Abstract Semisimplicity

Introduction

In linear algebra semi-simplicity emerges in various contexts. They all derive from the most general theory of semisimple modules, as presented for example in BourbakiN. Bourbaki, Elements of Mathematics : algebra. chap. 8. § 3.. For example, semisimplicity of an endomorphism \(f\) of a linear space over a field \(K\) is none other than that of the structure of \(K[X]\)-module induced by \(f,\) and semisimplicity of a linear representation of a group (or monoïd) \(G\) is that of the structure of \(K[G]\)-module induced by the representation.

The theory of semisimple modules can be seen as a generalization of basic results concerning direct sums of subspaces of a linear space: these still remain for modules, provided that semisimple modules are considered. For example, the fact that every linear space is a direct sum of lines (even in infinite dimension), in other words has a basis, is a special case of the result every semisimple module is a direct sum of simple modules.

But looking more closely, we can see that the whole theory does not even make use of the module structure, but only of properties of the inclusion ordering between submodules. The notions and results that relate to semisimplicity (direct sum, semisimplicity, simplicity, isotypic components, multiplicity) survive in an "abstract" framework where modules are replaced by points of a lattice equipped with appropriate "axioms". Such a lattice generalizes the case of a lattice of submodules, and we will call it completely modular.

Throughout this post,

• \((E,≤)\) designates a poset;
• We will often illustrate our definitions and results with the case of left \(A\)-modules, we will refer to as modules for short. In this case \(A\) will be any module given once and for all.

Preliminaries

We recall that the poset \((E,≤)\) is a lattice if for every \(x,y\in E\) the set \(\{x,y\}\) admits a least upper bound (join, written \(x+y)\) and a greatest lower bound (meet, written \(x\wedge y).\) This amounts to saying that any non empty finite subset has a meet and a join.

We say that the poset \((E,≤)\) is complete if every subset has a join. Considering the empty subset and the whole set \(E\) we get: every complete poset \(E\) has a least element (written \(0),\) and a greatest element (written \(1),\) and in particular is non-empty.

Let \(F\subset E.\) The set \(G\) of lower bounds of \(F\) ha a join \(g.\) Every member of \(F\) is an upper bound of \(G\) hence is greater or equal to \(g.\) Therefore the latter is a lower bound of \(F.\) It is the greatest lower bound, since every lower bound of \(F\) lies in \(G\) hence is less than or equal to \(g.\) Hence we have \(g=\inf F.\)

In a complete lattice \(E,\) given a family \((x_i)_{i\in I},\) we will write

• \(\bigsum_{i\in I}x_i\) for the join of the set \(\{x_i\,,\>i\in I\}\);
• \(\biginf_{i\in I}x_i\) for the meet of the set \(\{x_i\,,\>i\in I\}.\)

The set of submodules of a given module, endowed with the inclusion ordering, is a complete lattice. Indeed, if \((M_i)_{i\in I}\) is a family of submodules of a given module \(M\) then the module \(\bigsum_{i\in I}M_i\) (set of sums of families with finite-support) is the join, for the inclusion ordering, of the family \((M_i)_{i\in I}.\) The meet is \(\Inter_{i\in I}M_i.\)

Let \((E,≤)\) be a lattice having a minimum \(0\) and a maximum \(1\) (for instance a complete lattice). Then \((E,+)\) is a commutative monoïd with unit \(0,\) and \((E,\wedge)\) is a commutative monoïd with unit \(1.\)

Let \(I\) be a set, and \(\Sig\) any set of subsets of \(I.\) We recall that \(\Sig\) is of finite character We translate word for word Bourbaki's terminology. See: N. Bourbaki, Éléments de mathématique : Théorie des ensembles, Fasc. de résultats, § 6, n^o 11 if, for every set \(J\subset I,\) property \(J\in\Sig\) is logically equivalent to every finite subset of \(J\) belongs to \(\Sig\). Then the set \(\Sig,\) ordered by inclusion, if non-empty, is inductive, hence, by Zorn's lemma, has a maximal element. Indeed, if \(\Sig'\subset\Sig\) is totally ordered by inclusion, by denoting \(J\) the union of sets in \(\Sig',\) for every finite subset \(X\) of \(J,\) all members of \(X\) belong to a same set \(K\in\Sig'.\) But \(\Sig'\subset\Sig,\) hence \(K\in\Sig,\) so \(X\in\Sig.\) Since the finite set \(X\subset J\) is arbitrary and \(\Sig\) has finite character, we deduce that \(J\in\Sig.\)

