Let P be a poset. For, we say a covers b, denoted by; if and there doesn’t exist such that. If P has the minimum (resp. maximum) element, then we denote it by 0 (resp. 1) and say that P is a poset with 0 (resp. 1). Let P be a finite poset with 0. By a rank function on P, we mean a function r from P to the set of all the integers such that and whenever. Observe the rank function is unique if it exists. P is said to be ranked whenever P has a rank function.

Let P be a finite ranked poset with 0 and 1. The polynomial is called the characteristic polynomial of P, where is the function on P and r is the rank function of P. A poset P is said to be a lattice if both and exist for any two elements. and are called the join and meet of a and b, respectively. Let P be a finite lattice with 0. By an atom in P, we mean an element in P covering 0. We say P is atomic if any element in is the join of atoms. A finite atomic lattice P is said to be a geometric lattice if P admits a rank function satisfying,. Notations and terminologies about posets and lattices will be adopted from books [1] [2] .

The special lattices of rough algebras were discussed in [3] . The lattices generated by orbits of subspaces under finite (singular) classical groups were discussed in [4] [5] . Wang et al. [6] -[8] constructed some sublattices of the lattices in [4] . The subspaces of a d-bounded distance-regular have similar properties to those of a vector space. Gao et al. [9] -[11] constructed some lattices and posets by subspaces in a d-bounded distance-regular graph. In this paper, we continue this research, and construct some new sublattices of the lattices in [4] , discussing their geometricity and computing their characteristic polynomials.

Let be a finite field with q elements, where q is a prime power. For a positive integer, let be the n-dimensional row vector space over. Let. For a fixed-dimensional subspace of, let.

If we define the partial order on by ordinary inclusion (resp. reverse inclusion), then is a poset, denoted by (resp.). In the present paper we show that both and are finite atomic lattices, discuss their geometricity and compute their characteristic polynomials.

In this section we prove that the lattice is a finite geometric lattice, and compute its characteristic polynomial. We begin with a useful proposition.

Proposition 2.1. ([12] , Lemma 9.3.2 and [13] , Corollaries 1.8 and 1.9). For, the following hold:

1) The number of k-dimensional subspaces contained in a given m-dimensional subspace of is

.

2) The number of m-dimensional subspaces containing a given k-dimensional subspace of is

.

3) Let P be a fixed m-dimensional subspaces of. Then the number of k-dimensional subspaces Q of satisfying is

.

Theorem 2.2. is a geometric lattice.

Proof. For any two elements,

Therefore is a finite lattice. Note that is the unique minimum element. Let be the set of all the -dimensional subspaces of, where. Then is the set of all the atoms in. In order to prove is atomic, it suffices to show that every element of is a join of some atoms. The result is trivial for. Suppose that the result is true for. Let. By Proposition 2.1 and, the number of -dimensional subspaces of contained in at least is

.

Therefore there exist two different l-dimensional subspaces of such that. By inductionis a join of some atoms. Hence is a finite atomic lattice. For any , define. It is routine to check that is the rank function on. For any, we have

Hence is a geometric lattice.

Lemma 2.3. For any, suppose that, and. Then the function of is

Proof. The function of is

By Proposition 2.1, we have

.

Thus, the assertion follows.

Theorem 2.4. The characteristic polynomial of is

Proof. By Proposition 2.1 and Lemma 2.3, we have

In this section we prove that the lattice is a finite atomic lattice, classify its geometricity and compute its characteristic polynomial.

Theorem 3.1. The following hold:

1) is a finite atomic lattice.

2) is geometric if and only if.

Proof. 1) For any two elements, and

Therefore is a finite lattice. Note that is the unique minimum element. Let be the set of all the j-dimensional subspaces of, where. Then is the set of all the atoms in. In order to prove is atomic, it suffices to show that every element of is a join of some atoms. The result is trivial for. Suppose that the result is true for. Let. By Proposition 2.1, the number of subspaces of containing is equal to

.

Then there exist two different subspaces such that. By induction is a join of some atoms. Therefore is a finite atomic lattice.

2) For any, we define. It is routine to check that is the rank function on. It is obvious that is a geometric lattice. Now assume that. Let P be a -dimensional subspace of and. By Proposition 2.1, the number of 2-dimensional subspaces of containing P is equal to

Therefore, there exist two different 2-dimensional subspaces such that. So,. Hence, which implies that is not a geometric lattice when.

Lemma 3.2. For any, suppose that, and. Then the function of is

Proof. The function of is

Proposition 2.1 implies that

.

Theorem 3.3. The characteristic polynomial of is

.

Proof. By Proposition 2.1, we have