text:
Page:United States Statutes at Large Volume 121.djvu/374
[121 STAT. 353]
PUBLIC LAW 110-000—MMMM. DD, 2007
PUBLIC LAW 110–53—AUG. 3, 2007
121 STAT. 353
dkrause on GSDDPC44 with PUBLAW
the development and implementation of such legislation, regulations, policies, and guidelines; and ''(D) in providing advice on proposals to retain or enhance a particular governmental power, consider whether the department, agency, or element of the executive branch has established— ''(i) that the need for the power is balanced with the need to protect privacy and civil liberties; ''(ii) that there is adequate supervision of the use by the executive branch of the power to ensure protection of privacy and civil liberties; and ''(iii) that there are adequate guidelines and oversight to properly confine its use. ''(2) OVERSIGHT.—The Board shall continually review— ''(A) the regulations, policies, and procedures, and the implementation of the regulations, policies, and procedures, of the departments, agencies, and elements of the executive branch relating to efforts to protect the Nation from terrorism to ensure that privacy and civil liberties are protected; ''(B) the information sharing practices of the departments, agencies, and elements of the executive branch relating to efforts to protect the Nation from terrorism to determine whether they appropriately protect privacy and civil liberties and adhere to the information sharing guidelines issued or developed under subsections (d) and (f) of section 1016 and to other governing laws, regulations, and policies regarding privacy and civil liberties; and ''(C) other actions by the executive branch relating to efforts to protect the Nation from terrorism to determine whether such actions— ''(i) appropriately protect privacy and civil liberties; and ''(ii) are consistent with governing laws, regulations, and policies regarding privacy and civil liberties. ''(3) RELATIONSHIP WITH PRIVACY AND CIVIL LIBERTIES OFFICERS.—The Board shall— ''(A) receive and review reports and other information from privacy officers and civil liberties officers under section 1062; ''(B) when appropriate, make recommendations to such privacy officers and civil liberties officers regarding their activities; and ''(C) when appropriate, coordinate the activities of such privacy officers and civil liberties officers on relevant interagency matters. ''(4) TESTIMONY.—The members of the Board shall appear and testify before Congress upon request. ''(e) REPORTS.— ''(1) IN GENERAL.—The Board shall— ''(A) receive and review reports from privacy officers and civil liberties officers under section 1062; and ''(B) periodically submit, not less than semiannually, reports— ''(i)(I) to the appropriate committees of Congress, including the Committee on the Judiciary of the Senate, the Committee on the Judiciary of the House
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Q: Unexpected behavior in keys() redis python module using NULL & ETX sequence I'm using redis with the python module to save byte sequence (as keys)
When i'm trying to get keys contains: "\x00??\x03" (? = any ASCII character)
for example:
str = "\x00\x01\x00\x03" (as byte sequence)
db.keys(pattern=''+str+'')
gives me all the keys.
Is this a normal behavior?

redpajama_set_name: RedPajamaStackExchange


text:
\section{Introduction}

\noindent A distributive lattice is a partially order set with join and meet operations which distribute over each other. The prototypical example of a such structure are the sets where join and meet are the usual union and intersection. Further examples include the Lindenbaum algebra of logics that support conjunction and disjunction, every Heyting algebra, and Young's lattice whose join and meet of two partitions are given by the union and intersection of the corresponding Young diagrams. This survey mainly aims to provide a complete proof that, if $L$ is a complete lowly finite distributive lattice, $M$ a subset of $L$ containing its join-irreducible elements, $f:L \rightarrow G$ a valuation on $L$ to a module $G$, and $a$ an element of $M$ which is not join-irreducible, then
\begin{equation} \label{EqM}
\sum_{b \in [\mathsf{0},a] \cap M} \mu_M(b,a)f(b) = 0.
\end{equation}

\noindent The proof is carried out in several stages. We first have to consider the general case of posets in Section~\ref{SecPo}. A proof of Zorn's lemma and an introduction to the Möbius algebra $\mathrm{M\ddot{o}b}(L)$ of a lowly finite poset $L$ are namely provided. The lemma bearing his name was proved by Zorn in 1935 \cite{Zo}. Although we can easily find diverse proofs of that lemma in the literature, new ones are still proposed other time like that of Lewin in 1991 \cite{Le}. Ours is inspired by the lecture notes of Debussche \cite[§~2.II]{De}. The Möbius algebra was discovered in 1967 by Solomon who defined it for finite posets \cite[§~2]{So}. We establish the Möbius inversion formula, and prove that $\displaystyle \bigg\{\sum_{b \in [\mathsf{0},a]} \mu_L(b,a)b\ \bigg|\ a \in L\bigg\}$ is a complete set of orthogonal idempotents in $\mathrm{M\ddot{o}b}(L)$.

\smallskip

\noindent We study the special case of lattices in Section~\ref{SecLa}. After viewing some essential generalities, we focus on the distributive lattices, and establish diverse properties like the distributivity of a lattice $L$ if and only if, for all $a,b,c \in L$, $c \vee a = c \vee b$ and $c \wedge a = c \wedge b$ imply $a = b$.

\smallskip

\noindent Those properties are necessary to investigate the valuation algebra in Section~\ref{SecVa}. It is the central part of this survey, and principally inspired from the articles of Geissinger \cite{Ge1}, \cite[§~3]{Ge2} and that of Zaslavsky \cite[§~2]{Za}. In that section is particularly proved that if $M$ is a subset of a complete lowly finite distributive lattice $L$ containing its join-irreducible elements, and $a$ an element of $M$ which is not join-irreducible, then $\displaystyle \sum_{b \in M \cap [\mathsf{0},a]} \mu_M(b,a)b$ belongs to the submodule $\langle a \wedge b + a \vee b - a - b\ |\ a,b \in L \rangle$ of $\mathbb{Z}L$. It is the property that allows to deduce Equation~\ref{EqM}.

\smallskip

\noindent Thereafter, we use Equation~\ref{EqM} to deduce the fundamental theorem of dissection theory in Section~\ref{SecDi}, that is, if $\mathscr{A}$ is a subspace arrangement in a topological space $T$ with $|\chi(T)| < \infty$, and $L$ a meet-refinement of $L_{\mathscr{A}}$, then $\displaystyle \sum_{C \in C_{\mathscr{A}}}\chi(C) = \sum_{X \in L} \mu_L(X,T) \chi(X)$. In its original form of 1977 \cite[Theorem~1.2]{Za}, Zaslavsky expressed it for CW complexes. The number of chambers is consequently $\displaystyle \#C_{\mathscr{A}} = \frac{1}{c} \sum_{X \in L} \mu_L(X,T) \chi(X)$ if $c$ is the Euler characteristic of every chamber. Deshpande showed a similar result in 2014 for the special case of a submanifold arrangement with chambers having the same Euler characteristic $(-1)^l$ \cite[Theorem~4.6]{Des}.

\smallskip

\noindent We finally compute the $\mathrm{f}$-polynomial of submanifold arrangements from the dissection theorem of Zaslavsky in Section~\ref{SecCo}. Face counting of topological space has doubtless its origins in the formulas for the numbers of the $i$-dimensional faces if planes in $\mathbb{R}^3$ fall into $k$ parallel families in general position established by Steiner in 1826 \cite{Ste}. About 150 years later, Alexanderson and Wetzel computed the same numbers but for an arbitrary set of planes \cite{AlWe}, and Zaslavsky for hyperplane arrangements in a Euclidean space of any dimension \cite[Theorem~A]{Z}. One of our formulas is a generalization of those results as it considers a submanifold arrangement $\mathscr{A}$ such that $\chi(X) = (-1)^{\dim X}$ for every $X \in L_{\mathscr{A}} \cup F_{\mathscr{A}}$, and states that $\mathrm{f}_{\mathscr{A}}(x) = (-1)^{\mathrm{rk}\,\mathscr{A}} \mathrm{M}_{\mathscr{A}}(-x,-1)$ where $\mathrm{M}_{\mathscr{A}}$ is the Möbius polynomial of $\mathscr{A}$. Moreover, Pakula computed the number of chambers of a pseudosphere arrangement with simple complements in 2003 \cite[Corollary~1]{Pa}. Another formula is a generalization of his result considering a submanifold arrangement $\mathscr{A}$ such that
$$\forall C \in F_{\mathscr{A}}:\, \chi(C) = (-1)^{\dim C} \quad \text{and} \quad \forall X \in L_{\mathscr{A}}:\, \chi(X) = \begin{cases}
	2 & \text{if}\ \dim X \equiv 0 \mod 2 \\
	0 & \text{otherwise}
\end{cases},$$
and states
$$\mathrm{f}_{\mathscr{A}}(x) = (-1)^{n-\mathrm{rk}\,\mathscr{A}} \big(\mathrm{M}_{\mathscr{A}}(x,-1) + \gamma_n \mathrm{M}_{\mathscr{A}}(-x,-1)\big)\ \text{with}\  \gamma_n := \begin{cases}
	1 & \text{if}\ \dim X \equiv 0 \mod 2 \\
	-1 & \text{otherwise}
\end{cases}.$$

\section{Poset} \label{SecPo}

\noindent We begin with the general case of posets. The Zorn's lemma is especially proved, and the Möbius algebra described as it plays a key role in this survey.

\begin{definition}
A \textbf{partial order} is a binary relation $\preceq$ over a set $L$ such that, for $a,b,c \in L$,
\begin{itemize}
\item $a \preceq a$,
\item if $a \preceq b$ and $a \succeq b$, then $a=b$,
\item if $a \preceq b$ and $b \preceq c$, then $a \preceq c$.
\end{itemize}
The set $L$ with a partial order is called a \textbf{partially ordered set} or \textbf{poset}, and two elements $a,b \in L$ are said comparable if $a \preceq b$ or $a \succeq b$.
\end{definition}

\begin{definition}
	A poset $L$ has an uppest resp. lowest element $\mathsf{1}$ resp. $\mathsf{0} \in L$ if, for every $a \in L$, one has $a \preceq \mathsf{1}$ resp. $a \succeq \mathsf{0}$. The poset is said \textbf{complete} if it has an uppest and a lowest element.
\end{definition}

\subsection{Zorn's Lemma}

\begin{definition}
A subset $C$ of a poset $P$ is a \textbf{chain} if any two elements in $C$ are comparable.
\end{definition}

\noindent Denote by $\mathcal{C}_L$ the set formed by the chains of a poset $L$. A subset $S$ of $L$ has an upper resp. lower bound if there exists $u$ resp. $l \in L$ such that $s \preceq u$ resp. $l \preceq s$ for each $s \in S$. The upper resp. lower bound $u$ resp. $l$ is said strict if $u$ resp. $l \notin S$.

\begin{definition}
A poset $L$ is said \textbf{inductive} if every chain included in $L$ has an upper bound.
\end{definition}

\noindent For an inductive poset $L$, and $C \in \mathcal{C}_L$, let $C_{\prec}$ be the set formed by the strict upper bound of $C$, and denote by $\mathcal{E}_L$ the set $\{C \in \mathcal{C}_L\ |\ C_{\prec} \neq \emptyset\}$. The axiom of choice allows to deduce the existence of a function $\mathrm{c}:2^L \setminus \{\emptyset\} \rightarrow L$ such that, for every $A \in 2^L \setminus \{\emptyset\}$, we have $\mathrm{c}(A) \in A$. Define the function $\mathrm{m}: \mathcal{E}_L \rightarrow L$, for $C \in \mathcal{E}_L$, by $\mathrm{m}(C) := \mathrm{c}(C_{\prec})$.

\begin{definition}
Let $S,A$ be subsets of a poset $L$. The set $S$ is called a \textbf{segment} of $A$ if
$$S \subseteq A \quad \text{and} \quad \forall s \in S,\, \forall a \in A:\ s \succeq a \Rightarrow a \in S.$$
\end{definition}

\begin{definition}
An upper resp. lower bound $u$ resp. $l$ of the subset $S$ of a poset $L$ is called a \textbf{join} resp. \textbf{meet} if $u \preceq a$ resp. $b \preceq l$ for each upper resp. lower bound $a$ resp. $b$ of $S$.
\end{definition}

\begin{definition}
A chain $C$ of an inductive poset $L$ is called a \textbf{good set} if, for every segment $S$ of $C$ with $S \neq C$, we have $S_{\prec} \cap C \neq \emptyset$ and $\mathrm{m}(S)$ is the meet of $S_{\prec} \cap C$.
\end{definition}

\noindent For elements $a,b$ of a poset, by $a \prec b$ we mean that $a \preceq b$ and $a \neq b$.

\begin{lemma} \label{LeGo}
Let $A,B$ be nonempty good sets of an inductive poset $L$. Then, either $A$ is a segment of $B$ or vice versa.
\end{lemma}

\begin{proof}
Note first that $\emptyset$ is a chain of $L$. As $L$ is inductive, $\emptyset$ has then an upper bound in $L$ which is necessary a strict upper bound, hence $\emptyset \in \mathcal{E}_L$. Moreover, since $\emptyset$ is obviously a segment of both $A$ and $B$ which are good sets, then $\mathrm{m}(\emptyset) \in \emptyset_{\prec} \cap A \cap B$ and $A \cap B \neq \emptyset$.