Totally Modular Lattices

A lattice \(E\) is called modular if it satisfies the following property, called modularity Stated by Dedekind.: \[\forall a,b,c\in E\,,\quad a≤c\quad\impl\quad (a+b)\wedge c\>=\>a+(b\wedge c)\,.\]

1. a One always has \((a+b)\wedge c≥a+(b\wedge c)\) (hence only direction \(≤\) matters in the definition of modularity). Indeed, if \(a≤c,\) then
   • \(b\wedge c≤c,\) d'où : \(a+(b\wedge c)≤a+c=c\) ;
   • \(b\wedge c≤b,\) d'où : \(a+(b\wedge c)≤a+b.\)
   We deduce \(a+(b\wedge c)≤(a+b)\wedge c.\)
2. b The lattice of submodules of a given module is precisely modular: one has to verify that if \(M,N,P\) are three submodules of given module such that \(M\subset P,\) then \[(M+N)\inter P\subset M+(N\inter P).\] If \(x\in (M+N)\inter P,\) on one hand \(x=m+n\) where \((m,n)\in M\times N,\) and on the other hand \(x\in P.\) Then \(n=x-m\in P+M=P,\) hence \[x=m+n\in M+(N\inter P).\]

Let \(E\) be a complete lattice. An element \(x\in E\) is said to be of finite type Usually called a compact element. Since we have in mind to generalize the case of a submodule lattice, we rather say of finite type, see example below. if the folowing condition is satisfied: for every family \((x_i)_{i\in I}\) of elements of \(E,\) if \(x≤\bigsum_{i\in I}x_i\) then there exists a finite set \(J\subset I\) such that \(x≤\bigsum_{i\in J}x_i.\)

In the lattice of submodules of a module, a module is of finite type in the previous sense if, and only if, it is of finite type in the usual sense, that is if there exists a finite generating family. Indeed,

• Let \((M_i)_{i\in I}\) be a family of submodules of a given module, and \(M\) a module generated by a finite family \((e_k)_{k\in K}\) and contained in \(\bigsum_{i\in I}M_i.\) For all \(k\in K\) there exists a finite set \(I_k\subset I\) such that \(e_k\) belongs to \(\bigsum_{i\in I_k} M_i.\) The set \(J=\Union_{k\in K}I_k\) is then finite, and all of the \(e_k\) are members of \(\bigsum_{i\in J}M_i\) hence \(M\) is contained in \(\bigsum_{i\in J}M_i.\)
• Conversely, suppose \(M\) is a module of finite type in the sense of above. The module \(M\) is a union, hence also a sum, of the monogenic submodules \(Ax,\) \(x\in M.\) Therefore it is the sum of a finite number of them, that is generated by a finite family.

A lattice \((E,≤)\) will be called totally modular if it satisfies both following conditions:

• \(({\rm CM}_1)\)\(E\) is a complete modular lattice
• \(({\rm CM}_2)\)Every non-zero element of \(E\) is bounded below by an element of finite type.

Every lattice of submodules is a totally modular lattice. Indeed, property (CM\(_1\)) has already been verified in and , and every non-zero module \(M\) has a an element \(x≠0\) so it contains the non-zero monogenic (hence of finite type) submodule \(Ax.\)

Let \(E\) be a totally modular lattice, \((x_i)_{i\in I}\) a family of elements of \(E,\) and \(x\in E.\) If \(x\wedge\bigsum_{i\in I}x_i≠0,\) there exists a finite set \(J\subset I\) such that \(x\wedge\bigsum_{i\in J}x_i≠0.\) Indeed, there exists an element \(t≠0\) of finite type such that \(t≤x\wedge\bigsum_{i\in I}x_i,\) and \(t\) is bounded above by \(x\) and by a finite sum \(\bigsum_{i\in J}x_i,\) hence by \(x\wedge\bigsum_{i\in J}x_i.\)

