AD-A254 748

CMS Technical Summary Report #92-13

ON ASCERTAINING INDUCTIVELY THEDIMENSION OF THE JOINT KERNEL OFCERTAIN COMMUTING LINEAR OPERATORS

Carl de Boor, Amos Ronand Zuowei Shen

Center for the Mathematical SciencesUniversity of Wisconsin-Madison610 Walnut Street DTICMadison, Wisconsin 53705 A I62i 19tU

June 1992

(Received June 15, 1992) 92-23580

92 8 2 5 01 4 Approved for public release

Distribution unlimited

Sponsored by

U. S. Army Research Office National Science FoundationP. 0. Box 12211 1800 G Street, N.W.Research Triangle Park Washington, DC 20550

UNIVERSITY OF WISCONSIN-MADISONCENTER FOR THE MATHEMATICAL SCIENCES

On ascertaining inductively the dimension of the joint kernel

of certain commuting linear operators

Carl de Boor, Amos Ron, and Zuowei Shen

Technical Summary Report #92-13June, 1992

Abstract: Given an index set X, a collection B of subsets of X (all of the same cardinality),and a collection {ez}:EX of commuting linear maps on some linear space, the family of linearoperators whose joint kernel K = K(IB) is sought consists of all eA := IEA '- with A any subsetof X which intersects every B E 18. The goal is to establish conditions, on lB and £, which ensurethat

dim K(B) = Z dim K({B}),BEE

or, at least, one or the other of the two inequalities contained in this equality. Concrete instancesof this problem arise in box spline theory, and specific conditions on t were given by Dahmen andMicchelli for the case that ]B consists of the bases of a matroid.

We give a new approach to this problem, and establish the inequalities and the equality undervarious rather weak conditions on 1B and t. These conditions involve the solvability of certain linearsystems of the form lb? = Ob, b E B, with B E 18, and the existence of 'placeable' elements of X,i.e., of x E X for which every B E B not containing x has all but one element in common withsome B' E 1B containing x.

AMS (MOS) Subject Classifications: primary: 47A50; secondary: 05B353, 41A63, 35G05, 47D03.

Key Words and phrases: dimension of nullspaces, solvability, matroidal structure, placeable, boxsplines.

" This work was supported by the Unitcd States Army under Contract DAAL03-G-90-0090,and by the National Science Foundation under grants DMS-9000053, DMS-9102857.

On ascertaining inductively the dimension of the joint kernel

of certain commuting linear operators

1. Introduction

Given a linear space S (over some field), we attempt to determine the dimension of spaces ofthe form

K :-fnEL kerl,

with L a (finite) sequence of linear endomorphisms of S, i.e., a sequence in L(S), chosen in a mannerdescribed below.

We start with a finite set X of atoms, and associate each z E X with a map 4z E L(S). Noassumption is made in advance concerning the individual t., z E X, but it is assumed throughoutthis paper that the atomic maps [tz)=E *commute with one another:

t4t4=4y4, z, yEX.

This means, in particular, that the product

1A:= lI1' ACX,zEA

is well-defined, without any need for ordering X.The joint kernel K C S whose dimension we attempt to determine can be described in terms

of a subset lB of the power set 2 X; the latter consists of all subsets of X. In general, the set MD canbe chosen in quite an arbitrary manner, and, in particular, there is no assumption that

X(IB):= U BBEB

covers all of X. Once 11 is selected, the joint kernel K = K(IB) is defined as

K(1B) : n f kertA,AEA(B)

where2x D A := A(B) := {A C X:V{B E B} AnB #0}

is the set of all subsets of X which meet every B E B. A particularly simple situation arises when1B consists of the bases of some matroid. Consistent with this, we call each minimal element (under 1 Forinclusion) of A a cocircuit even when B doesn't have matroidal structure, and denote the set ofall cocircuits by

Amin. ed o

When B is empty, K(M) = {0}. More interestingly, when M consists of a single set B C X,K is simply the joint kernel of the atomic maps {tz}zEB. It was the ingenious idea of Dahmenand Micchelli [DM3] to study the relation between the dimension of K and the dimensions of the Ion/"block spacts" K({B}), B E 1B. Their work was stimulated by two non-trivial examples that occur Lty Cod08in box spline theory (cf. [B], [DM1], [DM2], [BeR], [DM3]), one of which we now describe.IDist ilSPol., .

Example 1.1. Assume that S is the space V(IR) of complex-valued distributions (entire functionsor formal power series will do as well), and let each t. be a differential operator of the form4 = D - A, where v_ E IR'\{0}, A= E C, and Dy, Y E 1R", is the directional derivative in they-direction. Define

lB := {Y C X: {v,}yy is a basis for IRs}.

It is easy to verify that in this case, for each B E IM, K({B}) is spanned by one exponential, hence,in particular, dim K({B}) = 1. It is much harder to prove here that

(1.2) dim K = #IB.

This result was proved first for the case A = 0 in [DM1] (see also [HS]), and for the general case in[BeR] and [DM3]. Note that (1.2) can also be written as

(1.3) dim K = E dim K({B}),BEB

and it was Dahmen and Micchelli's observation in [DM3] that this latter formulation holds in moregeneral settings which started research into these matters. We will get to a more systematic discus-sion of the literature later on in this introduction. We mention at this point that the significanceof (1.3) in approximation theory lies in the fact that, whenever {Vx}.x _ 2V, K determinesthe exponential-polynomial space in the span of the integer translates of the corresponding boxspline (cf. the above-cited references for details). However, the above-mentioned connection is nomore valid when {Vz}xEX ! 2Z', and the analogous problem (of determining the dimension ofthe exponential-polynomials in that span) is hopelessly complicated. It was our desire to settlethis more general problem that partly motivated the research that led to the present paper. Moredetails can be found in §4. 0

In the above example, each B E 113, being a basis for IRS, is of cardinality s. We retain suchan assumption throughout this paper, i.e., assume that

(1.4) #B =s, VB E M,

for some positive integer s, and call it the rank of M1. Also, because of this example (and again inconsistency with matroid theory), we term the elements of IM bases. Our ultimate goal is to prove(1.3) which, however, cannot be proved in general without further assumptions, as simple examplesshow (see Example 2.1). All methods now in the literature, as well as our approach here, separatethe discussion of (1.3) into proving the inequality :5 (i.e., upper bounds) and the inequality (i.e.,lower bounds). Assumptions to be made for the derivation of (1.3) fall into two essentially differentcategories, those involving B and those involving t.

(i) IB-conditions. In addition to (1.4), we assume in the paper one or more of the following:(1) B is matroidal (i.e., M1 is the collection of bases for a matroid defined on a subset of X); (2) IBis order-closed; (3) B is minimum-closed; (4) B is fair; (5) X contains a replaceable element; (6) Bsatisfies the IE-condition (i.e., 0 E IE); (7) X contains a placeable element. All these conditions willbe defincd in the sequel; still, as an easy reference for the reader, we record the relations observed inthe paper between these various conditions in the following diagram, in which each arrow indicatesa proper (i.e., non-reversible) implication, and, in addition, the absence of an otherwise possiblearrow indicates that the corresponding implication does not hold in general.

(4) -- (5)/

(1) - (2) - (3)

(6) (7)

2

It is probably inherent in the problem (and this is confirmed by [DDM: Theorem 6.2] and byProposition 3.22) that, by imposing 1B-conditions (of any type), one can infer only upper bounds(i.e., prove the inequality < in (1.3)), and lower bounds must incorporate knowledge on the oper-ators involved, which is the second category of assumptions we impose:

(ii) 1-conditions, namely, assumptions on the operators {I=}= x. We employ three suchassumptions. One is the solvability of certain atomic systems (cf. 3.2), a second is directness of(IB,t) (cf. 7.1), and a third is 9-additivity, from [S] (cf. 8.1).

The known methods employed for the derivation of (1.3) can be divided into inductive andnon-inductive. The inductive methods partition MB into two (or more) disjoint subsets

)B = B1 U B2 ,

study the relations between dim K and dim K((13I) + dim K(MB2 ), and proceed to the considerationof each Bi, j = 1,2. This results in a binary (or higher-order) tree decomposition of lM. The onlytwo non-inductive results that we are aware of are the complex-variable proof in [BeR] that showsthat in Example 1.1 one has

dim K > #IB,

and the polynomial ideal argument in [BR] that shows that in Example 1.1 one has

(1.5) dim K(IB') #B'

for an arbitrary subset M13' C IB (as matter of fact, the argument in [BeR] also implies (1.5), butno formal statement to that extent is made there). The latter result (1.5) is of particular interestbecause it proves lower bounds while the matching upper bounds might be invalid; moreover, theselower bounds require no lB-conditions. We are unaware of non-inductive methods for the derivationof upper bounds; (The proof in [BR] that shows equality to hold in (1.5) in case M1 is order-closedis only seemingly non-inductive, since it invokes a result from [DR] which is proved there by aninductive method.) As for inductive arguments, all those that we are aware of (including, thus,those of the present paper) require some B-conditions and, moreover, the lB-conditions which areknown to suffice for lower bounds imply matching upper bounds as a by-product.

The two basic operations in matroid theory are deletion and restriction, and these operationsplay a major role in our more general context as well. Precisely, for a given Y E X, we delete y,from X to obtain

IB\y := {B E lB : y B),

and restrict B to y to obtainMl := {B E lB : y E B},

and in this way form a partition of 11 into two sets. Note that

(1.6) 11 C B' ==: K(11) _ K(IB'),

and hence both spaces K(B\,) and K(IB3y) are subspaces of K = K(IB).In principle, we study the exactness of sequences of the form

0 -? -+K--? - 0,

where the unknown terms should be related to the space of deletion and the space of restriction.Thanks to (1.6), we have (at least) two options to consider:

(1.7) 0 - K(3\,) K- ? 0,

3

and

(1.8)0 - K(My) - o.

The next step to be made then is the selection of the appropriate map j or i and of the correspond-ing space now missing in the above sequences. Before we discuss such completions of the abovesequences, we require some further notations and definitions.

Guided by Example 1.1, we refer to the elements of {Y C X : 3{B E IB C_ Y) as thespanning subsets of X, and to the collection

IFI = E(MCI)

of all maximally nonspanning subsets as hyperplanes. We also need the family

I=(IB):- U 2B

BEna

of all independent subsets of X. We say that B is matroidal whenever 1(B) defines a matroidon X(IB), which means (cf. [W]) that (B) A 0 and, for any 11, 12 E I((B) with #11 = #12 + 1,there is y E 11 for which 12 U {y} is still independent. Finally, for Y g X, we set

(1.9) By := {B E B: B C Y}.

This is consistent with the notation Ml\, introduced earlier if \y is interpreted as X\{y}.