\noindent For $a \in A \cap B$, the sets $S_{a,A} := \{s \in A\ |\ s \prec a\}$ and $S_{a,B} := \{s \in B\ |\ s \prec a\}$ are clearly segments of $A$ and $B$ respectively. Set $C := \{a \in A \cap B\ |\ S_{a,A} = S_{a,B}\}$, and let $b \in C$, $c \in A$, with $b \succ c$. We have $c \in S_{b,A} = S_{b,B}$, then $c \in B$ which implies $c \in A \cap B$. If $d \in S_{c,A}$, then $d \prec c \prec b$ implies $d \in S_{b,A} = S_{b,B}$, hence $b \in S_{c,B}$ and $S_{c,A} \subseteq S_{c,B}$. Similarly, we have $S_{c,B} \subseteq S_{c,A}$, then $c \in C$. Therefore, $C$ is a segment of $A$ and $B$.

\noindent Suppose now that $C \neq A$ and $C \neq B$. As $A,B$ are good sets, then $\mathrm{m}(C) \in A \cap B$. Remark that $C \sqcup \big\{\mathrm{m}(C)\big\} = S_{\mathrm{m}(C),A} = S_{\mathrm{m}(C),B}$, then $\mathrm{m}(C) \in C$ which is absurd. Hence $C = A$ or $C = B$, in other words, $A$ is a segment of $B$ or vice versa.
\end{proof}

\noindent Denoting by $\mathcal{G}_L$ the set formed by the good sets of an inductive poset $L$, set $\displaystyle U_L := \bigcup_{A \in \mathcal{G}_L}A$.

\begin{theorem}
If $L$ is an inductive poset, then $U_L$ is a good set.
\end{theorem}

\begin{proof}
For $a,b \in U_L$, there exist good sets $S_a, S_b$ such that $a \in S_a$ and $b \in S_b$. Using Lemma~\ref{LeGo}, we get either $S_a \subseteq S_b$ or $S_b \subseteq S_a$. That means either $a \preceq b$ or $a \succeq b$, and $U_L$ is consequently a chain.

\noindent Let $A \in \mathcal{G}_L$, $a \in A$, and $b \in U_L$ with $a \succeq b$. There is $B \in \mathcal{G}_L$ with $b \in B$. From Lemma~\ref{LeGo},
\begin{itemize}
\item if $A$ is a segment of $B$, then $A$ is a segment and $b \in A$,
\item if $B$ is a segment of $A$, then $B \subseteq A$ and $b \in A$.
\end{itemize}
In any case, we have $b \in A$, then $A$ is a segment of $U_L$.

\noindent Consider a segment $S$ of $U_L$ such that $S \neq U_L$. Since $U_L$ is a chain, necessarily $U_L \setminus S \subseteq S_{\prec}$. Let $a \in U_L \setminus S$, and $A \in \mathcal{G}_L$ such that $a \in A$. As $A$ is a segment of $U_L$, then $S \varsubsetneq A$ and $S$ is a segment of $A$. Moreover, $\mathrm{m}(S)$ is the meet of $S_{\prec} \cap A$. If there exists $b \in S_{\prec} \cap U_L$ such that $b \prec \mathrm{m}(S)$, we would get $b \in A$, which is absurd. Therefore, $\mathrm{m}(S)$ is the meet of $S_{\prec} \cap U_L$, and $U_L$ is a good set.
\end{proof}

\begin{definition}
An element $a$ of a poset $L$ is said \textbf{maximal} if there does not exist an element $b \in L \setminus \{a\}$ such that $b \succ a$.
\end{definition}

\begin{corollary}[Zorn's Lemma] \label{CoZo}
Every inductive poset $L$ has a maximal element.
\end{corollary}

\begin{proof}
Recall that, since $U_L$ is a chain, it consequently possesses an upper bound. Suppose $U_{L \prec} \neq \emptyset$, and let $u \in U_{L \prec}$. Then $U_L \sqcup \{u\}$ is a good set which is absurd. Hence, $U_L$ has a unique upper bound, contained in $U_L$, which is a maximal element of $L$.
\end{proof}

\subsection{Möbius Algebra}

\noindent For two elements $a,b$ of a poset $L$ such that $a \preceq b$, denote by $[a,b]$ the set $\{c \in L\ |\ a \preceq c \preceq b\}$.

\begin{definition}
A poset $L$ is \textbf{locally finite} if, for all $a,b \in L$ such that $a \preceq b$, $[a,b]$ is finite.
\end{definition}

\noindent For a locally finite poset $L$, denote by $\mathsf{Inc}(L)$ the module of the functions $f:L^2 \rightarrow \mathbb{Z}$ with the property that, if $x,y \in L$, then $f(x,y) = 0$ if $x \npreceq y$.

\begin{definition}
The \textbf{incidence algebra} $\mathrm{Inc}(L)$ of a locally finite poset $L$ is the module of functions $f:L^2 \rightarrow \mathbb{Z}$, having the property $f(a,b) = 0$ if $a \npreceq b$, with distributive multiplication $h = f \cdot g$ defined, for $f,g \in \mathrm{Inc}(L)$, by
$$h(a,b):=0\ \text{if}\ a \npreceq b \quad \text{and} \quad h(a,b) := \sum_{c \in [a,b]}f(a,c)g(c,b)\ \text{otherwise}.$$
Its multiplicative identity is the Kronecker delta $\delta:L^2 \rightarrow \mathbb{Z}$ with $\delta(a,b) := \begin{cases}
1 & \text{if}\ a=b, \\
0 & \text{otherwise}
\end{cases}$.
\end{definition}

\begin{definition}
For a locally finite poset $L$, the \textbf{zeta function} $\zeta_L$ and the \textbf{Möbius function} $\mu_L$ in the incidence algebra $\mathrm{Inc}(L)$ are defined, for $a,b \in L$ with $a \preceq b$, by
$$\zeta_L(a,b):= 1 \quad \text{and} \quad \mu_L(a,b) := \begin{cases}
1 & \text{if}\ a=b \\
\displaystyle -\sum_{\substack{c \in [a,b] \\ c \neq b}} \mu_L(a,c) = -\sum_{\substack{c \in [a,b] \\ c \neq a}} \mu_L(c,b) & \text{otherwise}
\end{cases}.$$
\end{definition}

\begin{lemma} \label{LeZe}
For a locally finite poset $L$, the zeta function is the multiplicative inverse of the Möbius function in the incidence algebra $\mathrm{Inc}(L)$.
\end{lemma}

\begin{proof}
For $a,b \in L$ with $a \preceq b$, we have $\zeta_L \cdot \mu_L(a,a) = \mu_L \cdot \zeta_L(a,a) = 1 = \delta(a,a)$, but also
$$\zeta_L \cdot \mu_L(a,b) = \sum_{c \in [a,b]} \mu_L(c,b) = 0 = \delta(a,b) \quad \text{and} \quad \mu_L \cdot \zeta_L(a,b) = \sum_{c \in [a,b]} \mu_L(a,c) = 0 = \delta(a,b).$$
\end{proof}

\noindent The proof of this proposition is inspired from the original proof of Rota \cite[Proposition~2]{Ro}.

\begin{proposition}[Möbius Inversion Formula]
Let $L$ be a locally finite poset, $a,b \in L$ with $a \preceq b$, and $f,g$ two functions from $L$ onto a module $M$ over $\mathbb{Z}$. Then,
$$\forall x \in [a,b]:\,g(x) = \sum_{c \in [a,x]} f(c) \quad \Longleftrightarrow \quad \forall x \in [a,b]:\,f(x) = \sum_{c \in [a,x]} g(c) \mu_L(c,x).$$
\end{proposition}

\begin{proof}
Assume first that, for every $x \in [a,b]$, $\displaystyle g(x) = \sum_{c \in [a,x]} f(c)$. Using Lemma~\ref{LeZe}, we get
\begin{align*}
\sum_{c \in [a,x]} g(c) \mu_L(c,x) & = \sum_{c \in [a,x]} \sum_{d \in [a,c]} f(d) \mu_L(c,x) = \sum_{c \in [a,x]} \sum_{d \in [a,c]} f(d) \zeta_L(d,c) \mu_L(c,x) \\
& = \sum_{d \in [a,c]} \sum_{c \in [a,x]} f(d) \zeta_L(d,c) \mu_L(c,x) = \sum_{d \in [a,c]} f(d) \sum_{c \in [a,x]} \zeta_L(d,c) \mu_L(c,x) \\
& = \sum_{d \in [a,c]} f(d)\, \zeta_L \cdot \mu_L(d,x) = \sum_{d \in [a,c]} f(d) \delta(d,x) \\
& = f(x).
\end{align*}
Similarly, if $\displaystyle f(x) = \sum_{c \in [a,x]} g(c) \mu_L(c,x)$ for every $x \in [a,b]$, we obtain
\begin{align*}
\sum_{c \in [a,x]} f(c) & = \sum_{c \in [a,x]} \sum_{d \in [a,c]} g(d) \mu_L(d,c) = \sum_{c \in [a,x]} \sum_{d \in [a,c]} g(d) \mu_L(d,c) \zeta_L(c,x) = \sum_{d \in [a,c]} g(d) \delta(d,x) \\
& = g(x).
\end{align*}
\end{proof}

\begin{definition}
We say that a poset $L$ is \textbf{lowly finite} if the set $\{b \in L\ |\ b \preceq a\}$ is finite for any $a \in L$.
\end{definition}

\noindent For a lowly finite poset $L$ and $a \in L$, let $u_L(a)$ be the element $\displaystyle \sum_{\substack{c \in L \\ c \preceq a}} \mu_L(c,a)c$ of the module $\mathbb{Z}L$.

\begin{definition}
The \textbf{Möbius Algebra} $\mathrm{M\ddot{o}b}(L)$ of a lowly finite poset $L$ is the module $\mathbb{Z}L$ with distributive multiplication defined, for $a,b \in L$, by
$$a \cdot b := \sum_{\substack{c \in L \\ c \preceq a,\, c \preceq b}} u_L(c).$$
\end{definition}

\noindent Remark that the Möbius algebra was initial defined for finite posets \cite[§~2]{So}. For a lowly finite poset $L$ with a lowest element, define the algebra $\mathrm{A}_L := \langle \alpha_a\ |\ a \in L\rangle$ over $\mathbb{Z}$ with multiplication
$$\alpha_a \alpha_b := \begin{cases}
	\alpha_a & \text{if}\ a=b,\\
	0 & \text{otherwise}
\end{cases}.$$
To each $a \in L$, associate an element $a' \in \mathrm{A}_L$ by setting $\displaystyle a' := \sum_{b \in [\mathsf{0},a]} \alpha_b$.

\begin{lemma} \label{LeBas}
For a lowly finite poset $L$ with a lowest element, the set $\{a'\ |\ a \in L\}$ forms a basis of the algebra $\mathrm{A}_L$.
\end{lemma}

\begin{proof}
From the Möbius inversion formula, we get $\displaystyle \alpha_a = \sum_{b \in [\mathsf{0},a]} \mu(b,a) b'$. The set $\{a'\ |\ a \in L\}$ consequently generates $\mathrm{A}_L$. Suppose that there exists a finite set $I \subseteq L$ and an integer set $\{i_a\}_{a \in I}$ such that $\displaystyle \sum_{a \in I} i_a a' = 0$. If $b$ is a maximal element of $I$, then $\displaystyle \alpha_b \sum_{a \in I} i_a a' = i_b \alpha_b = 0$, hence $i_b = 0$. Inductively, we deduce that $i_a = 0$ for every $a \in I$. The set $\{a'\ |\ a \in L\}$ is therefore independent.
\end{proof}

\noindent The following results were initially established by Greene for finite lattice \cite[§~1]{Gre}.

\begin{theorem}
For a lowly finite poset $L$ with a lowest element, the map $\phi: L \rightarrow \mathrm{A}_L,\, a \mapsto a'$ extends to an algebra isomorphism from $\mathrm{M\ddot{o}b}(L)$ to $\mathrm{A}_L$.
\end{theorem}

\begin{proof}
	The map $\phi$ clearly becomes a module homomorphism by linear extension, and an isomorphism by Lemma~\ref{LeBas}. Moreover, for $a,b \in L$,
	\begin{align*}
		\phi(a \cdot b) & = \phi\Big(\sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b]} u_L(c)\Big) = \sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b]} \phi\Big(\sum_{d \in [\mathsf{0},c]} \mu(d,c)d\Big) \\
		& = \sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b]} \sum_{d \in [\mathsf{0},c]} \mu(d,c)d' = \sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b]} \alpha_c,
	\end{align*}
	and $\displaystyle \phi(a) \phi(b) = a'b' = \sum_{c \in [\mathsf{0},a]} \alpha_c \times \sum_{d \in [\mathsf{0},b]} \alpha_d = \sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b]} \alpha_c$. Then $\phi(a \cdot b) = \phi(a) \phi(b)$, and $\phi$ is consequently an algebra isomorphism.
\end{proof}

\begin{corollary} \label{CoBaI}
For a lowly finite poset $L$ with a lowest element, the set $\{u_L(a)\ |\ a \in L\}$ is a complete set of orthogonal idempotents in $\mathrm{M\ddot{o}b}(L)$.
\end{corollary}

\begin{proof}
	Since $\displaystyle \phi\big(u_L(a)\big) = \sum_{b \in [\mathsf{0},a]} \mu_L(b,a)b' = \alpha_a$, then $\{u_L(a)\ |\ a \in L\}$ is a basis of $\mathrm{M\ddot{o}b}(L)$. Moreover $\phi\big(u_L(a) \cdot u_L(a)\big) = \alpha_a = \phi\big(u_L(a)\big)$, so the $u_L(a)$'s are idempotents.\\ Finally $\phi\big(u_L(a) \cdot u_L(b)\big) = \alpha_a \alpha_b = 0$ if $a \neq b$, hence the $u_L(a)$'s are orthogonal.
\end{proof}

\begin{corollary} \label{CoHoA}
Let $L$ be a lowly finite poset with a lowest element, and $M$ a subset of $L$ containing $\mathsf{0}$. Then, the linear map $\mathrm{j}: \mathrm{M\ddot{o}b}(L) \rightarrow \mathrm{M\ddot{o}b}(M)$, which on the basis $\{u_L(a)\ |\ a \in L\}$ has the values
$$\mathrm{j}\big(u_L(a)\big) := \begin{cases}
u_M(a) & \text{if}\ a \in M,\\
0 & \text{otherwise},
\end{cases}$$
is an algebra homomorphism.
\end{corollary}

\begin{proof}
Using Corollary~\ref{CoBaI}, $\mathrm{j}\big(u_L(a) \cdot u_L(a)\big) = \mathrm{j}\big(u_L(a)\big) = u_M(a) = \mathrm{j}\big(u_L(a)\big) \cdot \mathrm{j}\big(u_L(a)\big)$ if $a \in M$. Otherwise, $\mathrm{j}\big(u_L(a) \cdot u_L(a)\big) = 0 = \mathrm{j}\big(u_L(a)\big) \cdot \mathrm{j}\big(u_L(a)\big)$. For $a,b \in L$ with $a \neq b$, $\mathrm{j}\big(u_L(a) \cdot u_L(b)\big) = 0 = \mathrm{j}\big(u_L(a)\big) \cdot \mathrm{j}\big(u_L(b)\big)$.
\end{proof}

