Draft lesson.

Distributive sublattices, duality and simultaneous splittings

Original text public domain (CC0).

Distributivity of a finite labeled family extends to the sublattice it generates. That extension makes duality transparent and permits a finite direct-sum decomposition adapted to the family when the ambient object is semisimple. For families assembled from chains, it is enough to test one member from each chain. We prove all these statements with their finite-subfamily quantifiers intact.

Let \(\mathcal A\) be abelian and let \((X_s)_{s\in S}\) be finite subobjects of \(X\). The conventions, finite meet and join constructions, signed intersection complexes and definition of a distributive family are those of Intersection complexes and the full distributivity criterion. We use its complete Theorem 3.2, its three-member criterion (4.1), and its ambient-independence and identity-cone proofs. The size conventions are those of Universes and small categories. All decompositions below specify actual subobjects and their maps into the ambient object.

1. Adjoining subobjects preserves distributivity

Assume the original labeled family is distributive. We may adjoin a new label whose subobject is zero, the whole ambient object, a duplicate member, the sum of two members, or the intersection of two members. The resulting labeled family is distributive in every case.

Zero and ambient. For a new distinguished ambient member, the deleted family is the original one, its intersection family is that same family, and every required sum/intersection identity is tautological. For a new zero member, its intersection family consists of zero subobjects in the zero ambient object. Every one of those intersection complexes is zero, and all the sum/intersection identities are zero. The full recursive theorem therefore applies, including an empty original family.

A duplicate. Let the new member equal \(X_t\). The original recursive theorem supplies a distributive family \((X_t\cap X_s)_{s\ne t}\) inside \(X_t\). Adjoin its full ambient member \(X_t\) by the case just proved. This is exactly the intersection family \((X_t\cap X_s)_{s\in S}\) needed for the duplicate. For a subset \(U\subseteq S\) not containing \(t\), the original recursive theorem gives

If \(t\in U\), both sides are \(X_t\), since the sum contains it and one right-hand term is \(X_t\). Thus all subsets satisfy (1.1), and the theorem proves the duplicate case. We have also established that intersecting the entire family with any of its own members gives a distributive family, including the member itself.

A sum. Adjoin \(V=X_a+X_b\). If the labels coincide, this is the duplicate case. Otherwise argue by induction on \(m=|S|\). For \(m=2\), the deleted family is the pair \((X_a,X_b)\), its intersection family with \(V\) is the same pair, and \(V\cap\sum_{s\in U}X_s=\sum_{s\in U}X_s\) for every subset \(U\) of the pair. The full recursive theorem proves this base case.

For \(m\ge3\), choose \(t\notin\{a,b\}\). The family with \(X_t\) deleted is distributive, and the induction hypothesis permits adjoining \(V\) to it. This is the deleted family for the enlarged family with distinguished member \(X_t\). The three-member subfamily \((X_t,X_a,X_b)\) is distributive, hence

The original recursion supplies the distributive family \((X_t\cap X_s)_{s\ne t}\). It has \(m-1\) labels, so the induction hypothesis permits adjoining the sum on the right of (1.2). This is precisely the enlarged intersection family. Finally take any subset \(U\) of the enlarged deleted labels. If it does not contain the new label, its identity is the original one. If it contains the new label, replace \(V\) in its sum by \(X_a+X_b\). Applying the original all-subset identity (1.1) at \(X_t\) gives the sum of its intersections with \(X_a,X_b\) and the remaining selected members. By (1.2) this is \((X_t\cap V)\) plus those remaining intersections. Repetitions disappear by idempotence of sum. Thus every required subset identity holds, and the full recursive theorem proves the induction step.

An intersection. Adjoin \(W=X_a\cap X_b\). We already proved that \(Z_s=X_a\cap X_s\), for all \(s\in S\), form a distributive family. This family contains \(Z_b=W\). Intersecting that entire family with its own member \(W\) gives the distributive family \((W\cap X_s)_{s\in S}\). For every subset \(U\subseteq S\), apply (1.1) first at \(X_a\) in the original family and then at \(W\) in the \(Z\)-family. Since \(W\le X_a\), this gives

The deleted original family, this whole intersection family, and every subset identity are now checked. The full recursive theorem proves the intersection case. These arguments establish the labeled statements as well as the statements about distinct subobjects; duplicate labels cause no exception.

