On forbidden poset problems in the linear lattice

In this note, we determine the maximum size of a $\{V_{k}, \Lambda_{l}\}$-free family in the lattice of vector subspaces of a finite vector space both in the non-induced case as well as the induced case, for a large range of parameters $k$ and $l$. These results generalize earlier work by Shahriari and Yu. We also prove a general LYM-type lemma for the linear lattice which resolves a conjecture of Shahriari and Yu.

[n] q ! = n i=1 [i] q . Then, it is easy to check that The general study of forbidden poset problems in the linear lattice was initiated by Ghassan and Shahriari [7]. For any collection of finite posets P, let La q (n, P) be the maximum size of a family of subspaces of V (viewed as a poset under inclusion) which does not contain any P ∈ P as a subposet, and let La * q (n, P) be the maximum size of a family of subspaces of V which does not contain any P ∈ P as an induced subposet.
We write simply La q (n, P ) and La * q (n, P ) if P = {P } for some poset P . We denote by Σ q (n, k) the sum of the k-largest q-binomial coefficients of the form n i q . Let V and Λ be the posets on three elements x, y, z defined by the relations x, y > z and x, y < z, respectively. In 1983, Katona and Tarján [8] proved the following result.
Shahriari and Yu [11] showed that in the linear lattice we have the following. Theorem 1.3 (Shahriari and Yu [11]).
The extremal construction is either V , except in the case n = 3 and q = 2, in which we have two other constructions shown in Figure 2.
We prove the following induced version of Theorem 1.3.
The extremal construction is either V , except in the case n = 3 and q = 2, in which we have two other constructions shown in Figure 2.
Let V k denote the poset with elements x 1 , x 2 , . . . , x k , y such that x 1 , x 2 , . . . , x k > y, and let Λ k denote the same poset but with all relations reversed. In the case when k or l is at least 3 only asymptotic results are known for La(n, {V k , Λ l }).
In the linear lattice, on the other hand, one can prove exact results for larger k and l as well. Shahriari and Yu [11] proved the following. Theorem 1.5 (Shahriari and Yu [11]). Let n be an even integer, and k, l be two integers such that k, l ≤ q. Then La q (n, {V k , Λ l }) = n n 2 q , and the only {V k , Λ l }-free family of maximum size is V n 2 q .
We extend Theorem 1.5 by weakening the conditions on k and l.
Theorem 1.6. Let n be an even integer, and k, l be two integers such that k, l ≤ q n 2 . Then In the induced case, we have the following two results.
Theorem 1.7. Let n be an even integer and let k, l be two integers such that k, l ≤ q, then , and the only maximum size {V k , Λ l }-free family is V  Equality occurs only for a family consisting of the union of two consecutive levels in the Boolean lattice of largest size.
We denote by Y k the poset with elements x 1 , x 2 , . . . , x k , y, z such that x 1 ≤ x 2 ≤ · · · ≤ x k ≤ y, z and Y ′ k the same poset but with all relations reversed. In the proof of Theorem 1.9, De Bonis, Katona and Swanepoel actually proved a stronger result by determining La(n, {Y 2 , Y ′ 2 }). Later pairs of posets {Y k , Y ′ k } were investigated for their own sake. Methuku and Tompkins [10] obtained the following theorem. Theorem 1.10 (Methuku and Tompkins [10]). Let k ≥ 2 and n ≥ k + 1, then Martin, Methuku, Uzzell and Walker [9] and Tompkins and Wang [13] proved the induced version of Theorem 1.10 independently.
Theorem 1.11 (Martin et al. [9], Tompkins and Wang [13]). Let k ≥ 2 and n ≥ k + 1, then In the vector space setting, Shahriari and Yu [11] proved a version of Theorem 1.9 holds. Namely, they proved Theorem 1.12 (Shahriari and Yu [11]). Len n ≥ 3 be an integer and q be a power of a prime, then Equality occurs only for a family consisting of the union of two consecutive levels in the linear lattice of maximum size.
Furthermore, they posed a conjecture for the case when {Y k , Y ′ k } is forbidden. For any poset P , let |P | be the size of P and h(P ) be the length of the largest chain in P . Burcsi and Nagy [1] and Grósz, Methuku and Tompkins [6] proved the following theorems for any poset P (another result in this direction was obtain by Chen and Li [2]). Theorem 1.13 (Burcsi and Nagy [1]). For any poset P , when n is sufficiently large, we have Theorem 1.14 (Grósz, Methuku and Tompkins [6]). For any poset P , when n is sufficiently large, we have for any fixed k.
We will prove that a version of these theorems holds in the vector space case as well.