\section{Lattice} \label{SecLa}

\noindent We study the special but important case of lattices. After viewing some generalities, we focus on distributive ones, and establish diverse properties which are necessary to investigate the valuation algebra in the next section.

\begin{definition}
	A poset $L$ is a join-semilattice resp. meet-semilattice if each $2$-element subset $\{a,b\} \subseteq L$ has a join resp. meet denoted by $a \vee b$ resp. $a \wedge b$. It is called a \textbf{lattice} if $L$ is both a join- and meet-semilattice, moreover $\vee$ and $\wedge$ become binary operations on $L$.
\end{definition}

\begin{proposition} \label{LeLow}
If a lattice $L$ is lowly finite, then it has a lowest element.
\end{proposition}

\begin{proof}
	For any $a \in L$, the principal ideal $\mathrm{id}(a)$ has a lowest element which is $\displaystyle \mathsf{0}_a := \bigwedge_{x \in \mathrm{id}(a)}x$. Consider $b \in L \setminus \{a\}$ and the lowest element $\mathsf{0}_b$ of $\mathrm{id}(b)$. The fact $\mathsf{0}_a \wedge \mathsf{0}_b \neq \mathsf{0}_a$ would contradict the fact that $\mathsf{0}_a$ is the lowest element of $\mathrm{id}(a)$. Hence, $L$ has a lowest element $\mathsf{0}$.
\end{proof}

\subsection{Generalities on Lattice}

\begin{definition}
	A \textbf{sublattice} of a lattice $L$ is a nonempty subset $M \subseteq L$ such that, for all $a,b \in M$, we have $a \vee b \in M$ and $a \wedge b \in M$.
\end{definition}

\begin{definition}
	A \textbf{lattice homomorphism} is a function $\varphi: L_1 \rightarrow L_2$ between two lattices $L_1$ and $L_2$ such that, for all $a,b \in L_1$,
	$$\varphi(a \vee b) = \varphi(a) \vee \varphi(b) \quad \text{and} \quad \varphi(a \wedge b) = \varphi(a) \wedge \varphi(b).$$
\end{definition}

\begin{definition}
	An \textbf{ideal} of a lattice $L$ is a sublattice $I \subseteq L$ such that, for any $a \in I$ and $b \in L$, we have $a \wedge b \in I$. If in addition $I \neq L$ and, for any $a \wedge b \in I$, either $a \in I$ or $b \in I$, then $I$ is a \textbf{prime ideal}.
\end{definition}

\begin{definition}
	Dually, a \textbf{filter} of a lattice $L$ is a sublattice $F \subseteq L$ such that, for any $a \in F$ and $b \in L$, we have $a \vee b \in F$. If in addition $F \neq L$ and, for any $a \vee b \in F$, either $a \in F$ or $b \in F$, then $F$ is a \textbf{prime filter}.
\end{definition}

\begin{proposition} \label{LeDu}
	A subset $M$ of a lattice $L$ is a prime ideal if and only if the subset $L \setminus M$ is a prime filter.
\end{proposition}

\begin{proof}
	Assume that $M$ is a prime ideal:
	\begin{itemize}
		\item If $a,b \in L \setminus M$, clearly $a \wedge b \in L \setminus M$ and $a \vee b \in L \setminus M$ since $(a \vee b) \wedge b \in L \setminus M$, then $L \setminus M$ is a sublattice.
		\item If $a \in M$ and $b \in L \setminus M$, once again $a \vee b \in L \setminus M$ since $(a \vee b) \wedge b \in L \setminus M$, then $L \setminus M$ is a filter.
		\item If $a \vee b \in L \setminus M$, it is clear that both $a,b$ cannot be all in $M$, then $L \setminus M$ is prime.
	\end{itemize}
	One similarly proves that if $M$ is a prime filter, then $L \setminus M$ is a prime ideal.
\end{proof}

\begin{definition}
Let $L$ be a lattice, and $a \in L$. The \textbf{principal ideal} generated by $a$ is the ideal $\mathrm{id}(a) := \{b \in L\ |\ b \preceq a\}$, dually the \textbf{principal filter} generated by $a$ is the filter $\mathrm{fil}(a) := \{b \in L\ |\ b \succeq a\}$.
\end{definition}

\begin{definition}
	An element $a$ of a lattice $L$ is \textbf{join-irreducible} if, for any subset $S \subseteq L$, $\displaystyle a = \bigvee_{b \in S}b$ implies $a \in S$. Denote by $\mathrm{ji}(L)$ the set formed by the join-irreducible elements of $L$.
\end{definition}

\begin{lemma} \label{LeJi}
	Let $L$ be a lattice, and $a \in L$. Then, $a \in \mathrm{ji}(L)$ if and only if $\displaystyle a \neq \bigvee_{\substack{b \in L \\ b \prec a}} b$.
\end{lemma}

\begin{proof}
	If $a \in \mathrm{ji}(L)$, as $a \notin \{b \in L\ |\ b \prec a\}$, then $\displaystyle a \neq \bigvee_{\substack{b \in L \\ b \prec a}} b$.

	\noindent Assume now that $\displaystyle a \neq \bigvee_{\substack{b \in L \\ b \prec a}} b$, and let $S \subseteq L$ such that $\displaystyle a = \bigvee_{b \in S}b$. Since $b \preceq a$ for every $b \in S$, the only possibility is $a \in S$, and consequently $a \in \mathrm{ji}(L)$.
\end{proof}

\noindent The proof of the following proposition takes inspiration from that of Bhatta and Ramananda \cite[Proposition~2.2]{BhRa}.

\begin{proposition} \label{PrJi}
	Let $L$ be a lowly finite lattice, and $a \in L$. Then, $\displaystyle a = \bigvee_{b \, \in \, \mathrm{id}(a) \cap \mathrm{ji}(L)}b$.
\end{proposition}

\begin{proof}
	It is obvious if $a \in \mathrm{ji}(L)$. Now, assume that $a \in L \setminus \mathrm{ji}(L)$ and $\displaystyle a \neq \bigvee_{b \, \in \, \mathrm{id}(a) \cap \mathrm{ji}(L)}b$. The set $\displaystyle S = \Big\{x \in L\ \Big|\ x \neq \bigvee_{b \, \in \, \mathrm{id}(x) \cap \mathrm{ji}(L)}b\Big\}$ is nonempty and has a minimal element $c$ as $L$ is lowly finite. Since $\displaystyle c \neq \bigvee_{b \, \in \, \mathrm{id}(c) \cap \mathrm{ji}(L)}b$, then $c \notin \mathrm{ji}(L)$, and it follows from Lemma~\ref{LeJi} that $\displaystyle c = \bigvee_{\substack{b \in L \\ b \prec c}} b$. Clearly, $c$ is an upper bound of the set $\displaystyle X = \bigcup_{\substack{b \in L \\ b \prec c}} \mathrm{id}(b) \cap \mathrm{ji}(L)$. If $u$ is another upper bound of $X$, then $u$ is an upper bound of $\mathrm{id}(x) \cap \mathrm{ji}(L)$ for every $x \in L$ with $x \prec c$. As $c$ is minimal in $S$, then $\displaystyle x = \bigvee_{b \, \in \, \mathrm{id}(x) \cap \mathrm{ji}(L)} b$ if $x \prec c$, hence $u$ is an upper bound of $\{b \in L\ |\ b \prec c\}$ implying $u \succeq c$. Observe that
	$\displaystyle X = \bigcup_{\substack{b \in L \\ b \prec c}}\big(\mathrm{id}(b) \cap \mathrm{ji}(L)\big) = \mathrm{ji}(L) \cap \bigcup_{\substack{b \in L \\ b \prec c}} \mathrm{id}(b) = \mathrm{ji}(L) \cap \mathrm{id}(c)$. Therefore, $c$ is a minimal upper bound for $\mathrm{id}(c) \cap \mathrm{ji}(L)$ which is a contradiction.
\end{proof}

\noindent For two elements $a,b$ of a lattice $L$ such that $a \preceq b$, let $\mathrm{j}_{a,b}: [a \wedge b,\, b] \rightarrow [a,\, a \vee b]$ and $\mathrm{m}_{a,b}: [a,\, a \vee b] \rightarrow [a \wedge b,\, b]$ be functions respectively defined by $$\mathrm{j}_{a,b}(x):= a \vee x \quad \text{and} \quad \mathrm{m}_{a,b}(x):= x \wedge b.$$

\begin{definition}
A lattice $L$ is \textbf{modular} if, for all $a,b \in L$, $x \in [a \wedge b,\, b]$, and $y \in [a,\, a \vee b]$, we have $$x = \mathrm{m}_{a,b}\, \mathrm{j}_{a,b}(x) \quad \text{and} \quad y = \mathrm{j}_{a,b}\, \mathrm{m}_{a,b}(y).$$
\end{definition}

\begin{proposition} \label{PrMo}
A lattice $L$ is modular if and only if, for all $a,b,z \in L$, we have
$$(a \vee z) \wedge (a \vee b) = a \vee \big(z \wedge (a \vee b)\big) \quad \text{and} \quad (a \wedge z) \vee (a \wedge b) = a \wedge \big(z \vee (a \wedge b)\big).$$
\end{proposition}

\begin{proof}
Assume first that $L$ is modular. We have $a \preceq (a \vee z) \wedge (a \vee b) \preceq a \vee b$. Letting $u = (a \vee z) \wedge (a \vee b)$, we get $$u = \mathrm{j}_{a,b}\, \mathrm{m}_{a,b}(u) = a \vee \big((a \vee z) \wedge (a \vee b) \wedge b\big) = a \vee \big((a \vee z) \wedge b\big).$$
Since it is true for all $a,b,z \in L$, interchanging $z$ and $b$, we obtain $u = a \vee \big(z \wedge (a \vee b)\big)$. Likewise, we have $a \wedge b \preceq (z \wedge b) \vee (a \wedge b) \preceq b$. Letting $v = (z \wedge b) \vee (a \wedge b)$, we get $$v = \mathrm{m}_{a,b}\, \mathrm{j}_{a,b}(v) = b \wedge \big((z \wedge b) \vee (a \wedge b) \vee a\big) = b \wedge \big((b \wedge z) \vee a\big).$$
Since it is true for all $a,b,z \in L$, interchanging $z$ and $a$, we obtain $v = b \wedge \big(z \vee (a \wedge b)\big)$.

