ARS MATHEMATICA CONTEMPORANEA Also available at http://amc-journal.eu ISSN 1855-3966 (printed edn.), ISSN 1855-3974 (electronic edn.) ARS MATHEMATICA CONTEMPORANEA 10 (2016) 383-392 The endomorphisms of Grassmann graphs* Li-Ping Huang School of Mathematics, Changsha University of Science and Technology, Changsha, 410004, China Benjian Lv Kaishun Wang Sch. Math. Sci. & Lab. Math. Com. Sys., Beijing Normal University, Beijing, 100875, China Received 17 December 2014, accepted 17 December 2015, published online 15 March 2016 A graph G is a core if every endomorphism of G is an automorphism. A graph is called a pseudo-core if every its endomorphism is either an automorphism or a colouring. Suppose that Jq(n, m) is a Grassmann graph over a finite field with q elements. We show that every Grassmann graph is a pseudo-core. Moreover, J2(4,2) is not a core and Jq (2k + 1, 2) (k > 2) is a core. Keywords: Grassmann graph, core, pseudo-core, endomorphism, maximal clique. Math. Subj. Class.: 05C60, 05C69 1 Introduction Throughout this paper, all graphs are finite undirected graphs without loops or multiple edges. For a graph G, we let V(G) denote the vertex set of G. If xy is an edge of G, then x and y are said to be adjacent, and denoted by x ~ y. Let G and H be two graphs. A homomorphism p from G to H is a mapping p : V(G) ^ V(H) such that p(x) ~ p(y) whenever x ~ y. If H is the complete graph Kr, then p is a r-colouring of G (colouring for short). An isomorphism from G to H is a bijection p : V(G) ^ V(H) such that x ~ y ^ p(x) ~ p(y). Graphs G and H are called isomorphic if there is an isomorphism from * Projects 11371072, 11301270, 11271047, 11371204, 11501036 supported by NSFC. Supported by the Fundamental Research Funds for the Central University of China, Youth Scholar Program of Beijing Normal University (2014NT31) and China Postdoctoral Science Foundation (2015M570958). t Corresponding author E-mail address: lipingmath@163.com (Li-Ping Huang), bjlv@bnu.edu.cn (Benjian Lv), wangks@bnu.edu.cn (Kaishun Wang) Abstract ©® This work is licensed under http://creativecommons.org/licenses/by/3.0/ 383 Ars Math. Contemp. 10 (2016) 255-268 G to H, and denoted by G = H .A homomorphism (resp. isomorphism) from G to itself is called an endomorphism (resp. automorphism) of G. Recall that a graph G is a core if every endomorphism of G is an automorphism. A subgraph H of G is a core of G if it is a core and there exists a homomorphism from G to H. Every graph has a core, which is an induced subgraph and is unique up to isomorphism [5]. A graph is called core-complete if it is a core or its core is complete. A graph G is called a pseudo-core if every endomorphism of G is either an automorphism or a colouring. Every core is a pseudo-core. Any pseudo-core is core-complete but not vice versa. For more information, see [2, 6, 9]. For a graph G, an important and difficult problem is to distinguish whether G is a core [2, 5, 6, 7, 11, 15]. If G is not a core or we don't know whether it is a core, then we need to judge whether it is a pseudo-core because the concept of pseudo-core is the most close to the core. Recently, Godsil and Royle [6] discussed some properties of pseudo-cores. Cameron and Kazanidis [2] discussed the core-complete graph and the cores of symmetric graphs. The literature [10] showed that every bilinear forms graph is a pseudo-core which is not a core. One of the latest result is from [9], where it was proved that every alternating forms graph is a pseudo-core. Moreover, Orel [13, 12] proved that each symmetric bilinear forms graph (whose diameter is greater than 2) is a core and each Hermitian forms graph is a core. Suppose that Fq is the finite field with q elements, where q is a power of a prime. Let V be an n-dimensional row vector space over Fq and let be the set of all m-dimensional subspaces of V. The Grassmann graph Jq (n, m) has the vertex set , and two vertices are adjacent if their intersection is of dimension m - 1. If m = 1, we have a complete graph and hence it is a core. Since Jq(n, m) = Jq(n, n - m), we always assume that 4 < 2m < n in our discussion unless specified otherwise. The number of vertices of Jq (n, m) is the Gaussian binomial coefficient: nq 1 -1. (1.1) qi-1 For Jq(n, m), the distance of two vertices X and Y is d(X, Y) := m — dim(X n Y). Any Grassmann graph is distance-transitive [1, Theorem 9.3.3] and connected. By [6, Corollary 4.2], every distance-transitive graph is core-complete, thus every Grassmann graph is core-complete. The Grassmann graph plays an important role in geometry, graph theory, association schemes and coding theory. Recall that an independent set of a graph G is a set of vertices that induces an edgeless graph. The size of the largest independent set is called the independence number of G, denoted by a(G). The chromatic number x(G) of G is the least value of k for which G can be k-colouring. A clique of a graph G is a complete subgraph of G. A clique C is maximal if there is no clique of G which properly contains C as a subset. A maximum clique of G is a clique with the maximum size. The clique number of G is the number of vertices in a maximum clique, denoted by w(G). By [6, p.273], if G is a distance-transitive graph and x(G) > w(G), then G is a core. Unluckily, applying the eigenvalues or the known results of graph theory for Grassmann graph, to prove the inequality x(G) > w(G) is difficult. Thus, it is a difficult problem to verify a Grassmann graph being a core. However, there are some Grassmann graphs which are not cores (see Section 4). Therefore, we need to judge whether a Grassmann graph is a pseudo-core. So far, this is an open problem. We solve this problem as follows: L.-P. Huang, B. Lv, K. Wang: The endomorphisms of Grassmann graphs 385 Theorem 1.1. Every Grassmann graph Jq (n, m) is a pseudo-core. The paper is organized as follows. In Section 2, we give some properties of the maximal cliques of Grassmann graphs. In Section 3, we shall prove Theorem 1.1. In Section 4, we discuss cores on Grassmann graphs. We shall show that J2(4,2) is not a core, Jq (2k +1, 2) (k > 2) is a core. 2 Maximal cliques of Grassmann graph In this section we shall discuss some properties of the maximal cliques of Grassmann graphs. We will denote by |X| the cardinal number of a set X. Suppose that V is an n-dimensional row vector space over Fq. For two vector subspaces S and T of V, the join S V T is the minimal dimensional vector subspace containing S and T. We have the dimensional formula (cf. [8, Lemma 2.1] or [16]): dim(S V T) = dim(S) + dim(T) - dim(S n T). (2.1) Throughout this section, suppose that 4 < 2m < n. For every (m - 1)-dimensional subspace P of V, let [P}m denote the set of all m-dimensional subspaces containing P, which is called a star. For every (m + 1)-dimensional subspace Q of V, let (Q]m denote the set of all m-dimensional subspaces of Q, which is called a top. By [4], every maximal clique of Jq (n, m) is a star or a top. For more information, see [14]. By [16, Corollary 1.9], qn-m+1 _ 1 qm+1 _ 1 |[P}m| = --:-, |(Q]m| = --r-. (2.2) q — 1 q — 1 If n > 2m, then every maximum clique of Jq(n, m) is a