The rest of this paper is organized as follows. In the next section, we present some preliminary results. Then we will prove Theorems 1.4, 1.7 and 1.8 in Section 3. In the last section, we prove a general LYM-type lemma and use this lemma to prove the vector space analogues of Theorems 1.10, 1.11, 1.13 and 1.14. We note that a recent manuscript of Gerbner [5] independently initiates a general study of LYM-type properties of the linear lattice and implies some similar results.

Preliminary results
In this section, let F be a {V k , Λ l }-free family of subspaces of V , and let F s = F ∩ V s q . Now, we define the bipartite graph Let F ′ s be any subset of F s , and Before beginning the proof, we need some preliminary results. Lemma 2.1 and Corollary 2.2 are motivated by an idea from [8].
Lemma 2.1. Let n be an even integer, and let k, l be two integers such that k, l ≤ q n 2 . Then, La q (n, {V k , Λ l }) can be realized with a family G of subspaces G satisfying dim(G) ≤ n 2 .
Proof: We first prove that for s ≥ n 2 + 1, the bipartite graph F s ∪ ( V s−1 \ F s−1 ), E contains a matching such that every element of F s is contained in some edge. To prove this, it is enough to check the condition of Hall's theorem, that is , and q ≥ 2. Applying Hall's theorem, let M be a matching which saturates every vertex in F s , and let F * s−1 be the set of neighbors of F s contained in edges of M . Clearly, Call the resulting family G. Clearly, ✷ Since linear lattices are symmetric, one can use the same idea to prove the following corollary.
Corollary 2.2. Let n be an even integer, and k, l be two integers such that k, l ≤ q n 2 . Then, La q (n, {V k , Λ l }) can be realized with a family G of subspaces G satisfying dim(G) ≥ n 2 .
The next technical lemma will be needed for determining the structure of the extremal families.
. We may now show by a simple averaging argument that there exists an F ∈ F ⌈ n 2 ⌉+1 such that F has at least q ⌈ n 2 ⌉ subspaces in F ⌈ n 2 ⌉ . Indeed, the number of relations between F ⌈ n 2 ⌉+1 and F ⌈ n 2 ⌉ is at least Thus, on average an element of F ⌈ n 2 ⌉+1 contains at least q ⌈ n 2 ⌉ subspaces in F ⌈ n 2 ⌉ . ✷ In the same way one can show the following.
Now, we can prove Theorem 1.6.
Proof: Combining Lemma 2.1 and Corollary 2.2, it is easy to see that Suppose not. If there is a subspace F ∈ F of dimension larger than n 2 , then we may assume, without loss of generality, that |F| = n n 2 q and for every F ∈ F, n 2 ≤ dim(F ) ≤ n 2 + 1 and F n 2 +1 = ∅. By Lemma 2.3, F contains a copy of Λ l , a contradiction. The case when F contains only subspaces of dimension at most n 2 is handled similarly by Lemma 2.4. ✷ 3 Proofs of Theorems 1.4, 1.7 and 1.8 . . , f r are not necessarily distinct). Then, we have the following proposition.
Proof: The subspace F is either small (suppose F = F i in this case) or contains some small subspace F i . In both cases we have f i ⊆ F i ⊆ F . ✷ Now, we define a family M (A) collecting all (s−1)-subspaces of A which do not contain any of the f i .
By Proposition 3.1, we have that the following properties of M (A) hold.
Let F ′ s be any subset of F s , and Lemma 3.3. Let n be an odd integer, and k, l be two integers such that k, l ≤ q. Then, La * q (n, {V k , Λ l }) can be realized with a family G of subspaces G satisfying dim(G) ≤ n+1 2 . Proof: We first show that for s ≥ n+3 2 , the bipartite graph F s ∪ V s−1 q \ F s−1 , E contains a matching such that every element of F s is contained in some edge. By Hall's theorem, it is enough to prove On the other hand, every (s − 1)-dimensional subspace in N s−1 (F ′ s ) has at most [n − s + 1] q superspaces in F ′ s . Then, by (iii) from Proposition 3.2, we have since s ≥ n+3 2 , l ≤ q and q ≥ 2. Let M be a matching which saturates every vertex in F s , and F * s−1 be matched under M . Clearly, |F * s−1 | = |F s |, and F * s−1 ∩ F s−1 = ∅ by (ii) from Proposition 3.2. Now, let t = t(F) be the largest integer s satisfying F s = ∅ in our family F. We repeatedly replace F by (F \ F t ) ∪ F * t−1 until t ≤ n+1 2 . Call the resulting family G. Clearly, |G| = |F|. Then it is enough to show that F is induced {V k , Λ l }-free in every step.
By contradiction, assume that at some step contains V k or Λ l . We distinguish two cases.