\smallskip

\noindent Assume now that $(a \vee z) \wedge (a \vee b) = a \vee \big(z \wedge (a \vee b)\big)$ and $(a \wedge z) \vee (a \wedge b) = a \wedge \big(z \vee (a \wedge b)\big)$ for all $a,b,z \in L$. If $a \preceq z \preceq a \vee b$, then
$$z = (a \vee b) \wedge z = (a \vee b) \wedge (a \vee z) = a \vee \big(b \wedge (a \vee z)\big).$$
Since $a \vee z = z$, then $z = a \vee (z \wedge b) = \mathrm{j}_{a,b}\, \mathrm{m}_{a,b}(z)$. Likewise, if $a \wedge b \preceq z \preceq b$, then
$$z = (a \wedge b) \vee z = (b \wedge a) \vee (z \wedge b) = b \wedge \big(a \vee (z \wedge b)\big).$$
And since $z \wedge b = z$, then $z = b \vee (a \wedge z) = \mathrm{m}_{a,b}\, \mathrm{j}_{a,b}(z)$.
\end{proof}

\subsection{Distributive Lattice}

\begin{proposition}
	Let $L$ be a poset, and $a,b,c \in L$. The condition
	$$a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c) \text{\: is equivalent to\: } a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c).$$
\end{proposition}

\begin{proof}
	Assume that $a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c)$. Then,
	\begin{align*}
		(a \vee b) \wedge (a \vee c) & = \big((a \vee b) \wedge a\big) \vee \big((a \vee b) \wedge c\big) \\
		& = a \vee \big((a \vee b) \wedge c\big) \\
		& = a \vee (a \wedge c) \vee (b \wedge c) \\
		& = a \vee (b \wedge c).
	\end{align*}
	Similarly, if we assume $a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c)$, then we obtain
	$$(a \wedge b) \vee (a \wedge c) = \big((a \wedge b) \vee a\big) \wedge \big((a \wedge b) \vee c\big) = a \wedge (b \vee c).$$
\end{proof}

\begin{definition}
	A lattice $L$ is \textbf{distributive} if, for all $a,b,c \in L$, $a \wedge (b \vee c) = (a \wedge b) \vee (a \wedge c)$.
\end{definition}

\noindent Denote by $\mathcal{I}_L$ the poset formed by the ideals of a lattice $L$ with inclusion as partial order. It is a lattice such that, for $I,J \in \mathcal{I}_L$, $\displaystyle I \vee J := \bigcap_{\substack{K \in \mathcal{I}_L \\ I,J \subseteq K}}K$ and $\displaystyle I \wedge J := \bigcup_{\substack{K \in \mathcal{I}_L \\ K \subseteq I \cap J}}K$.

\begin{theorem} \label{ThPr}
Let $L$ be a distributive lattice, $I$ an ideal of $L$, and $F$ a filter of $L$ such that $I \cap F = \emptyset$. Then, there exists a prime ideal $P$ of $L$ such that $I \subseteq P$ and $P \cap F = \emptyset$.
\end{theorem}

\begin{proof}
Set $\displaystyle \mathcal{X}_{I,F} := \{M \in \mathcal{I}_L\ |\ I \subseteq M,\, M \cap F = \emptyset\}$. It is a poset with inclusion as partial order, and is nonempty since $I \in \mathcal{X}_{I,F}$. Consider a chain $\mathcal{E} \in \mathcal{C}_{\mathcal{X}_{I,F}}$, and let $\displaystyle E = \bigcup_{C \in \mathcal{E}} E$. If $a,b \in E$, then $a \in A$ and $b \in B$ for some $A,B \in \mathcal{E}$. Since $\mathcal{E}$ is a chain, either $A \subseteq B$ or $A \supseteq B$ hold, so let assume $A \subseteq B$. Then, $a \in B$, and $a \vee b \in B \subseteq E$, as $B$ is an ideal. Moreover, if $c \in L$, then $a \wedge c \in A \subseteq E$, as $A$ is also an ideal. We deduce that $E \in \mathcal{I}_L$. Besides, $I \subseteq E$ and $E \cap F = \emptyset_{\prec}$ obviously hold. Hence, $E$ is an upper bound of $\mathcal{E}$ in $\mathcal{X}_{I,F}$. Therefore, $\mathcal{X}_{I,F}$ is an inductive poset, and Zorn's lemma allows to state that it has a maximal element $P$.

\noindent Suppose that $P$ is not prime. Then, there exists $a,b \in L$ such that $a,b \notin P$ but $a \wedge b \in P$. The maximality of $P$ yields $\big(P \vee \mathrm{id}(a)\big) \cap F \neq \emptyset$ and $\big(P \vee \mathrm{id}(b)\big) \cap F \neq \emptyset$. Thus, there are $p,q \in P$ such that $p \vee a \in F$, $q \vee b \in F$, and $(p \vee a) \wedge (q \vee b) \in F$ since $F$ is a filter. Expanding by distributivity, we obtain
$$(p \vee a) \wedge (q \vee b) = \big((p \vee a) \wedge q\big) \vee \big((p \vee a) \wedge b\big) = (p \wedge q) \vee (a \wedge q) \vee (p \wedge b) \vee (a \wedge b)$$
which belongs to $P$. That means $P \cap F \neq \emptyset$ or a contradiction.
\end{proof}

\begin{corollary} \label{CoPr}
Let $L$ be a distributive lattice, $I \in \mathcal{I}_L$, and $a \in L$ such that $a \notin I$. Then, there exists a prime ideal $P$ of $L$ such that $I \subseteq P$ and $a \notin P$.
\end{corollary}

\begin{proof}
Remark that $I \neq \mathrm{fil}(a) = \emptyset$, otherwise, if $b \in I \neq \mathrm{fil}(a)$, then $b \wedge a = a \in I$, which is absurd. Now, for the proof, we apply Theorem~\ref{ThPr} to $I$ and $F = \mathrm{fil}(a)$.
\end{proof}

\begin{corollary} \label{CoSep}
Let $L$ be a distributive lattice, and $a,b \in L$ such that $a \neq b$. Then, $L$ has a prime ideal containing exactly one of $a$ and $b$.
\end{corollary}

\begin{proof}
If $a$ and $b$ are not comparable or $b \prec a$, then $a \notin \mathrm{id}(b)$. It remains to apply Corollary~\ref{CoPr} to $I = \mathrm{id}(b)$.
\end{proof}

\begin{theorem} \label{ThEq}
A lattice $L$ is distributive if and only if, for all $a,b,c \in L$, $c \vee a = c \vee b$ and $c \wedge a = c \wedge b$ imply $a = b$.
\end{theorem}

\begin{proof}
Suppose first that $L$ is distributive and that there exist $a,b,c \in L$ such that $a \vee c = b \vee c$ and $a \wedge c = b \wedge c$. Then,
$$a = a \vee (a \wedge c) = a \vee (b \wedge c) = (a \vee b) \wedge (a \vee c) = (a \vee b) \wedge (b \vee c) = b \vee (a \wedge c),$$
which implies $a \preceq b$, and similarly we have $b \preceq a$.

\smallskip

\noindent Suppose now that $a \vee c = b \vee c$ and $a \wedge c = b \wedge c$ imply $a = b$. If $x \in [a \wedge b,\,b]$,
\begin{itemize}
\item as $x \preceq b \wedge (a \vee x)$ then $a \vee x \preceq a \vee \big(b \wedge (a \vee x)\big)$, as $a \vee x \succeq b \wedge (a \vee x)$ then $a \vee x \succeq a \vee \big(b \wedge (a \vee x)\big)$, hence $a \vee x = a \vee \big(b \wedge (a \vee x)\big)$ on one side,
\item on the other side, $a \wedge x = a \wedge b \wedge x = a \wedge b \wedge (a \vee x)$.
\end{itemize}
By canceling $a$, we obtain $x = \mathrm{m}_{a,b}\, \mathrm{j}_{a,b}(x)$. If $y \in [a,\,a \vee b]$, as $y \succeq a \vee (b \wedge y)$ then $b \wedge y \succeq b \wedge \big(a \vee (b \wedge y)\big)$, as $b \wedge y \preceq a \vee (b \wedge y)$ then $b \wedge y \preceq b \wedge \big(a \vee (b \wedge y)\big)$, hence $b \wedge y = b \wedge \big(a \vee (b \wedge y)\big)$ on one side, and $b \vee y = a \vee b \vee y = a \vee b \vee (b \wedge y)$ on the other side. By canceling $b$, we obtain $y = \mathrm{j}_{a,b}\, \mathrm{m}_{a,b}(y)$. Therefore, $L$ is modular.

\noindent Let $a^* = a \wedge (b \vee c)$, $b^* = b \wedge (c \vee a)$, and $c^* = c \wedge (a \vee b)$. Then, $a^* \wedge b^* = a \wedge (c \vee a) \wedge b \wedge (b \vee c) = a \wedge b$, $a^* \wedge c^* = a \wedge c$, and $b^* \wedge c^* = b \wedge c$. Set $d = (a \vee b) \wedge (b \vee c) \wedge (c \vee a)$. Using twice Proposition~\ref{PrMo}, we get
\begin{align*}
a^* \vee b^* & = a^* \vee \big(b \wedge (a \vee c)\big) = (a^* \vee b) \wedge (a \vee c) \\
& = \Big(\big((b \vee c) \wedge a\big) \vee b\Big) \wedge (a \vee c) = (b \vee c) \wedge (a \vee b) \wedge (a \vee c) \\
& = d.
\end{align*}
By symmetry, we also have $a^* \vee c^* = b^* \vee c^* = d$. Hence,
\begin{itemize}
\item $c^* \vee a^* \vee (b \wedge c) = c^* \vee b^* \vee (a \wedge c) = d$,
\item and $c^* \wedge \big(a^* \vee (b \wedge c)\big) = (c^* \wedge a^*) \vee (b \wedge c) = (c^* \wedge b^*) \vee (a \wedge c) = c^* \wedge \big(b^* \vee (a \wedge c)\big)$.
\end{itemize}
By canceling $c^*$, we obtain $a^* \vee (b \wedge c) = b^* \vee (a \wedge c)$, whence
$$a^* \vee (b \wedge c) = a^* \vee (b \wedge c) \vee b^* \vee (a \wedge c) = a^* \vee b^* = d.$$
It follows that $(a \vee b) \wedge c = c^* = c^* \wedge d = c^* \wedge \big(a^* \vee (b \wedge c)\big) = (a \wedge c) \vee (b \wedge c)$, hence $L$ is consequently distributive.
\end{proof}

\section{Valuation on Lattice} \label{SecVa}

\noindent This section is the central part of this survey. After defining the valuation algebra and showing some important properties, we prove that if $M$ is a subset of a complete lowly finite distributive lattice $L$ containing its join-irreducible elements, and $a$ an element of $M$ which is not join-irreducible, then $\displaystyle \sum_{b \in M \cap [\mathsf{0},a]} \mu_M(b,a)b$ belongs to the submodule $\langle a \wedge b + a \vee b - a - b\ |\ a,b \in L \rangle$ of $\mathbb{Z}L$. It would not have been possible to write the first two subsections without the articles of Geissinger \cite{Ge1}, \cite[§~3]{Ge2}, and the third without that of Zaslavsky \cite[§~2]{Za}.

\begin{definition}
A \textbf{valuation} on a lattice $L$ is a function $f$ from $L$ to a module $G$ such that, for all $a,b \in L$,
$$f(a \wedge b) + f(a \vee b) = f(a) + f(b).$$
\end{definition}

\subsection{Valuation Module}

\begin{definition}
The \textbf{valuation module} of a lattice $L$ is the module $\mathrm{Val}(L) := \mathbb{Z}L/\mathrm{N}(L)$, where $\mathrm{N}(L)$ is the submodule $\langle a \wedge b + a \vee b - a - b\ |\ a,b \in L \rangle$ of the module $\mathbb{Z}L$.
\end{definition}

\begin{proposition} \label{PrHom}
Let $\mathrm{i}:L \rightarrow \mathrm{Val}(L)$ be the natural induced map for a lattice $L$. Then, $\mathrm{i}$ is a valuation, and, for every valuation $f:L \rightarrow G$, there exists a unique module homomorphism $h: \mathrm{Val}(L) \rightarrow G$ such that the following diagram is commutative
\begin{center}
	\begin{tikzcd}
		L \arrow[rd, "f"] \arrow[r, "\mathrm{i}"] & \mathrm{Val}(L) \arrow[d, "h"] \\
		& G
	\end{tikzcd}
\end{center}
\end{proposition}

\begin{proof}
It is clear that $\mathrm{i}$ is a valuation as $\mathrm{i}(a \wedge b) + \mathrm{i}(a \vee b) - \mathrm{i}(a) - \mathrm{i}(b) = a \wedge b + a \vee b - a - b = 0$. Besides, we get the homomorphism $h$ by setting
$$\forall a \in L:\, h(a) := f(a) \quad \text{and} \quad \forall x,y \in \mathrm{Val}(L):\, h(x+y) = h(x) + h(y).$$
\end{proof}

\begin{proposition}
For lattices $L_1, L_2$ with natural induced maps $\mathrm{i}_1, \mathrm{i}_2$ respectively, a lattice homomorphism $\varphi: L_1 \rightarrow L_2$ induces a unique module homomorphism $\psi: \mathrm{Val}(L_1) \rightarrow \mathrm{Val}(L_2)$ such that, for every $a \in L_1$, $\psi \mathrm{i}_1(a) = \mathrm{i}_2 \varphi(a)$.
\end{proposition}

\begin{proof}
We obtain the homomorphism $\psi$ by setting
$$\forall a \in L_1:\, \psi(a) := \varphi(a) \quad \text{and} \quad \forall x,y \in \mathrm{Val}(L_1):\, \psi(x+y) = \psi(x) + \psi(y).$$
\end{proof}

\begin{proposition} \label{PrIn}
For any prime ideal or prime filter $M$ of a lattice $L$ with natural induced map $\mathrm{i}$, each element of $\mathrm{i}(M)$ is linearly independent of those in $\mathrm{i}(L \setminus M)$ and vise versa.
\end{proposition}

\begin{proof}
Assume that $M$ is a prime ideal, and consider the indicator function $\mathrm{1}_M:L \rightarrow \mathbb{Z}$ defined as $\mathrm{1}_M(a) := \begin{cases}
	1 & \text{if } a \in M \\
	0 & \text{otherwise}
\end{cases}$. For $a,b \in L$,
\begin{itemize}
\item if $a,b \in M$, we clearly have $\mathrm{1}_M(a \wedge b) + \mathrm{1}_M(a \vee b) = \mathrm{1}_M(a) + \mathrm{1}_M(b) = 2$,
\item if $a \in M$ and $b \notin M$, since $(a \vee b) \wedge b = b \notin M$, then $a \vee b \notin M$ and $\mathrm{1}_M(a \wedge b) + \mathrm{1}_M(a \vee b) = \mathrm{1}_M(a) + \mathrm{1}_M(b) = 1$,
\item if $a,b \notin M$, then $a \wedge b \notin M$, the fact $(a \vee b) \wedge b = b \notin M$ implies $a \vee b \notin M$, and $\mathrm{1}_M(a \wedge b) + \mathrm{1}_M(a \vee b) = \mathrm{1}_M(a) + \mathrm{1}_M(b) = 0$.
\end{itemize}
Therefore, $\mathrm{1}_M$ is a valuation on $L$. One similarly proves that if $M$ is prime filter, then $\mathrm{1}_M$ is also a valuation on $L$. We know from Proposition~\ref{PrHom} that there exists a unique homomorphism $h: \mathrm{Val}(L) \rightarrow \mathbb{Z}$ such that the diagram
\begin{tikzcd}
		L \arrow[rd, "\mathrm{1}_M"] \arrow[r, "\mathrm{i}"] & \mathrm{Val}(L) \arrow[d, "h"] \\
		& \mathbb{Z}
\end{tikzcd}
is commutative. As $h\mathrm{i}(a) = 1$, for every $a \in M$, and $\big\langle \mathrm{i}(b)\ \big|\ b \in L \setminus M \big\rangle \subseteq \ker h$, each element of $\mathrm{i}(M)$ is then linearly independent of those in $\mathrm{i}(L \setminus M)$. Likewise, Proposition~\ref{LeDu} allows to state that $\mathrm{1}_{L \setminus M}$ is a valuation, then one also proves that each element of $\mathrm{i}(L \setminus M)$ is linearly independent of those in $\mathrm{i}(M)$.
\end{proof}

\begin{proposition} \label{PrInj}
The natural induced map $\mathrm{i}:L \rightarrow \mathrm{Val}(L)$ of a lattice $L$ is an injection if and only if $L$ is distributive.
\end{proposition}

\begin{proof}
If $L$ is distributive, we know from Corollary~\ref{CoSep} that any two different elements $a,b \in L$ can be separated by a prime ideal, hence Proposition~\ref{PrIn} allows to deduce that $\mathrm{i}(a)$ and $\mathrm{i}(b)$ are independent in $\mathrm{Val}(L)$.

