On irreducible algebraic sets over linearly ordered semilattices II

TL;DR

This paper investigates the structure of solution sets of equations over linearly ordered semilattices, identifying irreducible components and calculating their average number across all equations with a given number of variables.

Contribution

It provides a method to find irreducible components of equations over linearly ordered semilattices and computes the average number of such components for equations with n variables.

Findings

Identified irreducible components of solution sets for equations over linearly ordered semilattices.

Calculated the average number of irreducible components for all equations with n variables.

Abstract

Equations over linearly ordered semilattices are studied. For any equation $t(X)=s(X)$ we find irreducible components of its solution set and compute the average number of irreducible components of all equations in $n$ variables.

Equations

$$a_i\le a_j\iff a_i{a}_j=a_i.$$

$$Eq(n)=\bigcup _{(k_1,k_2)\in K_n}Eq(k_1,k_2,n),$$

$$K_n=\{(k_1,k_2)\mid k_1+k_2\le n\}\setminus \{(0,n),(n,0)\}.$$

$$\#Eq(k_1,k_2,n)=\left(\genfrac{}{}{0ex}{}{n}{k_1}\right)\left(\genfrac{}{}{0ex}{}{n-k_1}{k_2}\right).$$

$$\#Eq(n)=3^n-2.$$

$$Y=Y_{1}\cup Y_{2}\cup\ldots\cup Y_{m},\quad Y_{i}\nsubseteq Y_{j}\mbox{ for all $i\neq j$},$$

$$t(X)\sim_{Y}s(X)\Leftrightarrow t(P)=s(P)\mbox{ for any point $P\in Y$}.$$

$$Y_1=\mathrm{V}(\{x_1\le x_2\le x_3\}),Y_2=\mathrm{V}(\{x_3\le x_2\le x_1\})$$

$$\Gamma (Y_1)=⟨x_1,x_2,x_3\mid x_1\le x_2\le x_3⟩\cong L_3,$$

$$\Gamma (Y_2)=⟨x_1,x_2,x_3\mid x_3\le x_2\le x_1⟩\cong L_3.$$

$$\overline{\mathrm{Irr}}(3)=\frac{6+2(2+2+2+2+2+2+3+3+3+4+4+4)}{25}=\frac{72}{25}=2.88$$

$$Y_1=\mathrm{V}(x_1\le x_2\le x_3),Y_2=\mathrm{V}(x_1\le x_3\le x_2),Y_3=\mathrm{V}(x_2=x_3\le x_1)$$

$${\sigma }_1=\left(\begin{array}{ccc}1& 2& 3\\ 1& 2& 3\end{array}\right),{\sigma }_2=\left(\begin{array}{ccc}1& 2& 3\\ 1& 3& 2\end{array}\right),{\sigma }_3=\left(\begin{array}{ccc}1& 2& 3\\ 2& 3& 1\end{array}\right).$$

$$Y_{\sigma }=\mathrm{V}(\bigcup _{i=1}^{n-1}\{x_{\sigma (i)}\le x_{\sigma (i+1)}\})$$

$$Y_{\sigma }=\mathrm{V}(\{x_{\sigma (1)}=x_{\sigma (2)}\}\bigcup _{i=2}^{n-1}\{x_{\sigma (i)}\le x_{\sigma (i+1)}\})$$

$$\Gamma(Y_{\sigma})\cong\begin{cases}L_{n},\mbox{ if $\chi({\sigma})=1$}\\ L_{n-1},\mbox{ if $\chi({\sigma})=2$}\end{cases}$$

$$x_{\sigma (1)}\le x_{\sigma (2)}\le \dots x_{\sigma (n)}\text{mbox}if\chi (Y_{\sigma })=1$$

$$x_{\sigma (1)}=x_{\sigma (2)}\le \dots x_{\sigma (n)}\text{mbox}if\chi (Y_{\sigma })=2.$$

