Projections of k-simplex onto the subsimplices of arbitrary type are derivations

TL;DR

This paper proves that projections of arbitrary k-simplices onto certain subsimplices are derivations, expanding understanding of geometric projections in simplicial complexes.

Contribution

It introduces a novel proof that such projections are derivations, providing new insights into the structure of simplices and their subsimplices.

Findings

Projections of k-simplices onto subsimplices are derivations.

The proof applies to arbitrary types of subsimplices.

Enhances theoretical understanding of simplicial projections.

Abstract

The aim of this paper is to prove that there is a projection of an arbitrary k-simplex onto (m - l)-subsimplex, where 1 < l < m < k - 1, which is a derivation.

Equations

$$\alpha (0)=\cdots =\alpha (i_0-1)=a_0,\alpha (i_0)=\cdots =\alpha (i_0+i_1-1)=a_1,\cdots$$

$$\alpha (i_0+\cdots +i_{k-2})=\cdots =\alpha (i_0+\cdots +i_{k-1}-1)=a_{k-1}$$

$$h_0=\min \{i_0,j_0\},$$

$$h_s=\min \left\{\sum _{p=0}^s{i}_p-\sum _{p=0}^{s-1}{h}_p,\sum _{p=0}^s{j}_p-\sum _{p=0}^{s-1}{h}_p\right\},\forall s=1,\dots k-1.$$

$$\sum _{p=0}^s{h}_p=\min \left\{\sum _{p=0}^s{i}_p,\sum _{p=0}^s{j}_p\right\},\forall s=1,\dots k-1.$$

$$\sum _{p=0}^{k-1}{h}_p=\min \left\{\sum _{p=0}^{k-1}{i}_p,\sum _{p=0}^{k-1}{j}_p\right\}=\min \{n,n\}=n.$$

$$\beta (a_0)=\cdots =\beta (a_{n_0})=a_0^{\prime }.$$

$$\beta (a_{n_0+1})=\cdots =\beta (a_{n_1})=a_1^{\prime }.$$

$$\beta (a_p)=a_0^{\prime },\forall p,0\le p\le n_0,$$

$$\beta (a_p)=a_1^{\prime },\forall p,n_0+1\le p\le n_1,$$

$$............................................................$$

$$\beta (a_p)=a_r^{\prime },\forall p,n_{r-1}+1\le p\le n_r=k-1.$$

$$\alpha \beta =(a_0^{\prime }{)}_{{\sum }_{p=0}^{n_0}{i}_p}(a_1^{\prime }{)}_{{\sum }_{p=n_0+1}^{n_1}{i}_p}\cdots (a_r^{\prime }{)}_{{\sum }_{p=n_{r-1}+1}^{n_r}{i}_p}.$$

$$\beta (a_{k-1})=\cdots =\beta (a_{n_{r-1}+1})=a_{k-1}^{\prime }.$$

$$\alpha \beta =(a_0^{\prime }{)}_{{\sum }_{p=0}^{n_0}{i}_p}\cdots (a_u^{\prime }{)}_{{\sum }_{p=n_{u-1}+1}^{n_u}{i}_p}\cdots (a_{k-1}^{\prime }{)}_{{\sum }_{p=n_{r-1}+1}^{k-1}{i}_p}.$$

$${\partial }_{m-\ell }^{k-1-0}={\partial }_{m-\ell }^{k-1}:{\sigma }^{(n)}\{a_0,\dots ,a_{k-1}\}\to {\sigma }^{(n)}\{a_{\ell },\dots ,a_m\},$$

$${\partial }_{m-\ell }^{k-1}(\alpha )=(a_{\ell }{)}_{{\sum }_{p=0}^{\ell }{i}_p}(a_{\ell +1}{)}_{i_{\ell +1}}\cdots (a_{m-1}{)}_{i_{m-1}}(a_m{)}_{{\sum }_{p=m}^{k-1}{i}_p}.$$

$${\partial }_{m-\ell }^{k-1}(\alpha +\beta )={\partial }_{m-\ell }^{k-1}(\alpha )+{\partial }_{m-\ell }^{k-1}(\beta ),$$

$$h_0=\min \{i_0,j_0\},$$

$$h_s=\min \left\{\sum _{p=0}^s{i}_p-\sum _{p=0}^{s-1}{h}_p,\sum _{p=0}^s{j}_p-\sum _{p=0}^{s-1}{h}_p\right\}\text{mbox}fors=1,\dots k-1.$$

$${\partial }_{m-\ell }^{k-1}(\alpha )=(a_{\ell }{)}_{{\sum }_{p=0}^{\ell }{i}_p}(a_{\ell +1}{)}_{i_{\ell +1}}\cdots (a_{m-1}{)}_{i_{m-1}}(a_m{)}_{{\sum }_{p=m}^{k-1}{i}_p},$$

$${\partial }_{m-\ell }^{k-1}(\beta )=(a_{\ell }{)}_{{\sum }_{p=0}^{\ell }{j}_p}(a_{\ell +1}{)}_{j_{\ell +1}}\cdots (a_{m-1}{)}_{j_{m-1}}(a_m{)}_{{\sum }_{p=m}^{k-1}{j}_p},$$

$${\partial }_{m-\ell }^{k-1}(\alpha +\beta )=(a_{\ell }{)}_{{\sum }_{p=0}^{\ell }{h}_p}(a_{\ell +1}{)}_{h_{\ell +1}}\cdots (a_{m-1}{)}_{h_{m-1}}(a_m{)}_{{\sum }_{p=m}^{k-1}{h}_p}.$$