The DM map. Assuming that 11 is matroidal, Dahmen and Micchelli offer in [DM3] the followingchoice for the missing map and space in (1.7). They choose the map j as

(1.10) j: K - X K(1 1 X\A) :f - (Af)AEAni.(O\,AEiAmjn(B\ u)

and employ it, in [DM3: Theorem 3.1], to prove the upper bound

(1.11) dimK < dim K({B})BEB

for a matroidal B. They also assert (cf. [DM3: Theorem 3.3]) equality in (1.11) under additionalf-conditions, and one of the by-products of the present paper (cf. §5) is the bridging of an apparentgap in the proof of the supporting Lemma 3.2 of [DM3]. Shen in [S] introduces a condition, called's-additivity' (cf. §8), on an abelian semi-group G of linear maps on S and, using the DM map,shows his condition to be necessary and sufficient for equality in (1.11) to hold for all maps t fromX into G and all matroidal B with rank s. Jia, Riemenschneider and Shen [JRSI] refine andextend the results of [S], from a matroidal M1 to an "order-closed" B, a notion introduced in [BR].Further, [JRSI] prove that if G is a semi-group of differential (resp. difference) operators, generatedby polynomials in s indeterminates over some algebraically closed field, and the linear space S isa space of formal power series (resp., sequences) in s indeterminates over the same field, then Gis s-additive (cf. Corollary 3.5 and Theorem 4.4 in (JRSI]). More recently, Dahmen, Dress andMicchelli [DDM], using hornological algebra and a replaceability condition (cf. §2), derived (1.3),i.e., equality in (1.11), for the matroidal and order-closed structure under certain f-conditions (cf.[DDM: Theorems 6.2, 6.5]). Those conditions are stronger than the basic solvability condition 3.2assumed in the present paper.

4

It is the DM map that naturally gives rise to the notion of "replaceability". More about thismap and the exactness of the corresponding short sequence is given in §2 and §8.

The atomic map. It is quite surprising that this simple idea was not used before. Here we choosei in (1.8) to be the restriction of 4. to K, and thus obtain, for any y E X, the following shortsequence

(1.12) 0 --+ K(lB) '-+ K K(IB\,) -. 0.

We will readily observe in §2 that K(BI ) _ keri and yK(IB) _ K(IB\,), hence (1.12) is a shortsequence in the homological sense. However, we can infer neither upper bounds nor lower boundsfrom this sequence, since in general, the sequence is inexact in two different locations: first, we donot expect in general to have K(IB y) = ker i, and further, we do not expect in general i to be onto.The derivation of upper bounds relies on the first exactness, the derivation of lower bounds relieson the second exactness. It is the desire to prove that K(BiM) = ker i that leads to the notion ofplaceability (§2) and the further desire to prove the ontoness of i that leads to the IE-condition (cf.§3).

The rest of this paper is organized as follows. Section 2 is devoted to the derivation of upperbounds using either of the above two approaches. Its main result is Theorem 2.16. The equality(1.3) is obtained (via the atomic map) in §3, which contains the main result of this paper (Theorem3.17). An example relevant to box spline theory is studied in §4, and the application of Theorem3.17 to matroidal and minimum-closed structures (together with some improvements) are discussedin §5 and §6, respectively (see, in particular, Theorems 5.2 and 6.4), with an application of theresults on matroids and minimum-closed sets presented in §7. The DM map is revisited in §8, whichis the counterpart of §3.

Our joint venture that led eventually to the present paper was initiated by the reading of[DDMI. We take this opportunity to thank the authors of [DDM] for making a preprint of theirpaper available to us.

2. Replaceability and placeability

We describe in this section lB-conditions which allow us to obtain upper bounds on dim Kin terms of dim K(IB\,) and dim K(IBy) for a suitably chosen Il E X. We emphasize that not-conditions are imposed here, hence these bounds are valid for an arbitrary S and arbitraryI : X - L(S). One might wonder whether it may be possible to establish realistic upper bounds ondim K without any 1B-conditions, especially since (1.5) shows that this might be the case for lowerbounds. The following example hints at the difficulties in obtaining such upper bounds withoutIB-conditions.

Example 2.1. Let X, B and {r}Zx be as in Example 1.1. We assume that {vi}zEX are heldfixed, select an arbitrary IB' C 1B, and consider the possible influence of the choice of the constantsA := {\}, on dim K(B') (such considerations are intimately related to the notions of "algebraicmultiplicity" and "geometric multiplicity" of a zero of an analytic ideal, cf., e.g., [AGV]). Ideally, wewould like dim K(IB') to be independent of the choice of A, as is the case for certain IB'. [BR] showsthat for an arbitrary IB' C B and for a generic choice of A, K(IB') is spanned by #M' exponentials,hence its dimension is #B'. On the other hand, if we choose B' to consist of parwise disjointbases, then Ami consists of all sets containing exactly one element from each B E IB', hence, withthe choice A = 0, all z, K(B') necessarily equals the space of all s-variate polynomials of degree< k := #M' (since it trivially contains the latter polynomial set, yet can contain no nontrivialhomogeneous polynomial of degree k), and hence dim K(IB') = (k+:-) > k = #113' (unless s = 1).

5

Now, suppose that we choose 1B' as above and want to derive lower bounds and upper boundson dim K(IB') without specifying the choice of A. In view of the above discussion, the best possiblelower bound is (1.5), and this is a realistic bound since it generically coincides with the correctdimension. In contrast, we cannot provide an upper bound better than dim K(IB') _5 (k+;-),

which, generically, is a gross overestimate of the correct dimension, and deviates from the desiredestimate (1.3). 0

The example shows, in particular, that the computation of dim K for a general IB mightrequire detailed knowledge of the interplay between the atomic maps involved. In contrast, wecompute dim K in this paper under mild general assumptions on the atomic maps. It is thereforeunderstandable that we must employ in our course suitable IB-conditions.

Since the DM map and the atomic map require different lB-conditions, we separate the dis-cussion accordingly. In these discussions, we use intensively the following simple fact which followsfrom the observation that, for any Y C X and any A E A(DB\y), Y U A E A (where, as mentionedbefore, \Y := X\Y).

Proposition 2.2. For any X, IB and t, and any Y C X, ly maps K into K(IB\,).

2.1. The DM map, Jia's intersection condition, and replaceability

We consider the DM map j defined in (1.10) and the corresponding short sequence (1.7).Because of Proposition 2.2, j is well-defined, and further, one observes that kerj = K(IB\y).

We find it useful in this section to index the target of j by H E Iff rather than by A EAmin(]B\y). This is possible, because

Amin(B\y) = {X\(y U H): y V H E III U {X\H : y E H E 1I} = {X\(yU H) : H E Il}.

Further, since lJyuH = 0 in case y E H E III, the only nontrivial components of the elements in thetarget

x K(BMuH)HEH

of the DM map are those belonging tolll\y := (H E III: y /H}.

Therefore we infer from (1.7), (1.10) the following inequality:

(2.3) dimK< dimK(B\y)+ E dimK(MByuH).HEIH\,

The arguments so far are valid for a general B, and hence (2.3) holds in general. It correspondsto writing B as the union

(2.4) B-=B\YU U ByuH,HEH\,

but, offhand, there is no reason to believe that this is a partition of 1B, since we might find thesame basis B in two different UMyuH (this is the case, e.g., for the B' in Example 2.1). In case theunion in (2.4) is not disjoint, (2.3) will not lead to the desired upper bound (1.11) on dim K.

This means that we are led to require the following intersection condition

(2.5) V{H, H' E 11\y} ByuH n ByuH, = 0, unless H = H',

first suggested by Jia, in [J].

6

Lemma 2.6. The intersection condition (2.5) is satisfied (for y) if and only if, for every B E BIY,there is at most one H E El\, containing B\y.

Proof. We observe that any B E ElyuH n lByuH' must contain y, i.e., is in lI.. Thus thecondition B E ElyuH n ElyuH,, H # H', is equivalent to the condition that (B\y) is contained inthe two different hyperplanes H and H'. .I

The intersection condition (2.5), as we will prove in a moment, is equivalent to having y"replaceable" in MD, in the sense of the following definition.

Definition 2.7. y is replaceable in l (or, l-replaceable) if for every B E l and everyB' E l there is some z E B' so that (B\y) U z E l.

For example, if #X(B) _ s + 1, then every z E X(El) is replaceable. Thus, the simplest lBwithout a replaceable element is {12,34}.

Proposition 2.8. y is Bl-replaceable if and only if (2.5) holds (for y).

Proof. '€:=': Let B E Illy and B' E l. Since B\y is not spanning, there exists H E lH\ycontaining B\y. We claim that, necessarily,

H = H' := Ex X : (B\y) U x IB}.

For, H C H' since H contains B\y but contains no basis (in particular no basis of the form(B\y) U z). On the other hand, if there were z E H'\H, then (B\y) U z would be not spanning,hence would be contained in some hyperplane H", and this hyperplane could not be H, since Hdoes not contain z. This would give us two distinct hyperplanes both containing B\y, hence neitherone containing y, and this would contradict (2.5), by Lemma 2.6. But now, knowing that H' is ahyperplane, we know that it cannot contain B', hence there is some x E B'\H' and, by the verydefinition of H', (B\y) U z E IB for each such z.

'==*': If B fails to satisfy (2.5) (for y), then there exist two distinct hyperplanes H, H' not

containing y for which there is some B E luHn M Euun'. B is necessarily of the form (B\y) U ywith B\y C H n H'. Since H # H', the union H U H' properly contains H and H', hence spans,

i.e., contains a basis B'. For z E B', (B\y) U z is a subset of either H or H' (since (B\Y) E H n H'and z E H U H'), hence cannot span. This means that y E B is not replaceable by any x E B'. 0

We note that this proposition ib close to [DDM: Lemma 6.4]. We also note, for later use, thefollowing characterization of B being matroidai (this is a standard result; cf., e.g., [NV: Theorem1]).

Proposition 2.9. The collection B # 0 is matroidal if and only if every y E X(B) is B-replaceable.

Proof. '==>': Let y E B E B and B' E B. Since #(B\,) < #B' and both sets areindependent, the assumption that B is matroidal implies that there must be z E B' so that(B\y) U z is independent, hence a basis.

'==': Let P,Q E I with #P < #Q. Then there are P',Q' with P U P', Q U Q' in M. We

order P' in any manner and replace sequentially each p' E P' C (P U P') E l by an element fromthe basis Q u Q'. At the end, we obtain a basis of the form P U P", with P" C Q U Q'. Since#P" = #P' > #Q', we must have P" n Q ? 0, and any set of the form P U q for some q E P" n Q

is independent. o

Corollary 2.10. For any independent (s-1)-set C, BIC is matroidal.

Proof. Every z E X(Blc) is either in C or else completes C to a basis, hence, either way,is replaceable. 0

7

To summarize: if y is IB-replaceable, then the union (2.4) is disjoint and therefore the estimate(2.3) provides a first inductive step toward the final desired upper bound (1.11). However, ourprimary aim in this paper is the application of the atomic map to which we now turn our attention.