star. If n = 2m, then every maximal clique of Jq (n, m) is a maximum clique. By (2.2) we have w(Jq(n,m))= [n"m+1] ifn > 2m. (2.3) Since n > 2m, we have |[P}m| > |(Q]m|, and |[P}m| > |(Q]m| if n > 2m. (2.4) Lemma 2.1. If [P}m n (Q]m = 0, then the size of [P}m n (Q]m is q + 1. Proof. Since [P}m n (Q]m = 0, one gets P C Q. It follows that [P}m n (Q]m consists of all m-dimensional subspaces containing P in Q. By [16, Corollary 1.9], the desired result follows. □ Lemma 2.2. ([8, Corollary 4.4]) Let M1 and M2 be two distinct stars (tops). Then |Mi n m2| < 1. Lemma2.3. Suppose [A}m = [B}m. Then [A}mn [B}m = 0 ifandonlyif dim(AnB) = m — 2. In this case, [A}m n [B}m = {A V B}. Proof. Since dim(A) = dim(B) = m — 1 and A = B, one gets dim(A V B) > m. If [A}m n [B}m = 0, then by Lemma 2.2, there exists a vertex C of Jq (n, m) such that {C} = [A}mn[B}m. Itfollows from (2.1) and A, B c C that C = AVB anddim(AnB) = m—2. Conversely, if dim(A n B) = m — 2, then Lemma 2.2 and (2.1) imply that C := A V B is a vertex of Jq(n, m) and hence {C} = [A}m n [B}m. □ 386 Ars Math. Contemp. 10 (2016) 255-268 Lemma 2.4. Suppose (P ]m = (Q]m. Then (P ]m n(Q]m = 0 if and only if dim(P n Q) = m. In this case, (P]m n (Q]m = {P n Q}. Proof. By dim(P) = dim(Q) = m +1 and P = Q, we have dim(P n Q) < m. If (P]m n (Q]m = 0, then Lemma 2.2 implies that there exists a vertex C of Jq(n, m) such that {c} = (P]m n(Q]m. Since C c Pn Q, we get that C = Pn Q and dim(Pn Q) = m. Conversely, if dim(P n Q) = m, then by P n Q G (P]m n (Q]m and Lemma 2.2, we have {P n Q} = (P]m n(Q]m. □ In the following, let f be an endomorphism of Jq(n, m) and Im(f) be the image of f. Lemma 2.5. If M is a maximal clique, then there exists a unique maximal clique containing f (M). Proof. Suppose there exist two distinct maximal cliques M' and M" containing f (M). Then f (M) CM' n M''. By Lemmas 2.1 and 2.2, |M' n M''| < q +1. Since |M| = |f (M) |, by (2.2) we have |f (M) | > q + 1, a contradiction. □ Lemma 2.6. Let M be a star and N be a top such that |f (M) n f (N)| > q +1. Then f(N) C f (M). Proof. LetN' be the maximal clique containing f (N). Then |f (M) nN'| >q +1. One gets f (M) = N' by Lemmas 2.1 and 2.2. □ Lemma 2.7. Suppose there exist two distinct stars [A)m and [B)m such that [A)m n [B)m = {X}, f([A)m) = f([B)m). If f ([A)m) is a star, then f is a colouring of Jq(n, m). Proof. Write M := f ([A)m). Then f ([B)m) = M and f (X) G M. Assume that M is a star. If Im(f) = M, then f is a colouring of Jq(n, m). Now we prove Im(f) = M as follows. Suppose that Y is any vertex with Y ~ X. Since G := Jq(n, m) is connected, it suffices to show that there exist two distinct stars [C)m and [D)m such that {Y} = [C)m n [D)m and f([C)m) = f([D)m) = M. In fact, if we can prove this point, then we can imply that f (Z) G M for all Z G V(G). We prove it as follows. Since X G (X V Y]m n [A)m n [B)m, using Lemma 2.2 we get |(X V Y]m n [A)m n [B)m| = 1. By Lemma 2.1 we obtain |(X V Y]m n [A)m| = |(X V Y]m n [B)m | = q +1. It follows that |(X V Y]m n ([A)m U [B)m)| = 2q +1. Observe that f((XvY]mn([A)mu [B)m)) c f((XvY]m)nf([A)mu[B)m) c f((XvY]m)nM. Since the restriction of f on a clique is injective, one gets |f ((X V Y]m) n M| > 2q + 1 > q + 1. L.