$${\partial }_{m-\ell }^{k-1}(\alpha )+{\partial }_{m-\ell }^{k-1}(\beta )=(a_{\ell }{)}_{h^{\prime }}(a_{\ell +1}{)}_{h_{\ell +1}}\cdots (a_{m-1}{)}_{h_{m-1}}(a_m{)}_{h^{\prime \prime }},$$

$$h^{\prime }=\min \left\{\sum _{p=0}^{\ell }{i}_p,\sum _{p=0}^{\ell }{j}_p\right\}\text{mbox}and{h}^{\prime \prime }=\min \left\{\sum _{p=m}^{k-1}{i}_p,\sum _{p=m}^{k-1}{j}_p\right\}$$

$$\min \left\{\sum _{p=0}^{\ell }{i}_p,\sum _{p=0}^{\ell }{j}_p\right\}=\sum _{p=0}^{\ell }{h}_p\text{mbox}and\min \left\{\sum _{p=m}^{k-1}{i}_p,\sum _{p=m}^{k-1}{j}_p\right\}=\sum _{p=m}^{k-1}{h}_p,$$

$$S_{m-\ell }^{k-1}=\left\{\alpha |\alpha \in {\sigma }^{(n)}\{a_0,\dots ,a_{k-1}\},\alpha (a_0)=\cdots =\alpha (a_{\ell })\ge a_{\ell +1},\alpha (a_m)\le a_m\right\}.$$

$$(\alpha +\beta )(a_p)=\alpha (a_p)+\beta (a_p)=\alpha (a_q)+\beta (a_q)=(\alpha +\beta )(a_q),$$

$$(\alpha +\beta )(a_p)=\alpha (a_p)+\beta (a_p)\ge a_{\ell +1}\text{mbox}forallp=0,\dots ,\ell .$$

$$(\alpha +\beta )(a_m)=\alpha (a_m)+\beta (a_m)\le a_m.$$

Full text

PROJECTIONS OF $\mathbf{k}$ – SIMPLEX ONTO THE SUBSIMPLICES OF ARBITRARY TYPE ARE DERIVATIONS

Dimitrinka Vladeva

Abstract: The aim of this paper is to prove that there is a projection of an arbitrary $k$ –simplex onto $(m-\ell)$ – subsimplex, where $1\leq\ell<m\leq k-1$, which is a derivation.

Keywords: endomorphism semiring of a finite chain, differential algebra, simplicial complex, derivations in semirings.

1   Introduction and preliminaries

Differential algebra is an area of algebra in the study of algebraic structures equipped with one or finitely many derivations that are additive maps satisfied Leibniz product rule. A simple examples are the usual derivatives on various algebras consisting of differentiable functions in sense of analysis. In 1950, Ritt [5] and in 1973, Kolchin [3] wrote their classical books on differential algebra. During the last few decades there has been a great deal of works concerning derivations in fields and rings, in Lie rings, in skew polynomial rings and other algebraic structures.

About derivations in semirings we know:

— the definition in Golan’s book [1],

— properties of derivations in strings and some simplicial complexes of strings – [9],

— the projections on the strings or on the triangles of 2, 3 and 4 simplices which are derivations – [11] – [14],

— the projections on the triangles $\triangle^{(n)}\{a_{0},a_{1},a_{m}\}$, where $2\leq m\leq k-1$, of an arbitrary simplex $\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$ which are derivations – [15],

— the projections on the all basic strings of an arbitrary simplex $\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$ which are derivations – [16].

The theorem proved here generalise most of these results.

The endomorphism semirings of a finite semilattice are well-established, see [2], [7], [8] and [17]. Concerning the background of simplicial complexes, algebraic topology and combinatorics a reader can refer to [4], and [6].

An algebra $R=(R,+,.)$ with two binary operations $+$ and $\cdot$ on $R$, is called a semiring if:

1. $\;(R,+)$ is a commutative semigroup,

2. $\;(R,\cdot)$ is a semigroup,

3. distributive laws hold $x\cdot(y+z)=x\cdot y+x\cdot z$ and $(x+y)\cdot z=x\cdot z+y\cdot z$ for any $x,y,z\in R$.

For a join-semilattice $\left(\mathcal{M},\vee\right)$ set $\mathcal{E}_{\mathcal{M}}$ of the endomorphisms of $\mathcal{M}$ is a semiring with respect to the addition and multiplication defined by:

$\bullet\;h=f+g\;\mbox{when}\;h(x)=f(x)\vee g(x)\;\mbox{for all}\;x\in\mathcal{M}$,

$\bullet\;h=f\cdot g\;\mbox{when}\;h(x)=f\left(g(x)\right)\;\mbox{for all}\;x\in\mathcal{M}$.

This semiring is called the endomorphism semiring of $\mathcal{M}$.

In this article, all semilattices are finite chains. Following [7] we fix a finite chain $\mathcal{C}_{n}=\;\left(\{0,1,\ldots,n-1\}\,,\,\vee\right)\;$ and denote the endomorphism semiring of this chain with $\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$. We do not assume that $\alpha(0)=0$ for arbitrary $\alpha\in\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$. So, there is not a zero in endomorphism semiring $\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$.

In [10] there is a new treatment of the subsemirings of endomorphism semiring $\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$ of a finite chain. For arbitrary elements $a_{0},a_{1},\ldots,a_{k-1}\in\mathrm{C}_{n}$, where $k\leq n$ and $a_{0}<a_{1}<\ldots<a_{k-1}$ we denote $A=\{a_{0},a_{1},\ldots,a_{k-1}\}$. Now, consider endomorphisms $\alpha\in\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$ with $Im(\alpha)\subseteq A$. The set of the all such endomorphisms $\alpha$ is a maximal simplex. We denote this simplex by $\sigma^{(n)}_{k}(A)=\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$.