2.2. The atomic map and placeability

Considering (1.12), we observe that, by Proposition 2.2, the map i is well-defined (i.e., mapsinto K(IB\,)). Further, we always have that K(IBy) C ker i = kerfl nK: the inclusion K(BI) C Kis due to BI, C 1B, while the inclusion K(B1 y) C kerty follows from the fact that {y} is a cocircuitin Bhy.

The atomic map provides the necessary inductive step towards an upper bound if, in additionto the above, we also know that K(]BIy) D ker i. For this, we introduce the following notion ofplaceability:

Definition 2.11. We say that Y is placeable in B ifYUC E B for some C C B. If Y is placeablein every B E B, then we say that Y is placeable (in B), or, B-placeable.

For example, if #X < s + 1, then every X E X(B) is placeable. Thus, the simplest B withouta placeable element is {12,34}. Further, y is replaceable in 1B iff for each B E B, B\y is B-placeable. On the other hand, there may be some replaceable atom even though none of the atomsare placeable, as is the case for B = {123, 126, 129,345, 678}, in which 9 is trivially replaceable,while none of the atoms in 345 can be placed in 678 and vice versa, and 1, 2, or 9 cannot be placedin either.

Further, if B is matroidal, then every independent element, i.e., every y E X(B), is placeable(by Proposition 2.13 below), but the converse does not hold, as the following example shows.

Example 2.12. Let X = 12345678 := {1,...,8} and let B consist of all 3-sets in X excluding123, 124, 567, and 568. Then, every z E X is B-placeable, but no x E X is B-replaceable: given1 < x < 4, x in B = 56x cannot be replaced by any element from 578, and a similar argumentapplies to z > 4. In particular, 1B is neither matroidal (by Proposition 2.9), nor is it minimum-closed (by Proposition 6.6, and for whatever ordering we choose to impose on X), hence cannot beorder-closed.

The lack a replaceable atom makes this example inappropriate for an application of the DMmap. On the other hand, we will verify (cf. Example 6.8) that B here satisfies the JE-condition,and this guarantees a successful binary decomposition of B via the atomic map. 0

The example demonstrates, in particular, the fact that total placeability (i.e., having everyy E X(B) placeable) falls short of implying that B is matroidal. In this regard, it is useful to notethe following two propositions.

Proposition 2.13.(i) If B is matroidal, then every independent set is placeable.

(ii) If every independent (s-1)-set is placeable, then B is matroidal.

Proof. (i): This is a standard matroid argument. Let C E I and B E B. We are to provethat B U C contains some B' E MIc. This is certainly so in case #C = a. In the contrary case,B contains some (#C + 1)-set C', and, B being matroidal, this implies that, for some Y E C',C U y E I. Downward induction on #C then completes the proof.

(ii): Since every independent (s-1)-set is placeable if and only if every element of X(B) isreplaceable, Proposition 2.9 supplies the proof. 03

8

Proposition 2.14. I, for every Y C X, every y E X(BMy) is ]Biy-placeable, then MB is matroidal.

Proof. In view of Proposition 2.9, it suffices to show that every y E X(]B) is replaceable.Let B, B' E lB and let y E B. We need to find x E B' such that (B\y) U X E 13, and may assumewithout loss that y V B' (otherwise, choose x to be y). We prove the existence of such an atom :by (downward) induction on #Y, with Y :=B B', there being nothing to prove when #Y = s.Also, when #Y = s - 1, we choose z as the single element of B'\B. So, assume #Y < 3 - 1.Then B\(y U Y) is not empty. Let b be one of its elements. By assumption, b is ]ly-placeable,hence we can place b in B', i.e., there is some B" E IBIy for which B"\B' = {b}. This implies that#(B n B") > #Y. Thus, by induction, there exists x E B" for which (B\y)U x E B. This z differsfrom b (since b is in B\y, hence cannot complete this latter set to a basis), and thus X is in B'. C3

The proposition makes clear that total placeability, while being preserved under deletion (un-less, of course, we delete the atom in question), cannot be preserved under restriction. Indeed, wesee that, in Example 2.12, 2 fails to be ]B11-placeable into 134.

The next lemma prepares for the main result of this section.

Lemma 2.15. Let Y C X, and set Y := {Z c X :Z n Y 0 0} = UYEy(2X)I1. Then, A(Bly )

A(IB) U Y, with equality if and only if Y is B placeable. In the latter case,

K(BIy) = Kf n ker t.pEY

Proof. The containment A(]Bly) 2) A(MB) U Y is straightforward.Assume that Y is not B-placeable. Then there exists B E B for which B U Y fails to contain

an element of lBIy, hence X\(B U Y) E A(Mly), yet X\(B U Y) is neither in A(M) (since itscomplement contains B) nor in Y (since it is disjoint from Y).

Assume that Y is B-placeable, and let A V A(B) U Y. Since A V A(B), X\A contains someB E 1B, and since A 0 Y, X\A must contain Y. Since Y is B-placeable, there is B1y - B' CB U Y C X\A, hence A V A(MBy). 0

Theorem 2.16. If y is B-placeable, then

(2.17) dim K < dim K(B1 ) + dim K(B\),

with equality if and only if t4 maps K onto K(MB\).

Proof. Since y is lB3-placeable, Lemma 2.15 implies that K(Ml ) = K fn kerry, hence thatkeri in (1.12) coincides with K(BI.). Thus, (1.12) is exact at K and (2.17) follows. Equality in(2.17) holds if and only if (1.12) is also exact at K(B\y), i.e., if and only if ty maps K onto K(B\,).

0

Use of the atomic map and the placeability notion seems to be more applicable and powerfulthan the alternative idea of the DM map and the notion of replaceability. For example, using theformer approach we obtain in the next two sections the equality (1.3) under t-conditions which areweaker than those employed in [DM3] and [DDM], and weaker than the s-additivity used in [S]and [JRS1]. Further, the replaceability of y E X is a necessary (albeit not sufficient) condition forthe exactness of the short sequence employed in the DM map, while, in contrast, the short sequence(1.12) can be exact even for nonplaceable y's. Indeed, if we choose A in Example 2.1 in such away that K is spanned by (pure) exponentials (which is the generic choice), then, for an arbitraryMB' C 1B and an arbitrary y E X, the sequence (1 12) (with B' replacing B) can be easily shown tobe exact (since then K(IB') is spanned by eigenvectors of I,).

9

3. E-condition and special solvability

In this section we discuss conditions on the map t : X --* L(S) and on lB under which there is

equality in the inequality (2.17).We expect equality in (2.17) in case 4y maps K onto K(I\y), i.e., in case the equation

t? = f

has solutions in K for any f E K(B\y). For this reason, our e-conditions are connected to thesolvability of systems of the form

(3.1) (C,(): t.=? cE C,

with C C X and W a map into S and defined (at least) on C.

Definition. We call the system (C, V)(i) special, or, more explicitly, 11-special if V, E K(IB\,), all c E C;

(ii) compatible if tcpb = 10P, for all c, b E C;(i;i) independent, resp. basic, if C E I, resp. C E lB.

It goes without saying that such compatibility is a necessary condition for the solvability of

such a system.

Solvability condition 3.2. Any special compatible basic system is solvable.

As an example, in Example 1.1 one can easily verify that any compatible basic system is

solvable. However, our solvability condition requires only the solvability of "special" systems,hence might hold even when some more general compatible basic systems fail to admit solutions.

For example, we can allow S to be finite-dimensional, e.g., to be K itself, which is not possible with

other approaches in the literature.For our subsequent purposes, it will be important to know that the solution of the special

compatible basic system is in K, but this fact comes for free:

Lemma 3.3. Any solution of a special basic system (.B, V) lies in K.

Proof. Let f be a solution, and let A E A(IB). Then A contains some element b of our

B, therefore t A = tA\bebf = t A\bPb • Since A E A(B), it follows that (A\b) E A(IB\b), and thus,because we assume that Wb E K(IB\b), we obtain 0Af 0. 0

3.1. The set IE

The solvability condition is all we need for the derivation of (1.3) in case the "effective" rank

is 1 (by Theorem 5.2, since BIc is matroidal for any independent (s-1)-set C, by Corollary 2.10).

For "effective" rank higher than 1, we use Lemma 3.3 to show that y maps K onto K(B\y), by

extending the equation t? = V, with Vy E K(IB\y) to a special compatible basic system (B,V),

but the existence of such an extension is not trivial. Our proof that this is possible is by induction,

and requires that IB satisfies the E-condition, by which we mean that

0E IE,

with IE = IE(IB) the following peculiar subset of I.

10

Definition 3.4. Let E = IE(]B) be the collection of all those C E I which either are in IB, or elsethere is some b E X\C, called a ]B-extender for C, which satisfies the following two conditions:(i) Cub EE;(ii) if JB\b 0 0, then C E IE(IB\b).

The recursion required in this definition does terminate after finitely many steps. For, the firstbranch leads to a set of higher cardinality, hence this branch terminates after exactly s - #C steps.The second branch keeps the cardinality of C the same but decreases the number of bases, thus isguaranteed to terminate since #IB is finite.

Note also that IE(B) is not (in general) monotone in MB. For, while MB C IM ' implies thatI(]B) _ I(1B'), this resulting increase in independent sets could lead to a nontrivial B'b wherebefore we had B\b = 0 (hence C E IE merely because C U b E JE), without guaranteeing that Clies in IE(B 'b) (since, before, we didn't need to know whether or not C E IE(IB\b)). On the otherhand, if we make an appropriate assumption, such as that, for all Y, By - 0 = IB'. - 0, inorder to avoid this objection, then we get the trivial conclusion that M = B'.

As an example, an independent (s-1)-set C is in IE if and only if {b E X : C U b E B} E A,as the proof of the following connection between IE and the lB-placeable subsets of X makes clear.

Proposition 3.5. Every C E IE is lB-placeable, and the converse is true if #C = s - 1. Inparticular, if s = 2, then y E IE if and only if y is B-placeable.

Proof. We prove the first claim by (downward) induction on #C and induction on #IB,it being trivial if #C = s or #13 = 1.

Assume that #C < s, and let B E 1B. Since C E IE, it has an extender, b say. In particular,C u b E IE. If b E B, we apply our induction hypothesis to C u b to conclude that C U b is placeablein B, a fortiori C is placeable there. If b V B, then B E M\b. Since IB\b is a (non-empty) propersubset of B, and since we know that we still have C E IE(B\b), we can conclude by induction thatC is placeable in B.