From now \((E, ≤)\) denotes a totally modular lattice

Direct Sums in a Totally Modular Lattice

Let \((x_i)_{i\in I}\) be a family of members of \(E.\) We say that the sum \(\bigsum_{i\in I}x_i\) is direct (or equivalently that the family \((x_i)_{i\in I}\) is in direct sum, or independant) if for every \(j\in I,\) we have: \[x_j\wedge\bigsum_{i\in I\setminus\{j\}}x_i=0\,.\]

• Every family whose indexing set has \(0\) or \(1\) element is in direct sum.
• A family \((x,y)\) with two elements is direct if and only if, \(x\wedge y=0.\) In this case we will say that \(x\) and \(y\) are in direct sum, or independant.
• If a family is in direct sum then any subfamily is itself in direct sum.
• According to reamark , a family is in direct sum if, and only if, every finite subfamily is in direct sum.

A sum \(\bigsum_{i\in I}x_i\) when direct is denoted \(\bigoplus_{i\in I}x_i\) (and \(x\oplus y\) for two elements). We say that a family \((x_i)_{i\in I}\) is complementary in \(z\) if \(z=\bigoplus_{i\in I}x_i.\) We say that \(x\) is a complement or a direct summand of \(y\) in \(z\) if \(z=x\oplus y.\)

Let \(x'\) be such a complement, that is \(x=x'\oplus (x\wedge y).\)

• From \(x'\wedge y≤x'\) and \(x'\wedge y≤x\wedge y\) we deduce: \(x'\wedge y≤x'\wedge(x\wedge y)=0.\)
• We also have: \(x+y=x'+(x\wedge y)+y=x'+y.\)

Fix \(j\in I.\) By taking \(a=x_j,\) \(b=\bigsum_{i≠j}x_i,\) and \(c=y_j\) in modularity property, we obtain \[\eqalign{y_j&=\Big(\bigoplus_{i\in I}y_i\Big)\wedge y_j≤\Big(\bigsum_{i\in I}x_i\Big)\wedge y_j\cr &=x_j+\Big(\big(\bigsum_{i≠j}x_i\big)\wedge y_j\Big)\qquad\hbox{(modularity)}\cr &≤x_j+\Big(\big(\bigoplus_{i≠j}y_i\big)\wedge y_j\Big)=x_j+0=x_j}\]

• We have: \(x\wedge(x'\wedge y)≤x\wedge x'=0.\)
• By applying modularity to the triple \((x,x',y)\) we have: \[x+(x'\wedge y)=(x+x')\wedge y=z\wedge y=y\,.\]

Since \(x≤x+y\) we get: \[\eqalign{(y+z)\wedge x&=(y+z)\wedge (x+y)\wedge x\cr &=(y+(z\wedge(x+y)))\wedge x\quad\hbox{(by applying modularity to \((y,z,x+y)\))}\cr &=(y+0)\wedge x\cr &=y\wedge x.}\]

According to previous theorem we have \[x≤y+z\ssi (y+z)\wedge x=x\ssi y\wedge x=x\ssi x≤y\,,\]

According to previous theorem we have: \[y\wedge x=0\ssi (y+z)\wedge x=0\,.\]

One has to prove that \(\bigoplus_{i\in J} x_i\wedge\bigoplus_{i\in K} x_i=0.\) According to remark , we can assume \(J\) is finite. We then proceed by induction on \(n=\card(J)\) (with universal quantification on \(K).\) If \(\card J\in\{0,1\}\) the result is obvious. If it's true when \(\card J=n≥1\) consider a subset \(J\) of size \(n+1\) not meeting \(K,\) and some \(j\in J.\) By induction assumption (applied with \(K\union\{j\}\) instead of \(K),\) we have \[\Big(\bigoplus_{i\in J\setminus\{j\}} x_i\Big)\wedge\Big(x_j+\bigoplus_{i\in K} x_i\Big)=0\,.\] Moreover we have \(x_j\wedge\bigoplus_{i\in K} x_i=0\) according to case \(n=1.\) According to exchange property applied to triple \(\Big(\bigoplus_{i\in K} x_i\,,\>x_j\,,\bigoplus_{i\in J\setminus\{j\}} x_i\Big),\) we deduce that \(\bigoplus_{i\in K} x_i\) is in direct sum with \(x_j+\bigoplus_{i\in J\setminus\{j\}} x_i\) that is with \(\bigoplus_{i\in J} x_i.\)