The endomorphisms $\alpha\in\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$ such that

[TABLE]

[TABLE]

we denote by $\alpha=(a_{0})_{i_{0}}(a_{1})_{i_{1}}\ldots(a_{k-1})_{i_{k-1}}$, where $\displaystyle\sum_{p=0}^{k-1}i_{p}=n$.

2   Addition and multiplication of endomorphisms

Here we show how to signify the sum of two endomorphisms.

Lemma 1. Let $\alpha,\beta\in\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$, where $\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$, $\beta=(a_{0})_{j_{0}}\cdots(a_{k-1})_{j_{k-1}}$ and $\displaystyle\sum_{p=0}^{k-1}i_{p}=\sum_{p=0}^{k-1}j_{p}=n$. Let $\gamma=(a_{0})_{h_{0}}\cdots(a_{k-1})_{h_{k-1}}$, where

[TABLE]

[TABLE]

Then $\gamma=\alpha+\beta$.

Proof. From (1) follows

[TABLE]

Now for $s=k-1$ we find

[TABLE]

Hence, $\gamma\in\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$.

Let $t=0,1,\ldots,k-1$. Let for some $s$ have $\displaystyle\sum_{p=0}^{s}i_{p}<t\leq\sum_{p=0}^{s+1}i_{p}$. Then it follows $\alpha(t)=a_{s+1}$.

Let us suppose that $\displaystyle t\leq\sum_{p=0}^{s}j_{p}$. Then it follows $\displaystyle\sum_{p=0}^{s}h_{p}=\sum_{p=0}^{s}i_{p}$. Since $\displaystyle t\leq\sum_{p=0}^{s+1}i_{p}$ and $\displaystyle t\leq\sum_{p=0}^{s}j_{p}\leq\sum_{p=0}^{s+1}j_{p}$, it follows $\displaystyle t\leq\sum_{p=0}^{s+1}h_{p}$. So $\displaystyle\sum_{p=0}^{s}h_{p}<t\leq\sum_{p=0}^{s+1}h_{p}$, that is $\gamma(t)=a_{s+1}=\alpha(t)$.

On the other hand, from $\displaystyle t\leq\sum_{p=0}^{s}j_{p}$ it follows that $\beta(t)\leq a_{s}$.

Thus $\gamma(t)=\max\{\alpha(t),\beta(t)\}$.

Let us suppose that $\displaystyle\sum_{p=0}^{s}j_{p}<t\leq\sum_{p=0}^{s+1}j_{p}$. Then $\displaystyle\sum_{p=0}^{s}h_{p}<t\leq\sum_{p=0}^{s+1}h_{p}$ and now $\gamma(t)=\beta(t)=\alpha(t)$.

Let us suppose that $\displaystyle\sum_{p=0}^{s+1}j_{p}<t$. Then $\displaystyle\sum_{p=0}^{s+1}j_{p}<\sum_{p=0}^{s+1}i_{p}$, so, $\displaystyle\sum_{p=0}^{s+1}h_{p}=\sum_{p=0}^{s+1}j_{p}$. Now $\gamma(t)=\beta(t)\geq a_{s+1}$ and also $\gamma(t)=\max\{\alpha(t),\beta(t)\}$.

Hence, we prove that $\gamma(t)=\alpha(t)+\beta(t)$.

For multiplication of two endomorphisms $\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$ and $\beta=(a_{0})_{j_{0}}\cdots(a_{k-1})_{j_{k-1}}$, where $\displaystyle\sum_{p=0}^{k-1}i_{p}=\sum_{p=0}^{k-1}j_{p}=n$, we consider the following notations.

Let $\beta(a_{0})=a^{\prime}_{0}$ and $n_{0}$ be the greatest number such that

[TABLE]

Let $a_{n_{0}+1}$ be the next element in $\{a_{0},\ldots,a_{k-1}\}$ after $a_{n_{0}}$ and $\beta(a_{n_{0}+1})=a^{\prime}_{1}$. Let $n_{1}$ be the greatest number such that

[TABLE]

So, we define a partition of the set $\{a_{0},\ldots,a_{k-1}\}$ with properties

[TABLE]

[TABLE]

[TABLE]

[TABLE]

Since $(\alpha\beta)(t)=\beta(\alpha(t))$ for $t=0,\ldots,k-1$, it follows

[TABLE]

Let $\beta(a_{k-1})=a^{\prime}_{k-1}$. Since $n_{r}=k-1$, it follows that $a^{\prime}_{k-1}=a^{\prime}_{r}$. So, $n_{r-1}+1$ is the smallest number such that

[TABLE]

Then, from (2), it follows

[TABLE]

In (3) we note that $1\leq u\leq k-1$, $\;n_{u-1}+1$ is the smallest number and ${n_{u}}$ is the greatest number such that $\beta(a_{n_{u-1}+1})=\cdots=\beta(a_{n_{u}})=a^{\prime}_{u}$.

3   Projection onto $(m-\ell)$ – simplex

Let us consider the map

[TABLE]

where $0\leq\ell<m\leq k-1$ and $a_{\ell},\ldots,a_{m}$ are consecutive elements of the set $\{a_{0},\ldots,a_{k-1}\}$, so that for any endomorphism $\alpha\in\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$, $\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$, where $\displaystyle\sum_{p=0}^{k-1}i_{p}=n$,

[TABLE]

Lemma 2. For any endomorphisms $\alpha,\beta\in\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$, it follows

[TABLE]

where $0\leq\ell<m\leq k-1$.