\noindent If $L$ is not distributive, then, by Theorem~\ref{ThEq}, it contains distinct elements $a,b,c$ with $c \vee a = c \vee b$ and $c \wedge a = c \wedge b$. Hence, $\mathrm{i}(a) + \mathrm{i}(c) = \mathrm{i}(c \vee a) + \mathrm{i}(c \wedge a) = \mathrm{i}(c \vee b) + \mathrm{i}(c \wedge b) = \mathrm{i}(b) + \mathrm{i}(c)$, and $\mathrm{i}(a) = \mathrm{i}(b)$.
\end{proof}

\begin{proposition} \label{PrMDi}
	Let $L$ be a distributive lattice, and $a_1, \dots, a_n, b \in L$ with $\displaystyle b \notin \big[\bigwedge_{i \in [n]}a_i,\, \bigvee_{i \in [n]}a_i\big]$. Then, $b$ is linearly independent of $\{a_1, \dots, a_n\}$ in $\mathrm{Val}(L)$.
\end{proposition}

\begin{proof}
	If $\displaystyle b \notin \mathrm{id}\Big(\bigvee_{i \in [n]}a_i\Big)$, then there exists a prime ideal $P$ such that $\{a_1, \dots, a_n\} \subseteq P$ and $b \notin P$ by Corollary~\ref{CoPr}, and $b$ is linearly independent of $\{a_1, \dots, a_n\}$ by Proposition~\ref{PrIn}.

	\noindent If $\displaystyle b \in \mathrm{id}\Big(\bigvee_{i \in [n]}a_i\Big)$, then $\displaystyle b \notin \mathrm{fil}\Big(\bigwedge_{i \in [n]}a_i\Big)$, otherwise $\displaystyle b \in \big[\bigwedge_{i \in [n]}a_i,\, \bigvee_{i \in [n]}a_i\big]$ which is a contradiction. Hence, $\displaystyle \mathrm{id}(b) \cap \mathrm{fil}\Big(\bigwedge_{i \in [n]}a_i\Big) = \emptyset$, and there exists a prime ideal $P$ such that $\mathrm{id}(b) \subseteq P$ and $\displaystyle P \cap \mathrm{fil}\Big(\bigwedge_{i \in [n]}a_i\Big) = \emptyset$ by Theorem~\ref{ThPr}. As $\displaystyle \{a_1, \dots, a_n\} \subseteq \mathrm{fil}\Big(\bigwedge_{i \in [n]}a_i\Big)$, we once again obtain the independence of $b$ by Proposition~\ref{PrIn}.
\end{proof}

\noindent As the lattice $L$ with either the operation $\vee$ or $\wedge$ form a semigroup, the module $\mathbb{Z}L$ may consequently be considered as an algebra with either $\vee$ or $\wedge$ as multiplication. Besides, if $L$ is distributive, Proposition~\ref{PrInj} allows to identify $L$ with $\mathrm{i}(L)$.

\begin{proposition} \label{PrHL}
If $L$ is a distributive lattice, then $\mathrm{N}(L)$ is an ideal of the algebra $\mathbb{Z}L$ for both $\vee$ and $\wedge$ as multiplication.
\end{proposition}

\begin{proof}
For $a,b,c \in L$, we have
\begin{align*}
(a \wedge b + a \vee b - a - b) \wedge c & = (a \wedge b) \wedge c + (a \vee b) \wedge c - a \wedge c - b \wedge c \\
& = (a \wedge c) \wedge (b \wedge c) + (a \wedge c) \vee (b \wedge c) - a \wedge c - b \wedge c
\end{align*}
which belongs to $\mathrm{N}(L)$. Then, by linearly extension, we get $(a \wedge b + a \vee b - a - b) \wedge t \in \mathrm{N}(L)$ for any $t \in \mathbb{Z}L$. Similarly, we have $$(a \wedge b + a \vee b - a - b) \vee c = (a \vee c) \wedge (b \vee c) + (a \vee c) \vee (b \vee c) - a \vee c - b \vee c \in \mathrm{N}(L).$$
\end{proof}

\subsection{Valuation Algebra}

\noindent If the lattice $L$ is distributive, Proposition~\ref{PrHL} allows to state that the valuation module $\mathrm{Val}(L)$ becomes a commutative algebra for either $\vee$ or $\wedge$ as multiplication.

\begin{definition}
The $\textbf{valuation algebra}$ is the algebra $\big(\mathrm{Val}(L), \vee\big)$ or $\big(\mathrm{Val}(L), \wedge\big)$ for a distributive lattice $L$.
\end{definition}

\begin{lemma} \label{LeTau}
Let $L$ be a complete distributive lattice, and define the map $\tau: \mathrm{Val}(L) \rightarrow \mathrm{Val}(L)$ by $\tau(x):= \mathsf{1} + \mathsf{0} - x$. Then, for $a,b \in L$, we have $\tau(a \vee b) = \tau(a) \wedge \tau(b)$.
\end{lemma}

\begin{proof}
We have $\mathsf{1} + \mathsf{0} - a \vee b = \mathsf{1} + \mathsf{0} + a \wedge b - a - b = (\mathsf{1} + \mathsf{0} - a) \wedge (\mathsf{1} + \mathsf{0} - b)$.
\end{proof}

\begin{proposition} \label{PrDi}
Let $L$ be a complete distributive lattice, $n \in \mathbb{N}^*$, and $a_1, \dots, a_n \in L$. Then, we have $\displaystyle \mathsf{1} - \bigvee_{i \in [n]}a_i = \bigwedge_{i \in [n]}(\mathsf{1} - a_i)$, that is $$\displaystyle \bigvee_{i \in [n]}a_i = \sum_{k=1}^n (-1)^{k-1} \sum_{\substack{I \subseteq [n] \\ \#I=k}} \bigwedge_{i \in I}a_i.$$
\end{proposition}

\begin{proof}
Using Lemma~\ref{LeTau} and $\mathsf{0} \wedge (\mathsf{1} - a_i) = \mathsf{0}$, we obtain
$$\tau\Big(\bigvee_{i \in [n]}a_i\Big) = \mathsf{0} + \mathsf{1} - \bigvee_{i \in [n]}a_i = \bigwedge_{i \in [n]} \tau(a_i) = \bigwedge_{i \in [n]}(\mathsf{0} + \mathsf{1} - a_i) = \mathsf{0} + \bigwedge_{i \in [n]}(\mathsf{1} - a_i).$$
Then $\displaystyle \mathsf{1} - \bigvee_{i \in [n]}a_i = \bigwedge_{i \in [n]} \tau(a_i) = \bigwedge_{i \in [n]}(\mathsf{1} - a_i) = \mathsf{1} + \sum_{k=1}^n (-1)^k \sum_{\substack{I \subseteq [n] \\ \#I=k}} \bigwedge_{i \in I}a_i$.
\end{proof}

\begin{corollary} \label{CoDi}
Let $L$ be a complete distributive lattice, $n \in \mathbb{N}^*$, $a_1, \dots, a_n \in L$, and $f$ a valuation on $L$. Then,
$$f\Big(\bigvee_{i \in [n]}a_i\Big) = \sum_{k=1}^n (-1)^{k-1} \sum_{\substack{I \subseteq [n] \\ \#I=k}} f\Big(\bigwedge_{i \in I}a_i\Big).$$
\end{corollary}

\begin{proof}
If $f$ is a valuation to module $G$, we know from Proposition~\ref{PrHom} that a unique module homomorphism $h:\mathrm{Val}(L) \rightarrow G$ such that $h\mathrm{i}=f$ exists. Then, using Proposition~\ref{PrDi}, we obtain
$$f\Big(\bigvee_{i \in [n]}a_i\Big) = h\Big(\bigvee_{i \in [n]}a_i\Big) = \sum_{k=1}^n (-1)^{k-1} \sum_{\substack{I \subseteq [n] \\ \#I=k}} h\Big(\bigwedge_{i \in I}a_i\Big) = \sum_{k=1}^n (-1)^{k-1} \sum_{\substack{I \subseteq [n] \\ \#I=k}} f\Big(\bigwedge_{i \in I}a_i\Big).$$
\end{proof}

\begin{theorem} \label{ThAb}
Let $L$ be a complete lowly finite distributive lattice. Then, $\mathrm{Val}(L)$ is equal to $\mathbb{Z}\mathrm{ji}(L)$ as modules.
\end{theorem}

\begin{proof}
We obviously have $\mathsf{0} \in \mathrm{ji}(L)$. Let $a \in L$, and assume that every $b \in L$ such that $a \succ b$ is a linear combination in $\mathrm{Val}(L)$ of a finite number of elements in $\mathrm{ji}(L)$. We know from Proposition~\ref{PrJi} that there exists a subset $\{b_1, \dots, b_n\}$ of $\mathrm{ji}(L)$ such that $\displaystyle a = \bigvee_{i \in [n]}b_i$. Using Proposition~\ref{PrDi}, we get $\displaystyle a = \sum_{k=1}^n (-1)^{k-1} \sum_{\substack{I \subseteq [n] \\ \#I=k}} \bigwedge_{i \in I}b_i$ with $\displaystyle a \succ \bigwedge_{i \in I}b_i$ for each $I \subseteq [n]$. Thus $\mathrm{ji}(L)$ generates $\mathrm{Val}(L)$.

\noindent Assume now that every subset with cardinality $n-1$ in $\mathrm{ji}(L)$ is independent, and consider a subset of $n$ elements $\{a_1, \dots, a_n\} \subseteq \mathrm{ji}(L)$. We can suppose that $a_n$ is a maximal element in that set. Since $\displaystyle a_n \neq \bigvee_{i \in [n-1]}a_i$, then $\displaystyle a_n \notin \big[\bigwedge_{i \in [n-1]}a_i,\, \bigvee_{i \in [n-1]}a_i\big]$. We deduce from Proposition~\ref{PrMDi} that $\{a_1, \dots, a_n\}$ is independent. Hence $\mathrm{ji}(L)$ is an independent set in $\mathrm{Val}(L)$.
\end{proof}

\begin{corollary}
If $L$ is a complete lowly finite distributive lattice, then every valuation of $L$ is determined by its values on $\mathrm{ji}(L)$ which can be assigned arbitrarily.
\end{corollary}

\begin{proof}
If $f$ is a valuation to a module $G$, we know from Proposition~\ref{PrHom} that a unique module homomorphism $h:\mathrm{Val}(L) \rightarrow G$ such that $h\mathrm{i}=f$ exists. We know from Theorem~\ref{ThAb} that, if $a \in L$, there exist subsets $\{\lambda_1, \dots, \lambda_n\} \subseteq \mathbb{Z}$ and $\{a_1, \dots, a_n\} \subseteq \mathrm{ji}(L)$ such that $\displaystyle a = \sum_{i \in [n]} \lambda_i a_i$. Then, $\displaystyle f(a) = h(a) = h\Big(\sum_{i \in [n]} \lambda_i a_i\Big) = \sum_{i \in [n]} \lambda_i h(a_i) = \sum_{i \in [n]} \lambda_i f(a_i)$.
\end{proof}

\noindent For a poset $L$, and $a,b \in L$, we write $a \lessdot b$ if $a \prec b$ and $\{c \in L\ |\ a \prec c \prec b\} = \emptyset$.

\begin{proposition}
Let $L$ be a distributive lattice, and $a \in \mathrm{ji}(L)$ such that $a$ is not minimal. Then, there exists a unique element $a^* \in L$ such that $a^* \lessdot a$.
\end{proposition}

\begin{proof}
Suppose that there exist two different elements $b,c \in L$ such that $b \lessdot a$ and $c \lessdot a$. Then, $b \vee c \succeq b$, $b \vee c \succeq c$, and $b \vee c \notin \{b,c\}$. The only possibility is $b \vee c = a$ which contradicts the join-irreducibility of $a$.
\end{proof}