2. The generated sublattice and its characterization

Let \(L\) be the bounded sublattice generated by the \(X_s\)'s: include \(0,X\), and close under finite sums and intersections. Explicitly start with the finite set \(L_0=\{0,X\}\cup\{X_s\}\), and set

Every \(L_r\) is finite. The union is closed under either operation because two members occur together at some finite stage. Every closed family containing the generators contains all these stages by induction, proving minimality. Equivalently its members are values of finite expressions in the generators and bounds. This construction concerns only a set of finite expressions and does not require the class of all subobjects of \(X\) to be a set.

Theorem 2.1. The original family is distributive if and only if

Proof. Suppose the original family is distributive. Adjoin its bounds by Section 1. For any finite list of members of \(L\), choose finite expressions representing them. Collect the finitely many intermediate expression values and adjoin them in an order in which the two inputs of each operation have already been adjoined. Section 1 proves at every step that the enlarged labeled family is distributive. It eventually contains the selected \(U,V,W\). The three-member criterion of the prerequisite gives (2.2).

Conversely assume (2.2). Induct on the number of labels in a family whose members lie in \(L\), to prove that every such finite family is distributive. The empty family satisfies the definition. For a nonempty family choose a distinguished \(X_t\). The deleted family has fewer labels and inherits the lattice identity, so is distributive by induction. The intersection members \(X_t\cap X_s\), \(s\ne t\), also lie in \(L\); their generated lattice is contained in \(L\), so induction gives their distributivity. The prerequisite's ambient-independence theorem allows us to regard it inside \(X_t\). Iterating (2.2) over any finite sum gives all the required identities (1.1); the empty subset gives zero on both sides. The full recursive criterion proves the family distributive. Apply this to the original labeled family. \(\square\)

The same characterization holds for the sublattice formed using only binary sums and intersections without adjoining bounds: identities involving \(0\) or \(X\) are tautological by absorption, so adjoining these two bounds introduces no new nontrivial triple identity. This includes the empty-generator case.

When (2.2) holds, \(L\) is finite. Every expression can be rewritten as a finite sum of finite intersections of generators: sums combine the lists, and intersecting two such sums distributes into their pairwise intersections, which combine the corresponding generator subsets. The empty intersection is \(X\), and the empty sum is zero. Thus every member has the form

There are at most \(2^{2^{|S|}}\) such choices. This is a finite upper bound, with redundancies allowed, and proves finiteness without presuming it in the construction (2.1).

3. Opposite kernels and the dual lattice identity

The opposite \(\mathcal A^{\mathrm{op}}\) is abelian: finite biproducts remain biproducts, kernels and cokernels exchange, and the specified coimage–image isomorphism reverses into the corresponding one. Given \(A\le X\), reverse the inclusion to an epimorphism \(X^{\mathrm{op}}\to A^{\mathrm{op}}\). Its kernel subobject is

with arrow the reverse of the actual quotient \(X\to X/A\). This is the kernel because a cokernel in \(\mathcal A\) becomes a kernel in its opposite.

The rule \(\Theta\) is an order-reversing bijection between subobjects of \(X\) and of \(X^{\mathrm{op}}\). For surjectivity, any monomorphism \(Q^{\mathrm{op}}\to X^{\mathrm{op}}\) reverses to an epimorphism \(q:X\to Q\); its kernel \(A\) makes \(q\) the cokernel of that kernel, so its reversed subobject is exactly (3.1), up to its compatible isomorphism. This also gives the inverse rule. If \(A\le B\), quotient factorization gives an epimorphism \(X/A\to X/B\), so \(\Theta(B)\le\Theta(A)\). Conversely an inclusion between those reversed quotient subobjects gives their quotient factorization; its kernel inclusion recovers \(A\le B\). Thus both orders and the inverse are verified. Greatest lower and least upper bounds therefore exchange, giving

We also need both lattice distributive laws to agree. In any lattice satisfying (2.2), its absorption and that identity give

This is the join-over-meet law. Reversing the roles of the two operations in this same derivation proves the reverse implication; their absorption laws are dual. Hence distributivity is self-dual for lattices, with no extra modularity assumption.

By (3.2), \(\Theta\) carries the generated bounded lattice of a finite family to the generated bounded lattice of its opposite-kernel family. Equations (2.2) and (3.3) transfer its distributive identity in both directions. Theorem 2.1, applied separately in \(\mathcal A\) and \(\mathcal A^{\mathrm{op}}\), now proves

This argument uses the actual reversed quotient kernels. It does not replace them by an unspecified dual-object functor.

4. Finitely many chains and all transversal choices

Suppose \(S=\bigsqcup_{i\in I}S_i\), where \(I\) is finite and every \(S_i\) is finite and nonempty. Assume the subobjects labeled in each block are pairwise comparable. A full transversal chooses one \(s_i\in S_i\) for each block.