Proof. Let $\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$ and $\beta=(a_{0})_{j_{0}}\cdots(a_{k-1})_{j_{k-1}}$. On the basis of Lemma 1 we can conclude that $\alpha+\beta=(a_{0})_{h_{0}}\cdots(a_{k-1})_{h_{k-1}}$, where

[TABLE]

[TABLE]

Now we consider

[TABLE]

[TABLE]

[TABLE]

Now, it follows

[TABLE]

where

[TABLE]

and $h_{\ell+1},\ldots,h_{m-1}$ are well defined. Since

[TABLE]

it follows $\partial_{{m-\ell}}^{k-1}(\alpha)+\partial_{{m-\ell}}^{k-1}(\beta)=\partial_{{m-\ell}}^{k-1}(\alpha+\beta)$.

Let us consider the set below:

[TABLE]

For $\displaystyle\alpha,\beta\in S_{{m-\ell}}^{k-1}$ we find

[TABLE]

where $0\leq p\leq\ell$, $0\leq q\leq\ell$ and

[TABLE]

Moreover

[TABLE]

Also it follows

[TABLE]

where $0\leq p\leq\ell$, $0\leq q\leq\ell$ and

[TABLE]

Moreover

[TABLE]

So, we prove

Lemma 3. The set $S_{{m-\ell}}^{k-1}$ is a subsemiring of the semiring $\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$.

Let us consider the set below:

[TABLE]

For $\displaystyle\alpha,\beta\in R_{{m-\ell}}^{k-1}$ we find

[TABLE]

[TABLE]

Thus, we prove

Lemma 4. The set $R_{{m-\ell}}^{k-1}$ is a subsemiring of the semiring $\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$.

Let $\mathcal{D}_{{m-\ell}}^{k-1}=S_{{m-\ell}}^{k-1}\cup R_{{m-\ell}}^{k-1}$. Note that $S_{{m-\ell}}^{k-1}\cap R_{{m-\ell}}^{k-1}=\varnothing$.

Lemma 5. The set $\mathcal{D}_{{m-\ell}}^{k-1}$ is a subsemiring of the semiring $\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$.

Proof. Let $\alpha\in S_{{m-\ell}}^{k-1}$ and $\beta\in R_{{m-\ell}}^{k-1}$.

Since $(\alpha+\beta)(a_{p})=\alpha(a_{p})+\beta(a_{p})$, $\alpha(a_{p})\geq a_{\ell+1}$ and $\beta(a_{p})\leq\beta(a_{\ell})\leq a_{\ell}$, it follows $(\alpha+\beta)(a_{p})=\alpha(a_{p})\geq a_{\ell+1}$. As $\alpha(a_{p})=\alpha(a_{q})$ we have $(\alpha+\beta)(a_{p})=(\alpha+\beta)(a_{q})$, where $0\leq p\leq\ell$, $0\leq q\leq\ell$. Moreover, $(\alpha+\beta)(a_{m})=\alpha(a_{m})+\beta(a_{{m}})\leq a_{m}$. Hence $\alpha+\beta\in S_{{m-\ell}}^{k-1}$.

Let $0\leq p\leq\ell$. As $\beta(a_{p})\leq\beta(a_{\ell})\leq a_{\ell}$, it follows $\beta(a_{p})=a_{s}$, where $s\leq\ell$. Analogously, for any $q$, $0\leq q\leq\ell$, it follows $\beta(a_{q})=a_{t}$, where $t\leq\ell$. Then

[TABLE]

On the other hand, $(\beta\alpha)(a_{p})=\alpha(\beta(a_{p}))=\alpha(a_{s})\geq a_{\ell+1}$. Moreover, $(\beta\alpha)(a_{m})=\alpha(\beta(a_{m}))\leq\alpha(a_{{m}})\leq a_{m}$. Hence $\beta\alpha\in S_{{m-\ell}}^{k-1}$.

Let $\alpha(a_{p})=a_{\ell+1}$, where $0\leq p\leq\ell$, and $\beta(a_{\ell+1})\geq a_{\ell+1}$. Then

[TABLE]

As $(\alpha\beta)(a_{q})=\beta(\alpha(a_{q}))=\beta(a_{\ell+1})$ where $0\leq q\leq\ell$, it follows $(\alpha\beta)(a_{p})=(\alpha\beta)(a_{q})$. Moreover, $(\alpha\beta)(a_{m})=\beta(\alpha(a_{m}))\leq\beta(a_{{m}})\leq a_{m}$. Hence $\alpha\beta\in S_{{m-\ell}}^{k-1}$.

Let $\alpha(a_{p})=a_{\ell+1}$, where $0\leq p\leq\ell$, and $\beta(a_{\ell+1})\leq a_{\ell}$. Then

[TABLE]

[TABLE]

Hence $\alpha\beta\in R_{m-\ell}^{k-1}$.

Let $\alpha(a_{p})=a_{r}$, where $r>\ell+1$ and $0\leq p\leq\ell$. Let $\beta(a_{r})\geq a_{\ell+1}$. Then

[TABLE]

As $(\alpha\beta)(a_{q})=\beta(\alpha(a_{q}))=\beta(a_{r})$, it follows $(\alpha\beta)(a_{p})=(\alpha\beta)(a_{q})$, where

$0\leq p\leq\ell$, $0\leq q\leq\ell$. Moreover, $(\alpha\beta)(a_{m})=\beta(\alpha(a_{m}))\leq\beta(a_{{m}})\leq a_{m}$. Hence $\alpha\beta\in S_{{m-\ell}}^{k-1}$.

Let $\alpha(a_{p})=a_{r}$, where $r>\ell+1$ and $0\leq p\leq\ell$. Let $\beta(a_{r})\leq a_{\ell}$. Then

[TABLE]

Moreover, $(\alpha\beta)(a_{m})=\beta(\alpha(a_{m}))\leq\beta(a_{{m}})\leq a_{m}$. Hence $\alpha\beta\in R_{{m-\ell}}^{k-1}$.