\noindent Let $L$ be a distributive lattice having a lowest element $\mathsf{0}$. Define $$e_{\mathsf{0}} := \mathsf{0} \in \mathrm{Val}(L) \quad \text{and} \quad e_a := a - a^* \in \mathrm{Val}(L)\ \text{for each}\ a \in \mathrm{ji}(L) \setminus \{\mathsf{0}\}.$$

\begin{theorem}
Let $L$ be a complete lowly finite distributive lattice. Then, $\big\{e_a\ |\ a \in \mathrm{ji}(L)\big\}$ is an orthogonal idempotent basis of $\mathrm{Val}(L)$.
\end{theorem}

\begin{proof}
For $a,b \in \mathrm{ji}(L) \setminus \{\mathsf{0}\}$ with $a \neq b$, we have $e_{\mathsf{0}} \wedge e_{\mathsf{0}} = e_{\mathsf{0}}$ and $e_a \wedge e_{\mathsf{0}} = a \wedge \mathsf{0} - a^* \wedge \mathsf{0} = 0$,
$e_a \wedge e_a = a \wedge a - a \wedge a^* - a^* \wedge a + a^* \wedge a^* = a - a^* - a^* + a^* = e_a$, and \begin{align*}
	e_a \wedge e_b & = a \wedge b - a \wedge b^* - a^* \wedge b + a^* \wedge b^* \\
	& = \begin{cases}
		b - b^* - b + b^* & \text{if}\ a^* = b \\
		a^* \wedge b^* - a^* \wedge b^* - a^* \wedge b^* + a^* \wedge b^* & \text{otherwise}
	\end{cases} \\
	& = 0.
\end{align*}
Then, $\big\{e_a\ |\ a \in \mathrm{ji}(L)\big\}$ is orthogonal idempotent. Assume now that every subset with cardinality $n-1$ in $\big\{e_a\ |\ a \in \mathrm{ji}(L)\big\}$ is independent, and consider a subset of $n$ elements $\{e_{a_1}, \dots, e_{a_n}\}$. We can suppose that $a_n$ is a maximal element in the set $\{a_1, \dots, a_n\}$. Since $\displaystyle a_n \neq \bigvee_{i \in [n-1]}a_i  \vee \bigvee_{i \in [n]}a_{i}^*$, then $\displaystyle a_n \notin \big[\bigwedge_{i \in [n-1]}a_i \wedge \bigwedge_{i \in [n]}a_{i}^*,\, \bigvee_{i \in [n-1]}a_i \vee \bigvee_{i \in [n]}a_{i}^*\big]$. We deduce from Proposition~\ref{PrMDi} that $a_n$ is independent of $\{a_1, \dots, a_{n-1}, a_{1}^*, \dots, a_{n}^*\}$. Hence $e_{a_n}$ is independent of $\{e_{a_1}, \dots, e_{a_{n-1}}\}$, and $\{e_{a_1}, \dots, e_{a_n}\}$ is consequently an independent set in $\mathrm{Val}(L)$. Finally, since there is a natural bijection $a \mapsto e_a$ between $\mathrm{ji}(L)$ and $\big\{e_a\ |\ a \in \mathrm{ji}(L)\big\}$, by Theorem~\ref{ThAb} the latter is also a basis of $\mathrm{Val}(L)$.
\end{proof}

\subsection{Identities on Valuation Algebra}

\begin{theorem} \label{ThPrMo}
Let $L$ be a complete lowly finite distributive lattice. Then,
$$\forall x \in L:\, x = \sum_{\substack{a,b \in \mathrm{ji}(L) \\ b \preceq a \preceq x}} \mu_{\mathrm{ji}(L)}(b,a)b.$$
\end{theorem}

\begin{proof}
If $a \in \mathrm{ji}(L)$, then $a = e_a + a^*$, particularly $\mathsf{0} = e_{\mathsf{0}}$. Now, consider any $x \in L \setminus \mathrm{ji}(L)$, and assume that, for every $b \in L$ such that $b \prec x$, we have $\displaystyle b = \sum_{\substack{d \in \mathrm{ji}(L) \\ d \preceq b}} e_d$. There exist $b,c \in L \setminus \{x\}$ such that $x = b \vee c$. Note that $\displaystyle b \wedge c = \sum_{\substack{d \in \mathrm{ji}(L) \\ d \preceq b \wedge c}}e_d$ as the $e_d$'s are orthogonal idempotent. Hence, $\displaystyle b \vee c = b + c - b \wedge c = \sum_{d\, \in \, \mathrm{ji}(L) \cap (\mathrm{id}(b) \cup \mathrm{id}(c))}e_d$. Besides, remark that, for any $y \in \mathrm{ji}(L) \cap \mathrm{id}(x)$, there exist $b,c \in L \setminus \{x\}$ such that $y \preceq b$ and $b \vee c = x$. Therefore, $\displaystyle x = \sum_{\substack{d \in \mathrm{ji}(L) \\ d \preceq x}}e_d$.

\noindent Let $\mathrm{b}$ be the natural bijection $a \mapsto e_a$ between $\mathrm{ji}(L)$ and $\big\{e_a\ |\ a \in \mathrm{ji}(L)\big\}$. For $a \in \mathrm{ji}(L)$, we have $\displaystyle \mathrm{i}(a) = \sum_{d \, \in \, \mathrm{ji}(L) \cap [\mathsf{0},a]}\mathrm{b}(d)$. Then, using the Möbius inversion formula, we obtain $$\displaystyle \mathrm{b}(a) = \sum_{d \, \in \, \mathrm{ji}(L) \cap [\mathsf{0},a]} \mu_{\mathrm{ji}(L)}(d,a) \mathrm{i}(d) \quad \text{or} \quad e_a = \sum_{\substack{d \in \mathrm{ji}(L) \\ d \preceq a}} \mu_{\mathrm{ji}(L)}(d,a) d.$$

\noindent We obtain the result by combining $\displaystyle x = \sum_{\substack{a \in \mathrm{ji}(L) \\ a \preceq x}}e_a$ with $\displaystyle e_a = \sum_{\substack{d \in \mathrm{ji}(L) \\ d \preceq a}} \mu_{\mathrm{ji}(L)}(d,a) d$.
\end{proof}

\begin{lemma} \label{LeMob}
If $L$ is a lowly finite distributive lattice, then $\big(\mathbb{Z}L, \wedge\big)$ is naturally isomorphic to the Möbius algebra $\big(\mathrm{M\ddot{o}b}(L),\cdot\big)$.
\end{lemma}

\begin{proof}
For $a \in L$, we have $\displaystyle u_L(a) = \sum_{c \in [\mathsf{0},a]} \mu_L(c,a)c$. The Möbius inversion formula consequently allows to state that $\displaystyle a = \sum_{c \in [\mathsf{0},a]} u_L(c)$. Then, for $a,b \in L$, we have
$$a \cdot b = \sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b]} u_L(c) = \sum_{c \in [\mathsf{0},a \wedge b]} u_L(c) = a \wedge b.$$
\end{proof}

\begin{lemma} \label{LeIso}
If $L$ is a complete lowly finite distributive lattice, then $\big(\mathrm{M\ddot{o}b}(L)/\mathrm{N}(L),\cdot\big)$ is isomorphic to the Möbius algebra $\Big(\mathrm{M\ddot{o}b}\big(\mathrm{ji}(L)\big),\cdot\Big)$.
\end{lemma}

\begin{proof}
By Lemma~\ref{LeMob}, we get $\mathrm{M\ddot{o}b}(L)/\mathrm{N}(L) \simeq \mathbb{Z}L/\mathrm{N}(L) \simeq \mathrm{Val}(L)$. We know from Theorem~\ref{ThAb} that $\mathrm{Val}(L)$ is isomorphic to $\mathbb{Z}\mathrm{ji}(L)$ as modules. Now, as algebras, $\big(\mathrm{Val}(L), \wedge\big)$ is naturally isomorphic to $\Big(\mathrm{M\ddot{o}b}\big(\mathrm{ji}(L)\big),\cdot\Big)$ since, for $a,b \in \mathrm{ji}(L)$, Theorem~\ref{ThPrMo} allows to state that
$$a \cdot b = \sum_{c \in [\mathsf{0},a] \cap [\mathsf{0},b] \cap \mathrm{ji}(L)} u_{\mathrm{ji}(L)}(c) = \sum_{c \in [\mathsf{0},a \wedge b] \cap \mathrm{ji}(L)} u_{\mathrm{ji}(L)}(c) = a \wedge b.$$
\end{proof}

\noindent The following theorem is the main result of this survey. Zaslavsky originally proved it for finite distributive lattice \cite[Theorem~2.1]{Za}.

\begin{theorem} \label{ThUma}
Let $L$ be a complete lowly finite distributive lattice, and $M$ a subset of $L$ such that $\mathrm{ji}(L) \subseteq M$. If $a \in M \setminus \mathrm{ji}(L)$, then $$u_M(a) \in \mathrm{N}(L).$$
\end{theorem}

\begin{proof}
Consider the linear maps $\mathrm{j}: \mathrm{M\ddot{o}b}(L) \rightarrow \mathrm{M\ddot{o}b}\big(\mathrm{ji}(L)\big)$, $\mathrm{j}_1: \mathrm{M\ddot{o}b}(L) \rightarrow \mathrm{M\ddot{o}b}(M)$, and $\mathrm{j}_2: \mathrm{M\ddot{o}b}(M) \rightarrow \mathrm{M\ddot{o}b}\big(\mathrm{ji}(L)\big)$ which on the basis $\{u_L(a)\ |\ a \in L\}$, and $\{u_M(a)\ |\ a \in M\}$ respectively have the values
\begin{align*}
& \mathrm{j}\big(u_L(a)\big) := \begin{cases}
	u_{\mathrm{ji}(L)}(a) & \text{if}\ a \in \mathrm{ji}(L),\\
	0 & \text{otherwise}
\end{cases}, \quad \mathrm{j}_1\big(u_L(a)\big) := \begin{cases}
	u_M(a) & \text{if}\ a \in M,\\
	0 & \text{otherwise}
\end{cases}, \\
& \text{and}\ \mathrm{j}_2\big(u_M(a)\big) := \begin{cases}
	u_{\mathrm{ji}(L)}(a) & \text{if}\ a \in \mathrm{ji}(L),\\
	0 & \text{otherwise}
\end{cases}.
\end{align*}
Then, $\mathrm{j}$, $\mathrm{j}_1$, and $\mathrm{j}_2$ are algebra homomorphisms by Corollary~\ref{CoHoA}. Moreover, as the diagram
\begin{center}
\begin{tikzcd}
	\mathrm{M\ddot{o}b}(L) \arrow[rd, "\mathrm{j}"] \arrow[r, "\mathrm{j}_1"] & \mathrm{M\ddot{o}b}(M) \arrow[d, "\mathrm{j}_2"] \\
	& \mathrm{M\ddot{o}b}\big(\mathrm{ji}(L)\big)
\end{tikzcd}
\end{center}
is commutative, then $u_M(a) \in \ker \mathrm{j}_2 \subseteq \ker \mathrm{j}$ if $a \in M \setminus \mathrm{ji}(L)$.

\noindent Finally, since $\mathrm{M\ddot{o}b}\big(\mathrm{ji}(L)\big) \simeq \mathrm{M\ddot{o}b}(L)/\ker \mathrm{j}$ \cite[II-Theorem~6.12]{BuSa}, we obtain $\ker \mathrm{j} = \mathrm{N}(L)$ using Lemma~\ref{LeIso}, and consequently $u_M(a) \in \mathrm{N}(L)$ if $a \in M \setminus \mathrm{ji}(L)$.
\end{proof}

\begin{corollary} \label{CoUma}
Let $L$ be a complete lowly finite distributive lattice, $M$ a subset of $L$ such that $\mathrm{ji}(L) \subseteq M$, and $f:L \rightarrow G$ a valuation on $L$. If $a \in M \setminus \mathrm{ji}(L)$, then
$$\sum_{b \in [\mathsf{0},a] \cap M} \mu_M(b,a)f(b) = 0.$$
\end{corollary}

\begin{proof}
Let $h:\mathrm{Val}(L) \rightarrow G$ be the module homomorphism associated to $f$ as in Proposition~\ref{PrHom}. We already know from Lemma~\ref{LeMob} that $\mathrm{Val}(L) \simeq \mathrm{M\ddot{o}b}(L)/\mathrm{N}(L)$. By Theorem~\ref{ThUma}, we then obtain
\begin{align*}
\sum_{b \in [\mathsf{0},a] \cap M} \mu_M(b,a)b & = 0 \\
h\Big(\sum_{b \in [\mathsf{0},a] \cap M} \mu_M(b,a)b\Big) & = h(0) \\
\sum_{b \in [\mathsf{0},a] \cap M} \mu_M(b,a)h(b) & = 0 \\
\sum_{b \in [\mathsf{0},a] \cap M} \mu_M(b,a)f(b) & = 0.
\end{align*}
\end{proof}

\section{Dissection Theory} \label{SecDi}

\noindent We use Corollary~\ref{CoUma} to prove the fundamental theorem of dissection theory.

\begin{definition}
Let us call \textbf{subspace arrangement} in a topological space $T$ a finite set of subspaces in $T$.
\end{definition}

\noindent For a subspace arrangement $\mathscr{A}$ in $T$, let $\displaystyle L_{\mathscr{A}} := \Big\{\bigcap_{H \in \mathscr{B}}H \in 2^T \setminus \{\emptyset\}\ \Big|\ \mathscr{B} \subseteq \mathscr{A}\Big\}$ be the poset with partial order $\preceq$ defined, for $A,B \in L_{\mathscr{A}}$, by $A \preceq B$ if and only if $A \subseteq B$.