-P. Huang, B. Lv, K. Wang: The endomorphisms of Grassmann graphs 387 Thus, Lemma 2.6 implies that ^((X V Y]m) CM. (2.5) So ) € M. Write C := X n Y. Since every vertex of [C)m \ {X} is adjacent to X, by our claim we have y([C)m) = M. Pick a vertex Z such that Z — Y and the distance from X is 2. Write D = Yn Z. Since Y € [D)m n (X V Y]m, by Lemma 2.1 we have |[D)m n (X V Y]m| = q + 1. It follows from (2.5) that |y([D)m) n M| > q +1. Thus Lemma 2.2 implies that y([D)m) = M. Since {Y} = [C)m n [D)m, [C)m and [D)m are the desired stars. □ 3 Proof of Theorem 1.1 For the proof of Theorem 1.1, we only need to consider the case 4 < 2m < n. We divide the proof of Theorem 1.1 into two cases: n > 2m and n = 2m. Lemma 3.1. If n > 2m, then every Grassmann graph Jq (n, m) is a pseudo-core. Proof. Suppose that n > 2m > 4. Then by (2.4), every maximum clique of Jq (n, m) is a star. Let ^ be an endomorphism of Jq(n, m). Then the restriction of ^ on any clique is injective, so ^ transfers stars to stars. Suppose ^ is not a colouring. It suffices to show that ^ is an automorphism. Write Gr := Jq (n, r), where 1 < r < m - 1. By Lemma 2.7, the images under ^ of any two distinct and intersecting stars are distinct. Hence by Lemma 2.3, ^ induces an endomorphism ym-i of Gm-1 such that y([A)m) = [ym-i(A))m. Let X be any vertex of Jq (n, m). Then there exist two vertices X' and X" of Gm-1 such that X = X 'VX ".Then [X ')m n[X ")m = {X } and y(X) € m-1 is not a colouring of Gm-i for m > 3. For any two vertices A1 and A3 of Gm-i at distance 2, we claim that ^m-i(Ai) = ^m-i(A3). There exists an A2 € V(Gm-i) such that Ai - - A3. Write Yi := Ai V A2 and Y2 := A2 V A3. Then Yi - Y>, so ^(Yi) = ^(Y>). By (3.1), ^(Yi) = ^m-i(Ai) V ^m-i(A2), ^(Y2) = ^m-i(A2) V ^m-1 (A3). Thus our claim is valid. Otherwise, one has ^(Y1) = y(Y2), a contradiction. Pick a star N of Gm-1. Since the diameter of Gm-1 is at least two, there exists a vertex A4 € V (Gm-1) \ N that is adjacent to some vertex in N .If B € N such that A4 is not adjacent to B, then d(A4, B) = 2. By our claim, ^>m-1(A4) = y(B) and hence ^m-1(A4) € ^m-1(N). Therefore, y>m-1 is not a colouring. By induction, we may obtain induced endomorphism ^>r of Gr for each r. Furthermore, y(X) = ^ (Xfcl) V Vk2 (Xfc2) V • • • V Vks (Xfcs), (3.2) 388 Ars Math. Contemp. 10 (2016) 255-268 where X = Xkl V Xkl V ■ ■ ■ V Xks G V(Gm) and 1 < dim(Xki) = k, < m - 1. In order to show that ^ is an automorphism, it suffices to show that ^ is injective. Assume that X and Y are any two distinct vertices in Gm with d(X, Y) = s. Thus dim(X n Y) = m — s. If s = 1, then y(X) = ^(Y). Now suppose s > 2. There are 1-dimensional row vectors X,, Y,, i = 1,... .s, such that X, Y can be written as X = (Xn Y)VXiV---VXs, Y = (XnY)VYiV---VYs.LetZ = (XnY)VXiV---VXs_iVYs G V(Gm). By X - Z, dim(^(X) V y>(Z)) = m + 1. Applying (3.2), one has that y(X) = ^m_s(X n Y) V ^i(Xi) v ■ ■ ■ v ^i(Xs), y>(Y) = ^m_s(X n Y) v ^i(Yi) v ■ ■ ■ v ^i(Ys) and ^(Z) = ^m_s(X n Y) V ^(Xi) V ■ ■ ■ V y>i(Xs_i) V ^(Ys). Therefore, we get y(X) V ^(Z) C y>(X) V ^(Y). It follows that y(X) = ^(Y). Otherwise, one has y(X) V ^(Z) C ^>(X), a contradiction to dim(^(X) V <^(Z)) = m + 1. Hence, ^ is an automorphism, as desired. By above discussion, Jq(n, m) is a pseudo-core when n > 2m. □ Lemma 3.2. If n = 2m, then every Grassmann graph Jq (n, m) is a pseudo-core. Proof. Suppose that n = 2m > 4. For a subspace W of V, the dual subspace WL of W in V is defined by W^ = {v G V | wvt = 0, V w G W}, where vt is the transpose of v. For an endomorphism ^ of Jq (2m, m), define the map ^ : V(Jq(2m, m)) —> V(Jq(2m, m)), A i—> ^(A)^. Then ^ is an endomorphism of Jq (2m, m). Note that ^ is an automorphism (resp. colouring) whenever ^ is an automorphism (resp. colouring). For any maximal clique M of Jq(2m, m), <^(M) and <^(M) are of different types. Next we shall show that Jq(2m, m) is a pseudo-core. Case 1. There exist [A)m and (X]m such that [A)m n (X]m = 0 and