 Y ' ) k i = 1 and (A2)ki = 0}, and the matrix X' as X',i = {1 if and o n l y i f s E S~ U{i}} . Then since r o w i of A2 is e~, ( X A 2Y ' ) #~ = ( X X ' A 2 ) # i , and column a of A2 is not used. Therefore, we can assume that column a is ignored. This contradicts part 2 of our proof. Hence, P:rh ~ ~ , and we can achieve the desired submatr ix by permut ing the columns of Y only.

c. Y = I [ m + 1].

Any summing that Y does in columns 1 through m (which can only be of the form c o l u m n i + column m + 1 since Y = I[m] in that m • m submatr ix) can be done by premult iplying (by arguing as before). Thus, we can assume that Ym+l,i = 0 for i E rh. I claim that in the representation

Im C ) O

the only nonzero column of C is (em+a) t , and Y = I[m + 1]. We establish this with the following: i. Column k of A1 contains column m + 1 of Ax if and only if column k of A2 contains column m + 1 of A2 , ii. Therefore, Y can not sum , iii. Y = I[m + 1].

i. By construction, the nonzero rows of A1 and A2 differ only in certain lead l 's . If the matrices have the same number of nonzero rows, the only entry in which the matrices differ from each other is in the lead 1 of row n of Aa �9 If they have a different number of nonzero rows, row m + j of the two matrices will differ in the coordinate corresponding to the lead 1 of A1. Rows m + j + 1 through n of A1 will be identical to row m + j of A2 in all coordinates except their (the rows of A1 ) lead l 's .

In either case, no column of A1 which contains column m + 1 could contain a lead 1 otherwise the coordinate directly above it would be 0. However, column m + 1 is not 0 in this coordinate because A1 has at least two nonzero rows below row m (A1 r I[m;k] since this part of the chain is maximal by an earlier argument) .

The matr ix A1 has at least as many l ' s as A2, hence any column of A2 which contained column m + l would also represent a column of A1 which contained column m+ 1. The claim is true. Note that if these type of columns exist, they are consecutive, from k to m for some k.

ii. C can not sum .

What of the structure of C? We can assume that

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where D E Bm+a, for A1 has m + 1 nonzero columns. We would like to show that (em+l) t is the only nonzero column of D (and C). We can see that all nonzero columns of D are 1 in coordinate m + 1 (otherwise Y is not reduced; since A2 and B are reduced, we can assume that Y is). Also, if (era+ 1 )t E C(D), then the rest of the columns of D must be 0 t (otherwise again Y would not be reduced).

Suppose (em+l)t ~ C(D), that is that it is not one of the columns of D. Without loss of generality, assume that the nonzero columns of D are columns 1 through r. Then, for each d, m + 1 _< d _< m + r, (A2)#,,~+1 + (A2)#d, ~ ( A s Y ) # a for some dt where dt E rh. When such is the case, no Zaretskii map from C( A1 ) into C( As Y) is possible. The support for the previous s ta tement follows.

In C(A1 ) suppose we have columns 1 through i, none of which contains column m + 1 (of A1); columns i + 1 through m, each containing column m + 1; and colmnn m + 1. (Note that the second category may be void.) Then, in a discussion similar to previous ones, the first category of columns and column m + 1 will lie on level one of C(A1 ) each belonging to chains which have m elements above them. Columns from the second category each lie on level two and belong to chains with exactly m - 1 elements above them.

If for all d (as before) (A~)#m+l + (A2)#a , _C ( A 2 Y ) # a , then each of these columns will have at most m - 1 zeros which can be filled by elements of C(A2 Y) (in the other coordinates which are 0, every element of the column space is also 0). Thus, each of these columns will belong to chains with at most m - 1 elements above them. Then the basis columns of C(A2 Y) will be part i t ioned in such a manner that columns 1 through i do not contain any other nonzero column of A2 Y; columns i + 1 through m may or may not contain one or more of columns m + 1 through m + r but have at most m - 1 elements above them in any chain they are in; and columns m + 1 through m + r which also have at most m - 1 elements above them in any chain they are in.

Hence, any Zaretskii map from C( A1 ) into C( As Y) must map columns 1 through i and m + 1 of A1 to the set consisting of columns 1 through i of As Y. This is not possible. Therefore, no summing by Y is possible.

iii. Since Y can not ignore, Y = I[m + 1].

Since A1 = X1BY1 = XI(XA2Y)Y, = XI(XA2)Y, = Xl (XA2Vl ) , we can discuss Y 1 in the same manner in which we discussed Y and conclude that Y 1 = I[m+ 1]. Likewise, by an argument similar to the one which shows that X can not ignore (by treating XI X as one matrix), we can show that X1 can not ignore. Thus, we are left with investigating the occasion where X and X 1 sum.

