Extremal graph for intersecting odd cycles

An extremal graph for a graph $H$ on $n$ vertices is a graph on $n$ vertices with maximum number of edges that does not contain $H$ as a subgraph. Let $T_{n,r}$ be the Tur\'{a}n graph, which is the complete $r$-partite graph on $n$ vertices with part sizes that differ by at most one. The well-known Tur\'{a}n Theorem states that $T_{n,r}$ is the only extremal graph for complete graph $K_{r+1}$. Erd\"{o}s et al. (1995) determined the extremal graphs for intersecting triangles and Chen et al. (2003) determined the maximum number of edges of the extremal graphs for intersecting cliques. In this paper, we determine the extremal graphs for intersecting odd cycles.


Introduction
In this paper, all graphs considered are simple and finite. For a graph G and a vertex x ∈ V (G), the neighborhood of x in G is denoted by N G (x). Let N G [x] = {x}∪N G (x). The degree of x, denoted by deg G (x), is |N G (x)|. Let δ(G) and ∆(G) denote the minimum and maximum degrees of G, respectively. A matching M in G is a subgraph of G with δ(M) = ∆(M) = 1. The matching number of G, denoted by ν(G), is the maximum number of edges in a matching in G. Let e(G) be the number of edges of G.
For a graph G and S, T ⊂ V (G), let e G (S, T ) be the number of edges e = xy ∈ E(G) such that x ∈ S and y ∈ T , if S = T , we use e G (S) instead of e G (S, S); and we use e G (u, T ) instead of e G ({u}, T ) for convenience, the index G will be omitted if no confusion from the context. For a subset X ⊆ V (G) or X ⊆ E(G), let G[S] be the subgraph of G induced by X, that is G[X] = (X, E(X)) if X ⊆ V (G), or G[X] = (V (X), X) if X ⊆ E(G). A cycle of length q is called a q-cycle.
Given two graphs G and H, we say that G is H-free if G does not contain an H as a subgraph. The Turán function, denoted by ex(n, H), is the largest possible number of edges of an H-free graph on n vertices. That is, ex(n, H) = max{e(G) : |V (G)| = n, G is H-free}.
And for positive integers n ≥ r, the Turán graph, denoted by T n,r , is the complete r-partite graph on n vertices with part sizes that differ by at most one (also called the complete balanced r-partite graph). The well-known Turán Theorem states that ex(n, K r+1 ) = e(T n,r ) and T n,r is the only extremal graph for complete graph K r+1 .
A k-fan, denoted by F k , is a graph on 2k + 1 vertices consisting of k triangles which intersect in exactly one common vertex. In 1995, Erdös et al. [4] gave the value of ex(n, F k ) and determined the extremal graphs for F k as follows.
Moreover, when k is odd, the extremal graph must be a T n,2 with two vertex disjoint copies of K k embedding in one partite set. When k is even, the extremal graph must be a T n,2 with a graph having 2k − 1 vertices, k 2 − 3 2 k edges with maximum degree k − 1 embedded in one partite set.
In 2003, Chen et al. [2] proved that ex(n, F k,r ) = e(T n,2 ) + g(k), where F k,r is a graph consisting of k complete graphs of order r which intersect in exactly one common vertex and g(k) is the same as in Theorem 1.1. Recently, Gelbov [5] and Liu [6] gave the extremal graphs for blow-ups of paths [5], cycles and a large class of trees [6].
In this paper, motivated by the results in [2,4,5,6], we generalize Theorem 1.1 in another way. For a positive integer k and an odd integer q with q ≥ 5, let C k,q be the graph consisting of k q-cycles which intersect exactly in one common vertex, called the center of it. For n 4(k − 1) 2 , let F n,k be the family of graphs with each member is a Turán graph T n,2 with a complete bipartite graph K k−1,k−1 embedded into one class. Our main result is as follows.
Theorem 1.2. For an integer k 2 and an odd integer q 5, there exists n 1 (k, q) ∈ N such that for all n n 1 (k, q), we have ex(n, C k,q ) = e(T n,2 ) + (k − 1) 2 , and the only extremal graphs for C k,q are members of F n,k .