y([A)m), ¥>((X]m) are of the same type. By Lemma 2.1, the size of [A)m n (X]m is q +1. Then |y([A)m) n y((X]m)| > q +1. Since y([A)m) and y((X]m) are of the same type, by Lemma 2.2 one gets ^([A)m) = ^((X ]m). (3.3) Note that A C X. Pick any Y G [J+J satisfying A C Y and dim(X n Y) = m. Then (Y]m n [A)m = 0. By Lemma 2.1 we have |y((Y]m) n y([A)m)| > q +1. By Lemma 2.2 and (3.3) we obtain either y>((Y]m) = ^((X]m) or y>((Y]m) and ^((X]m) are of different types. Case 1.1. There exists a Y g [mV J such that <^((Y]m) and ^((X]m) are of different types. For any B G [m_T], we have that B C Y and B C X. Since |[B)m) n (X]m| = | [B)m) n (Y]m| = q + 1, we have similarly b([B)m) n ^((X]m)| > q +1 and |^([B)m) n ^((Y]m)| > q +1. Since y((Y]m) and ^((X]m) are of different types, Lemma 2.2 implies that y([B)m) = ^((X]m) or ^([B)m) = ^((Y]m) for any B G [^^i]. Since the size of [m^Yl is at least 3, by above discussion, there exist two subspaces Bi, B2 g [m^5!] such that ([Bi)m) is a star, then f is a colouring by Lemma 2.7. Suppose f([B1)m) is a top. Then f±([B1 )m) is a star. By Lemma 2.7 again, f^ is a colouring. Hence, f is also a colouring. Case 1.2. f ((Y]m) = f ((X]m) for any Y € [„V+J. Consider a star [C)m where C satisfies C c X and dim(C n A) = m - 2. Then (a V C) C X and dim(A V C) = m. For any T € [C)m, since (A V C) C (A V T) and m < dim(A V T) < m + 1, there exists a subspace W € [mV J such that (A V T) C W and dim(W n X) > m (because (A v C) c w n X). Since T € (W]m, f (T) € f ((W]m). By the condition, f ((W]m) = f ((X]m). Then f ((W]m) = f ([A)m) by (3.3). It follows that f (T) € f ([A)m) for all T € [C)m, and so f([C)m) C f([A)m). Hence, f([C)m) = f([A)m). Since [C)m n [A)m = 0, similar to the proof of Case 1.1, f is a colouring. Case 2. For any two maximal cliques of different types containing common vertices, their images under f are of different types. In this case, f maps the maximal cliques of the same type to the maximal cliques of the same type. Case 2.1. f maps stars to stars. In this case f maps tops to tops by Lemmas 2.1 and 2.2. If there exist two distinct stars M and M' such that MnM' = 0 and f (M) = f (M'), then f is a colouring by Lemma 2.7. Now suppose f (M) = f (M') for any two distinct stars M and M' with M n M' = 0. By Lemma 2.3, f induces an endomorphism fm-1 of Jq(2m, m - 1) such that f ([A)m) = [fm-1(A))m. By Lemma 3.1, Jq(2m, m - 1) is a pseudo-core. Thus, f m-1 is an automorphism or a colouring. We claim that fm-1 is an automorphism of Jq(2m, m - 1). For any C € [V] and B € L-J, since C € [B)m and f([B)m) = [fm_1(B))m, we have f (C) € [fm-1(B))m. Then fm-1(B) C f (C), which implies that fm-1((C]m-1) is a top of Jq(2m, m - 1). If m = 2, our claim is valid. Now suppose m > 3 and fm-1 is a colouring. Then Im(fm-1) is a star of Jq(2m, m - 1). Note that fm-1((C]m-1) C Im(fm-1) and |fm-1 ((C]m-1)| > q +1, contradicting to Lemma 2.1. Hence, our claim is valid. Therefore, f maps distinct stars onto distinct stars, and f is an automorphism. Case 2.2. f maps stars to tops. In this case f maps tops to stars by Lemmas 2.1 and 2.2. Note that f^ maps stars to stars. By Case 2.1, f^ is an automorphism. Hence, f is an automorphism. By above discussion, we have proved that every Grassmann graph Jq (2m, m) is a pseudo-core. □ By Lemmas 3.1 and 3.2, we have proved Theorem 1.1. 4 Cores on Grassmann graphs In this section, we shall show that J2 (4,2) is not a core and Jq(2k + 1, 2) (k > 2) is a core. It is well-known (cf. [3, Theorem 6.10 and Corollary 6.2]) that the chromatic number of G satisfies the following inequality: X(G) > max{w(G), |V(G)|/a(G)} 390 Ars Math. Contemp. 