Theorem 4.1. The entire labeled family is distributive if and only if every full transversal family \((X_{s_i})_{i\in I}\) is distributive.

Proof. The forward implication is inheritance by subfamilies. For the converse assume all full transversals are distributive and induct on the total number of labels. If every block is a singleton, the whole family is itself a transversal. If \(I\) is empty, the family is empty and distributive. Otherwise choose a nonsingleton block \(S_j\), and a label \(t\) whose subobject is minimal in that block. Such a label exists in a finite comparable family, with ties permitted. Deleting it leaves every block nonempty and reduces the total number of labels. All transversals of the deleted family are original transversals, so the induction hypothesis proves the deleted family distributive.

Consider the intersection family \((X_t\cap X_s)_{s\ne t}\) inside \(X_t\), keeping the same blocks with \(t\) removed. Each block remains a chain by monotonicity of intersection. In block \(j\), every member is the full ambient subobject \(X_t\), since minimality gives \(X_t\le X_s\). For a transversal of the intersection blocks, take its selected original labels in every other block. The original transversal consisting of these choices and \(t\) is distributive by hypothesis. Recursion at \(X_t\) proves that their intersection members in the other blocks form a distributive family. Adjoining the full ambient member \(X_t\) preserves distributivity by Section 1. Thus every transversal of the intersection-block family is distributive. It has one fewer total label, so the induction hypothesis proves the entire intersection family distributive.

It remains to check each subset identity at \(X_t\). Let \(U\subseteq S\setminus\{t\}\). If \(U\) contains a member of block \(j\), both sides of (1.1) are \(X_t\), since that selected subobject contains \(X_t\). If it does not, in each nonempty \(U\cap S_i\) choose a maximal subobject among the selected members. Their sum is exactly \(\sum_{s\in U}X_s\), and the sum of their intersections with \(X_t\) is exactly the sum of all intersections over \(U\), because intersection preserves the comparisons within a chain. These selected maxima form a partial transversal in the other blocks. Extend it to a full transversal by choosing \(t\) in block \(j\) and arbitrary members in unrepresented blocks, possible because every block is nonempty. That full transversal is distributive; its subfamily consisting of \(t\) and the selected maxima is distributive. The full recursive criterion for this subfamily gives (1.1) for the selected maxima, hence for the original \(U\). The empty subset also satisfies the identity. We have now proved the deleted-family condition, the whole intersection-family condition, and every required subset identity. The recursive criterion proves the entire original family distributive, completing the induction. \(\square\)

In particular every finite chain is distributive, and the union of any two finite chains is distributive: each of their full transversals has at most two members, so the complete two-member result in the prerequisite applies. Repeated subobjects and zero members remain covered by the proof.

5. Semisimple objects and an adapted direct-sum construction

Call \(X\) semisimple when every subobject \(Y\le X\) has a complementary subobject \(Z\) such that the sum of their specified inclusions is an isomorphism \(Y\oplus Z\to X\). This is equivalent to every monomorphism into \(X\) having a retraction, and to every epimorphism out of \(X\) having a section.

Here are the complete equivalences. A complement gives the retraction by projecting onto \(Y\). Conversely, if \(r:X\to Y\) retracts its inclusion \(i\), let \(z:Z\to X\) be \(\ker r\). The map \(1_X-ir\) is killed by \(r\), so factors as \(zp\). Then \(p i=0\), \(p z=1_Z\), \(ri=1_Y\), and \(rz=0\), by composing with the monomorphism \(z\) or with \(r\). Thus \((r,p)\) and \((i,z)\) are inverse biproduct maps. For an epimorphism \(q:X\to W\), a complement \(Z\) of its kernel makes \(q|_Z\) both monic and epic: its kernel is the zero intersection with that kernel, and its image is all of \(W\) since \(X\) is their sum. The abelian comparison makes it invertible, yielding a section of \(q\). Finally if every such \(q\) splits, apply this to \(X\to X/Y\) and a section \(s\). The map \(1-sq\) factors through its kernel \(Y\), giving a retraction of \(i_Y\) after restriction to \(Y\). This proves all three definitions agree.

Every subobject \(V\le X\) of a semisimple object is semisimple. For \(Y\le V\), a retraction \(X\to Y\) restricts to one from \(V\), so the characterization just proved applies. Every quotient of \(X\) is semisimple as well: its split quotient map identifies it with a subobject of \(X\).

Theorem 5.1. For a finite family of subobjects of a semisimple \(X\), distributivity is equivalent to the existence of a finite decomposition

under whose actual component inclusions each \(X_s\) is the sum of a subset of the components.