• Assume (i). For all \(k\in K,\) \((x_i)_{i\in I_k}\) is in direct sum as a subfamily of \((x_i)_{i\in I}.\) Moreover, according to lemma above, \(\bigoplus_{i\in I_k}x_i\wedge \bigsum_{l≠k}\bigoplus_{i\in I_l}x_i=\bigoplus_{i\in I_k}x_i \wedge\bigoplus_{i\in I\setminus I_k}x_i =0,\) that is: \(\big(\bigoplus_{i\in I_k}x_i\big)_{k\in K}\) is in direct sum. Hence we have (i) \(\impl\) (ii).
• Assume (ii). Let \(j\in I.\) One has to verify that \(x_j\wedge\bigsum_{i\in I\setminus\{j\}}x_i=0.\) Let \(k\in K\) such that \(j\in I_k.\) Since family \((x_i)_{i\in I_k}\) is in direct sum one has \[x_j\wedge\bigoplus_{i\in I_k\setminus\{j\}}x_i=0\,.\] Moreover since \(\Big(\bigoplus_{i\in I_k}x_i\Big)_{k\in K}\) is in direct sum we have \[\Big(x_j+\bigoplus_{i\in I_k\setminus\{j\}}x_i\Big)\wedge \bigsum_{i\in I\setminus I_k}x_i=0\,.\] By exchange property applied to triple \(\Big(x_j\,,\bigoplus_{i\in I_k\setminus\{j\}}x_i\,,\bigsum_{i\in I\setminus I_k}x_i\Big)\) we obtain the desired result.

Semisimplicity in Totally Modular Lattices

We say that an element \(x\in E\) is simple if \(x≠0\) and \(x\) is minimal. We say that \(x\) is indecomposable if \(x≠0\) an \(x\) is not a direct sum of two non-zero elements. We say that \(x\) is semisimple if every lower bound of \(x\) has a complement in \(x.\)

1. a Let \(x\in E.\) Then \(x\) is simple if, and only if, it is both indecomposable and semi-simple.
2. b Every simple element is of finite type. Indeed, such an element \(x\) is non-zero hence bounded below by a non-zero element \(t\) of finite type (axiom (CM2)), and simplicity of \(x\) implies \(t=x.\) Note that in the module case, a simple module is not only of finite type but monogenic. In an abstract totally modular lattice, being monogenic, unlike having finite type character, in not definable.

Let \(x,y\in E\) such that \(y\) is semisimple and \(x≤y.\) We have to show that every element \(x'≤x\) has a complement in \(x.\) By semisimplicity of \(y\) there exists a complement \(z\) of \(x'\) in \(y.\) According to prop. , \(x\wedge z\) is then a complement of \(x'\) in \(x.\)

We say that \(E\) is semisimple if every element \(x\in E\) is semisimple. According to the previous proposition, this is equivalent to \(1=\max(E)\) being semisimple.

Let \(x≠0\) be semisimple. According to axiom (CM2) \(x\) has a lower bound \(t≠0\) of finite type.