It remains to show that a B-placeable C of cardinality s - 1 is in M: If C is placeable, thenevery B E B must meet the set C' := {b E X : C U b E M}. This means that B\c = 0, henceC E IE follows by (downward) induction on #C', it being trivially true when #C' = 1. 0

In general, placeability does not imply membership in E. For example, not every IE containsthe empty set (cf. Proposition 3.11 below). As a concrete example, if X = 12345 := {1,2,3,4,5}and B = {123,234,245, 135}, then 5 can be placed into any basis, but doesn't make it into IE sinceno 2-set containing 5 is placeable and therefore no such set makes it into IE (in view of the lastproposition). As it turns out, the additional condition needed for a placeable C to be in IE is thatit be already in IE(BIc). This is a consequence of the following lemma.

Lemma 3.6. If Y C X contains some placeable C, then Y E IE if and only if Y E IE(BIc).

Proof. We begin with the observation that, for any Z C X\C,

(3.7) B\z=0 0- IBic\z =0.

Indeed, the necessity is trivial. As for the sufficiency, if B E B\z, then C, being placeable, can beplaced into B, and this provides an element of IBIC\Z.

'=: ': The proof is by (downward) induction on #Y and induction on #B, it being triviallytrue when #M < 1 or if #Y = a. Assume that #B > 1, and that #Y < s. Since we assumeY E lE, there exists a B-extender, say b, for Y. We now verify that b is also a Bic-extender forY: (i) Since Y U b is in IE and is larger than Y, induction on #Y ensures that Y U b E IE(BIc).(ii) If BjC\6 # 0, then, by (3.7), B\b # 0, hence Y E IE(B\b), therefore, by induction on #B,Y E IE(MBc\b).

11

'==': Since (3.7) is an equivalence, the argument just given also works with lB and IBIcinterchanged. E3

Note that, offhand, Lemma 3.6 implies nothing about the relationship between E and IE(IBjc).This reflects the fact that, in general, IE(IB) is not a monotone function of B.

Corollary 3.8. C E ME if and only if the following two conditions hold:(a) C is IB-placeable;(b) C E IE(]Blc).

Proof. Because of Proposition 3.5, it is sufficient to prove that, for a lB-placeable C,C E IE if and only if C E IE(IBIc). But this is just the special case Y = C in the lemma.

[]

Note that we could not use (a) and (b) of this corollary to define IE because the equivalenceproved in this corollary is a tautology whenever MB = IBic.

Lemma 3.9. Assume that lB = 1B1 c.(a) IfYuC E IE, then J E IE for anyY\C _ J C YUC.(b) If Y E IE, then JEIE for anyYC JCYUC.

Proof. (a): The proof is by (downward) induction on #J, there being nothing to provewhen #J = #(Y U C). If now #J < #(Y U C), then there exists z E (Y U C)\J. We verify thatany such z is an extender for J: (i) J U z E IE by induction hypothesis; (ii) any such x is necessarilyin C (since Y\C C J), hence B\. = 0.

(b): The proof is by induction on #13 and (downward) induction on #Y, it being triviallytrue when #1B = 1 or #Y = s. So, assume that #113 > 1 and #Y < s. We are to verify thatJ E IE. Since Y E IE, it has an extender, b say. Since Y U b is in IE and larger than Y, inductionon #Y implies that J U b E IE, hence we are done in case b E J. Otherwise, to verify that b is anextender for J, assume that MB\b 6 0. Then Y E IE(lB\b), hence J E IE(IB\b), by induction on #IB(applicable since J U b E IE implies that #IB\b < #]B, and since IB\b = (IB\b)IC). 0

Corollary 3.10. Assume that lB = lBc and Y C X. Then Y E IE if and only if Y U C E IE ifand only if Y\C E JE. In particular, 0 E IE(IBIC) if and only if C E lE(IBic).

Proof. Both implications '==.' are special cases of the lemma, as is the first '=', whilethe second '€==' follows from (b) of the lemma since Y U C = (Y\C) U C. 0

The final results in this subsection aim at providing efficient methods for an inductive verifi-cation of the IE-condition, i.e., the condition 0 E IE.

Proposition 3.11. Assume #IB > 1. If 0 E IE, then, for some b E IE, 0 E IE(Mlb) andO E IE(IB\b).

Conversely, if 0 E IE(lB\b) for some b E IE, then 0 E E.

Proof. For the sake of both claims here, we note that

(3.12) b E IE = b E IE(lib) == 0 E IE(lB1b).

Indeed, the first implication corresponds to the choice C - b in Corollary 3.8, and the secondimplication corresponds to the choice C = b in Corollary 3.10.

12

Proof of first daim: Let C C_ X be a maximal set for which lB = JBIC and note that #C < a(since #IB > 1). Since 0 E JE, C E ]E by Corollary 3.10, hence has a IB-extender, b. We nowverify that any such b does the job: B\b 0 0 (for, if IB\b were empty, then lB = MBcub, and thiswould contradict the maximality of C), and therefore, because b extends C, we have C E X(I\b),

or equivalently (by Corollary 3.10, since IB\b = (IB\b)Ic), 0 E IE(B\b). Further, since C U b E IE,the choice Y = b in Corollary 3.10 provides the conclusion that b E IE, and hence, by (3.12),0 E IE(Mlb).

Proof of second claim: Since b E E and 0 E IE(IB\b), b is a rn-extender for 0. I

A repeated application of the last proposition leads to the following partial unraveling of thecondition 0 E IE:

Corollary 3.13. 0 E JE if and only if X contains a sequence bl,. . ., b, for which bj Eall j, while #1B\b,,..., = 1.

We note that, by Proposition 5.1, 0 EIE in case lB is matroidal. This provides the followingstrengthening of the above corollary.

Corollary 3.14. 0 E IE if and only if X contains a sequence bl, . . ., b, for which b, E IE(B\b .b,_ ),

all j, while IB\b ..... b, is matroidal.

Remark 3.15 A complete unraveling of the condition in Proposition 3.11 produces a binary treewhose nodes are of the form lJy\z for certain Y, Z C X with Y n Z = 0. Further, each such nodeis either a leaf, in which case it contains exactly one B E IB, or else it is the disjoint union of itstwo children, IBI(bUY)\z and lljY\(zub), with b E X\(Y U Z) Bwy\z-placeable. Finally, IB is theroot of this tree.

Conversely, if we have such a tree, then every one of its nodes (including lB itself) satisfiesthe IE-condition, as can be seen by induction on #M3, as follows: It is trivially true if #B = 1.If #B > 1, and b is the placeable element used to split the root node, IB, then, by Corollary 3.8,b E IE if and only if b E IE(IBlb), and, by Corollary 3.10, this latter condition is equivalent to having0 E IE(lBib), and this condition holds by induction hypothesis. Thus b E IE, and, by Proposition3.11, this implies that 0 E IE since 0 E IE(IB\b) by induction hypothesis. E3

Without the requirement that the b used to split IBly\z be lBy\z-placeable, every lB wouldhave such a tree.

With or without the placeability requirement, the leaves of such a tree constitute the partitionof lB into its bases.

3.2. Dimension estimates

We now turn our attention to the main topic of this section, namely the connection betweenthe content of IE and the validity of (1.3). The central ingredient for our argument is the followingproposition for whose proof the set IE was tailor-made.

Proposition 3.16. If the solvability condition 3.2 holds, then, any special compatible system(C, W) with C E IE can be extended to a special compatible basic system, hence has solutions in K.

Proof. The proof is by (downward) induction on #C and induction on #lB. The statementis trivial if IB is empty or if C is a basis.

Let lB and C E IE\IB be given, and assume that we already know the claim for larger C E lEas well as for any set C' E IE(IB') with #lB' < #lB.

13

Let b be an extender for C. Then C U b is in E and larger than C. We claim that we cancorrespondingly find some Vb E K(B\b) so that the extended special system (C U b, W) is stillcompatible. For this, it is necessary and sufficient that jPb solve the system (C, lbw). There are twocases:

(i) if IB\b = 0, then lB = BIb, hence {b} E A, therefore kereb 2 K _D K(B\0 ) for all c E Cand, in particular, t b'Pc = 0 for all c E C, thus the trivial choice (Pb = 0 solves (C, IbW).

(ii) if IB\b 6 0, then we know that C E IE(IB\b), and the system (C, tbW) is compatible andIB\b-special (since (C, W) is compatible and IB-special, and because of Proposition 2.2). Also, sinceC U b E IE, it is contained in some B E B and this B is necessarily not in B\b. This implies thatIB\b is a proper subset of lB. It follows, by induction hypothesis, that (C, tbV) has a solution inK(B\b), and any such is suitable as Wb-

Since C U b is in IE and larger than C, induction now allows the conclusion that our present(extended) special and compatible system is part of a special compatible basic system. 0

We are now ready to state and prove the main result of this paper:

Theorem 3.17.(a) Assume that the solvability condition 3.2 holds. Then, for any y E IE, t maps K onto K(]B\,),

and

(3.18) dim K = dim K(Bl.) + dim K(IB\y).

(b) Assume that 0 E IE. Then

(3.19) dimK < E dim K({B}),BEE

with equality in case the solvability condition 3.2 holds.

Proof. (a): Since Y E IE, Proposition 3.16 implies that the linear equation t? = 'P withWy E K(I1\,) can be extended to a special compatible basic system (B,WP), and, by assumption,this is solvable, while, by Lemma 3.3, any solution of such a system is in K. This proves that t.maps K onto K(IB\,). On the other hand, since Y E IE, it is B-placeable (by Proposition 3.5).Now apply Theorem 2.16.

(b): We prove this part by induction on #113, it being trivially true when #IB = 1. Assumethat #IB > 1. Then, by Proposition 3.11, there exists b E IE for which 0 is contained in both]E(Bib) and IE(IB\b). In particular, neither IBib nor IB\b is empty, hence both are of cardinality< #MB, and induction therefore provides the inequalities

(3.20) dim K(Bjb) < Z dim K({B)), dim K(IB\b) _< 1 dim K({B}).BEB1l6 BEB\

On the other hand, since b is in IE, hence placeable, Theorem 2.16 implies that

(3.21) dim K < dim K(BIb) + dim K(IB\b).

Combining (3.20) and (3.21), we obtain (3.19).For the equality assertion, note that, as soon as the solvability condition 3.2 is assumed with

respect to B, it automatically holds with respect to any subset B' C B (since K(EB') C K).Therefore, if the solvability condition 3.2 holds, then, by (a) and by induction, we have equality in(3.20) and (3.21), hence obtain equality in (3.19). 0

The final claim in this section provides a partial converse of (b) in the above theorem.

14

Proposition 3.22. Let B be a basis in ]B that satisfies, for some ordering B = (bl,...,b'),

]B\. I. :=lb EB :b < a)E E(B\,), a EB.

dim K dim K({B),BEE

then any 11B-special compatible system (B, W) is solvable.