$$Y=\bigcup _{\chi (\sigma )\in \{1,2\}}{Y}_{\sigma }.$$

$$p_{\sigma (1)}\le p_{\sigma (1)}\le \dots \le p_{\sigma (n)},$$

$$x_{\sigma (1)}\in \mathrm{Var}(t)\cap \mathrm{Var}(s)\Rightarrow t(P)=s(P)=p_{\sigma (1)}\Rightarrow P\in \mathrm{V}(t(X)=s(X)).$$

$$p_{\sigma (i)}=a_i\text{mbox}if\chi (\sigma )=1,$$

$$p_{{\sigma}(i)}=\begin{cases}a_{i},\;2\leq i\leq n\\ a_{2},\;i=1\end{cases}\;\mbox{ if $\chi({\sigma})=2$.}$$

$$P_1=(a_1,a_2,a_3),P_2=(a_1,a_3,a_2),P_3=(a_3,a_2,a_2),$$

$$P_1\in Y_1\setminus (Y_2\cup Y_3),P_2\in Y_2\setminus (Y_1\cup Y_3),P_3\in Y_3\setminus (Y_1\cup Y_2)).$$

$$(n-k_1-k_2)(n-1)!$$

$$k_1{k}_2(n-2)!$$

$$\mathrm{Irr}(k_1,k_2,n)=(n-k_1-k_2)(n-1)!+k_1{k}_2(n-2)!$$

$$\begin{gathered} \overline{\mathrm{Irr}}(n)=\frac{{\sum }_{(k_1,k_2)\in K_n}\#Eq(k_1,k_2,n)\mathrm{Irr}(k_1,k_2,n)}{\#Eq(n)}= \\ \frac{{\sum }_{k_1=0}^{n-1}{\sum }_{k_2=0}^{n-k_1}\#Eq(k_1,k_2,n)\mathrm{Irr}(k_1,k_2,n)-\#Eq(0,n,n)\mathrm{Irr}(0,n,n)}{\#Eq(n)}. \end{gathered}$$

$$\mathrm{Irr}(0,n,n)=(n-0-n)(n-1)!+0n(n-2)!=0,$$

Full text

1 Introduction

This paper is devoted to the following problem. One can define a notion of an equation over a linearly ordered semilattice $L_{l}=\{a_{1},a_{2},\ldots,a_{l}\}$ (the formal definition of an equation is given below in the paper). A set $Y$ is algebraic if it is the solution set of some system of equations over $L_{l}$. Let us consider an equation $t(X)=s(X)$ in $n$ variables over $L_{l}$, and $Y$ be the solution set of $t(X)=s(X)$. One can find algebraic sets $Y_{1},Y_{2},\ldots,Y_{m}$ such that $Y=\bigcup_{i=1}^{m}Y_{i}$. One can decompose each $Y_{i}$ into a union of other algebraic sets, etc. This process terminates after a finite number of steps and gives a decomposition of $Y$ into a union of irreducible algebraic sets $Y_{i}$ (the sets $Y_{i}$ are called the *irreducible components *of $Y$). Roughly speaking, irreducible algebraic sets are “atoms” which form any algebraic set. The size and the number of such “atoms” are important characteristics of the semilattices $L_{l}$, since there are connections between irreducible algebraic sets and universal theory of linearly ordered semilattices (see [1]). Moreover, the number of irreducible components was involved in the estimation of lower bounds of algorithm complexity (see [2] for more details).

In this paper we assume $n\leq l$ (i.e. the order of the semilattice $L_{l}$ is not less than the number of variables in $t(X)=s(X)$) and study (Section 4) the properties of algebraic sets over $L_{l}$. Precisely, for any equation $t(X)=s(X)$ in $n$ variables we count the number of irreducible components (see (8)), and in Section 5 we count the average number $\overline{{\mathrm{Irr}}}(n)$ of irreducible components of the solution sets of equations in $n$ variables.

Remark that the current paper is the sequel of [3], where we solved the similar problems assuming $n>l$ (we discuss this case in Remark 2 below).