• We show that the set \(A\) of strict lower bounds of \(t\) is inductive. Let \(B\) be a totally ordered subset of \(A,\) and \(m=\sup(B)=\bigsum_{x\in B}x.\) We have to see that \(m\in A,\) that is \(m≠t.\) If we had \(m=t,\) since \(t\) is of finite type there would exists a finite set \(B'\subset B\) such that \(t≤\bigsum_{x\in B'}x=\sup (B').\) But since \(B'\) is totally ordered we would have: \(t≤\sup (B')\in B',\) in contradiction with elements of \(B'\) being strict lower bounds of \(t.\)
• Using Zorn's lemma we deduce that \(A\) has a maximal element, that is: \(t\) has a maximal strict lower bound \(t'.\) Since \(x\) is semisimple, so is \(t\) (see ) Hence there exists an element \(t''\) such that \(t=t'\oplus t''.\) We will see that \(t''\) answers the problem.
• We already have \(t''≤x.\) We have now to see that \(t''\) is simple. We already have \(t''≠0\) since \(t'≠t.\) It remains to show that every element \(s\) such that \(0\lt s≤t''\) equals \(t''.\) We have \(s\wedge t'≤t''\wedge t'=0≠s.\) Therefore \(s\) is not a lower bound of \(t',\) hence \(t'\lt t'+s.\) By maximality of \(t'\) we deduce \(t'+s=t.\) By modularity applied to the tripls \((s,t',t''),\) we deduce \[s=s+0=s+(t'\wedge t'')=(s+t')\wedge t''=t\wedge t''=t''\,.\]

Decomposition into Simple Elements

The following result is in some way the abstract version of the incomplete basis theorem in linear algebra.\[ \]

Let \(\Sig\) be the set of subsets \(K\subset I\) such that the sum \(\bigsum_{i\in K}x_i \) is direct and is independant from \(y.\) According to remark , the set \(\Sig\) has finite character, hence admits a maximal element \(J\) (see ; \(\Sig\) is not empty since \(\vide\in\Sig\)). By definition of \(\Sig,\) the sum \(\bigsum_{i\in J}x_i \) is direct and independant from \(y.\) Put \(z=y\oplus\Big(\bigoplus_{i\in J}x_i\Big);\) We have now to see that \(z=\bigsum_{i\in I}x_i.\) For all \(i\in I,\) by maximality of \(J,\) \(z\) and \(x_i\) are not independant, hence, since \(x_i\) is simple, \(x_i\wedge z=x_i,\) that is \(x_i≤z,\) Hence we have \(\bigsum_{i\in I}x_i≤z.\) The reverse inequality is obvious.

• (i) \(\impl\) (ii) : just take \(y=0\) in the previous theorem.
• Assume (ii). Let \(x'≤x.\) By taking \(y=x'\) is the previous lemma we obtain: \(x'\) has a complement in \(x.\) Hence we have (ii) \(\impl\) (iii).
• Assume (iii). Let \(y\) be the sum (join) of simple elements below \(x.\) We must verify that \(y=x.\) Since \(x\) is semisimple there exists \(y\in E\) such that \(x=y\oplus y'.\) Assume, by contradiction, \(y\lt x.\) Then \(y'≠0.\) According to \(y'\) is semi-simple, hence according to \(y'\) is bounded below by a simple element \(s.\) But then \(s≤y,\) hence \(s≤y\wedge y',\) in contradiction with \(y\) and \(y'\) being independant.

\(\bullet\) (i) \(\impl\) (ii) : just take \(y=0\) in the previous theorem.

\(\bullet\) Assume (ii). Let \(x'≤x.\) By taking \(y=x'\) is the previous lemma we obtain: \(x'\) has a complement in \(x.\) Hence we have (ii) \(\impl\) (iii).

\(\bullet\) Assume (iii). Let \(y\) be the sum (join) of simple elements below \(x.\) We must verify that \(y=x.\) Since \(x\) is semisimple there exists \(y\in E\) such that \(x=y\oplus y'.\) Assume, by contradiction, \(y\lt x.\) Then \(y'≠0.\) According to \(y'\) is semi-simple, hence according to \(y'\) is bounded below by a simple element \(s.\) But then \(s≤y,\) hence \(s≤y\wedge y',\) in contradiction with \(y\) and \(y'\) being independant.