Proof. We are to prove that, for any such B E B, any lB-special compatible system (B, V)is solvable. Since, by Lemma 3.3, any solution of such a system is necessarily in K, this is equivalentto proving that the map

P: S -S. ' 1 -- (tbf)beB

carries K onto the space t := 'Ob., where, for a, b, c E B, we define

{(Vb) E x K(IB\b) :V{b, c < a} £tcb = tb(o}.

bKa

Since P maps K into $ and K n kerP = K({B}), while

dimK= E dim K({B'})B'EM

by assumption, it is therefore sufficient to prove that

(3.23) dim 4t < 1 dim K{(B'}).B'EB\B

For this, we claim, and prove inductively, that

(3.24) dim <S dim (K(B\b)f nnker tc)b<a COb

The case a = b, is trivial, since fb, = K(B\b1 ), and (3.24) asserts that dim 4pb, _ dim K(]B\b,).Assume, thus, that (3.24) holds for a < c := bk-1, and consider the case a = bk. We note thatevery element V E 4). is of the form (0, W.), with 0 E $c, and W. satisfying suitable compatibilityconditions. Conversely, if 40 E t is extendible to V = (, °) E %, then it is easily checked thatany other such extension (0,, y) must satisfy

SE K( \) nn ker t.b<a

Therefore, it readily follows that

dim 4. :< dim$ 4+ dim (K(B\.) n n ker tb)

and (3.24) follows.

15

Further, if B\. 6 0, then the set I. = {b E B : b < a) is in IE(MB\.) by assumption, hence isIB\-placeable by Proposition 3.5, thus it follows from Lemma 2.15 that

K(IB\.) n nkerlb = K((M\.)II.),b<a

and this equality holds trivially when lB], = 0. Therefore, since t = Ob., we obtain from (3.24)and the last equation that

di, < E<i (Ma1r.: i Q')aEB aEB B'E(1B\D.) 1

Here, the second inequality is trivial for all a with lB\. = 0 and follows for all other a E B fromTheorem 3.17(b), since I. E IE(IB\) by assumption, hence in IE((IB\a) 1I.) (by Corollary 3.8),therefore 0 E IE((IB\).) (by Corollary 3.10 applied to C = .). Further, this double sum equals

" B'EB\B dim K({B'}) since IB\B is the disjoint union of the sets (B\.)11., a E B. 0

Note that all inequalities established during the proof must actually be equalities. Further,in terms of the binary tree obtained in Remark 3.15 from the IE-condition, the only bases whichoffhand satisfy the conditions of the proposition are those belonging to the largest matroidal node(of the tree) of the form IBl.

4. An example

In this section, we apply our results from the previous one to an example whose solution isimportant in box spline theory. Since Approximation Theory and in particular box spline theory isnot an issue in this paper, we discuss neither the connections nor the applications of this example tobox splines. However, the example has intrinsic importance for the discussion in this paper since itprovides a naturally arising instance when the condition 0 E IE holds (and hence (1.3) holds), whilethe seemingly more verifiable, but stronger, conditions (minimum-closed, order-closed, matroidal)are invalid.

The example goes as follows: M is an s-dimensional linear subspace of JRd which is spannedby integer vectors. We associate each x E X with a vector v. E Qd\ 0 and define the coverage ofz as the set

(4.1) Zz(M) := {a E M n7Zd : vz. a E Z\0}.

This set does not change if we replace v. by its orthogonal projection onto M, and this orthogonalprojection is again in Qd since M is spanned by integer vectors. We may, and do, therefore assumewithout loss of generality that v, E M for all z E X. More generally, the coverage of Y C X is,by definition, the union

ZY(M) := UYEyZY(M).

Further, we say that z E X weakly covers Z C 2Zd n M in case {0} 0 v,' .Z C 72Z, and call z anM-integer if it weakly covers all of M n Md. We note that z is an lRd-integer exactly when v_is a nontrivial integer vector. Finally, for ease of notation, we use the orthogonal complement ofY C X to mean the orthogonal complement of the corresponding set of vectors in Q:

Y :rgyv .

Based on these notions, we define the set 113 as follows. An illustration for this definition canbe found in Example 6.9.

16

Definition 4.2. The collection B = B(M) consists of those sets B C X of cardinality s for whichM n B1 = {0), and, in some ordering B = {bI,..., b), and for each j = 1,...,., B -covers (M n 7Zd)\B,±.

Since dim M = s = #B, the condition that M n B - = {0} implies that, for each B E IB,(Vb)bEB is a basis for M (recall our assumption that, for each x, v. E M). Each B E B is not justa set, but an ordered set, but we do not count two such B as different elements of lB if they onlydiffer in the ordering of the elements.

Further, if B = (b1,...,b,) and, for some j > 1, bj is an M-integer, then we can exchange itwith its neighbor to the left, i.e., then also B' := (Bi- 2 , bi, b-, 1, bj+I,... , b.) satisfies the conditionof the definition: Since (bi,..., bj) covers (M n 7Zd)\(bl,..., bj) 1 regardless of the order in whichwe write the terms in the sequence (bl,... , bi), the only issue is whether (Bj- 2 , bj) covers (M n2Zd)\(Bj_ 2 , bi)1'. However, MnTZd is the union of (Mnl2Zd)\Bj_ 2

1L with (MnTZd)flBj- 21, and the

first set is covered by Bj- 2 , while bj, being an M-integer, covers everything in M n 2Zd except thoseelements in bj ' . Thus, (Bj- 2 , bj) covers everything in M n7d except for (M n 2Zd) n Bj2 .L n bj1-,i.e., everything in (M n 7Zd)\(Bj_ 2, bi) "'.

A slightly different definition of IB could have been to merely require the s-set B E B to coverall non-trivial integers in M. If s < 3, this latter variant can be proved to be equivalent to the onewe had chosen above. However, it is possible to give examples for s = 4 of s-sets of rational vectorswhich cover all the (nontrivial) integers in a space of dimension 4, which nevertheless do not satisfythe terms of the definition for any ordering of its elements.

If M = R d , and each x is an M-integer (which is equivalent in this case to having vx E 2Zd),

then B consists exactly of all subsets B whose corresponding {Vz,-zEB form a basis for IRd. Inother words, the present setting generalizes the box spline setup described in Example 1.1. As amatter of fact, it is the present setup that one needs to study when the 'directions' of a box splineare permitted to be rational vectors.

Our technical goal is to prove the following two results:

Lemma 4.3. Every M-integer is in IE.

Lemma 4.4. 0 E JE.

This latter lemma, when combined with Theorem 3.17, provides us with

Theorem 4.5.

(4.6) dimK < E dim K({B}),BED

and equality holds whenever the solvability condition 3.2 holds.

The problem germane to box spline theory needs only the case M = R d , and t and S as inExample 1.1. For this case, we need only the upper bound result (4.6), with the matching lowerbound already being provided by (1.5) (recall that, for this choice of the operators, dim K({B}) =1). The fact that we have chosen to define the problem on linear subspaces M C IRa is technical:we need it for the inductive proof. We did not restrict attention to the specific t of Example 1.1simply because the results of §3 allow us to prove the above theorem without prescribing t.

We prove Lemma 4.4 simultaneously with Lemma 4.3:

Proof of Lemma 4.4 and Lemma 4.3. We use induction on s = dim M, the proof being obviousif dim M = 1, since for any rank-1 113, we always have 0 E IE, and further {y) is then in lB iff y isan M-integer. Assume thus that a = dim M > 1.

17

First note that the first element in any basis must be an M-integer: Indeed, with b, the firstelement in question, it must cover (M n 2Zd)\blr-, hence Z 2) Vb, • (M n 27d) $ {0}, the inequalitydue to the fact that Vb, is a rational vector and part of a basis for M. In particular, there areM-integers in X unless B is empty.

Further, we claim that any M-integer y can be placed into any B E IB: Let B := bl, ... , b,}and set vj := vj,, all j. Since vs E M\0, and (vj,..., v.) is a basis for M, there is a smallest r > 1for which

(4.7) span{vj, ... , vr} = SpanIVI,... i,--, VY}.

We contend that y can replace br in B. Since the role of br in B is to cover (M n 2Zd n B J- )\B1"r-1 *rv

we need to prove that y covers this set as well. Since y weakly covers all of M n Zd, we need onlyverify that it does not vanish on B1. \B - , i.e., that vy -a 5 0, Va E B-_1 \Bk. For this, observethat, by choice of r, v. E CVr + span(vj)j<, for some nonzero c, hence, for any a E B -_\B*,Vat. a = CVr"- a 5 0. Thus, indeed, B' := (B\br) U Y E B, and y is thus placeable. After placing yinto B, the discussion following the definition of B implies that we can place y as the first elementof B' (without changing the order of the rest).

Since y is placeable, it is, by Corollary 3.8, in IE if and only if it is in IE(IB1l), and, by Corollary3.10 (applied with C = {y} and Y = 0), this latter condition is equivalent to having 0 E IE(IBl).Thus, to prove that y E IE, it remains to show that 0 E IE(MBl). For this we apply inductionon s: we first define M' to be the subspace of M which is perpendicular to vy. Since y covers(M\M') n 2Zd, but certainly does not cover any a E (M n 2d), we conclude that B E IB(M') ifand only if (y, B) E lB. In particular, 0 E IE(MBl) if and only if 0 E IE(IB(M')), and the lattercondition holds by induction hypothesis since dim M' = s - 1.

To complete the inductive step (and thereby the proof of the two lemmata), it remains to showthat 0 E IE, which we prove by induction on the number of M-integers in X. Assume that thereis an M-integer y. Since y is in IE, it can serve as an extender for 0, provided 0 E IE(B\y) in caseIB\Y 0 0. But the latter proviso holds by our induction hypothesis (on the number of M-integers,of which X\y is guaranteed to contain at least one since IB\y 0, but fewer than does X). 0

5. Matroid structure and special solvability

In this section, we prove the dimension formula (1.3) under the assumption that B is matroidal.Recall from Proposition 2.9 that B is matroidal if and only if each independent z is replaceable

in B, and, from Proposition 2.13, that, if B is matroidal, then every independent element z is B-placeable, with the converse not true in general.

Proposition 5.1. B is matroidal if and only if JE = I.

Proof. '== ': It is sufficient to prove that, for any C C B E B, and any b E B\C,C E I(B\b) in case B\b # 0. For this, if B' E B\b, then, since B is matroidal, C is placeable in B',i.e., extendible to a basis using only elements of B' and, since b B', this implies that C E I(B\b).

'=:': If IE = I, then, by Proposition 3.5, every independent set is placeable, hence B ismatroidal by Proposition 2.13. 0

If B is matroidal, then, for any z E X, also BI. and B\, are matroidal.For matroidal IM, we have the following theorem.