2 Main definitions

Let $L_{l}=\{a_{1},a_{2},\ldots,a_{l}\}$ be the linearly ordered semilattice of $l$ elements and $a_{1}<a_{2}<\ldots<a_{l}$. The multiplication in $L_{l}$ is defined by $a_{i}\cdot a_{j}=a_{\min(i,j)}$. Obviously, the linear order on $L_{l}$ can be expressed by the multiplication as follows

[TABLE]

A term $t(X)$ in variables $X=\{x_{1},x_{2},\ldots,x_{n}\}$ is a commutative word in letters $x_{i}$.

Let ${\mathrm{Var}}(t)$ be the set of all variables occurring in a term $t(X)$. Following [1], an equation is an equality of terms $t(X)=s(X)$. Below we consider inequalities $t(X)\leq s(X)$ as equations, since $t(X)\leq s(X)$ is the short form of $t(X)s(X)=t(X)$. Notice that we consider equations as ordered pairs of terms, i.e. the expressions $t(X)=s(X)$, $s(X)=t(X)$ are different equations. Let $Eq(n)$ denote the set of all equations in $X=\{x_{1},x_{2},\ldots,x_{n}\}$ variables (we assume that each $t(X)=s(X)\in Eq(n)$ contains the occurrences of all variables $x_{1},x_{2},\ldots,x_{n}$). An equation $t(X)=s(X)\in Eq(n)$ is said to be a $(k_{1},k_{2})$-equation if $|{\mathrm{Var}}(t)\setminus{\mathrm{Var}}(s)|=k_{1}$ and $|{\mathrm{Var}}(s)\setminus{\mathrm{Var}}(t)|=k_{2}$. For example, $x_{1}x_{2}=x_{1}x_{3}x_{4}$ is a $(1,2)$-equation. Let $Eq(k_{1},k_{2},n)\subseteq Eq(n)$ be the set of all $(k_{1},k_{2})$-equations in $n$ variables. Obviously,

[TABLE]

where

[TABLE]

Each equation $t(X)=s(X)\in Eq(k_{1},k_{2},n)$ is uniquely defined by $k_{1}$ variables in the left part and by $k_{2}$ other variables in the right part (the residuary $n-k_{1}-k_{2}$ variables should occur in both parts of the equation). Thus,

[TABLE]

By (1), one can compute that

[TABLE]

**Remark 2.1. **

In this paper we consider only equations $t(X)=s(X)$ with $n\leq l$, i.e. the number of variables occurring in $t(X)=s(X)$ is not more than the order of the semilattice $L_{l}$. The case $n>l$ needs a completely different technic and was considered in [3]. All main results of the current paper do not hold for the case $n>l$.

A point $P\in L_{l}^{n}$ is a solution of an equation $t(X)=s(X)$ if $t(P),s(P)$ define the same element in the semilattice $L_{l}$. By the properties of linearly ordered semilattices, a point $P=(p_{1},p_{2},\ldots,p_{n})$ is a solution of $t(X)=s(X)$ iff there exist variables $x_{i}\in{\mathrm{Var}}(t)$, $x_{j}\in{\mathrm{Var}}(s)$ such that $p_{i}=p_{j}$ and $p_{i}\leq p_{k}$ for all $1\leq k\leq n$. The set of all solutions of an equation $t(X)=s(X)$ is denoted by ${\mathrm{V}}(t(X)=s(X))$.

An arbitrary set of equations is called a system. The set of all solutions ${\mathrm{V}}({\mathbf{S}})$ of a system ${\mathbf{S}}=\{t_{i}(X)=s_{i}(X)\mid i\in I\}$ is defined as $\bigcap_{i\in I}{\mathrm{V}}(t_{i}(X)=s_{i}(X))$. A set $Y\subseteq L_{l}^{n}$ is called algebraic over $L_{l}$ if there exists a system ${\mathbf{S}}$ in $n$ variables with ${\mathrm{V}}({\mathbf{S}})=Y$. An algebraic set $Y$ is irreducible if $Y$ is not a proper finite union of other algebraic sets.

**Proposition 2.2. ([3], Proposition 2.2) Any algebraic set $Y$ over $L_{l}$ is a finite union of irreducible sets

[TABLE]

and this decomposition is unique up to a permutation of components.