Lemma 6. For any endomorphisms $\alpha,\beta\in\mathcal{D}_{{{m-\ell}}}^{k-1}$, it follows

[TABLE]

where $0\leq\ell<m\leq k-1$.

Proof. Let $\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$ and $\beta=(a_{0})_{j_{0}}\cdots(a_{k-1})_{j_{k-1}}$, where

$\displaystyle\sum_{p=0}^{k-1}i_{p}=\sum_{p=0}^{k-1}j_{p}=n$.

Case 1. Let $\beta\in S_{{m-\ell}}^{k-1}$ and let assume that $\beta(a_{0})=\cdots=\beta(a_{\ell})=\cdots=\beta(a_{q_{0}})=a_{q}$, where $q_{0}\geq\ell$ be the greatest number with this property and $q\geq\ell+1$.

Case 1.1. Let us assume that the image of any element of the set $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ is smaller than or equal to $a_{m}$, so, we suppose that $\beta(a_{k-1})=a^{\prime}_{k-1}=a_{r}\leq a_{m}$.

Since $\displaystyle\sum_{p=0}^{q-1}j_{p}\leq a_{0},\ldots,a_{q_{0}}\leq\sum_{p=0}^{q}j_{p}-1$, it follows

[TABLE]

where $\displaystyle\sum_{p=0}^{q_{0}}i_{p}+\sum_{p=q+1}^{r}s_{p}=n$. Now, it follows $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\alpha\beta$. Since

[TABLE]

using that $\ell\leq q_{0}$ and $r\leq m$, it follows $\partial_{{m-\ell}}^{k-1}(\alpha)\beta=\alpha\beta$. It is clear that

[TABLE]

and since $\displaystyle a_{0}\geq\sum_{p=0}^{q-1}j_{p}\geq\sum_{p=0}^{\ell}j_{p}$, it follows $\alpha\partial_{{m-\ell}}^{k-1}(\beta)=\alpha\beta$.

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Case 1.2. Let us assume that the image of any element of the set $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ is greater than $a_{m}$. Then

[TABLE]

where $\displaystyle\sum_{p=0}^{q_{0}}i_{p}+\sum_{p=q+1}^{k-1}s_{p}=n$ Hence

[TABLE]

Using that $\displaystyle\partial_{{m-\ell}}^{k-1}(\alpha)=(a_{\ell})_{\sum_{p=0}^{\ell}i_{p}}(a_{\ell+1})_{i_{\ell+1}}\cdots(a_{m-1})_{i_{m-1}}(a_{m})_{\sum_{p=m}^{k-1}i_{p}}$ and $\ell\leq q_{0}$, it follows

[TABLE]

Note that the indices $\displaystyle s_{q+1},\ldots,\sum_{p=m}^{k-1}s_{p}$ are the same as in $\partial_{{m-\ell}}^{k-1}(\alpha\beta)$ because $\partial_{{m-\ell}}^{k-1}(\alpha)(a_{p})=\alpha(a_{p})$ for $p=q+1,\ldots,m-1$.

Since

[TABLE]

and using that $\partial_{{m-\ell}}^{k-1}(\beta)(a_{p})=\beta(a_{p})$ for $p=q+1,\ldots,m-1$, it follows

[TABLE]

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Case 1.3. Let us assume that only images of some of the elements of $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ are smaller than or equal to $a_{m}$. So, we suppose that $\beta(a_{k-1})=a^{\prime}_{k-1}=a_{r}>a_{m}$ and there exist a number $t$, $m+1\leq t<k-1$, such that $\beta(a_{t})<a_{m}$. Let $t$ be the smallest number with this property. Hence, $a^{\prime}_{t}<a_{m}\leq a^{\prime}_{t+1}$.

Now, considering (3), it follows

[TABLE]

where $a^{\prime}_{0}=a_{q}$, $n_{0}=q_{0}$ and $1\leq u\leq k-1$. Then

[TABLE]

Since $\displaystyle\partial_{{m-\ell}}^{k-1}(\alpha)=(a_{\ell})_{\sum_{p=0}^{\ell}i_{p}}(a_{\ell+1})_{i_{\ell+1}}\cdots(a_{m})_{\sum_{p=m}^{k-1}i_{p}}$, $\;\ell\leq q_{0}=n_{0}$ and $\displaystyle\partial_{{m-\ell}}^{k-1}(\alpha)(a_{u})=\alpha(a_{u})$ for $u=n_{0}+1,\ldots,n_{t}$, it follows

[TABLE]

Since $\displaystyle\partial_{{m-\ell}}^{k-1}(\beta)=(a_{\ell})_{\sum_{p=0}^{\ell}j_{p}}(a_{\ell+1})_{j_{\ell+1}}\cdots(a_{m-1})_{j_{m-1}}(a_{m})_{\sum_{p=m}^{k-1}j_{p}}$,

$\;\ell\leq q_{0}=n_{0}$ and $\displaystyle\partial_{{m-\ell}}^{k-1}(\beta)(a_{u})=\beta(a_{u})$ for $u=\ell+1,\ldots,m-1$, it follows

[TABLE]

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Case 2. Let $\beta\in R_{{m-\ell}}^{k-1}$.

Case 2.1. Let us assume that the image of any element of the set $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ is smaller than or equal to $a_{\ell}$. So, we suppose that $\beta(a_{k-1})=a^{\prime}_{k-1}=a_{r}$, where $r\leq\ell$. Thus $\alpha\beta=(a_{0})_{s_{0}}\cdots(a_{r})_{s_{r}}$ which implies $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=(a_{\ell})_{n}$. It is clear that

[TABLE]

and then $\partial_{{m-\ell}}^{k-1}(\alpha)\beta\leq(a_{\ell})_{n}$. Since

[TABLE]

and $\displaystyle a_{k-1}\leq\sum_{p=0}^{\ell}j_{p}-1$, it follows $\alpha\,\partial_{{m-\ell}}^{k-1}(\beta)=(a_{\ell})_{n}$.