The remaining of the paper is arranged as follows. Section 2 gives some lemmas. Section 3 gives the proof of Theorem 1.2.

Lemmas
The following two lemmas are useful to estimate the number of edges of a graph with restrict degree and matching number. The following stability result due to Erdös [3] and Simonovits [7] gives the rough structure of the extremal graphs for a graph H with χ(H) = r 3 and H = K r . 3,7]). Let H be a graph with χ(H) = r 3 and H = K r . Then, for every γ > 0, there exists δ > 0 and n 0 = n 0 (H, γ) ∈ N such that the following holds. If G is an H-free graph on n n 0 vertices with e(G) ex(n, H) − δn 2 , then there exists a partition of The following is a simple observation.
where ω(G) is the number of components of G.
Proof. Since ∆(G) ≤ 2, each component of G is a path or a cycle. Hence each component C of G has matching number at least |V (C)|−1 2 . This implies the desired result.
) r, then e(G) r 2 . Moreover, the equality holds if and only if G = K r,r .
Proof. Clearly, ∆(G) r. We claim that ν(G) r. Let u 1 v 1 , · · · , u ℓ v ℓ be a matching in G. Wlog, assume that the claim is true. Now we prove the result according to the following two cases. Case 1. ∆(G) < r. Then, by Lemma 2.1, we have e(G) f (ν, ∆) ≤ r(r − 1 + 1) = r 2 , and the equality holds only if ν = r and ∆ = r−1. We claim that the equality does not hold in this case.
Suppose to the contrary that e(G) = 9. Since ∆(G) ≤ 2 and G has no isolated vertex, |V (G)| ≥ e(G) = 9 (the equality holds if and only if G is 2-regular) and ω(G) ≤ ν(G) = 3. By Observation 2.3, Hence |V (G)| ≥ 9. Thus |V (G)| = 9 (and so G is 2-regular) and ω(G) = 3. Therefore, G consists of three vertex-disjoint triangles. But this contradicts the assumption that Moreover, the equality holds if and only if d i = r andd i = 0 for each i ∈ [1, r], that is G is a bipartite graph with partites N(x) = {x 1 , · · · , x r } and V (G)\N(x). To show that The following lemma states that the members F n,k are C k,q −free.
Lemma 2.5. Each member of F n,k is C k,q -free for all k ≥ 2, n 4(k − 1) 2 , and odd integer q ≥ 5.
Proof. Suppose to the contrary that there is a graph G ∈ F n,k containing a copy of C k,q . Let K be the copy of K k−1,k−1 in G. Then each odd cycle of C k,q must contain odd number of the edges of K. Let A = E(C k,q ) ∩ E(K). Then |A| ≥ k. We claim that the center of C k,q must lie in K. If not, then G[A] contains a matching of order at least k by the structure of C k,q , a contradiction with ν(K) = k − 1. Let Lemma 2.6. Let n 0 be an integer and let G be a graph on n ≥ n 0 + n 0 2 vertices with e(G) = e(T n,2 ) + j for some integer j > 0. Then G contains a subgraph G ′ on n ′ > n 0 vertices such that δ(G ′ ) δ(T n ′ ,2 ) and e(G ′ ) e(T n ′ ,2 ) + j + n − n ′ .
Proof. If δ(G) ≥ n 2 , then G is the desired graph and we have nothing to do. So assume that δ(G) < n e(T n,2 ) + j − n 2 + 1 = e(T n−1,2 ) + j + 1, since e(T n,2 ) − e(T n−1,2 ) = n 2 . We may continue this procedure until we get G ′ on n − i vertices with δ(G ′ ) δ(T n−i,2 ) for some i < n − n 0 , or until i = n − n 0 . For the latter case, G ′ has n 0 vertices but e(G ′ ) e(T n 0 ,2 ) + j + i > n − n 0 n 0 2 , which is impossible.
By the choice of n 1 and Lemma 2.6, we may assume δ(G) ≥ δ(T n,2 ) = n 2 , otherwise, we consider a subgraph G ′ with the desired minimum degree instead of G. Let V 0∪ V 1 be a partition of V (G) such that e(V 0 , V 1 ) is maximized. Lemma 2.2 implies that m = e(V 0 ) + e(V 1 ) < γn 2 . The following claim asserts that the partition is closed to be balanced.