The subsets $Y_{i}$ from the union (2) are called the irreducible components of $Y$.

Let $Y$ be an algebraic set over $L_{l}$ defined by a system ${\mathbf{S}}(X)$. One can define an equivalence relation $\sim_{Y}$ over the set of all terms in variables $X$ as follows

[TABLE]

The set of all $\sim_{Y}$-equivalence classes is called the coordinate semilattice of $Y$ and denoted by $\Gamma(Y)$ (see [1] for more details). The following statement describes the coordinate semilattices of irreducible algebraic sets.

**Proposition 2.3. ([3], Proposition 2.3) A set $Y$ is irreducible over $L_{l}$ iff $\Gamma(Y)$ is embedded into $L_{l}$

There are different algebraic sets over $L_{l}$ with isomorphic coordinate semilattices. Such sets are called isomorphic. For example, the following sets

[TABLE]

has the isomorphic coordinate semilattices

[TABLE]

[TABLE]

Thus, $Y_{1},Y_{2}$ are isomorphic.

3 Example

Let $n=3$, $l=3$. We have exactly $Eq(3)=3^{3}-2=25$ equations in three variables over $L_{3}$. The following table contains the information about such equations over $L_{3}$. The second column contains systems which define irreducible components of the solution set of an equation in the first column. A cell of the table contains $\uparrow$ if an information in this cell is similar to the cell above.

Table 1.

[TABLE]

Notice that ${\mathrm{V}}(x_{1}=x_{2}\leq x_{3})$ does not define an irreducible component for $Y={\mathrm{V}}(x_{1}x_{2}=x_{1}x_{3})$, since ${\mathrm{V}}(x_{1}=x_{2}\leq x_{3})$ is included into the solution set of another irreducible component ${\mathrm{V}}(x_{1}\leq x_{2}\leq x_{3})$. Similarly, ${\mathrm{V}}(x_{3}=x_{1}\leq x_{2})$ is not an irreducible component for $Y$, since it is contained in the irreducible component ${\mathrm{V}}(x_{1}\leq x_{3}\leq x_{2})$.

It turns out that the number of irreducible components does not depend on the semilattice order $l$. One can directly compute the average number of irreducible components of algebraic sets defined by equations in three variables:

[TABLE]

Recall that in Section 5 we obtain the general expression for $\overline{{\mathrm{Irr}}}(n)$ (10). Clearly, (10) will give (3) for $n=3$.

4 Decompositions of algebraic sets

Let $Y$ denote the solution set of an equation $t(X)=s(X)$ over the semilattice $L_{l}=\{a_{1},a_{2},\ldots,a_{l}\}$. The table above shows that any irreducible component sorts the variables $X$ into some order. The following definition formalizes this property of irreducible components.

Let ${\sigma}$ be a permutation of the set $\{1,2,\ldots,n\}$; ${\sigma}$ sorts the set $X$ as follows $\{x_{{\sigma}(1)},x_{{\sigma}(2)},\ldots,x_{{\sigma}(n)}\}$, i.e. ${\sigma}(i)$ is the $i$-th variable in the sorted set $X$. A permutation ${\sigma}$ is called a a permutation of the first (second) kind if $x_{{\sigma}(1)}\in{\mathrm{Var}}(t)\cap{\mathrm{Var}}(s)$ (respectively, $x_{{\sigma}(2)}\in{\mathrm{Var}}(t)\setminus{\mathrm{Var}}(s)$, $x_{{\sigma}(1)}\in{\mathrm{Var}}(s)\setminus{\mathrm{Var}}(t)$). Let $\chi({\sigma})\in\{1,2\}$ denote the kind of a permutation ${\sigma}$.

**Example 4.1. ** Let us consider an algebraic set $Y_{0}={\mathrm{V}}(x_{1}x_{2}=x_{1}x_{3})$. By the table above, $Y_{0}$ is the union of the following irreducible components

[TABLE]

The irreducible components $Y_{1},Y_{2},Y_{3}$ define the following permutations

[TABLE]

Moreover, ${\sigma}_{1},{\sigma}_{2}$ are permutations of the first kind, whereas ${\sigma}_{3}$ is of the second kind.