18

Theorem 5.2. If MB is matroidal and dim K < oo, then, the following are equivalent:(i) The solvability condition 3.2 holds.(ii) For all Y C X and y E Y, 4. maps K(lBy) onto K(IBy\,).

dimK = dim K({B}).BEE

Proof. (i)==*(ii): Let Y E Y C_ X. Since MB is matroidal, so is IBy, hence Y E IE(IBy) byProposition 5.1. Since, for every b E Y, K(IBy\b) C K(MB\b), the solvability condition 3.2 (which isassumed to hold with respect to MB) holds with respect to By, too. Therefore, Theorem 3.17 (part(a), with 13 there replaced by By) implies that ly maps K(By) onto K(IBy\y).

(ii)=*(iii): The proof is by induction on #B, it being trivially true when #M < 1. Assume#M > 1, and choose Y E X(B) so that B\, 0. Since y E IE (by Proposition 5.1), we concludefrom Theorem 3.17 that

dim K = dim K(B1I) + dim K(I~y),

and this equals EBEO1 , dim K({B}) + EBEIB, dim K({B}) by induction (applicable since both

1 1 and B\, have smaller cardinality than B), and this equals

E dimK({B})BEE

since B is the disjoint union of Bly and B\y.(iii)==.(i): This implication is a special case of Proposition 3.22, since B\b is matroidal for

any b, hence, by Proposition 5.1, every B E B satisfies the conditions imposed on B in Proposition3.22. 0

Corollary 5.3. If B is matroidal, dim K < oo, and s > 2, then any of the conditions (i)-(iii) inTheorem 5.2 is equivalent to the following condition:(iv) For every (some) r E (1..s), and every Y C X, every By-special compatible independent

system (C,'p) with #C = r has solutions in K(By).

Proof. (i) =* (iv): Given a By-special compatible independent system (C, p), since Byis matroidal, we have C E IE(1By) by Proposition 5.1. Hence, by Proposition 3.16, (C,W) can beextended to a By-special compatible basic system (B, V). Since every By-special system is alsoIB-special, assumption (i) implies the solvability of (B, V), and any of its solutions is necessarily inK(IBy), by Lemma 3.3. Hence, (C, W) has solutions in K(By).

(iv) == (ii): Given an independent Y E Y and 'p E K(By\y), we know, from Proposition 5.1and the fact that By is matroidal, that Y E IE. Therefore, the proof of Proposition 3.16 showsthat the equation (y, V) can be extended, step by step, to a By-special compatible basic system.Instead of performing all the s-1 steps of this extension process, we can stop after r-1 steps toobtain a By-special compatible independent system of r equations, which admits a solution inK(By), since (iv) is assumed. 13

Remark. We note that, in [DM3], (iii) is obtained under the (explicit) assumption that everycompatible independent system (C, V) is solvable and the (implicit) assumption that any B-specialcompatible independent system (C, W) actually has solutions in K. Since we are able to derive (iii)from (i), we are in effect avoiding the possibly hard task of verifying that certain systems not onlyare solvable, but have solutions in some subspace (like K). In fact, since we show that (i), (ii), and(iv) are all equivalent, we are incidentally closing the gap in the argument for (i)==(iii) in [DM3:Lemma 3.2] by showing that any B-special compatible independent system (C, W) has solutions inK if any 3-special compatible basic system is solvable.

19

6. Order-closed and minimum-closed

We now consider a weakening of the assumption that lB is matroidal. Any total order on Xinduces a corresponding partial order on ]B by the prescription

B := (b, < ... < b,) S_ B' := (b'l < ... < b') bj :_ b', =,...,...S

We recall from [BR] the following weakening of being matroidal.

Definition. We call lB order-closed in IM' if(i) M C M';(ii) IB' is matroidal;(iii) B' < B for some B' E 11' and B E ID implies that B' E ID.

We note that, for any Y C X, By is order-closed in B', if 1M is order-closed in M1'.

Lemma 6.1. Any order-closed ID has a unique minimal element, namely the unique minimalelement of the associated matroidal ID'.

Proof. Let B =: (b, < ... < b,) be a minimal element of B. If not B < B' for all B' E MB',then there would exist B' = (b, ....b) E ID' so that b < bj for some j. Assume without loss thatj is the smallest such index. Since 1M' is matroidal and both (bl,...,bj-,1 ) and (b,...,b ) are inI(B'), there would exist some k < j so that (b1,. . ., bj-1 , b') E I(IM') and, further, (bl,..., bi-1, b)could be completed to an element B" = (bj,...,bj-.,b',...) of MB' using elements from B. Sinceb' < b < bj, it would follow that B" < B, hence B" E 11 (since ID is order-closed), and this wouldcontradict the minimality of B.

In particular, given an ordering on X, any matroidal ID has a unique minimal element, whichwe will denote by

min I.

Since we often need only this consequence of order-closedness, we give it a special name.

Definition 6.2. We call ID minimum-closed in ID' if(i) M C M';(ii) ID' is matroidal;

(iii) for all Y C X, min B' E By.

Again, if ID is minimum-closed in ID', so is By in IM', for any Y C X.

Proposition 6.3. If(b, < ... < b,) := m ID, with ID minimum-closed in M', then, for any k > 0,Ik := (b,...,bk) is in E, as well as in IE(IB\b,+,) if I\b,+1 is not empty. In particular, 0 E IE.

Proof. The proof is by induction on #IB and (downward) induction on k < s, it beingtrivially true when #IB = 1 or k = s.

So assume that #IB > 1 and k < a. Then, by induction hypothesis, Ik U bk+l E IE. IflB\b&,i 0 0, let B' be its minimal element. Then Ik g B' since, otherwise, we could complete Ik toan element B" of the matroidal IB(M6 , by elements from B', and this would imply that B" < B',contradicting the minimality of B'. Therefore, Ik is the initial segment of the minimal element forIB~, ,, hence in IE(IB\b.,), by induction hypothesis (on #IM). This verifies that Ik E ]E. 3

For a minimum-closed IM, we have the following dimension formula.

20

Theorem 6.4. If IB is minimum-closed (in particular, if lB is order-closed) in IB', then

(6.5) dimK< K({B}),BEB

with equality in case the solvability condition 3.2 holds. Further, if equality holds, then any IB-special compatible system (min IB, V) is solvable.

Proof. By Proposition 6.3, 0 E IE, hence the claim here follows from part (b) of Theorem3.17, with the final statement true by the same Proposition 6.3 and Proposition 3.22. 0

In the rest of this section we make several observations relevant to minimum-closedness.

Proposition 6.6. If ]B is minimum-closed in B' and #IB > 1, then y := max{z E X(IB) : IB\0} is well-defined and replaceable.

Proof. Let B, B' E B and assume y E B. We need to find b E B' that replaces y. Ify E B', take b = y. Otherwise, since B' is matroidal and B,B' E B C B' , we can find b E B'for which B" := (B\y) U b E B'. Since Y := B" U y contains a basis (namely B) from M andB is minimum-closed, 3 must contain min B. However, min B = B" because y > b (by themaximality of y and the fact that B E B\b, hence B\b # 0). 0

Proposition 6.7. Let B C B ' for some matroidal IB'. Then, B is minimum-closed in B' if andonly if, for each Y C X of cardinality s + 1,

My19# 0 = minIB'EB.

Proof. The implication '==>' is trivial.'==': Let Y C X, and assume that B E By. Let B' := minMB'. We need to show that

B' E B, and for this we can assume without loss that Y = B U B' (since otherwise we can replaceY by its subset B U B'). We prove the desired result by induction on #Y. If #Y < s + 1, thenB' = min By E B, by assumption. Assume now that #Y > s + 1. Let y be the maximal elementin B\B' and C be the set of all elements in B which are larger than y. Then, C C B n B' by thechoice of y.

First we observe that we need only to prove that By\1 # 0. Indeed, we clearly have B' C(Y\y), and therefore B' = min IB'\ , and hence, by the induction hypothesis (applicable since Y\yhas one less element and our proof goes by induction on #Y), B' E 1B.

Since B' is matroidal, we can replace y E B by an element x E B'.We claim that x < y for any such x: if not, every element in the subset C U x of B' is larger

than y. Thus, B' contains at least #C + 1 elements which are larger than y, while B contains only#C elements larger than y, and this is impossible, since B' < B.

We now let B" := (B\y) U x. We claim that it suffices to prove that B" = min u . Indeed,B U x consists of a + 1 atoms, and contains a basis from B (viz. B), therefore, B" E IB by thehypothesis of the proposition. Since B" C Y\y, this proves that lB.\y # 0, and, by the aboveobservation, completes the proof of the proposition.

Thus, it remains to show that, with B" = B" := min ]BuX. If not, then B" < B". Since B"misses only y (from B U x), B" must miss then a larger element, and because we already provedthat z < y, this missed atom must belong to C. But then B" contains only #C - 1 atoms largerthan y, while B' contains at least #C atoms larger than y, hence B' cannot be smaller than B'.This contradicts the minimality of B', thereby completing the proof. 3

21

We now give examples to show that, in general, the implications

order-closed = minimum-closed = 0 E IE

proved and used in this section cannot be reversed even if we permit complete freedom in the choiceof the matroidal B' in which B is to be order-, resp., minimum-closed, and also permit completefreedom in the ordering.

The first example shows that a B satisfying the JE-condition need not be minimum-closed inany matroidal B' and in any ordering.

Example 6.8. Let X and B be as in Example 2.12. Since X contains no B-replaceable atom,Proposition 6.6 shows that B is never minimum-dosed regardless of the ordering we choose for X.On the other hand, we claim that 0 E IE(B). One binary tree that proves this claim goes as follows:we choose 3 (which was verified to be placeable). In B13 every atom is replaceable (since only one3-set that contains 3 is not a basis), hence -is a matroid, by Proposition 2.9. As for M\3, here 4 isplaceable (since it was so in the beginning). Again, B\3 14 is matroidal, and we need to look onlyat IB\ 3,4 , which is an order-closed subset for the ordering {1,2,7,8,5,6} and with B' containingall possible 3-subsets of X\{3,4}. 0

The next example is a strengthening of the preceding one, in that it shows that the resultson minimum-closedness are not general enough to solve the problem of §4; i.e., while Lemma 4.4asserts that 0 E IE, the stronger assertion "B is minimum-closed" is, in general, invalid for Bconsidered in §4.

Example 6.9. Let X = 123456 := {1,...,6}, and B = {12,23,13,14,25,36}. This is the Bobtained in §4 for the choice M = IRd, d = 2, and

V1 = (1,0), v2 = (0, 1), v3 = (1, 1), v 4 = (1/2, 1), v5 = (1, 1/2), v6 = (1/2, -1/2).

We now assume that B is minimum-closed in some matroidal B ' and with respect to some ordering< on X, and derive from this a contradiction.