B. O n l y o n e of X a n d X1 c a n s u m .

Let row m + i of A= be the first row that is summed by X1X. Note that since neither matr ix ignores, any sum done by one is kept /not lost by the other. Suppose row m + i has w ones. If row m + i of A2 has fewer ones than row m + j of A2, then it is not row m + j of A2 and A1 will have the same number of rows with w ones as As does because the matrices are equal there. If row m + i of A2 is summed in X 1 X As, then it can not be a row of this matr ix (otherwise the matr ix is not reduced). By our assumption, no row of A2 with fewer than w ones is summed, so at most h - 1 rows of A2 are permuted to obta in h rows of A1 (for some h). This is impossible. We can conclude that if row

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m + i of A2 is s u m m e d , then it has as m a n y l ' s as row m + j of A2. By cons t ruc t ion , A1 has h rows wi th w l ' s , and A2 has h + 1 rows wi th w l ' s . Using this fact and an a r g u m e n t s imi lar to the p reced ing one, row m + i is the only row of A2 which is s u m m e d in X1 X A 2 . If we consider the respec t ive lead l ' s in row m + i and row m + j, we can see tha t the re is only one o ther 1 in each of those co lumns and tha t tha t 1 lies in one of the first m rows. There fore , by p e r m u t a t i o n , we could replace row m + i wi th row m + j and as sume tha t it is row m + j of A tha t is s u m m e d .

T h e ma t r i ces A1 and A~ toge the r have one of two possible forms. E i the r

i. A1 = I [m;k l , . . . , k t ] and A2 = I [m;k l , . . . , k t + 1] , o r

ii. A1 = I [ m ; k l ) . . . , k j , . . . , k t ] and A~ = I [ m ; k l , . . . , k j + 1) .

(Note tha t in the l a t t e r case, the number s in the label for A1 f rom k j on, decrease by one.)

If A1 and A~ are as in case i, then the two mat r i ces differ in only one en t ry so we can choose the X ' s such tha t X1 or X but not bo th does the one s u m of rows of A2 necessary to ob t a in A1 and B is as desired.

If A1 and A2 are as in case ii, then row m + j of A~ is used to ob t a in rows m + j t h r o u g h n of Aa. In this case, each of these rows has exac t ly one more 1 than row m + j of A2. None of these rows conta ins any of the others , so tha t no row in this group is ob t a inab l e f rom any of the o thers . If one row of this group is p e r m u t e d to ob ta in ano the r row, then we have lost the fo rmer row. Therefore , all of these rows mus t be ob ta ined f rom row m + j by the ac t ion of exac t ly one ma t r ix . We can choose X and X 1 such tha t exact ly one of t h e m sums. T h e m a t r i x B is as desired. T h e chain is m a x i m a l .

S e c t i o n 5: F i n a l C o m m e n t s

We are left wi th some ques t ions and conjec tures .

1. Is this the longest chain possible? Up to J ( B 4 ), the answer is yes.

2. Given tha t a chain is m a x i m a l and ex tends f rom (O) to (I), wha t is the shor tes t chain possible? T h e chain (in J ( B n ) ) must have at least n + 2 e lements . This will compr i se classes whose row spans are f rom 1 to n + 1 and the class (1). How shor t i t m a y be be tween the (n + 1)-st level and (I), is not known. For J ( B n ) where 0 < n < 2, the set is a chain. In J ( B 3 ) , chain lengths m a y be e i ther 5 or 6. In J ( B 4 ), chain l eng ths of 7, 8, 9, 10, and 11 are possible.

3. W h a t types of number s can serve for lengths of m a x i m a l chains?

4. W h a t row spans are possible in Bn ?

T h e a u t h o r would like to t hank the referees for the i r c o m m e n t s and suggest ions;

and Professor Boris M. Schein for his i n s t ruc t ion and advice.

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R e f e r e n c e s

Breen, Michael A, "The Semigroup of Binary Relations" (dissertation), 1988. Howie, J.M. "An Introduction to Semigroup Theory," New York, Aca- demic Press, 1976.

Kim, Ki Hang. "Boolean Matrix Theory and its Applications" Marcel Dekker, Inc., New York, 1982.

Luce, R. Duncan, A Note on Boolean Matriz Theory. Proceedings of the American Mathematical Society. Number 3 (1952), pp. 382 - 388.

Plemmons, R.J. and West, M.T., On the Semigroup of Binary Rela- tions. Pacific Journal of Mathematics. 35, (1970). 743 - 753.

Schein, Boris M., Semigroups of Binary Relations. Proceedings of a Miniconference on Algebraic Semigroup Theory, Szeged, Hungary (1972), 17-24.

Vagner, V.V., Theory of Relations and Algebra of Partial Mappings, Theory of Semigroups and its Applications, Saratov University Press, 1, (1965). pp. 3- 178. [in Russian].

Zaretskii, K.A., The Semigroup of Binary Relations. Mat. Sbornik. Vol. 61, no. 3, (1963). pp. 291- 305. [in Russian 1.

Math Department, Alfred University, Alfred N.Y. 14802

Received November 8, 1989 and in final form April 10, 1990

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