Let $(P, \leq)$ be a poset. We write $a < b$ whenever $a\leq b$ and $a\neq b$. For any two elements $a, b\in P$, we say $a$ covers $b$, denoted by $b\lessdot a$, if $b < a$ and there exists no $c\in P$ such that $b < c < a$. If $P$ has the minimum (respectively maximum) element, then we denote it by 0 (respectively $\upharpoonleft$), and say that $P$ is a poset with 0 (respectively $\upharpoonleft$). A poset $P$ is said to be a lattice if both $a\vee b:=\rm{sup}\{a, b\}$ and $a\wedge b:=\rm{inf}\{a, b\}$ exist for any two elements $a, b\in P$. Let $P$ be a finite lattice with 0. For $a \in P$, if $0\lessdot a$, then $a$ is called an atom. A lattice $P$ with 0 is called an atomic lattice if $a\in P\backslash\{0\}$ is the least upper bound of some atoms. Let $P$ be a finite poset with 0. If there is a function $r$ from $P$ to set of all the nonnegative integers such that

(1) $r(0)=0, $

(2) $r(b)=r(a)+1$, if $a\lessdot b$.

Then $r$ is said to be the rank function on $P$. Note that the rank function on $P$ is unique if it exists.

Let $P$ be a finite atomic lattice. $P$ is said to be geometric lattice, if $P$ admits a rank function $r$ and for any two elements $a, b\in P, $

$ r(a\wedge b)+r(a\vee b)\leq r(a)+r(b). $

Let $P$ be a poset with 0 and $\upharpoonleft$, and let $P$ admit the rank function $r$. The polynomial

$\begin{eqnarray*} \chi(P, x)=\sum\limits_{a\in P}\mu(0, a)x^{r(\upharpoonleft)-r(a)}\end{eqnarray*} $

is called the characteristic polynomial of $P$.

Let $A$ be an $m$-subset of $[n]:=\{1, 2, \cdots, n\}$, and $f:A\rightarrow [n]$ be a map. Then the pair $(A, f)$ is said to be an $m$-partial map of $[n]$. In particular, we write $(A, f)=0$ if $A=\emptyset$.

Let $P=\{(A, f)\mid(A, f) \hbox{ be a partial map of [n]}\}\cup \{\upharpoonleft\}$. For any two elements $(A, f)\in P\backslash\{\upharpoonleft\}$, $(B, g)\in P\backslash\{\upharpoonleft\}$, we define that $\upharpoonleft$ includes $(A, f)$, and $(B, g)$ includes $(A, f)$ if $A\subseteq B$ and $g|_A=f$. Partially ordered $P$ by ordinary or reverse inclusion, two families of finite posets are obtained, denoted by $P_O$ and $P_R$, respectively.

In this paper we will prove that $P_O$ and $P_R$ are finite atomic lattices, discuss their geometricity and compute their characteristic polynomials.

The results on the lattices generated by transitive sets of subspaces under finite classical groups may be found in Huo, Liu and Wan [3-5]. In [1], Guo discussed the lattices associated with finite vector spaces and finite affine spaces. The lattices generated by the orbits of subspaces under finite classical groups have been obtained in a series of papers by Huo and Wan [6], Wang and Feng [8], Wang and Guo [9, 10], Guo and Nan [2, 7], Wang and Li [11].

In this section we will prove that $P_O$ is a finite atomic lattice and computes its characteristic polynomial. We begin with a useful lemma.

Lemma 2.1   $P_O$ is a finite lattice.

Proof  For any $(A, f)\in P_O\backslash\{\upharpoonleft\}$, it is easy to see that

$ \upharpoonleft=(A, f)\vee \upharpoonleft \hbox{ and } (A, f)=\upharpoonleft\wedge(A, f). $

For any $(A, f), (B, g)\in P_O\backslash\{\upharpoonleft\}$, we assert that

$ (A, f)\vee (B, g)=\left\{ \begin{array}{ll} (A\cup B, h), h|_A=f, h|_B=g, &f|_{A\cap B}=g|_{A\cap B}, \\\upharpoonleft, &f|_{A\cap B}\neq g|_{A\cap B}. \end{array} \right. $

Case 1   $f|_{A\cap B}=g|_{A\cap B}$. Let $(C, \varphi)$ be an upper bound of $(A, f)$ and $(B, g)$, then

$ A\subseteq C, B\subseteq C\hbox{ and } \varphi |_{A}=f=h|_{A}, \varphi |_{B}=g=h|_{B}. $

Thus, $A\cup B\subseteq C$ and $\varphi|_{A\cup B}=h$, i.e., $(A\cup B, h)\leq (C, \varphi)$. Hence $(A, f)\vee(B, g)=(A\cup B, h)$.