First, due to the symmetries in B, we can assume without loss of generality that 4 < 5 < 6.Then, we consider the following three subsets of X:(a): Y = 356. The only basis in By is 36. If 35 E B', then, as 5 < 6 implies 35 < 36, By wouldnot be minimum-closed in B'. Therefore, we must have 35 0 B'.(b): Y = 346. Repeating the argument in (a), with 4 replacing 5, we conclude that 34 0 B'.Consequently, since 34,35 0 B', and B' is matroidal, also 45 % B' (since, otherwise, 3 could notbe placed into the basis 45).(c): Y = 245. Since 25 E B C B', and 45 0 B', we must have 24 E B' (otherwise, 4 cannot beplaced into 25). Thus By = {25} while By = {24,251, and since 24 < 25, By is not minimum-closed in B'. [

The final example shows that the results concerning minimum-closed B are a true general-ization of their order-closed counterparts, by exhibiting a minimum-closed B which fails to beorder-closed in any matroidal B containing it and any ordering of X.

Example 6.10. Let X = 12345 :- {1, ... , 5}, and construct B from the collection B of all 3-setsin X by omitting the four 3-sets

125, 135, 245, 345.

The resulting IB is trivially minimum-closed in B0 with respect to the natural ordering, since eachof the four sets omitted contains the largest atom, 5, hence is the minimum basis in some MB' onlyin the trivial case when it equals Y.

22

Assume now that B is order-clos-i in some matroidal M' and with respect Co some ordering< on X. We show that this assumption is untenable.

First, we claim that, with this assumption, necessarily 125 E IBM' and prove this by contradic-tion. Indeed, if 125 % 1', then necessarily 135 E 1M' since otherwise no element from 123 E B C 18'could be used to replace 4 in 145 E B C IB'. With that, comparison of 135 E IBM'\IB with 145 E Mimplies that 4 < 3. On the other hand, the same argument shows that (still under the assumption125 0 B') also 245 is necessarily in B', and, now, comparison of 245 E ]B'\B with 235 E B showsthat 3 < 4, a contradiction.

Since 2 and 3 enter the definition of B symmetrically, as do 1 and 4, it follows that necessarilyall the four sets excluded from B must be in B'. In particular, both 135 and 245 must be in B',yet, as we just saw, this leads to the contradictory conclusions that 4 < 3 and 3 < 4. We havereached a contradiction. 0

Note that all the 3-sets actively involved in this example are in BIs. We can therefore think ofthis example as being of rank 2, with the atom 5 added only in order to make B minimum-closed.With this, B itself reduces to that simplest of pathological examples in the context of this paper,namely the set

12, 34

which must fail to be order-closed since the dimension theorem fails for it in general.

7. Dimension equalities without B-conditions

In this section, we consider a nice application of the material detailed in the last two sec-tions. This application is based on the following t-condition, which we show later on to imply oursolvability condition 3.2 under the additional assumption that (7.4) holds.

Definition 7.1. We call the pair (M,t) direct if, for every B E B and every z E X\B, 4, definesa (linear) automorphism on K({B}).

For example, the pair (B, t) of Example 1.1 is direct for a generic choice of the constants{Az}=. Note that t is a (linear) automorphism on K({B}) exactly when it is 1-1 on K({B}) since,in any case, for any f E K({B}) and any b E B, tb((J) = t(tbf) = 0, hence t maps K({B})into itself.

We chose the term "direct" since, for a direct (1, t), the sum E"BEE K({B}) is direct. Thisimplies at once that, for a direct (M, t),

(7.2) dim K > E dim K({B}),BEE

since, by (1.6), we always have the inclusion

(7.3) K _? E K((B}).BED

It is also clear that, for a matroidal B, equality holds in (7.2), since the converse inequality

(7.4) dimK _ E dim K({B})BEE

holds for such 18, by virtue of Theorem 3.17(b) and the fact that every matroidal B satisfies theE-condition. However, the following theorem seems to be less obvious:

23

Theorem 7.5. Assume that the pair (B0,t) is direct and lB0 is matroidal. Then the equality

dim K= E dimK({B})BEE

holds for an arbitrary lB C lBo.

In view of (7.2), we need only to prove (7.4). We present two different arguments for (7.4), eachof which proves (7.4) in a more general setup than required here. The first approach relaxes therequirement that lB0 be matroidal, and the second approach relaxes the 1-condition of directness.

Our first generalized version of Theorem 7.5 reads as follows:

Theorem 7.6. Theorem 7.5 holds even if we assume that Bo, in lieu of being matroidal, merelysatisfies

dim K(Bo) < 1 dim K({B}).BEIBo

Thus, this stronger version of Theorem 7.5 applies to any Mo saisfying the E-condition (inparticular, to order-closed or minimum-closed BO), as well as to any fair MB0 (cf. the next section).

Proof. Consider the map

P : K(1o) --+ x K({B}) : f (x\Bf)BEBo\B.

BEB0o\0

P is well-defined (i.e., into) by Proposition 2.2, and

kerP=K(IBo) nf kertx\B D K.BcBo\B

Further, P is onto since, for any B, B' E lBo,

IXBK(B) 0, B' B;lK({B'}), B' = B,

and K({B}) C K(]Bo) for all B E lBo. It follows that

dim kerP+ Z dimK({B})=dimK(Bo) Ej dim K({B}).BEEo\B BEBo

Hence,dim K < dim ker P < E dim K({B}),

BEE

and the desired equality now follows from (7.2). 0-

The other approach for the proof of Theorem 7.5 goes as follows. We introduce a new atom xand define t,, to be the zero map. Further, using X U z as the atom set, we attempt to find a newset IB' of bases of rank s + 1 that satisfies the following three conditions:(i) (B, x) es M1' ,= B E MB.

(ii) K({ B'}) = 0, for every B' E IBM'.(iii) dim K(IB') _< EBeB, dim K({B}).

24

Proposition 7.7. IfIB has a rank-(s+l) "extension" B' that satisfies the conditions (i-iii) specifiedabove, then (7.4) holds.

Proof. We observe that, since 4, = 0, K({(B,z)}) = K({B}), while K({B'}) = 0, forany other B' E B', because of assumption (ii). Thus, (iii) leads to

dim K(IB') E dim K({B}).BEB

The claim then follows from the fact that K C K(IB') which can be observed in the followingway. Given A E A(IB'), we have two possibilities to consider: (1) z E A. In such a case LA = 0and therefore it annihilates K. (2) z 0 A. Then, since A intersects every (B, z), B E IB, it mustintersect every B E IB, hence lies in A(IB). Thus, indeed, K C K(IB'). 0

Consequently, the inequality (7.4) requiied for the proof of Theorem 7.5 is established, as soonas we demonstrate the existence of a IBM ' which satisfies (i-iii), as we do in the next proposition.

Proposition 7.8. Assume that IBo is matroidal and the pair (Bo,t) is direct. For an arbitraryIB B0 , and a new atom z V X, define

IB' := {(B,z) : B E IB} U {(B, y) : B E IBo, y E X\B}.

Then B' satisfies conditions (i-iii) above, and hence (7.4) holds (by Proposition 7.7).

Proof. The fact that (ii) is satisfied follows the directness of (IBo, t). Condition (i) triviallyfollows from the definition of B'. The last condition, (iii), will follows from Theorem 6.4, as soonas we show that B' is minimum-closed in

1 :{(B,y): B B0 , Y E (XUz)\B},

in any ordering that makes z the maximal atom.For that, we first want to show that B' is matroidal. Here, we consider two bases (B, y),

(B', z) in B' (namely, B, B' E BO), choose a E (B, y) and search for a replacement for a in (B', z).If a = y, we can replace it by any atom in (B' U z)\B. Otherwise, a E B, and in this case weconsider two different possibilities. (a): (B\a) U y E Bo. Then we can write (B, y) = (B", a), withB" E B0 , and proceed as in the previous case. (b): (B\a) U y V Bo. Since B, B' E B0 , and B0 ismatroidal, there exists b E B', for which (B\a) U b E Bo. Since we assume that (B\a) U y V IBo,we must have b ? y, and hence this b is an appropriate replacement for a.

To prove that B' is minimum-closed in B, we first observe that all the bases in B\B ' containz. Now, let Y C X U z be of cardinality > a + 1 = rank B. If Y contains a basis (B, z) E ]B\B',then B E Bo, and choosing any y E Y\(B U z), we obtain a basis (B, y) E B'. Since z is maximalin our ordering, (B, y) < (B, z), and hence (B, z) is not the minimal basis of 1B on Y. Therefore,B' is minimum-closed in B, as claimed. 0

We want to unravel a little bit the three conditions (i-iii) required of the "extension" B'.Condition (i) determines an initial set of bases in '3', and the subsequent problem is to determineB' . Condition (ii) is an t-condition, and asserts that for every basis B' E B and every b E B',tb is 1-1 on K({B\b)). This is a weakening of the assumption that (B, ) be direct, since we maytry to construct B' in such a way that lB\, is small. In contrast, Condition (iii) (which should

be regarded as a B-condition on B', because upper bound assertions do not require t-conditions)pulls the situation in the opposite direction, since the B-conditions which are known to imply upperbounds usually assert a certain "richness" property of the underlying set of bases.

25

The following example illustrates further the conditions (i-iii).

Example 7.9. Let lBo be the collection of all a-sets in X, hence lB0 is matroidal. Assume that,in some ordering < on X, the following condition is satisfied: for every B E B0 and every Y E Xwith y > b for all b E B, 4. is 1-1 on K({B}). We claim that then the inequality (7.4) holds for anarbitrary B C lB0 (i.e., an arbitrary collection of s-sets).

To verify this, we show that, given lB C 1B0 , we can construct MB' that satisfies (i-ii) (and theninvoke Proposition 7.8). We define IB' to be the collection of all (s + 1)-sets in X U z (with z anew atom and I, = 0), and define

B':= {(B,z): BEIB}u{B': B' CX, #B'=s+ 1}.

Here, condition (i) trivially holds, and condition (ii) holds, since, for the largest atom b in every(a + 1)-set B' E lBk,, Lb is assumed to be 1-1 on K({B\b}). As for condition (iii), one verifies,as in the proof for Proposition 7.8, that, With z chosen to be the largest atom in X U z, IB' isminimum-closed in lBs. 0

We close this section with a proof that, in the presence of the upper bound (7.4), directnessimplies the solvability condition 3.2.

Proposition 7.10. If (ID, I) is direct and satisfies (7.4), then every IB-special basic compatiblesystem is solvable.