10 (2016) 383-392 By [15, Lemma 2.7.2], if G is a vertex-transitive graph, then x(G) »^f > .(G). a(G) (4.1) Lemma 4.1. Let G be a Grassmann graph. Then G is a core if and only if x(G) > w(G). In particular, if ^(^j1 is not an integer, then G is a core. Proof. By [6, Corollary 4.2], every distance-transitive graph is core-complete, thus G is core-complete. Then, x(G) > w(G) implies that G is a core. Conversely, if G is a core, then we must have x(G) > w(G). Otherwise, there exists an endomorphism f of G such that f (G) is a maximum clique of G, a contradiction to G being a core. Thus, G is a core if and only if x(G) > w(G). By [2, p.148, Remark], if the core of G is complete, then |V(G)| = w(G)a(G). Assume that is not an integer. Then |V(G)| = w(G)a(G). Therefore, the core of G is not complete and hence G is a core. □ Denote by F^-*" the set of m x n matrices over F0 and F" «udFq = F^*". Let G = Jq(n,m) where n > m. If X is a vertex of G, then X = [ai,..., am] is an m-dimensional subspace of the vector space F", where {ai,..., am} is a basis of X. Thus, X has a matrix representation € F^*" (cf. [8, 16]). For simpleness, the matrix representation of X € V(G) is also denoted by X. For matrix representations X, Y of two vertices X and Y, X ~ Y if and only if rank ^ X j = m +1. Note that if X is a matrix representation then X = PX (as matrix representation) for any m x m invertible matrix P over Fq. Then, V (G) has a matrix representation V(G) = {X : X € F^*", rank(X) = m} . Now, we give an example of Grassmann graph which is not a core as follows. Example 4.2. Let G = J2(4,2). Then G is not a core. Moreover, x(G) = w (G) = 7 and a(G) = 5. Proof. Applying the matrix representation of V(G), G = J2(4, 2) has 35 vertices as fol- lows: Ai = 1 0 0 1 0 0 0 0 ^ A 2 = 1 0 0 1 1 0 0 0 A3 A5 = 1 0 0 0 ), A6 = 1 0 0 0 , A 7 0 1 1 0 0 1 0 1 Ag = 1 0 0 1 0 1 0 1 ), A10 = 1 0 0 1 1 0 0 1 , A11 A 13 = 1 0 0 1 1 0 11 ^ A14 = 1 0 0 1 11 0 1 , A15 A 7 = 0 0 0 0 1 0 0 1 ), A18 = 1 0 0 0 1 0 0 1 , A19 A 21 = 0 0 0 1 1 0 0 1 ^ A22 = 11 0 0 1 0 0 1 , a23 A 25 = 0 0 1 1 1 0 0 1 ), A26 = 11 11 1 0 0 1 , A27 A 29 = 0 0 1 0 0 0 11 ), A30 = 1 0 0 0 0 0 0 1 , A31 A 33 = 1 0 1 0 0 0 0 1 ^ A 34 = 11 0 0 0 0 11 , A35 A4 = As = A12 : A16 = A20 = A24 = A28 = a32 = ai a 0 0 0 0 0 L.-P. Huang, B. Lv, K. Wang: The endomorphisms of Grassmann graphs 391 Suppose that £1 = {A1, A10, A12, A15, A17}, £2 = {A2, A6, A20, A19, A34}, £3 = {A3, Ag, A21, A22, A35}, £4 = {A5, A9, A18, A24, A29}, £5 = {A7, A14, A23, A27, A33}, £5 = {A4, A13, A25, A28, A30}, and £7 = {An, A16, A26, A31, A32}. It is easy to see that V(G) = £1 U£2 U • • • U £7 and £1,..., £7 are independent sets. Thus x(G) < 7. On the other hand, (4.1) implies that x(G) > w(G) = 7. Therefore, x(G) = w(G) = 7. It follows from Lemma 4.1 that G is not a core. By (4.1) again, we have a(G) = 5. □ We believe that Jq(2k, 2) (k > 2) is not a core for all q (which is a power of a prime). But this a difficult problem. Next, we give some examples of Grassmann graphs which are cores. Example 4.3. If k > 2, then Jq (2k + 1,2) is core. Proof. When k > 2, let G = Jq(2k + 1, 2). Applying (1.1) and (2.3) we have |V(G)| _ q2fc+1 - 1_ q2fc+1 - q 1 w(G) q2 - 1 q2 - 1 q + 1' Thus is not an integer for any q (which is a power of a prime). By Lemma 4.1, G is a core. □ Acknowledgement We are grateful to the referees for useful comments and suggestions. Projects 11371072, 11301270, 11271047, 11371204, 11501036 supported by NSFC. 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