Proof, construction. Suppose the family is distributive. Its bounded lattice \(L\) is finite and distributive by Section 2. A nonzero element \(J\in L\) is join-irreducible if \(J=U+V\), with \(U,V\le J\) in \(L\), implies \(J=U\) or \(J=V\). Every element \(U\) of the finite lattice is a sum of join-irreducibles below it: induct on the number of elements at or below \(U\); if it is neither zero nor join-irreducible, it is a sum of two proper lower elements, to which induction applies.

For a join-irreducible \(J\), the sum \(J^-\) of all elements of \(L\) strictly below \(J\) is itself strictly below \(J\). Otherwise a finite sum of proper lower elements would equal \(J\), contradicting join-irreducibility by induction on the number of summands. It is consequently the greatest proper lower element. The object \(J\le X\) is semisimple, so choose a subobject \(Y_J\le J\) complementing \(J^-\):

List the finitely many join-irreducibles in an order in which every strictly smaller one occurs earlier. A finite partial order has such an order: repeatedly remove a minimal element; a minimal element exists because following strict decreases in a finite set cannot continue indefinitely. Let \(J\) be the next member and let \(V\) be the sum of those listed previously. By (2.2),

For the inclusion from left to right, no previously listed \(K\) contains \(J\): if it did and were distinct, the ordering would have placed \(J\) earlier. Hence every \(J\cap K\) is a proper lower element, contained in \(J^-\). For the other inclusion, \(J^-\) is the sum of join-irreducibles below it by the preceding induction; all are strictly below \(J\), thus are previously listed and lie in \(V\). They also lie in \(J\). This proves both inclusions in (5.3), including the empty preceding list.

Equation (5.2) then gives \(Y_J\cap V=0\) and \(V+J=V+Y_J\), since \(J^-\le V\). The exact signed intersection sequence in Section 1 of the prerequisite shows that the sum map \(V\oplus Y_J\to V+Y_J\) is an isomorphism: its kernel is the zero intersection and it is epic by its image presentation. Induction over the list therefore yields

since the top element is the sum of all join-irreducibles. Moreover, inducting along that same order gives \(J=\sum_{K\le J}Y_K\): its proper lower element \(J^-\) is a sum of lower join-irreducibles, whose analogous identities are already established, and (5.2) adjoins \(Y_J\). Any \(U\in L\) is a sum of the \(J\le U\), so

The sum is direct as a subfamily of (5.4), with the specified component inclusions. This gives the required simultaneous decomposition for every original \(X_s\).

Proof, converse. For a decomposition (5.1), write \(X_T\) for the component sum with labels in \(T\subseteq A\). Its inclusion consists of the corresponding biproduct injections. Coordinate projections show \(X_T\cap X_U=X_{T\cap U}\): a map factoring through both has zero coordinates outside each subset, hence zero outside their intersection, and factors uniquely through that component sum. The least-upper-bound property gives \(X_T+X_U=X_{T\cup U}\); both containments also follow from the same inclusions. The finite-set identity \(T\cap(U\cup V)=(T\cap U)\cup(T\cap V)\) therefore proves (2.2) for every coordinate subobject. The original family and its generated lattice consist of such subobjects, so Theorem 2.1 proves it distributive. Zero components give the same coordinate statements, and if \(X=0\) the empty decomposition suffices. For an empty original family with nonzero \(X\), the one-component decomposition \(X\) suffices; the construction also gives it from the lattice \(\{0,X\}\). \(\square\)

The components \(Y_J\) need not be simple objects or have finite length. The theorem constructs finitely many blocks because the family is finite, while semisimplicity supplies the complements of the particular finite-lattice subobjects.

6. Four graded exercises with complete solutions

Exercise 1 (the actual opposite kernels). In abelian groups take \(X=\mathbb Z/4\mathbb Z\) and \(A=2\mathbb Z/4\mathbb Z\). Describe the three opposite subobjects associated with \(0,A,X\), their inclusion order, and their distributivity.

Solution. They are \(\Theta(0)=X^{\mathrm{op}}\), \(\Theta(A)=(X/A)^{\mathrm{op}}\simeq(\mathbb Z/2\mathbb Z)^{\mathrm{op}}\), and \(\Theta(X)=0\). The middle inclusion into \(X^{\mathrm{op}}\) is the reverse of the quotient \(X\to X/A\), whose element map in abelian groups is reduction modulo two. It is the kernel of the reverse of \(A\hookrightarrow X\), by the cokernel universal property of that inclusion. Their order is \(0\le\Theta(A)\le X^{\mathrm{op}}\), the reverse of the original order. A finite chain is distributive by Theorem 4.1 (one block with singleton transversals); equivalently its meet and join are minimum and maximum and obey (2.2). The duality proof therefore confirms distributivity of both labeled chains, with their actual opposite-category maps specified.