Just take the sum of all simple (os semisimple) elements below \(x.\)

Cardinality Results

The following theorem can be viewed as the abstract version of the linear algebra result a family of independant vectors cannot have greater size than any generating family.\[ \]

We proceed by induction on \(\card(I).\)
If \(\card(I)=0\) then \(z\oplus\bigoplus_{j\in J}y_j≤z,\) that is\(\bigoplus_{j\in J}y_j≤z.\) But then \(\bigoplus_{j\in J}y_j=z\wedge \bigoplus_{j\in J}y_j\) is null since the sum is direct. Since the \(y_j\)'s are non-zero, the only possible case is \(J=\vide,\) that is \(\card(J)=0.\)
Now let \(n\in\nmat^*\) and assume the result is given when \(\card(I)=n-1.\) Now assume we have \(z\oplus\bigoplus_{j\in J}y_j≤z+\bigsum_{i\in I}x_i\) with \(\card(I)=n,\) the \(y_j\)'s non-zero, and the \(x_i\)'s simple. Fix \(k\in I.\)
1^st case: For all \(j\in J,\) \(x_k\) is not bounded above by \(y_j+z+\bigsum_{i\in I\setminus\{k\}}x_i.\)
Since \(x_k\) is simple, it is in direct sum with the latter. But by assumption \(y_j≤z+\bigsum_{i\in I}x_i.\) Using Corollary , we obtain that \(y_j≤z+\bigsum_{i\in I\setminus\{k\}}x_i.\) Since \(j\in J\) is arbitrary, we deduce that \[z\oplus\bigoplus_{j\in J}y_j≤z+\bigsum_{i\in I\setminus\{k\}}x_i\,.\] Since \(\card\l(I\setminus\{k\}\r)=n-1,\) by induction hypothesis we get that \(J\) is finite and \[\card(J)≤n-1≤n=\card(I)\,.\] 2^nd case: There exists \(l\in J\) such that \(x_k≤y_l+z+\bigsum_{i\in I\setminus\{k\}}x_i.\)
Then \[ z\oplus\bigoplus_{j\in J\setminus\{l\}}y_j ≤z\oplus\bigoplus_{j\in J}y_j≤z+\bigsum_{i\in I}x_i≤y_l+z+\bigsum_{i\in I\setminus\{k\}}x_i\,.\] Since \(\card\l(I\setminus\{k\}\r)=n-1,\) by induction hypothesis (applied to \(y_l+z\) instead of \(z),\) we deduce that \(\card\l(J\setminus\{l\}\r)≤n-1,\) that is \(\card(J)≤n.\)

By exchanging families \((x_i)\) and \((y_j)\) it's enough to prove the inequality \(\card(J)≤\card(I).\)
1^st case: The set \(I\) is finite. Since the \(x_i\)'s are semisimple and the \(y_j\) non-zero, by applying previous theorem with \(z=0\) we obtain: \(J\) is finite and \(\card(J)≤\card(I).\)
2^nd case: The set \(I\) is infinite. For every \(i\in I,\) there exists a finite subset \(J(i)\) s.t. \(x_i\wedge \bigoplus_{j\in J(i)}y_j≠0\) (see Rem. ), and consequently, since \(x_i\) is simple, \(x_i≤\bigoplus_{j\in J(i)}y_j.\) The sum of the \(x_i\)'s equals \(x,\) which gives \(\Union_{i\in I}J(i)=J.\) Since \(I\) is infinite and the \(J(i)\)'s are finite, we conclude that \[\card(J)=\card\l(\Union_{i\in I}J(i)\r)≤\card(I).\]

Isomorphic Elements

To be able to extend to the abstract case the notion of isotypic module (semi-simple module of which all simple submodules are isomorphic), it is necessary to have a notion of isomorphic elements. However, it is probably not possible to define, from the order relation in a completely abstract modular lattice, a binary relation that would play the role of the isomorphism relation between submodules of a module. For this purpose we will enrich the lattice structure so as to have a notion of isomorphism between elements. Let's start by noticing (it's classic and easy) that two submodules of a module \(M\) having a common complement are isomorphic (the converse is false, as shown by the example of a hyperplane of an infinite dimensional vector space). Note also that an isomorphism between two modules induces an isomorphism of ordered sets between the two lattices of submodules.