Case 2   $f|_{A\cap B}\neq g|_{A\cap B}$. Assume that $(C, \varphi)$ is an upper bound of $(A, f)$ and $(B, g)$, i.e.,

$ (A, f)\leq(C, \varphi) \hbox{ and } (B, g)\leq(C, \varphi). $

Then $\varphi |_{A\cap B}=f|_{A\cap B}, \varphi |_{A\cap B}=g|_{A\cap B}$, a contradiction.

On the other hand, for any $(A, f), (B, g)\in P_O\backslash\{\upharpoonleft\}$, we assert that $(A, f)\wedge (B, g)=(D, h)$, where D is the maximum element of the set

$ \{C\subseteq A\cap B \mid f|_{C}=g|_{C}\} $

and

$ h=f|_{D}=g|_{D}. $

In fact, let $(C, \varphi)$ be a lower bound of $(A, f)$ and $(B, g)$. Then

$ C\subseteq A, C\subseteq B\;\mathrm{and}\; f|_C=g|_C. $

Thus $C$ belongs to $\{C\subseteq A\cap B \mid f|_{C}=g|_{C}\}$. Hence, $(A, f)\wedge (B, g)=(D, h)$.

Theorem 2.2  Let $n\geq 2$. Then $P_O$ is a finite atomic lattice, but not a geometric lattice.

Proof  Define $r_O(A, f)=|A|$ for any $(A, f)\in P_O\backslash\{\upharpoonleft\}$ and $r_O(\upharpoonleft)=n+1$. Then $r_O$ is the rank function on $P_O$.

Pick

$ \begin{equation}\label{shi1} A=\{1\}, f:A\rightarrow [n], 1\mapsto 1;\;\mathrm{and}\;g:A\rightarrow [n], 1\mapsto 2. \end{equation} $

(2.1)

Then $(A, f)$ and $(A, g)$ are the atoms of $P_O$, and $\upharpoonleft=(A, f)\vee (A, g).$

For any $(A, f)\in P_O\backslash\{\upharpoonleft\}$ with $A=\{a_1, a_2, \cdots, a_m\}$, we have

$ (A, f)=(\{a_1\}, f|_{\{a_1\}})\vee(\{a_2\}, f|_{\{a_2\}})\vee\cdots\vee(\{a_m\}, f|_{\{a_m\}}). $

Hence $P_O$ is a finite atomic lattice.

Pick $(A, f)\;\mathrm{and} \;(A, g)$ as in $(2.1)$. Then $(A, f)\vee (B, g)=\upharpoonleft$ and $(A, f)\wedge(B, g)=0$, which implies that

$ r_O((A, f)\vee(B, g))+r_O((A, f)\wedge(B, g))=n+1>2=r_O(A, f)+r_O(B, g). $

Therefore, the desired result follows.

Lemma 2.3  The Möbius function on $P_O$ is

$ \mu_O(x, y)=\left\{ \begin{array}{ll} 0, &x \nleq y, \\(-1)^{|B|-|A|}, &x=(A, f)\leq(B, g)=y\neq \upharpoonleft, \\-(1-n)^{n-|A|}, &x=(A, f) < y=\upharpoonleft, \\ 1, &x=y=\upharpoonleft. \end{array} \right. $

Proof  In order to prove that $\mu_O$ is the Möbius function on $P_O$, we only need to show that

$\begin{eqnarray*} \sum\limits_{x\leq z\leq y}\mu_O(x, z)=0\end{eqnarray*} $

for any $x, y\in P_O$ with $x < y$.

If $y=(B, g)\neq\upharpoonleft$, let $|B|-|A|=m$. Then

$\begin{eqnarray*} \sum\limits_{x\leq z\leq y}\mu_O(x, z)=\sum\limits_{k=0}^{m}C^k_m(-1)^k =(1-1)^m=0.\end{eqnarray*} $

If $y=\upharpoonleft$, let $|A|=m$. Then

$\begin{eqnarray*} \sum\limits_{x\leq z\leq\upharpoonleft}\mu_O(x, z)=\sum\limits_{k=0}^{n-m}C^k_{n-m}(-1)^{k}n^{k} +\mu(x, 1)=(1-n)^{n-m}+[-(1-n)^{n-|A|}]=0.\end{eqnarray*} $

Hence, the function $\mu_O$ is the Möbius function on $P_O$.