By (i) from Proposition 3.2, F ⊆ F 1 , a contradiction. ✷ Using the same idea in Section 2, one can similarly prove the following corollaries.
Corollary 3.6. Let n be an odd integer, and k, l be two integers such that k, l ≤ q. Then, La * q (n, {V k , Λ l }) can be realized with a family G of subspaces G satisfying n−1 Corollary 3.7. Let n be an even integer, and k, l be two integers such that k, l ≤ q. Then, La * q (n, {V k , Λ l }) can be realized with a family G of subspaces G satisfying dim(G) = n 2 .
Theorem 1.7 follows from Corollary 3.7 and the equality cases are again settled by applying Lemmas 2.3 and 2.4. (Once a family is contained in two levels there is no distinction between an induced and noninduced copy of V k or Λ l .) Now, we turn to prove Theorem 1.8. Before beginning the proof, we need the following lemma.
and the only families which attain equality are V 3 1 q and V 3 2 q .
Note that V 3 1 q = V 3 2 q = q 2 + q + 1, so we have |A| > |B ′ | and |B| > |A ′ |. Since F is Λ l -free, for every A ∈ A, the number of subspaces of A in B is at most l − 1, thus the number of subspaces of A in B ′ is at least (q + 1) − (l − 1) = q + 2 − l. Since |A| > |B ′ |, there exists a subspace B ∈ B ′ with at least q + 3 − l superspaces A 1 , A 2 , . . . , A q+3−l in A by the pigeonhole principle.
For 1 ≤ i, j ≤ q + 3 − l, A i and A j have only one common subspace B, since there is no butterfly in two consecutive levels of a linear lattice. So we have and similarly, we have since F is V k -free. Then, a contradiction when k, l ≤ q − √ 2 2 q. This completes the proof of the inequality. Furthermore, if |F| = q 2 + q + 1 and A, B = ∅, we will have |A| = |B ′ | instead of |A| > |B ′ |. Then there exists a subspace B ∈ B ′ with at least q + 2 − l superspaces in A, and so but this contradicts the condition k, l ≤ q − √ 2 2 q. ✷ Remark 3.9. Lemma 3.8 is true when the weaker condition q 2 + q + 1 < (q + 2 − l)(q + 1 − l) + (q + 2 − k)(q + 1 − k) + 2 holds.
Now, we are ready to prove Theorem 1.8.
Proof of Theorem 1.8: A maximal chain in a linear lattice of dimension n is a sequence of subspaces V 0 , V 1 , . . . , V n where {0} = V 0 ⊂ V 1 ⊂ · · · ⊂ V n = V . We denote by C the set of all maximal chains in a linear lattice. Now, we double count the number of pairs (F, C), where F ∈ F, C ∈ C such that F is in the chain C.
For every F ∈ F, there are [dim(F )] q ![n − dim(F )] q ! maximal chains though F . On the other hand, we consider a pair of subspaces (G 1 , G 2 ) such that dim(G 1 ) = n+3 2 , dim(G 2 ) = n−3 2 and G 2 ⊆ G 1 . Then the subfamily of F between G 1 and G 2 satisfies the condition of Lemma 3.8. Hence, the size of the subfamily can be bounded as q 2 + q + 1, and the number of chains between G 1 and G 2 though some F in the subfamily is (q 2 + q + 1)(q + 1).
Clearly, the number of maximal chains between {0} and G 2 (G 1 and V ) is [ n−3 2 ] q !, and the number of such pairs (

This completes the proof of La
Since equality must hold in the first inequality of (1), we have , then the size of subfamily between G 1 and G 2 is q 2 + q + 1 for any pair (G 1 , G 2 ) such that dim(G 1 ) = n+3 2 , dim(G 2 ) = n−3 2 and G 2 ⊆ G 1 .
We also need the following theorem from [3].
Theorem 3.11 (Chowdhury and Patkós [3]). Let F ⊂ V k q and x ≥ k be a real number such that |F| = x k q , then ∂F ≥ x k−1 q , where ∂F = {G ∈ V k−1 q : G ⊂ F ∈ F}.
Now we turn to the proof of Theorem 1.4. Clearly, the even case follows from Theorem 1.7. So we need to prove the case when n is odd.
Proof of Theorem 1.4: By Remark 3.12, when k = l = 2, the weaker condition for upper bound of |F| is q 2 − q − 1 > 0, this is true for q ≥ 2, and this completes the proof of La * q (n, {V, Λ}) = n n+1 2 q . Furthermore, the weaker condition for structure of F is q 2 − 3q + 1 > 0, and this inequality is true for q ≥ 3. Note that in this structure, there is a matching with 3 edges connecting 6 subspaces and a single isolated subspace. In Figure 2, g and G are the single isolated subspaces, respectively.