Furthermore, m = e(V 0 ) + e(V 1 ) ≥ (k − 1) 2 and if the equality holds then G contains a complete balanced bipartite graph with classes V 0 and V 1 .
Proof. Let |V 0 | = n 2 + a. Then |V 1 | = n 2 − a. Since Hence e(V 0 , V 1 ) = e(T n,2 ), that is V 0 , V 1 are balanced and so G contains a complete balanced bipartite graph with classes V 0 and V 1 .
In the following, let Proof. We prove it by contradiction. Wlog, assume that there is an Then the number of bad vertices in G is at most 2m For any vertex u ∈ V i (i = 0, 1), by the maximality of e(V 0 , V 1 ), we have Particularly, if u ∈ V i is good, then we have We find a copy of C k,q passing through all the vertices of A to get a contradiction.
For each ℓ ∈ [1, s], we find a sequence of vertices w 1 1ℓ , w 2 0ℓ , · · · , w q−3 0ℓ , w q−2 1ℓ with w j ξ(j)ℓ ∈ V ξ(j) \ A for 1 j q − 2, such that w q−3 0ℓ is good and xw 1 1ℓ · · · w q−3 0ℓ w q−2 1ℓ x ℓ x is a q-cycle. Furthermore, we require that w j ξ(j)ℓ (ℓ ∈ [1, s], j ∈ [1, q − 2]) are pairwise different. This is possible since together with all vertices in A, the total number of good vertices which we have found is at most |V (C k,q )| = k(q − 1) + 1 and each vertex u ∈ V i has at least good neighbors in V 1−i and the number of common neighbors of w q−3 0ℓ and x ℓ in V 1 is at least (since n 20k 2 q) Thus we have found a copy of C s,q centered at x and passing through the edges of {xx 1 , xx 2 , · · · , xx s }. Particularly, since G is C k,q −free, we have s k − 1. Thus Consequently, all the vertices of G are good, and for each vertex u ∈ V i , Next we will find a copy of C k−s,q centered at x disjoint from the copy of C s,q . For every u ℓ (ℓ ∈ [s + 1, t]), choose a common neighbor of x and u l , say w 1 1ℓ , in V 1 such that w 1 1ℓ = w 1 1ℓ ′ if ℓ = ℓ ′ . We can do this because the number of common neighbors of x and u ℓ in V 1 is at least For each ℓ ∈ [s+1, t], with the same reason as above, we find vertices w 3 1ℓ , w 4 0ℓ , · · · , w q−3 0ℓ one by one, then a common neighbor of w q−3 0ℓ and v ℓ , say w q−2 1ℓ , in V 1 such that u ℓ w 1 1ℓ xw 3 1ℓ · · · w q−2 1ℓ v ℓ u ℓ is a q-cycle. And for every l ∈ [t, k], begin with x, we can find vertices w 2 1ℓ , w 3 0ℓ , · · · , w q−3 1ℓ one by one, then a common neighbor, say w q−2 0ℓ , of w q−3 1ℓ and v ℓ in V 0 such that u ℓ xw 2 1ℓ w 3 0ℓ · · · w q−2 0ℓ v ℓ u ℓ is a q-cycle. Also, we may require that w j ξ(j)ℓ (ℓ ∈ [s + 1, k], j ∈ [1, q − 2]) are pairwise different. Thus the k − s q-cycles form a copy of C k−s,q centered at x and passing through all of the edges of contains a desired copy of C k,q with center x. This completes the proof.
Proof. If not, suppose that M 0 = {u 1 v 1 , · · · , u s v s } and M 1 = {u s+1 v s+1 , · · · , u k v k } are matchings in G 0 and G 1 , respectively. Wlog, assume that s 1. First we find a common neighbor, say x, of u s+1 , · · · , u k in V 0 \ V (M 0 ). This is possible since the number of such neighbors is at least Proof. If not, then by Claim 2, max{∆ 0 , ∆ 1 } k − 2. Thus by Lemma 2.1 and Claim 3, a contradiction with Claim 2.