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Case 2.2. Let us assume that the image of any element of the set $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ is smaller than or equal to $a_{m}$. So, we suppose that $\beta(a_{k-1})=a^{\prime}_{k-1}=a_{r}$, where $\ell<r\leq m$.

Let $\beta(a_{t})\leq a_{\ell}$, where $\ell\leq t<k-1$ and $t$ be the greatest number with this property. Hence, $a^{\prime}_{t}\leq a_{\ell}<a^{\prime}_{t+1}$. Then, from (3), it follows

[TABLE]

which implies

[TABLE]

Clearly $\displaystyle\partial_{{m-\ell}}^{k-1}(\alpha)=(a_{\ell})_{\sum_{p=0}^{\ell}i_{p}}(a_{\ell+1})_{i_{\ell+1}}\cdots(a_{m})_{\sum_{p=m}^{k-1}i_{p}}$, but, if $\beta(a_{\ell})=a_{\ell_{0}}<a_{\ell}$, it follows

[TABLE]

Note, that generally we observe $\partial_{{m-\ell}}^{k-1}(\alpha)\beta\leq\partial_{{m-\ell}}^{k-1}(\alpha\beta)$.

Since

[TABLE]

and using that $a^{\prime}_{t}\leq a_{\ell}$ and $a^{\prime}_{k-1}=a_{r}\leq a_{m}$, it follows

[TABLE]

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Case 2.3. Let us assume that the image of any element of the set $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ is greater than or equal to $a_{m}$. Then, considering reasonings similar to those in Case 2.2 and (3), it follows

[TABLE]

where $\ell\leq t\leq m$ and $t$ be the smallest number such that $\beta(a_{t})\leq a_{\ell}$.

Now, it follows

[TABLE]

Since $\partial_{{m-\ell}}^{k-1}(\alpha)=(a_{\ell})_{\sum_{p=0}^{\ell}i_{p}}(a_{\ell+1})_{i_{\ell+1}}\cdots(a_{m})_{\sum_{p=m}^{k-1}i_{p}}$, as in the previous case we find $\partial_{{m-\ell}}^{k-1}(\alpha)\beta\leq\partial_{{m-\ell}}^{k-1}(\alpha\beta)$.

Since $\partial_{{m-\ell}}^{k-1}(\beta)=(a_{\ell})_{\sum_{p=0}^{\ell}j_{p}}(a_{\ell+1})_{j_{\ell+1}}\cdots(a_{m-1})_{j_{m-1}}(a_{m})_{\sum_{p=m}^{k-1}j_{p}}$ and using that $a^{\prime}_{t}\leq a_{\ell}$ and $a^{\prime}_{m+1}\geq a_{m}$, it follows

[TABLE]

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Case 2.4. Let us assume that only images of some of the elements of $\{a_{m+1},\ldots,a_{k-1}\}$ under the endomorphism $\beta$ are smaller than or equal to $a_{m}$. So, we suppose that $\beta(a_{k-1})=a^{\prime}_{k-1}=a_{r}>a_{m}$ and there exist a number $t$, $m+1\leq t<k-1$, such that $\beta(a_{t})<a_{m}$. Let $t$ be the smallest number with this property. Hence, $a^{\prime}_{t}<a_{m}\leq a^{\prime}_{t+1}$. Let $q$ be the greatest number such that $\beta(a_{q})\leq a_{\ell}$, where $\ell\leq q<t$. Hence, $a^{\prime}_{q}\leq a_{\ell}<a^{\prime}_{q+1}$. Then, from (3), it follows

[TABLE]

Now, it follows

[TABLE]

Since $\displaystyle\partial_{{m-\ell}}^{k-1}(\alpha)=(a_{\ell})_{\sum_{p=0}^{\ell}i_{p}}(a_{\ell+1})_{i_{\ell+1}}\cdots(a_{m})_{\sum_{p=m}^{k-1}i_{p}}$, if $\beta(a_{\ell})=a_{\ell_{0}}<a_{\ell}$, it follows

[TABLE]

Generally we observe $\partial_{{m-\ell}}^{k-1}(\alpha)\beta\leq\partial_{{m-\ell}}^{k-1}(\alpha\beta)$.

It is clear that

[TABLE]

First multiplier in $\partial_{{m-\ell}}^{k-1}(\alpha\beta)$ in its representation as a product is $(a_{\ell})_{\sum_{p=0}^{n_{q}}i_{p}}$. Since

[TABLE]

it means that $\partial_{{m-\ell}}^{k-1}(\alpha\beta)(a^{\prime}_{p})=a_{\ell}$, where $p=0,\ldots,q$. Since, for $p=0,\ldots,q$, we have $\displaystyle a^{\prime}_{p}<\sum_{p=0}^{\ell}j_{p}$, it follows $\alpha\partial_{{m-\ell}}^{k-1}(\beta)(a^{\prime}_{p})=a_{\ell}$. So, first multiplier in the representation as a product of $\alpha\partial_{{m-\ell}}^{k-1}(\beta)$ is $(a_{\ell})_{\sum_{p=0}^{n_{q}}i_{p}}$.

On the other hand $\displaystyle\partial_{{m-\ell}}^{k-1}(\beta)(a_{u})=\beta(a_{u})$ for $u=q+1,\ldots,m=1$. Thus, it follows

[TABLE]

Hence $\partial_{{m-\ell}}^{k-1}(\alpha\beta)=\partial_{{m-\ell}}^{k-1}(\alpha)\beta+\alpha\partial_{{m-\ell}}^{k-1}(\beta)$.