Proof. Let IB' C lM. By Theorem 7.6, the assumptions imply that

dim K(ID') = E dim K({B}),BEW'

while, by directness, BE1E' K({B}) is direct and in K(IB'). Hence, altogether,

K(I') = ® K({B}).BEB'

For each B E ID, let PB be the projector on K onto K({B}) corresponding to this direct sumdecomposition of K. Since each 4y maps each of these summands K({B}) into itself, ty commuteswith each such PB. Hence, if the basic system (B', V) is special and compatible, then, for eachB E ID, the system

(7.11) tb? = PB(Pb, b E B'

is compatible and, further, PB'P6 = 0 in case b E B, since by assumption, the original system isspecial, hence

(Pb E K(I({\b) = e K{B}).BEB\b

In particular, PB'pb = 0 for all b E B', hence fB' := 0 solves (7.11) in case B = B'. In the contrarycase, at least one of the 1b involved is invertible on K({B}), hence the system has a solution inK({B}), namely

It follows that f E K given by the identity

Pa!=fB, BEMD

solves the original system. 0

26

8. A replaceability condition and s-additivity

In the last five sections we analysed the dimension of K with the aid of the atomic map, henceare now in a position to enlarge on the remarks at the end of §2 concerning the relative merits of thetwo approaches, via the DM map and via the atomic map, to the bounding of dim K. In view of theexamples discussed in this paper, the 1-conditions required for the application of the atomic map(e.g., placeability) are more likely to hold than their DM counterparts (replaceability). Secondly(and more importantly), the t-condition we use in the atomic approach (i.e., the solvability condition3.2) is weaker than the one we need for the implementation of the DM map (the s-additivity, seebelow). This means that as long as we have in hand lB-conditions which allow us to decompose 1Bthrough the atomic map (for example if 0 E IE), we can get no better results by using the DM map.This observation applies, in particular, to matroidal, order-closed, and minimum-closed structures.The notion of replaceability plays an important role in the discussion in [DDM: §6], and hencevarious results obtained there are related to those of this section. We mention, however, that themethod and the t-condition that we employ here differ from the ones used in [DDM].

In this section, we consider as an e-condition the notion of s-additivity, which was introduced in[S] and was successfully applied in [S] and [JRS1] for a matroidal and order-closed MB respectively.While we already derived, in §5 and §6, results stronger than their counterparts from [S] and [JRS1],the approach of [JRS1] can be extended to yield new dimension results which are not obtained inthe previous sections. This is due to the fact that the existence of replaceable atom (needed here)does not imply the existence of a placeabh, element. It thus requires a complementary discussionof estimates for K via the DM map and the notion of replaceability.

For this discussion, letG

denote the abelian semi-group generated by (the elements of) I(X). Since this discussion involvesthe joint kernel of an arbitrary sequence L in G, we also use the letter K for such a joint kernel,i.e., write

K(L) := nkern,IEL

and trust that the reader will have no difficulty distinguishing between K(L) for a sequence L inG and K(B) for a collection I8 of subsets of X.

Definition 8.1. We say that G is s-additive in case

dim K(L, gh) = dim K(L, g) + dim K(L, h)

for arbitrary (s-1)-sequences L and arbitrary elements g, h (in G).

Before making use of this condition, it is perhaps useful to compare it to the solvabilitycondition 3.2 placed on t in §3, as is done in the following proposition which also fully answers thequestion raised in [RJS].

Proposition 8.2. G is s-additive if and only if, for every matroidal 1B and every t: X - G withdim K < oo, any special compatible basic system is solvable.

Proof. It follows from [S: Theorem (2.4)] that G is s-additive if and only if (iii) of Theorem5.2 holds for an arbitrary matroidal 13 (of rank s) and t : X -. G. But, for each fixed matroidalB and 1: X -. G, (iii) is equivalent to (i) of Theorem 5.2 which says that any special compatiblebasic system is solvable. 3

27

We note that a comparison of s-additivity and the solvability condition 3.2 has also been madein [JRS2]. In particular, [JRS2: Theorem (2.11)] can be derived from Proposition 8.2 and Corollary5.3.

The following lemma will play an important role in the proof of the main induction step in thenext theorem. It is a variant of [JRSl: Theorem (2.1)], and employs the notation

KI(B) fn ker taAEAuiu(]B)

whenever the dependence on the particular map t needs stressing.

Lemma 8.3. Assume that y E X, H E M, and t: X -- G satisfy the following conditions:(i) dim K < oo;(ii) For arbitrary t': y U H - G,

dim Ke(BuH) = dim Ke'({B});

BE8(yUH)

(iii) y is lB-replaceable.Then, for each W E K(lBMuH), the system

ex\(@uH)? =(8.4) t\Y1)

(X\(yuH')? 0 VH' E IH\H

has solutions in K.

Proof. We apply [JRSl: Theorem (2.1)] to prove this lemma. For this, note that B C B',where B' := {B C X(]B) : #B = s} is matroidal. Then the conditions (i) and (ii) are exactly thesame as the conditions (i) and (ii) of that Theorem.

As to condition (iii), since y is B-replaceable,

H= {x : (B\y)u IB}

for each B E lByuH, as proved at the beginning of the proof of Proposition 2.8. This implies that,for all B E IByuH and for all z E X\H, (B\y) U x E B which is the condition (ii) of (JRSl:Theorem (2.1)].

The solvability of (8.4) therefore follows from that theorem. We now prove that each suchsolution f is necessarily in K. This means that we need to show that t x\H'f = 0, for all H' E HI.If H' 0 H, then already tX\(yuH')f = 0. Otherwise, H' = H, and there are two possibilities toconsider: (a) y E H. In this case ]ByuH = 0, and hence = 0, and thus tx\Hf = tX\(yuH)f =

= 0. (b) y 0 H. Here, we compute tx\Hf = tytX\(yuH)f =typ = 0, with the last equality sinceW E K(ByuH), and y is a cocircuit in ByUH. r0

We note that the proof, in Proposition 3.16, of the solvability of system (3.1) does not rely onany dimension identity involving subspaces of K. This, as we saw in §3, gives a new approach tothe study of the joint kernels and solves problems which cannot be easily solved by the other way.On the other hand, the proof of the solvability of the system (8.4) relies on condition (ii) which isa dimension identity for the subspace K(]BvuH) of K. This motivates the following definition.

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Definition 8.5. We say that B is fair if, for all Y C X with #By > 1, there exists a By-replaceable y for which ]By\y 0 0.

Note that lB is fair in case it satisfies some property which (i) is inherited by subsets (i.e.,holds with respect to any By, Y C X), and which (ii) implies, in case #B > 1, the existenceof a replaceable y E X whose corresponding B\, and MB,, are not empty. An instance of such aproperty is minimum-closedness (which is obviously inherited by subsets, and which satisfies (ii)by Proposition 6.6), and hence we have

Corollary 8.6. Every minimum-closed B is fair.

This last corollary is not extremely useful, since results on minimum-closed MR were alreadyestablished in §6 by other means, and the results below on fair structures will not improve uponthose from §6. It is more significant to note that a fair B need not be minimum-closed, sinceotherwise our main result here (Theorem 8.10) would become a weaker version of (the first partof) Theorem 6.4. The next example serves this purpose.

Example 8.7. Let M := lRd, d = 2, and let B be chosen as in §4, with respect to the present M. Itcan be checked then that B is fair. Precisely, given Y C X, if Y contains only integer vectors, thenBy is matroidal (as observed in the discussion prior to Lemma 4.3) and hence every y E X(By) isBy-replaceable. Otherwise, every non-integer Y E X(By) is By-replaceable: since such y appearssecond, hence last, in every basis B that contains it, its only contribution is to cover the non-zerointegers on the line which is not covered by the other atom in that basis, or equivalently, to covera non-zero integer a, o' on this line which is closest to the origin. Given another basis B', theremust be b E B' that covers a and this b can replace y in B. 0

On the other hand, Example 6.9 exhibits a special case of the above setup which is notminimum-closed, hence fair cannot imply minimum-closedness. As a matter of fact, since thatexample satisfies the JE-condition (as does every B of §4), we see that even the IE-condition com-bined with the assumption that B is fair does not imply minimum-closedness. Finally, the followingexample shows that B can be fair without satisfying the IE-condition. This means that the resultsin this section concerning dim K could not have been derived directly from their counterparts in§3.

Example 8.8. Let B = {123,124,125,246,147,367,467,567}. Then only 4 is placeable, and, inM\4 = {123,125,367,567}, only 3 and 5 are placeable, and, with x = 3 or 5, B\ 4 1, = {12x,x67}cannot be split any further by a placeable element. Thus, by Remark 3.15, B does not satisfy theIE-condition. On the other hand, B is fair, as one verifies directly. 0

Proposition 8.9. If 1B is fair, then

dimK < Z dimK({B}).BEE

Proof. We use induction on #B, it being trivially true in case #B < 1. So, assumethat #MB > 1. Then, there is, by assumption, a replaceable y E X(MB), and, by Proposition 2.8,this implies (2.5) which, in turn, implies that B is the disjoint union of the collections M.luH,H E I1, and IB\,. Further, since y E X(B), #B\y < #B, and since B\, # 0, by assumption, also#1ByuH < #B, H E IR. Therefore, induction together with (2.3) finishes the proof. 0

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Theorem 8.10. Suppose that 1B (of rank s) is fair and G is s-additive. Then, for arbitrary1: X -- G with dimK < oo,

dim K dimK({B}).BEE

Proof. The proof is by induction on #1B, it being trivially true when #IB _< 1. LetY E X(IB) be B-replaceable, with B\ # 0. The major induction step is to prove that the shortsequence

(8.11) 0 -- K(IB\,) -- x K(lByuH) -- 0HEll

is exact, with j defined by (1.10), but using H E IH rather than A E Amin(]S\,) to index thecomponents of j's target, as discussed at the beginning of §2.1.

Since kerj = K(IB\,), the sequence is exact at K. To prove that the sequence (8.11) is exact,it remains to show that j is onto. This follows by applying Lemma 8.3 to each H E Il. Lemma8.3 can be applied to each H E Iff, since, for each H E IH, (i) holds by assumption and (iii) holdsby the choice of y, while, finally, since y E X(]B), #IB\v < #IB and since IB\y # 0, #MBYuH < #Bfor each H, hence (ii) holds by induction hypothesis.

It follows from the exactness of the sequence (8.11) that

dim K = dim K(IB\y) + E dim K(1BuH).HEH

Since y is IB-replaceable, UHEH IBuluH is a disjoint union, by Proposition 2.8. Therefore, lB isthe disjoint union of IB\y and {I 3 yUH, H E III}. Applying the induction hypothesis to K(IB\y) and{K(M1,uH): H E 11, we have

dim K = dim K({B}).BEB

We remark that the exactness of (8.11) gives the exactness of the "Hom" of the sequence (6.31)of [DDM], and Theorem 8.10 holds if IB is 'strongly coherent' as defined in [DDM]. Interestedreaders should consult (DDM] for details.

As noted before, being fair is implied by minimum-closedness, thereby is also implied by order-closedness, and these implications are proper. Our result, thus, improves [JRS1: Theorem (2.3)],since the latter concerns order-closed sets. The argument we use, however, is essentially the one in[JRS1].

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