Theorem 2.4  The characteristic polynomial of $P_O$ is

$ \chi(P_O, x)=x(x-n)^n-(1-n)^n. $

Proof  By Lemma 2.3 we obtain

$ \begin{eqnarray*} \chi(P_O, x)&=&\sum\limits_{u\in P_O}\mu_O(0, u)x^{r_O(\upharpoonleft)-r_O(u)}\\ &=&\sum\limits_{m=0}^{n}(-1)^mC_n^mn^mx^{r_O(\upharpoonleft)-m}+\mu_O(0, \upharpoonleft)x^{r_O(\upharpoonleft)-r_O(\upharpoonleft)}\\ &=&x\sum\limits_{m=0}^{n}(-1)^mC_n^mn^mx^{n-m}+[-(1-n)^nx^0]\\ &=&x(x-n)^n-(1-n)^n. \end{eqnarray*} $

In this section we will prove that $P_R$ is a finite atomic lattice and compute its characteristic polynomials. Similar to the proof of Lemma 2.1, we obtain the following result.

Lemma 3.1   $P_R$ is a finite lattice.

Theorem 3.2  Let $n\geq 2$. Then $P_R$ is a finite atomic lattice, but not a geometric lattice.

Proof  Define $r_R(A, f)=n+1-|A|$ for any $(A, f)\in P_R\backslash\{\upharpoonleft\}$ and $r_R(\upharpoonleft)=0.$ Then $r_R$ is the rank function on $P_R$.

Pick $f:[n]\rightarrow [n], i\mapsto 1$ and $g:[n]\rightarrow [n], i\mapsto 2.$ Then $0=([n], f)\vee([n], g)$.For $(A, f)\in P_R\backslash\{\upharpoonleft\}$ with $A=\{a_1, a_2, \cdots, a_m\}$, pick $([n], g), ([n], h)\in P_R$, such that

$ \begin{equation*}\label{shi2} g:[n]\rightarrow [n], a_i\mapsto f(a_i), a\mapsto 1 \hbox{ if } a \notin A ;\quad h:[n]\rightarrow [n], a_i\mapsto f(a_i), a\mapsto 2 \hbox{ if } a \notin A. \end{equation*} $

(3.1)

Then $([n], g)$ and $([n], h)$ are atoms in $P_R$ and $(A, f)=([n], g)\vee ([n], h).$ Hence, $P_R$ is a finite atomic lattice.

Pick $f, g$ as in (3.1). Then

$ ([n], f)\vee ([n], g)=0, ([n], f)\wedge([n], g)=\upharpoonleft, $

which implies that

$ r_R(([n], f)\vee([n], g))+r_R([n], f)\wedge([n], g))=n+1>2=r_R([n], f)+r_R([n], g). $

Therefore, the desired result follows.

Lemma 3.3  The M$\ddot o$bius function on $P_R$ is

$ \mu_R(x, y)=\left\{ \begin{array}{ll} 0, &x\nleq y, \\(-1)^{|A|-|B|}, &\upharpoonleft\neq x=(A, f)\leq y=(B, g), \\-(1-n)^{n-|B|}, &\upharpoonleft=x < y=(B, g), \\ 1, &x=y=\upharpoonleft. \end{array} \right. $

Proof  If $x\neq\upharpoonleft$, let $|A|-|B|=m$. Then

$\begin{eqnarray*} \sum\limits_{x\leq z\leq y}\mu_R(x, z)=\sum\limits_{k=0}^{m}C^{m-k}_m(-1)^k =(1-1)^m=0.\end{eqnarray*} $

If $x=\upharpoonleft$, let $|B|=m$. Then

$\begin{eqnarray*} \sum\limits_{\upharpoonleft\leq z\leq y}\mu_R (\upharpoonleft, z)=\mu(\upharpoonleft, \upharpoonleft)+\sum\limits_{k=0}^{n-m}C^k_{n-m}[-(1-n)^{n-m-k}]n^{k} =1-[n+(1-n)]^{n-m}=0.\end{eqnarray*} $

Hence, the function $\mu_R$ is the M$\ddot o$bius function on $P_R$.

Theorem 3.4  The characteristic polynomial of $P_R$ is

$ \chi(P_R, x)=x^{n+1}-(nx-n+1)^n. $

Proof  By Lemma 3.3 we obtain

$ \begin{eqnarray*} \chi(P_R, x)&=&\sum\limits_{u\in P_R}\mu_R(\upharpoonleft, (A, f))x^{r_R(0)-r_R(u)}\\ &=&\sum\limits_{m=0}^{n}C_n^mn^m[-(1-n)^{n-m}]x^m+\mu_R(\upharpoonleft, \upharpoonleft)x^{r_R(0)-r_R(\upharpoonleft)}\\ &=&x^{n+1}-(nx+1-n)^{n}. \end{eqnarray*} $