Theorem. The map $\partial_{{m-\ell}}^{k-1}:\mathcal{D}_{{{m-\ell}}}^{k-1}\rightarrow\sigma^{(n)}\{a_{\ell},\ldots,a_{m}\}$ is a derivation. The maximal subsemiring of $\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$ closed under the derivation $\partial_{{m-\ell}}^{k-1}$ is a semiring $\mathcal{D}_{{{m-\ell}}}^{k-1}$.

Proof. From the lemmas we obtain that $\partial_{{m-\ell}}^{k-1}:\mathcal{D}_{{{m-\ell}}}^{k-1}\!\!\rightarrow\!\sigma^{(n)}\{a_{\ell},\ldots,a_{m}\}$ is a derivation.

If we suppose that $\beta\notin\mathcal{D}_{{{m-\ell}}}^{k-1}=S_{{m-\ell}}^{k-1}\cup R_{{m-\ell}}^{k-1}$, then it follows $\beta(a_{m})>a_{m}$. Let $t>m$ be the smallest number such that $\beta(a_{m})=a_{t}$, so $a^{\prime}_{t-1}\leq a_{m}<a_{t}$. Let $q$ be the greatest number such that $\beta(a_{q})\leq a_{\ell}$, where $\ell\leq q$. Hence, $a^{\prime}_{q}\leq a_{\ell}<a^{\prime}_{q+1}$. From (3) we have

[TABLE]

Hence, it follows

[TABLE]

Since

[TABLE]

it follows by reasonings similar to those from Case 2.4 of Lemma 6, that

[TABLE]

when $\beta(a_{\ell})=a_{\ell}$, or

[TABLE]

when $\beta(a_{\ell})=a_{\ell_{0}}<a_{\ell}$. In both cases, since $a_{t}>a_{m}$ we have that $\partial_{{m-\ell}}^{k-1}(\alpha)\,\beta$ is not smaller than or equal to $\partial_{{m-\ell}}^{k-1}(\alpha\beta)$ and this completes the proof.

Immediately from the theorem follows that $\sigma^{(n)}\{a_{\ell},\ldots,a_{m}\}\subset\mathcal{D}_{{{m-\ell}}}^{k-1}$. But $\sigma^{(n)}\{a_{\ell},\ldots,a_{m}\}$ is a left ideal of $\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$, so it is a left ideal of semiring $\mathcal{D}_{{{m-\ell}}}^{k-1}$. Analogously any subsimplex $\sigma^{(n)}\{a_{r},\ldots,a_{m}\}$, where $r=\ell+1,\ldots,m-1$ is a left ideal of $\mathcal{D}_{{{m-\ell}}}^{k-1}$.

4   Projections onto the smallest $m$ – simpleces

Here we consider the derivation from the previous section in the case when $\ell=0$. So, we explore the map

[TABLE]

where $0<m\leq k-1$, such that for any endomorphism

$\alpha\in\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$, $\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$, where $\displaystyle\sum_{p=0}^{k-1}i_{p}=n$,

[TABLE]

Now, it follows

[TABLE]

Immediately from the theorem, it follows that

[TABLE]

is a derivation and $\mathcal{D}_{{m}}^{k-1}$ is the maximal subsemiring of $\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$ closed under this derivation.

Analogously, for any integer $m_{1}$, where $0<m_{1}<m$, we consider the derivation

[TABLE]

From theorem, using the last derivation we can construct

[TABLE]

which is a derivation and $\mathcal{D}_{{m_{1}}}^{m}=\left\{\alpha\left|\;\alpha\in\sigma^{(n)}\{a_{0},\ldots,a_{m}\},\alpha(a_{m_{1}})\leq a_{m_{1}}\right.\right\}$ is the maximal subsemiring of $\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{m}\}$ closed under this derivation. So, we consider a new derivation

[TABLE]

and obtain that for any $\alpha\in\mathcal{D}_{{m}}^{k-1}\cap\mathcal{D}_{{m_{1}}}^{k-1}$, it follows $\partial_{{m_{1}}}^{k-1}(\alpha)=\partial_{m_{1}}^{m}\left(\partial_{m}^{k-1}(\alpha)\right)$. Hence, $\partial_{{m_{1}}}^{k-1}=\partial_{m_{1}}^{m}\cdot\,\partial_{m}^{k-1}$. Note that the last composition is possible for arbitrary number $m$, where $m_{1}<m<k-1$.

In the semiring

[TABLE]

we can represent any derivation as composition of two or more derivations.

In the case when $k=n$, i.e. $\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}=\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$ the semiring $\displaystyle\bigcap_{m=0}^{n-1}\mathcal{D}_{{m}}^{n-1}$ is well known – it is the subsemiring $\mathcal{ON}_{n}$ of over nilpotent endomorphisms, see [7]. Hence, semiring $\mathcal{ON}_{n}$ is closed under any derivation $\partial_{{m}}^{n-1}$, where $m=1,\ldots,n-2$. The order of this semiring is the n-th Catalan number $\displaystyle\frac{1}{n+1}\binom{2n}{n}$.

The semiring $\mathcal{ON}_{n}$ has a zero element $\overline{0}$ and we can consider the subsemiring $\mathcal{N}_{n}$ of nilpotent elements, i.e. the endomorphisms $\alpha\in\widehat{\mathcal{E}}_{\mathcal{C}_{n}}$ such that $\alpha^{m}=\overline{0}$ for some positive integer $m$. It is known, [7], that $\mathcal{N}_{n}$ is an ideal of $\mathcal{ON}_{n}$ and it is consists of all endomorphisms $\alpha$ such that $\alpha(t)<t$ for any $t=1,\ldots,n-1$. The order of the ideal $\mathcal{N}_{n}$ is the $(n-1)$-th Catalan number $\displaystyle\frac{1}{n}\binom{2n-2}{n-1}$.