A permutation ${\sigma}$ defines an algebraic set $Y_{\sigma}$ as follows:

[TABLE]

if $\chi({\sigma})=1$, and

[TABLE]

if $\chi({\sigma})=2$.

**Example 4.2. ** Let ${\sigma}_{1},{\sigma}_{2},{\sigma}_{3}$ be permutations from Example 4. Obviously, the sets $Y_{{\sigma}_{1}},Y_{{\sigma}_{2}},Y_{{\sigma}_{3}}$ defined by (4,5) coincide with the sets $Y_{1},Y_{2},Y_{3}$ respectively.

*Lemma 4.3. *** Let $\chi({\sigma})\in\{1,2\}$, then the set $Y_{\sigma}$ is irreducible and moreover

[TABLE]

Proof.

By the definition of a coordinate semilattice, $\Gamma(Y_{\sigma})$ is generated by the elements $\{x_{1},x_{2},\ldots,x_{n}\}$ and has the following defined relations

[TABLE]

and

[TABLE]

Thus, $\Gamma(Y_{\sigma})$ is a linearly ordered semilattice, and (6) holds. By Proposition 2, the set $Y_{\sigma}$ is irreducible. ∎

The following lemma gives the irreducible decomposition of an algebraic set $Y={\mathrm{V}}(t(X)=s(X))$.

*Lemma 4.4. *** An algebraic set $Y={\mathrm{V}}(t(X)=s(X))$ is a union

[TABLE]

Proof.

Suppose $P=(p_{1},p_{2},\ldots,p_{n})\in Y$. Let us sort $p_{i}$ in the ascending order

[TABLE]

where ${\sigma}$ is a permutation of the set $\{1,2,\ldots,n\}$. We have that ${\sigma}$ induces the sorting of the variable set $X$. Obviously, we may assume that $x_{\sigma}(1)\in{\mathrm{Var}}(t)$ (if $x_{{\sigma}(1)}\notin{\mathrm{Var}}(t)$, the properties of $L_{l}$ provides an existsence of a variable $x_{\sigma}(i)\in{\mathrm{Var}}(t)$ such that $p_{{\sigma}(i)}=p_{{\sigma}(1)}$; in this case one can swap the values ${\sigma}(1)$ and ${\sigma}(i)$).

For example, the point $P=(a_{2},a_{1},a_{1})\in{\mathrm{V}}(x_{1}x_{2}=x_{1}x_{3})$ defines ${\sigma}(1)=2$, ${\sigma}(2)=3$, ${\sigma}(3)=1$ (the permutation obtained equals ${\sigma}_{3}$ from Example 4, so the point $(a_{2},a_{1},a_{1})$ belongs to the set $Y_{3}$).

Since ${\sigma}$ is defined by the inequalities between the coordinates $p_{i}$, it follows $P\in Y_{\sigma}$.

Let us prove now $Y_{\sigma}\subseteq Y$ for each ${\sigma}$. Suppose $P=(p_{1},p_{2},\ldots,p_{n})\in Y_{\sigma}$. If $\chi(Y_{\sigma})=1$ then

[TABLE]

Otherwise ($\chi(Y_{\sigma})=2$), $t(P)=p_{{\sigma}(1)}$, $s(P)=p_{{\sigma}(2)}$, and (5) gives $p_{{\sigma}(1)}=p_{{\sigma}(2)}$. Therefore $P\in{\mathrm{V}}(t(X)=s(X))$. ∎

*Lemma 4.5. *** For distinct permutations $\sigma,\sigma^{\prime}$ we have $Y_{\sigma}\nsubseteq Y_{\sigma^{\prime}}$ in (7).

Proof.

Let ${\sigma}$ be a permutation of the first or second kind, and $P_{\sigma}$ denote the following point

[TABLE]

and

[TABLE]

For example, the permutations ${\sigma}_{1},{\sigma}_{2},{\sigma}_{3}$ from Example 4 define the points

[TABLE]

respectively.