We check that if \(z\) is a semisimple upper bound of \(\{x,y\}\) the following conditions are equivalent:

1. \(x\) and \(y\) have a common complement in \(z\);
2. \(x\) and \(y\) have a common complement in \(x+y.\)

Assume (i). Let \(t\) be a common complement of \(x\) and \(y\) in \(z.\) Since \(z\) is an upper bound of \(x+y,\) according to prop. , \(t\wedge(x+y)\) is a common complement of \(x\) and \(y\) in \(x+y.\)
Assume (ii). Let \(t\) be a common complement of \(x\) and \(y\) in \(x+y,\) and \(z\) a semisimple upper bound of \(\{x,y\},\) that is of \(x+y.\) By using semisimplicity of \(z,\) \(x+y\) has a complement \(u\) in \(z.\) Then \(t\oplus u\) is a common complement of \(x\) and \(y\) in \(z.\)

In a totally modular lattice \(E,\) we will write \([0,x]\) for the set of lower bounds of \(x.\) For instance, if \(E\) is the lattice of all submodules of \(M\) and \(M'\in E,\) the set \([0,M']\) is the set of all submodules of \(M'.\)

Let \((E,≤)\) be a totally modular lattice. We call notion of isomorphism on \(E\) every set \(G\subset E\times E\) satisfying the four following conditions:

• \(({\rm IS}_1)\)\(G\) is the graph of an equivalence relation;
• \(({\rm IS}_2)\)If \((x,y)\in G,\) there exists an isomorphism \(\phi\) of posets from \([0,x]\) onto \([0,y]\) s.t. for every \(x'≤x,\) \((x',\phi(x'))\in G\);
• \(({\rm IS}_3)\)If \(x,y\in E\) are two semisimple elements having a common complement, then \((x,y)\in G\);
• \(({\rm IS}_4)\)If two families \((x_i)_{i\in I}\) and \((y_i)_{i\in I}\) are in direct sum (finite or infinite) and if, for all \(i\in I,\) \(\l(x_i, y_i\r)\in G,\) then \(\Big(\bigoplus_{i\in I}x_i\,,\,\bigoplus_{i\in I}y_i\Big)\in G.\)

When \((x,y)\in G\) we say that \(x\) and \(y\) are isomorphic.

The lattice of submodules of some module \(M,\) equipped with the usual isomorphism relation between submodules, satisfies axioms \(({\rm IS}_1)\) to \(({\rm IS}_1)\).

From now we fix a triple \((E,≤,G)\) where \((E,≤)\) is a totally modular lattice and \(G\) a notion of isomorphism.

Since \(x\) is a sum of simple elements it is semisimple, hence \(z\) has a complement \(y\) in \(x.\) According to th. , there exists \(J\subset I\) s.t. the sum \(\bigsum_{i\in J} x_i\) is direct and is a complement of \(y\) in \(x.\) Then \(\bigsum_{i\in J} x_i\) is isomorphic to \(z,\) since they both have \(y\) has a complement in \(x\) (axiome IS\(_3).\)

According to (IS\(_2)\) if two elements \(x\) and \(y\) are isomorphic then the posets \([0,x]\) and \([0,y]\) are isomorphic. Then just notice that the notions of nullity (resp. simplicity, resp. indécomposability, resp. semisimplicity) of an element \(x\) can be expressed in terms of the sublattice \([0,x].\)

Let \(z≤x\) be simple. There exists a set \(J\subset I\) s.t. the sum \(\bigsum_{i\in J} x_i\) is direct and isomorphic to \(z\) (Prop. ). Then \(\bigsum_{i\in J} x_i\) is simple (Prop. ) therefore indecomposable, hence \(\card J=1.\)