\begin{definition}
Let $\mathscr{A}$ be a subspace arrangement in a topological space $T$. A \textbf{meet-refinement} of $L_{\mathscr{A}}$ is a finite poset $L \subseteq 2^T \setminus \{\emptyset\}$ with the same partial order as that defined for $L_{\mathscr{A}}$ such that $\displaystyle \bigcup_{X \in L}X = \bigcup_{H \in \mathscr{A}}H$ and
\begin{itemize}
\item any element in $L_{\mathscr{A}}$ is a union of elements in $L$,
\item any nonempty intersection of elements in $L$ is also a union of elements in $L$.
\end{itemize}
\end{definition}

\noindent Denote by $\mathrm{C}(X)$ the set formed by the connected components of a topological space $X$, and let $\mathscr{A}$ be a subspace arrangement of $T$. The set $\displaystyle L_{\mathscr{A}}^c := L_{\mathscr{A}} \sqcup \bigg\{\mathrm{C}\Big(\bigcap_{H \in \mathscr{B}}H\Big)\ \bigg|\ \mathscr{B} \subseteq \mathscr{A},\, \bigcap_{H \in \mathscr{B}}H \neq \emptyset\bigg\}$ is for instance a meet-refinement of $L_{\mathscr{A}}$.

\begin{definition}
Let $\mathscr{A}$ be a subspace arrangement in a topological space $T$, and denote by $C_{\mathscr{A}}$ the set formed by the connected components of $\displaystyle T \setminus \bigcup_{H \in \mathscr{A}}H$. An element of $C_{\mathscr{A}}$ is called a \textbf{chamber} of $\mathscr{A}$.
\end{definition}

\noindent Consider a subspace arrangement $\mathscr{A}$, and a meet-refinement $L$ of $L_{\mathscr{A}}$. Let $\mathrm{D}(L)$ be the finite distributive lattice of sets generated by $L \sqcup C_{\mathscr{A}}$ through unions and intersections, that is $$\mathrm{D}(L) := \Big\{\bigcup_{A \in M}A \sqcup \bigcup_{X \in D}X\ \Big|\ M \subseteq L,\, D \subseteq C_{\mathscr{A}}\Big\}.$$
In that case, for $A,B \in \mathrm{D}(L)$, we have $A \vee B = A \cup B$ and $A \wedge B = A \cap B$.

\begin{lemma} \label{LejiD}
Let $\mathscr{A}$ be a subspace arrangement in a topological space $T$, and $L$ a meet-refinement of $L_{\mathscr{A}}$. Then, $\mathrm{ji}\big(\mathrm{D}(L)\big) \subseteq \{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}}$.
\end{lemma}

\begin{proof}
Every element of $\mathrm{D}(L) \setminus (\{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}})$ is the union of at least two elements of $L \sqcup C_{\mathscr{A}}$. Then, none of them can be join-irreducible.
\end{proof}

\begin{theorem} \label{ThFuT}
Let $\mathscr{A}$ be a subspace arrangement in a topological space $T$, $L$ a meet-refinement of $L_{\mathscr{A}}$, and $f$ a valuation on $\mathrm{D}(L)$. Then,
$$\sum_{C \in C_{\mathscr{A}}}f(C) = \sum_{X \in L \sqcup \{\emptyset\}} \mu_{L \sqcup \{\emptyset\}}(X,T)f(X).$$
\end{theorem}

\begin{proof}
Note first that $T \in L$ but $T \notin \mathrm{ji}\big(\mathrm{D}(L)\big)$ as $\displaystyle T = \bigcup_{H \in \mathscr{A}}H \sqcup \bigcup_{C \in C_{\mathscr{A}}}C$. From Corollary~\ref{CoUma} and Lemma~\ref{LejiD}, we get
$$\sum_{A \in \{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}}} \mu_{\{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}}}(A,T)f(A) = 0.$$
The result is finally obtained after taking into account the following remarks:
\begin{itemize}
\item if $C \in C_{\mathscr{A}}$, then $\mu_{\{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}}}(C,T) = -\mu_{\{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}}}(C,C) = -1$,
\item if $X \in \{\emptyset\} \sqcup L$, then $[X,T] \cap C_{\mathscr{A}} = \emptyset$, hence $\mu_{\{\emptyset\} \sqcup L \sqcup C_{\mathscr{A}}}(X,T) = \mu_{\{\emptyset\} \sqcup L}(X,T)$.
\end{itemize}
\end{proof}

\begin{definition}
Let $T$ be a topological space, and denote by $H_n(T)$ the $n^{\text{th}}$ singular homology group of $T$ for $n \in \mathbb{N}$. The \textbf{Euler characteristic} of $T$ is
$$\chi(T) := \sum_{n \in \mathbb{N}} (-1)^n\, \mathrm{rank}\,H_n(T).$$
\end{definition}

\noindent We can now state the fundamental theorem of dissection theory.

\begin{corollary}[Fundamental Theorem of Dissection Theory] Let $\mathscr{A}$ be a subspace arrangement in a topological space $T$ with $|\chi(T)| < \infty$, and $L$ a meet-refinement of $L_{\mathscr{A}}$. Then, $$\sum_{C \in C_{\mathscr{A}}}\chi(C) = \sum_{X \in L} \mu_L(X,T) \chi(X).$$
\end{corollary}

\begin{proof}
It is known that $\chi(A) + \chi(B) = \chi(A \cup B) + \chi(A \cap B)$, for $A,B \subseteq T$, like stated at the end of \cite[§~12.4]{To}. The Euler characteristic is then a valuation on $\mathrm{D}(L)$. Moreover, $\chi(\emptyset)=0$ by definition. We consequently obtain the result by using Theorem~\ref{ThFuT} with $\chi$.
\end{proof}

\begin{example} \label{ExSp}
Consider the arrangement $\mathscr{A}$ of parametric $1$-spheres $\displaystyle H_1: \begin{cases}
		x = \cos(\frac{\pi}{4})  \\
		y = \sin(\frac{\pi}{4}) \cos(t), \\
		z = \sin(\frac{\pi}{4}) \sin(t)
\end{cases}$ $\displaystyle H_2: \begin{cases}
		x = -\cos(\frac{\pi}{8}) \\
		y = \sin(\frac{\pi}{8}) \cos(t), \\
		z = \sin(\frac{\pi}{8}) \sin(t)
\end{cases}$ $\displaystyle H_3: \begin{cases}
		x = \cos(\frac{\pi}{6}) \sin(t) \\
		y = \cos(\frac{\pi}{6}) \cos(t), \\
		z = \sin(\frac{\pi}{6})
\end{cases}$ $\displaystyle H_4: \begin{cases}
		x = \cos(\frac{\pi}{3}) \sin(t) \\
		y = \cos(\frac{\pi}{3}) \cos(t), \\
		z = -\sin(\frac{\pi}{3})
\end{cases}$ where $t \in [0,2\pi]$, in $\mathbb{S}^2$ represented on Figure~\ref{Ex2}. On one side, $C_{\mathscr{A}}$ has $6$ chambers having Euler characteristic $1$, and $1$ with Euler characteristic $0$, then $\displaystyle \sum_{C \in C_{\mathscr{A}}}\chi(C) = 6$. On the other side,
\begin{align*}
\sum_{X \in L_{\mathscr{A}}} \mu_{L_{\mathscr{A}}}(X,\mathbb{S}^2) \chi(X) = \, & \mu_{L_{\mathscr{A}}}(\mathbb{S}^2,\mathbb{S}^2)\chi(\mathbb{S}^2) + \sum_{i \in [4]} \mu_{L_{\mathscr{A}}}(H_i,\mathbb{S}^2)\chi(H_i) \\
& + \mu_{L_{\mathscr{A}}}(H_1 \cap H_3,\mathbb{S}^2)\chi(H_1 \cap H_3) + \mu_{L_{\mathscr{A}}}(H_2 \cap H_3,\mathbb{S}^2)\chi(H_2 \cap H_3) \\
= \, & 1 \times 2 + 4 \times (-1) \times 0 + 1 \times 2 + 1 \times 2 \\
= \, & 6.
\end{align*}

\begin{figure}[h]
		\centering
		\includegraphics[scale=0.45]{Sphere.png}
		\caption{$1$-Sphere Arrangement of Example~\ref{ExSp}}
		\label{Ex2}
\end{figure}
\end{example}

\begin{corollary} \label{CoNCh}
Let $\mathscr{A}$ be a subspace arrangement in a topological space $T$ with $|\chi(T)| < \infty$, and $L$ a meet-refinement of $L_{\mathscr{A}}$. Suppose that every chamber of $\mathscr{A}$ has the same Euler characteristic $c \neq 0$. Then, $$\#C_{\mathscr{A}} = \frac{1}{c} \sum_{X \in L} \mu_L(X,T) \chi(X).$$
\end{corollary}

\begin{proof}
It is obviously a consequence of the fundamental theorem of dissection theory where $\chi(C) = c$ for $C \in C_{\mathscr{A}}$.
\end{proof}

\section{Face Counting for Submanifold Arrangement} \label{SecCo}

\noindent We use the fundamental theorem of dissection theory to compute the $\mathrm{f}$-polynomial of submanifold arrangements having specific face properties.

\begin{definition}
	Let $\mathscr{A}$ be a subspace arrangement in topological space $T$, and $X \in L_{\mathscr{A}}$. The \textbf{induced subspace arrangement} on $X$ is the subspace arrangement in $X$ defined by
	$$\mathscr{A}_X := \big\{H \cap X\ \big|\ H \in \mathscr{A},\, H \cap X \notin \{\emptyset, X\}\big\}.$$
	Let $\displaystyle F_{\mathscr{A}} := \bigsqcup_{X \in L_{\mathscr{A}}}C_{\mathscr{A}_X}$, and call an element of $F_{\mathscr{A}}$ a \textbf{face} of $\mathscr{A}$.
\end{definition}

\begin{definition}
Recall that a \textbf{$n$-dimensional manifold} or \textbf{$n$-manifold} is a topological space with the property that each point has a neighborhood that is homeomorphic to $\mathbb{R}^n$, and a \textbf{submanifold} of a $n$-manifold $T$ is a $k$-manifold included in $T$ where $k \in [0,n]$.
\end{definition}

\begin{definition}
Let us call \textbf{submanifold arrangement} in the $n$-manifold $T$ a finite set of submanifolds $\mathscr{A}$ in $T$ such that every element of $L_{\mathscr{A}} \cup F_{\mathscr{A}}$ is a submanifold of $T$.
\end{definition}

\begin{example} \label{ExR2}
Consider the arrangement $\mathscr{A}$ of $1$-manifolds $H_1: y = 6 \sin(x)$, $H_2: y = x + \cos(x)$, $\displaystyle H_3: \frac{x^2}{64} + \frac{y^2}{25} = 1$ in $\mathbb{R}^2$ represented on Figure~\ref{Ex1}. We see that
\begin{align*}
	\sum_{X \in L_{\mathscr{A}}} \mu_{L_{\mathscr{A}}}(X,\mathbb{R}^2) \chi(X) = \, & \mu_{L_{\mathscr{A}}}(\mathbb{R}^2,\mathbb{R}^2)\chi(\mathbb{R}^2) + \mu_{L_{\mathscr{A}}}(H_1,\mathbb{R}^2)\chi(H_1) + \mu_{L_{\mathscr{A}}}(H_2,\mathbb{R}^2)\chi(H_2) \\
	& + \mu_{L_{\mathscr{A}}}(H_3,\mathbb{R}^2)\chi(H_3) + \mu_{L_{\mathscr{A}}}(H_1 \cap H_2,\mathbb{R}^2)\chi(H_1 \cap H_2) \\
	& + \mu_{L_{\mathscr{A}}}(H_1 \cap H_3,\mathbb{R}^2)\chi(H_1 \cap H_3) + \mu_{L_{\mathscr{A}}}(H_2 \cap H_3,\mathbb{R}^2)\chi(H_2 \cap H_3) \\
	= \, & 1 \times 1 + (-1) \times (-1) + (-1) \times (-1) + (-1) \times 0 + 1 \times 3 + 1 \times 10 + 1 \times 2 \\
	= \, & 18
\end{align*}
is the number of chamber in $C_{\mathscr{A}}$.