Since $P_{\sigma}$ preserves the order of variables, we have $P_{\sigma}\in Y_{\sigma}$.

Let us show now $P_{\sigma}\notin Y_{{\sigma}^{\prime}}$ for every ${\sigma}^{\prime}\neq{\sigma}$ (for example, each of the points $P_{1},P_{2},P_{3}$ above belong to a unique irreducible component from Example 4:

[TABLE]

There exists indexes $i<j$ such that $i={\sigma}({\alpha})$, $j={\sigma}(\beta)$, $i={\sigma}^{\prime}({\alpha}^{\prime})$, $j={\sigma}^{\prime}(\beta^{\prime})$, with ${\alpha}<\beta$, ${\alpha}^{\prime}>\beta^{\prime}$. Hence the inequality $x_{i}\leq x_{j}$ holds in $Y_{\sigma}$, and $x_{j}\leq x_{i}$ holds in $Y_{{\sigma}^{\prime}}$. Let us consider the following two cases:

If $\chi({\sigma})=1$, then $p_{i}<p_{j}$ in $P_{\sigma}$, and we immediately obtain $P_{\sigma}\notin Y_{{\sigma}^{\prime}}$. 2. 2.

Suppose $\chi({\sigma})=2$. One should assume that $p_{i}=p_{j}=a_{2}$ (if $p_{i}<p_{j}$ we immediately obtain $P_{\sigma}\notin Y_{{\sigma}^{\prime}}$). Then ${\alpha}=1$, $\beta=2$ and $i={\sigma}(1)$, $j={\sigma}(2)$ (one can similarly consider the case $i={\sigma}(2)$, $j={\sigma}(1)$). Hence $x_{i}\in{\mathrm{Var}}(t)\setminus{\mathrm{Var}}(s)$, $x_{j}\in{\mathrm{Var}}(s)\setminus{\mathrm{Var}}(t)$. By the definition of a permutation of the second kind, ${\sigma}^{\prime}(1)=k\neq j$, and the inequality $x_{k}\leq x_{j}$ holds in $Y_{{\sigma}^{\prime}}$. Let $\gamma$ be the index such that ${\sigma}(\gamma)=k$. Since ${\alpha}=1$, $\beta=2$, we have $\gamma>2$. Then $p_{k}=a_{\gamma}$, and $p_{j}<p_{k}$ for $P_{\sigma}$. Thus, $P\notin Y_{{\sigma}^{\prime}}$.

∎

According to Lemmas 6, 7, 4, we obtain the following statement.

*Theorem 4.6. *** The union (7) is the irreducible decomposition of the set $Y={\mathrm{V}}(t(X)=s(X))$. The number of irreducible components is equal to the number of permutations of the first and second kind.

5 Average number of irreducible components

One can directly compute that any $(k_{1},k_{2})$-equation admits

[TABLE]

permutations of the first kind and

[TABLE]

permutations of the second kind.

By Theorem 4, for a $(k_{1},k_{2})$-equation $t(X)=s(X)$ the number of its irreducible components equals

[TABLE]

The average number of irreducible components of algebraic sets defined by equations from $Eq(n)$ is

[TABLE]

Since

[TABLE]

we obtain

[TABLE]

Below we compute $\overline{{\mathrm{Irr}}}$ using the following denotations:

$A\stackrel{{\scriptstyle(1)}}{{=}}B$: an expression $B$ is obtained from $A$ by the binomial identity

[TABLE] 2. 2.

$A\stackrel{{\scriptstyle(2)}}{{=}}B$: an expression $B$ is obtained from $A$ by the following identity of binomial coefficients

[TABLE]

Let us demonstrate the proof of (9):

[TABLE]

Let us compute $\overline{{\mathrm{Irr}}}(n)$. We have that

[TABLE]

where

[TABLE]

[TABLE]

[TABLE]

[TABLE]

and

[TABLE]

Finally, we obtain that

[TABLE]

and

[TABLE]

Notice that the final answer does not depend on $l$ if $l\leq n$. In particular, (10) gives

[TABLE]

for $n=3$, and (11) obviously coincides with (3).