However, these constructions do not extend beyond the case n > 3 for q = 2. Otherwise, by Corollary 3.6, we may suppose that for every F ∈ F, n−1 2 ≤ dim(F ) ≤ n+1 2 . If one of F n−1 2 and F n+1 2 is empty, we can find a contradiction by Lemma 2.3 or 2.4. Now, both of them are not empty, by Theorem 3.11 and the assumption that |F| = n n+1 2 q , we can find two subspaces (say d ⊂ A) in F. Suppose dim(Q) = n+3 2 and dim(P ) = n−3 2 such that P ⊂ d ⊂ A ⊂ Q. We apply Fact 3.10, then we have 7 (q 2 + q + 1) subspaces in F between P and Q. Since d and A are not in the same level, without loss of generality, we can suppose that A, B, C, d, e, f, g ∈ F (as in Figure 2).
Since n ≥ 5, we have [ n−1 2 ] q ≥ 3 superspaces for every subspace of dimension n+1 2 . Then E has two other n+3 2 -dimensional superspaces Q ′ and Q ′′ . By applying Fact 3.10, the construction of subspaces in F between P and Q ′ (P and Q ′′ ) is not a level, since b / ∈ F and e ∈ F. Moreover, B ⊂ Q ′ (Q ′′ ), otherwise B, E, Q and Q ′ (Q ′′ ) form a butterfly. Also note that B, e ∈ F and e ⊂ B, so e is an isolated subspace in the construction between P and Q ′ (P and Q ′′ ). Thus, there exist E ′ ⊂ Q ′ and E ′′ ⊂ Q ′′ in F such that g ⊂ E ′ , E ′′ .
Note that E ′ = E ′′ , otherwise E, E ′ , Q ′ and Q ′′ form a butterfly. Then g, E ′ and E ′′ form an induced V in F, a contradiction. We now present a general LYM-type lemma. The proof comes from adopting the methods from [6] to a vector space setting. and set H π = {H π : H ∈ H}.
Observe that if for two distinct H 1 , H 2 ∈ H we have F = H π 1 1 and F = H π 2 2 , then π 1 = π 2 . It follows that for each F ∈ F, there are mappings π such that F ∈ H π . Thus, on the one hand, the number of pairs (F, π) is F ∈F (q r − 1)(q r − q) . . . (q r − q r−1 )(q n−r − 1)(q n−r − q) . . . (q n−r − q n−r−1 )N dim(F ) (H), or equivalently, Now suppose we fix a mapping π. Since H and H π are isomorphic as posets with respect to the subspace relation, we have at most α(H, P ) many F ∈ F such that F ∈ H π . Since the total number of mappings π is (q n − 1)(q n − q) . . . (q n − q n−1 ), we have an upper bound on the number of pairs (F, π) of (q n − 1)(q n − q) . . . (q n − q n−1 )α(H, P ), or equivalently, [n] q !(q − 1) n α(H, P ).
Combining (2) and (3), we have and rearranging yields the desired inequality.  [11]). Let k ≥ 1 and n ≥ k + 1, then We will show that even the induced version of this conjecture holds.  Otherwise, if {0}, V ∈ F, we can assume the result is true for k − 1. Note that the base case k = 1 is proved by Theorem 1.4. Then it follows that |F| ≤ Σ q (n, k − 1) + 2 ≤ Σ q (n, k), since F is induced {Y k−1 , Y ′ k−1 }-free. Now we may assume that {0} ∈ F and V / ∈ F. Let G = F \{{0}}. If |G| ≤ Σ q (n, k)−1, then |F| ≤ Σ q (n, k). So we may assume |G| = Σ q (n, k), and G is a subfamily of the k (or k + 1) largest levels in the linear lattice. If n = k modulo 2, then G is uniquely determined (i.e., the largest k levels), and it is easy to find an induced Y k in F. Indeed, we can find an induced Y k−1 in G, which together with {0} form an induced Y k . If n = k modulo 2, then G is a subfamily of the k + 1 largest levels L 1 , L 2 , . . . , L k+1 . If G ∩ L 1 = ∅, then we can find an induced Y k as in the previous case. Otherwise, let L 1 ∈ G ∩ L 1 , then find an induced Y k−2 in L 2 ∪ · · · L k+1 such that L 1 is a subspace of every subspace in this Y k−2 .
By Lemma 4.1, the vector space versions of Theorems 1.13 and 1.14 follows from the above two lemmas, respectively. Remark 4.8. In proving the vector space version of Theorem 1.14 from the Lemma 4.7, we proceed exactly as in the corresponding proof in [6] replacing each binomial coefficient with the corresponding q-binomial, and verify that all the estimates still hold.