Decomposition into Isotypical Components

• (i) \(\ssi\) (ii) \(\ssi\) (iii) follows from theorem and corollary .
• Assume (i). Let \(y,z≤x.\) Since \(x\) is semisimple, so are \(y\) and \(z.\) There exists families \((y_i)_{i\in I}\) and \((z_j)_{j\in J}\) of simple elements such that \(y=\bigoplus_{i\in I}y_i\) and \(z=\bigoplus_{j\in J}z_j.\) Since the \(y_i\)'s and \(z_j\)'s are lower bounds of \(x,\) they all are isomorphic. Moreover one of the two sets \(I\) and \(J\) (the one with smallest cardinal) is in bijection with a subset of the other. Axiom (IS\(_4)\) then implies that \(y\) or \(z\) is isomorphic with a lower bound of the other. Hence we have (i) \(\impl\) (iv).
• Assume (iv). Let \(y,z≤x\) be simple. One of them (for instance \(y)\) is isomorphic to a lower bound of the other (say some \(z'≤z).\) We have \(y≠0\) hence \(z'≠0.\) Since \(z\) is simple, \(z=z'\) is isomorphic to \(y.\) Hence we have (iv) \(\impl\) (i).

(obvious) every lower bound of an isotypical element is isotypical.

From now we denote by \(\mskip 2mu\Sig\) the set of all isomorphism classes of simple elements of \(E.\) Let \(S\in\Sig.\) We say that an isotypical element \(x\) is of type \(S\) if all simple lower bounds of \(x\) belong to \(S.\)

Every isotypical element has a unique type, except 0 which is of type \(S\) for every \(S\in\Sig.\)

Let \(S\in\Sig.\) Any simple lower bound of \(x_S\wedge\bigsum_{S'\in\Sig\setminus\{S\}}\!S'\) is both of type \(S\) and, according to cor. , of a type \(S'≠S,\) which is absurd. We deduce \(x_S\wedge\bigsum_{S'\in\Sig\setminus\{S\}}\!S'=0.\)

Just take the sum of all simple lower bounds of \(x\) that have \(S\) for isomorphism class, that is \(\bigsum_{s\in S,\, s≤x}\!\!s.\)

\(\ci(y,S)\) is a lower bound of both \(y\) and \(\ci(x,S).\) Hence \(\ci(y,S)≤y\wedge \ci(x,S).\) Moreover \(y\wedge \ci(x,S)\) is an isotypical lower bound of \(y\) of type \(S.\) The greatest of them is \(\ci(y,S),\) so \(y\wedge \ci(x,S)≤\ci(y,S)\)

We already know (Prop. ) that every sum of isotypical elements of distinct types is direct. Moreover \(\bigoplus_{S\in\Sig}\ci(x,S)\) is a semisimple lower bound of \(x.\) We deduce the inequality \[\bigoplus_{S\in\Sig}\ci(x,S)≤x'.\tag{1}\] If \(x'\) is the sum of a family \((x_S)_{S\in\Sig}\) where each\(x_S\) is isotypical of type \(S,\) the inequality \((1)\) can be rewritten \[\bigoplus_{S\in\Sig}\ci(x,S)≤\bigsum_{S\in\Sig}x_S\,.\] Theorem then gives \(\ci(x,S)=x_S\) for all \(S\in\Sig,\) which proves the second part of the theorem. Finally since there exists at least one such family \((x_S)\) (by using th. and grouping isomorphic simple elements) we also deduce the inequality \(\bigoplus_{S\in\Sig}\ci(x,S)=x'.\)

Multiplicity

Let \(x\in E\) and \(S\in\Sig.\) The isotypical component \(\ci(x,S)\) can be splitted in a direct sum \(\bigoplus_{i\in I}s_i\) where the \(s_i\)'s are simple and belong to \(S.\) According to , the cardinal (finite or infinite) of \(I\) does not depend of the chosen decomposition. We call it the multiplicity of \(S\) in \(x\) and denote it by \(\mu_S(x)\)

Let \(x\in E.\) The family \((\mu_S(x))_{S\in\Sig}\) is called the signature of \(x.\)

If \(x,y\in E\) are two semisimple elements with same signature, according to th. and and axiom (IS4), \(x\) and \(y\) are isomorphic. Conversely, if \(x\) and \(y\) are isomorphic then according to axiom (IS2) they have same signatures. Hence we obtain:

In the general case of elements which are not necessarily semisimple, the signature only provides a charaterization of semisimple parts: \(x\) and \(y\) have the same signature if, and only if, their semisimple parts are isomorphic.