Corollary. The ideal $\mathcal{N}_{n}$ is closed under arbitrary derivation $\partial_{m}^{n-1}$, where $m=1,\ldots,n-2$.

Proof. Let $\alpha=(a_{0})_{i_{0}}\cdots(a_{n})_{i_{n}}\in\mathcal{N}_{n}$. Then

[TABLE]

For $\displaystyle 0\leq t\leq\sum_{p=0}^{m-1}i_{p}-1$, it follows $\partial_{m}^{n-1}(\alpha)=\alpha$, so, $\left(\partial_{m}^{n-1}(\alpha)\right)(t)<t$.

For $\displaystyle\sum_{p=0}^{m-1}i_{p}\leq t\leq n-1$, it follows $\left(\partial_{m}^{n-1}(\alpha)\right)(t)=a_{n-m}\leq\alpha(t)<t$.

Hence $\displaystyle\partial_{m}^{n-1}(\alpha)\in\mathcal{N}_{n}$.

5   Projections onto the greatest $k-\ell$ – simpleces

Here we consider the derivation from the section 3 in the case when $m=k-1$. We explore the map

[TABLE]

where $0\leq\ell<k-1$, such that for any $\alpha\in\sigma^{(n)}\{a_{0},\ldots,a_{k-1}\}$,$\alpha=(a_{0})_{i_{0}}\cdots(a_{k-1})_{i_{k-1}}$, where $\displaystyle\sum_{p=0}^{k-1}i_{p}=n$,

[TABLE]

So, we consider

[TABLE]

[TABLE]

and $\mathcal{D}_{{k-\ell-1}}^{k-1}=S_{{k-\ell-1}}^{k-1}\cup R_{{k-\ell-1}}^{k-1}$. Immediately from (4) and (5) it follows

[TABLE]

where $1\leq\ell<\ell_{1}<k-1$ and

[TABLE]

for arbitrary $1\leq\ell,\ell_{1}<k-1$.

Thus, using (6) and (7), we obtain

[TABLE]

Immediately from the theorem it follows that

[TABLE]

is a derivation and $\mathcal{D}_{{k-\ell-1}}^{k-1}$ is the maximal subsemiring of $\sigma^{(n)}\{a_{0},a_{1},\ldots,a_{k-1}\}$ closed under this derivation.

Analogously, for any integer $\ell_{1}$, where $\ell<\ell_{1}<k-1$, we consider the derivation

[TABLE]

Here $\mathcal{D}_{{k-\ell_{1}-1}}^{k-1}=S_{{k-\ell_{1}-1}}^{k-1}\cup R_{{k-\ell_{1}-1}}^{k-1}$, where

[TABLE]

[TABLE]

At last for arbitrary integers $\ell$ and $\ell_{1}$, where $\ell<\ell_{1}<k-1$, using theorem, we can construct the derivation

[TABLE]

Here $\mathcal{D}_{{k-\ell_{1}-1}}^{k-\ell-1}=S_{{k-\ell_{1}-1}}^{k-\ell-1}\cup R_{{k-\ell_{1}-1}}^{k-\ell-1}$, where

[TABLE]

[TABLE]

For derivation

[TABLE]

we obtain that for any endomorphism $\alpha\in\mathcal{D}_{{{k-\ell-1}}}^{k-1}\cap\mathcal{D}_{{{k-\ell_{1}-1}}}^{k-1}$, it follows $\partial_{{k-\ell_{1}-1}}^{k-1}(\alpha)=\partial_{{k-\ell_{1}-1}}^{k-\ell-1}\left(\partial_{{k-\ell-1}}^{k-1}(\alpha)\right)$. Hence,

[TABLE]

for any number $\ell_{1}$, where $\ell<\ell_{1}\leq k-1$.

In the semiring $\displaystyle\bigcap_{\ell=1}^{k-1}\mathcal{D}_{{{k-\ell-1}}}^{k-1}$ we can represent any derivation as composition of two (or more) derivations.

From (6) we find

[TABLE]

Since

[TABLE]

it follows that $\displaystyle\bigcap_{\ell=1}^{k-1}S_{{k-\ell-1}}^{k-1}=\{(a_{k-1})_{n}\}$.

On the other hand, for $\mathcal{D}_{{m}}^{k-1}$ from section 4 and $R_{{k-\ell-1}}^{k-1}$ from (5) we find that $R_{{k-\ell-1}}^{k-1}=\mathcal{D}_{{\ell}}^{k-1}$. Hence, using (7), it follows

[TABLE]

[TABLE]

So, as in the previous section, in the same semiring $\mathcal{D}$ we can represent any derivation of type $\partial_{{k-\ell-1}}^{k-1}$ as a composition of derivations.

In the case when $k=n$, the order of $\mathcal{D}=\mathcal{ON}_{n}$ is the $n$-th Catalan number. The impotant role of the Catalan numbers appears also in the next reasonings.

For $k=n$ from (4) it follows

[TABLE]

where $0\leq\ell<n-1$. As in (8) we find $S_{{1}}^{n-1}=\{(n-1)_{n}\}$.

Proposition. For any $1\leq p\leq n-2$ the order of semiring $S_{p}^{n-1}$ is

[TABLE]

where $C_{p}$ is the $p$-th Catalan number.

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Author:   Dimitrinka Vladeva, assoc. prof., Department "Mathematics and physics", LTU, Sofia, e-mail: d$\_$vladeva@abv.bg