Exercise 2 (two chains and the necessary hypothesis). Prove that the union of two finite chains of subobjects in any abelian category is distributive. Show why the transversal conclusion would fail without comparability inside a block.

Solution. Each full transversal of the two chains has two labels, so is distributive by the two-member result in the prerequisite. Theorem 4.1 then proves the whole union distributive, regardless of repeated values or the ambient object. If a chain is empty, discard that block; the remaining one-block result applies, and the completely empty family is distributive.

For failure without comparability, place the three lines \(\mathbb Q(1,0),\mathbb Q(0,1),\mathbb Q(1,1)\) in a single block. Every full transversal is a singleton and hence distributive. The union has the nonzero degree-one class \((-1,-1,1)\) computed in Exercise 2 of the prerequisite, so is not distributive. The lines are incomparable, exactly the hypothesis excluded by the theorem.

Exercise 3 (a distributive family without a splitting). Show that the singleton \(2\mathbb Z\le\mathbb Z\) is distributive but cannot be a coordinate subobject in any finite direct-sum decomposition of \(\mathbb Z\). Identify the failed semisimple condition.

Solution. Every singleton is distributive by the signed intersection-complex calculation in the prerequisite. If \(2\mathbb Z\) were a sum of selected components of a direct-sum decomposition of \(\mathbb Z\), projection onto those components would give a retraction \(r:\mathbb Z\to2\mathbb Z\) of the actual inclusion. A group homomorphism has \(r(1)=2m\) for an integer \(m\). The retraction condition would require \(r(2)=2\), whereas linearity gives \(r(2)=4m\). This is impossible. Thus the inclusion has no retraction and the object is not semisimple. Its quotient \(\mathbb Z\to\mathbb Z/2\mathbb Z\) also has no section, because an element of order two would have to map to a nonzero element of order two in the torsion-free group \(\mathbb Z\). This checks both equivalent failed splitting conditions of Section 5.

Exercise 4 (the finite blocks and the obstruction). In \(\mathbb Q^3\), take the three coordinate planes \(X_2=\{x=0\}\), \(X_3=\{y=0\}\), \(X_4=\{z=0\}\). Find their generated lattice, join-irreducibles and adapted decomposition. Explain why adjoining \(X_1=\{x+y+z=0\}\) cannot leave the family adapted to a finite direct-sum decomposition.

Solution. Their pair intersections give the three coordinate axes, their triple intersection is zero, and their pair sums give the full space. Their bounded generated lattice is exactly the eight spans of subsets of \(\{e_1,e_2,e_3\}\). Intersections and sums correspond to intersection and union of those subsets, so Theorem 2.1 proves distributivity. Its nonzero join-irreducibles are exactly the three axes: each has only zero properly below it, while every larger coordinate space is a sum of two proper coordinate subspaces. Thus \(J^-=0\) for each axis and the construction chooses \(Y_J=J\), giving \(\mathbb Q^3=\mathbb Q e_1\oplus\mathbb Q e_2\oplus\mathbb Q e_3\). Each plane is the sum of its two contained axes.

The four planes with the new diagonal one are distinct. A decomposition of the three-dimensional space has at most three nonzero summands, since their positive integer dimensions sum to three. With one summand there is no two-dimensional coordinate subobject; with dimensions \(2,1\) there is only one; with dimensions \(1,1,1\) there are three. Zero summands add none. Hence four distinct planes cannot all be coordinate subobjects of a decomposition. The exact \(H_2=\mathbb Q\) calculation in Counterexample 4.2 of the prerequisite gives the corresponding distributivity obstruction. The ambient vector space is semisimple: a basis of any subspace extends to a basis of \(\mathbb Q^3\), whose remaining basis vectors span a complement. A finite independent list can be extended by selecting vectors outside its span until it spans the three-dimensional space. Thus the failed hypothesis for the four-plane family is distributivity, while the ambient splitting hypothesis holds.

References

• Intersection complexes and the full distributivity criterion, Theorem 3.2, equation (4.1), Sections 1 and 5, for complete subfamily recursion, three-member tests, finite sum/intersection maps and ambient independence.
• The Stacks Project, Section 12.19, Filtrations, for the finite-chain and induced-intersection conventions. Every transversal, generated-lattice, opposite-kernel and simultaneous-splitting argument used here is proved above.