\begin{figure}[h]
	\centering
	\includegraphics[scale=0.8]{Submanifolds.png}
	\caption{Submanifold Arrangement of Example~\ref{ExR2}}
	\label{Ex1}
\end{figure}
\end{example}

\begin{definition}
Let $\mathscr{A}$ be a submanifold arrangement in a $n$-manifold $T$, and $x$ a variable. For $k \in [0,n]$, denote by $\mathrm{f}_k(\mathscr{A})$ the number of $k$-dimensional faces of $\mathscr{A}$. The \textbf{$\mathrm{f}$-polynomial} of $\mathscr{A}$ is
$$\mathrm{f}_{\mathscr{A}}(x) := \sum_{k \in [0,n]} \mathrm{f}_k(\mathscr{A}) x^{n-k}.$$
\end{definition}

\begin{proposition} \label{PrSub}
Let $\mathscr{A}$ be a submanifold arrangement in a $n$-manifold $T$ with $|\chi(T)| < \infty$. Suppose that
\begin{align*}
& \forall k \in [0,n],\, \forall X \in L_{\mathscr{A}},\, \dim X = k:\ \chi(X) = l_k, \\
& \forall k \in [0,n],\, \forall C \in F_{\mathscr{A}},\, \dim C = k:\ \chi(C) = c_k \neq 0.
\end{align*}
Then,
$$\mathrm{f}_{\mathscr{A}}(x) = \sum_{i \in [0,n]}  \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \sum_{k \in [0,i]} \sum_{\substack{X \in L_{\mathscr{A}_Y} \\ \dim X = k}} \frac{l_k}{c_i} \mu_{L_{\mathscr{A}}}(X,Y) x^{n-k}.$$
\end{proposition}

\begin{proof}
Using the fundamental theorem of dissection theory, we get
\begin{align*}
\mathrm{f}_i(\mathscr{A}) & = \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \#C_{L_{\mathscr{A}_Y}} \\
& = \frac{1}{c_i} \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \sum_{X \in L_{\mathscr{A}_Y}} \mu_{L_{\mathscr{A}_Y}}(X,Y) \chi(X)	\\
& = \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \sum_{k \in [0,i]} \sum_{\substack{X \in L_{\mathscr{A}_Y} \\ \dim X = k}} \frac{l_k}{c_i} \mu_{L_{\mathscr{A}_Y}}(X,Y) \\
& = \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \sum_{k \in [0,i]} \sum_{\substack{X \in L_{\mathscr{A}_Y} \\ \dim X = k}} \frac{l_k}{c_i} \mu_{L_{\mathscr{A}}}(X,Y).
\end{align*}
\end{proof}

\begin{definition}
Let $\mathscr{A}$ be a submanifold arrangement in a $n$-manifold $T$. The \textbf{rank} of $X \in L_{\mathscr{A}}$ is $\mathrm{rk}\,X := n - \dim X$, and that of $\mathscr{A}$ is $\mathrm{rk}\,\mathscr{A} := \max \{\mathrm{rk}\,X\ |\ X \in L_{\mathscr{A}}\}$.
\end{definition}

\begin{definition}
Let $\mathscr{A}$ be a submanifold arrangement in a $n$-manifold $T$, and $x,y$ two variables. The \textbf{Möbius Polynomial} of $\mathscr{A}$ is
$$\mathrm{M}_{\mathscr{A}}(x,y) := \sum_{X,Y \in L_{\mathscr{A}}} \mu_{L_{\mathscr{A}}}(X,Y) x^{\mathrm{rk}\,X} y^{\mathrm{rk}\,\mathscr{A} - \mathrm{rk}\,Y}.$$
\end{definition}

\begin{corollary}
Let $\mathscr{A}$ be a submanifold arrangement in a $n$-manifold $T$ with $|\chi(T)| < \infty$. Suppose that $\chi(X) = (-1)^{\dim X}$ for every $X \in L_{\mathscr{A}} \cup F_{\mathscr{A}}$. Then,
$$\mathrm{f}_{\mathscr{A}}(x) = (-1)^{\mathrm{rk}\,\mathscr{A}} \mathrm{M}_{\mathscr{A}}(-x,-1).$$
\end{corollary}

\begin{proof}
From Proposition~\ref{PrSub}, we obtain
\begin{align*}
\mathrm{f}_{\mathscr{A}}(x) & = \sum_{i \in [0,n]}  \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \sum_{k \in [0,i]} \sum_{\substack{X \in L_{\mathscr{A}_Y} \\ \dim X = k}} (-1)^{k-i} \mu_{L_{\mathscr{A}}}(X,Y) x^{n-k} \\
& = \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} (-1)^{\dim X - \dim Y} \mu_{L_{\mathscr{A}}}(X,Y) x^{n- \dim X} \\
& = \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} (-1)^{n - \dim Y} \mu_{L_{\mathscr{A}}}(X,Y) (-1)^{\dim X - n} x^{n- \dim X} \\
& = \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} (-1)^{\mathrm{rk}\,Y} \mu_{L_{\mathscr{A}}}(X,Y) (-x)^{\mathrm{rk}\,X} \\
& = (-1)^{\mathrm{rk}\,\mathscr{A}} \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}}  \mu_{L_{\mathscr{A}}}(X,Y) (-x)^{\mathrm{rk}\,X} (-1)^{\mathrm{rk}\,Y - \mathrm{rk}\,\mathscr{A}} \\
& = (-1)^{\mathrm{rk}\,\mathscr{A}} \mathrm{M}_{\mathscr{A}}(-x,-1).
\end{align*}
\end{proof}

\begin{corollary}
Let $\mathscr{A}$ be a submanifold arrangement in a $n$-manifold $T$ with $|\chi(T)| < \infty$. Suppose that
$$\forall C \in F_{\mathscr{A}}:\, \chi(C) = (-1)^{\dim C} \quad \text{and} \quad \forall X \in L_{\mathscr{A}}:\, \chi(X) = \begin{cases}
	2 & \text{if}\ \dim X \equiv 0 \mod 2 \\
	0 & \text{otherwise}
\end{cases}.$$
Moreover, define $\displaystyle \gamma_n := \begin{cases}
	1 & \text{if}\ \dim X \equiv 0 \mod 2 \\
   -1 & \text{otherwise}
\end{cases}$. Then,
$$\mathrm{f}_{\mathscr{A}}(x) = (-1)^{n-\mathrm{rk}\,\mathscr{A}} \big(\mathrm{M}_{\mathscr{A}}(x,-1) + \gamma_n \mathrm{M}_{\mathscr{A}}(-x,-1)\big).$$
\end{corollary}

\begin{proof}
	From Proposition~\ref{PrSub}, we obtain
	\begin{align*}
		\mathrm{f}_{\mathscr{A}}(x) & = \sum_{i \in [0,n]}  \sum_{\substack{Y \in L_{\mathscr{A}} \\ \dim Y = i}} \sum_{k \in [0,i]} \sum_{\substack{X \in L_{\mathscr{A}_Y} \\ \dim X = k}} (-1)^{-i} \chi(X) \mu_{L_{\mathscr{A}}}(X,Y) x^{n-k} \\
		& = \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} (-1)^{-\dim Y} \chi(X) \mu_{L_{\mathscr{A}}}(X,Y) x^{n- \dim X} \\
		& = (-1)^n \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} \chi(X) \mu_{L_{\mathscr{A}}}(X,Y) x^{\mathrm{rk}\,X} (-1)^{\mathrm{rk}\,Y} \\
		& = (-1)^{n - \mathrm{rk}\,\mathscr{A}} \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} \chi(X) \mu_{L_{\mathscr{A}}}(X,Y) x^{\mathrm{rk}\,X} (-1)^{\mathrm{rk}\,\mathscr{A} -\mathrm{rk}\,Y} \\
		& = (-1)^{n - \mathrm{rk}\,\mathscr{A}} \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} \mu_{L_{\mathscr{A}}}(X,Y) x^{\mathrm{rk}\,X} (-1)^{\mathrm{rk}\,\mathscr{A} -\mathrm{rk}\,Y} \\
		& \quad + (-1)^{n - \mathrm{rk}\,\mathscr{A}} \gamma_n \sum_{Y \in L_{\mathscr{A}}} \sum_{X \in L_{\mathscr{A}_Y}} \mu_{L_{\mathscr{A}}}(X,Y) (-x)^{\mathrm{rk}\,X} (-1)^{\mathrm{rk}\,\mathscr{A} -\mathrm{rk}\,Y} \\
		& = (-1)^{n-\mathrm{rk}\,\mathscr{A}} \mathrm{M}_{\mathscr{A}}(x,-1) + (-1)^{n-\mathrm{rk}\,\mathscr{A}} \gamma_n \mathrm{M}_{\mathscr{A}}(-x,-1).
	\end{align*}
\end{proof}

\bibliographystyle{abbrvnat}

redpajama_set_name: RedPajamaArXiv


text:
Carlos Ramirez was born in Cuba, but his family moved to the United States for political reasons when he was a child. His family's story is like that of many immigrants, and he recalls those early days in New York:
"My parents came to this country with nothing but the clothes in their suitcases. They gave up a comfortable life that was deteriorating in hopes of a safe future. As an adult I can now admire their courage and determination. In their late thirties they landed in a country where they didn't speak the language, had no financial resources and a small child."
Ramirez always drew and painted, but as the only child of an immigrant family, his parents were concerned that he have a structured profession rather than take a chance on a career as an artist. In college, he focused on drawing and drafting in the hopes of pursuing an architectural degree, but at the last minute, due to the prodding of his father, he applied to law school.
He received a full scholarship to the University of Miami Law School, where he excelled but soon realized his heart was not in it. In his last year, he left the program to follow a vision of putting a fashion collection together. "I applied all of my drawing, sketching, and design skills to this new endeavor," he notes. At age 23, he established the label Liancarlo, and today he is a highly recognized designer in the fashion world. The label has dressed dozens of celebrities, and his work has been featured in countless fashion magazines.
Throughout the years of growing his business, he became a designer by day, painter by night, taking workshops with such greats as painter Wolf Kahn. He made time to pursue his painting with the same detailed study and commitment that he brought to building a business.
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Line, color, and shape are at the core of both his designs and paintings. They are not separate but have cohabited for many years, one being a natural outgrowth of the other. Ramirez continues to paint daily out of his cottage studio in Coral Gables, Florida.
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Driving Into Siena
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Summer Hike

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Oriol Alamany is a very special nature photographer who combines wildlife photography with a particular photo technique to achieve jaw dropping results. Since decades ago, this renowned photographer (with prizes in the Wildlife Photographer of the Year in three occasions, for example), travels all over the world to show what he sees and what he feels in unique images.
In the DBF, Alamany will talk and show us images of the Svalvard Islands, located around 80ºN, that is, fully in the Arctic Circle. These are islands with a special spell that once lured restless explorers and now lure photographers and nature lovers -who are also restless- to admire their landscape and wildlife. Oriol Alamany is one of them. Fascinated by the beauty of the Svalvard, he has traveled to these islands five times and relates that all of them have been a memorable experience. The Arctic sea ice, the seabirds and a wildlife adapted to extreme conditions have hipnotized this passionate photographer.


text:
Publisher: Bonfirraro Editore
Il partito dei magistrati
Storia di una lunga deriva istituzionale
by Mauro Mellini
This book by Mauro Mellini, moves away from the usual points of view according to which they are judged, with a severity which, however, it is always less than that due, events and people of justice in our country. Instead of looking at the alarming state of dysfunction, chronic and now that leaves little room for hope, Mellini, on time and with almost pique, examines the progressive distortion of the institutional machinery of justice, whose power, over the years has gone by widening over the river bed that is just right, with alteration of the balance between the various powers and functions of the state. A real "deviance" judicial apparatus, with a tendency to usurp the legislative and the executive, which manifests itself with the consequent assumption of addresses, purposes and inevitably political connotations. Justice becomes the government, it becomes party. And, consequently, it becomes less correct. A deviance that has disparate origins and whose responsibilities are to be assigned to categories and different political forces. This work by Mellini, titled so provocative as realistic, it will be useful, therefore, to reflect, but also, it is hoped, to act in the way and for the purposes desired by many.
14.5 cm. x 21 cm., 208 pages
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Research experts in nutrition, diet and physical activity from across the country have come together to form the NIHR Diet and Activity Research Translation (DART) Collaboration. DART aims to tackle various challenges associated with sedentary lifestyle, such as obesity, diabetes, cancer and cardiovascular disease.
NIHR Imperial BRC researchers have been investigating effects of diet and nutrition on health over many years, including cancer epidemiology and prevention (Informatics & Biobanking Theme) and metabonomics and microbiome research (Cancer; Gut Health; Metabolic Medicine & Endocrinology Themes), with cross-cutting support of the Molecular Phenomics Theme. There are also a number of engineering projects that involve a nutritional component. Some recent examples in this area include research from Dr Isabel Garcia-Perez and Prof Gary Frost, who previously identified four distinct diet types based on metabolic profiles of participants. Cross-Theme collaborative work led by Dr Marc Dumas identified a metabolic marker that could predict steatosis, while also providing evidence for importance of microbial diversity in disease prevention.
Imperial College also established Network of Excellence in Nutrition & Food in 2017, aiming to bringing together researchers from different Faculties with multi-disciplinary expertise, to focus on the major health problems associated with diet. Prof Gary Frost, Chair in Nutrition and Dietetics at Imperial College, and NIHR Imperial BRC Gut Health Theme Lead, is head of this Imperial network. In the newly formed NIHR DART Collaboration, Prof Frost is leading education for the network.


Acronis Home User Backup ver 10.0. Message="processing, please wait." never ends.
Discussion in 'Acronis True Image Product Line' started by MplsBob, Jul 27, 2008.
I am using the Acronis home user backup version 10.
I started doing a backup and indicated I wanted to backup my "computer". In response to this I got a message saying "Processing, please wait." I have been waiting for over an hour for it to end.
I have never seen this message before.
This is the first problem I have experienced with Acronis backup version 10 since I got it over a year ago.
Do you start the windows installation of TrueImage? Or from rescue-cd?
And you want to backup the boot partition? Or annother partition?
I am using a Windows installation of TrueImage.
I want to backup the only partition on my C: drive.
I had only selected that I wanted to do a backup of my "computer".
Please start from your Ti-rescue-cd and try the restore from there.
Please make sure you use the latest build (4942) of Acronis True Image 10.0 Home. To get access to updates you should first register your software.
If the latest build doesn't solve the problem, please download the latest version of Acronis drivers and install it with disabled logging.
Please also try checking your hard drive for errors by running "chkdsk /r" command on all partitions.


Stationery set for dog lovers!
Isn't this the cutest letterpress stationery set? If you and your furry friend are in need of custom note cards, look no further. These note cards were printed for a couple with a beloved dachshund. The black chevron envelope liner adds a bit of modernity to the design.