The Chromatic Number of the q-Kneser Graph for Large q

We obtain a new weak Hilton-Milner type result for intersecting families of kspaces in F2k q , which improves several known results. In particular the chromatic number of the q-Kneser graph qKn:k was previously known for n > 2k (except for n = 2k+1 and q = 2) or k < q log q−q. Our result determines the chromatic number of qK2k:k for q > 5, so that the only remaining open cases are (n, k) = (2k, k) with q ∈ {2, 3, 4} and (n, k) = (2k + 1, k) with q = 2. Mathematics Subject Classifications: 51E20, 05C69, 05B25, 05D05, 06E30

We want to point out that the case n = 2k is trivial as K 2k:k is bipartite.A natural generalization is the q-analog of the Kneser graph: the q-Kneser graph qK n:k .Here we take the k-spaces of F n q as vertices and two vertices are adjacent if they intersect trivially.Let n k denote the number of k-spaces in F n q .Note that for 0 k n we have Due to two previous results by Blokhuis et al. [1] for n > 2k and Blokhuis et al. [2] for n = 2k, we know the following: Theorem 2 (Blokhuis, Brouwer, Chowdhury, Frankl, Mussche, Patkós, Szőnyi (2010), Blokhuis, Brouwer, Szőnyi (2012)).If k 3 and either q 3 and n 2k + 1, or q = 2 and n 2k + 2, then the chromatic number of qK n:k is n−k+1

1
. If either k < q log q − q or k 3, then the chromatic number of qK 2k:k is q k + q k−1 .
We complete this result for n = 2k and q 5. Theorem 3. Let q 5. Then the chromatic number of qK 2k:k is q k + q k−1 for n = 2k.
The key ingredient of the n = 2k proof by Blokhuis et al. [2] is a weak Hilton-Milner type result (see [11] for the Hilton-Milner theorem for the classical set case, a variation of the famous Erdős-Ko-Rado theorem [8]).In the following, we use projective notation, so we call 1-spaces points, 2-spaces lines, and (n − 1)-spaces hyperplanes.Call the set of all k-spaces on fixed point a dictator (also known as point-pencil).The dual of a dictator consists of all k-spaces in a fixed hyperplane.Due to work by Hsieh [12], Frankl and Wilson [10], and Godsil and Newman [17], we know that the largest independent sets of qK n:k are dictators and, if n = 2k, duals of dictators, that is the family of all k-spaces in a hyperplane.Blokhuis et al. showed the following: Theorem 4 (Blokhuis, Brouwer, Szőnyi (2011)).Let k < q log q − q and let Y be an independent set of qK 2k:k .If Y is not contained in a dictator or its dual, then Using a result by Tokushige on cross-intersecting families in vector spaces [20] and some properties of the spectrum of the Grassmann scheme, we improve this as follows: Theorem 5. Let q 3 and k 5 and let Y be an independent set of qK 2k:k .If Y is not contained in a dictator or its dual, then As this does not cover k = 4, we also provide the following: Theorem 6.Let q 4 and let Y be an independent set of qK 2k:k .If Y is not contained in a dictator or its dual, then While our second bound is slightly worse than Theorem 4 for k and q large, it is clearly better for q small compared to k.It is easy to construct independent sets of k-spaces of size vaguely k 1 2k−2 k−2 , so our result is close to a proper stability result.Recently, Cameron-Liebler k-space classes (also known as Boolean degree 1 functions) received some attention [3,9,18].In particular, Metsch showed the following [16]: Theorem 7 (Metsch (2017)).Let q q 0 for some universal constant q 0 and let k < q log q − q − 1.Let Y be a non-trivial Boolean degree 1 function on k-spaces of F 2k q , then |Y | q The condition on k comes from Theorem 4, so with Theorem 5 we can improve this to the following: Theorem 8. Let q q 0 for some universal constant q 0 .Let Y be a non-trivial Boolean degree 1 function on k-spaces of F 2k q , then |Y | q 5 2k−1 k−1 .Note that a tedious calculation shows that we can choose q 0 = 89 if we follow the argument in [16] without optimizing any of the used constants.We believe that all Boolean degree 1 functions for k > 2 are trivial, so most likely might be still far from the truth.
Our paper is organized as follows: In Section 2, we write down some basic background on the Grassmann scheme, so that we can exploit the spectrum of qK n:k .In Section 3 we prove Theorem 5, and then we finish our investigation with the mentioned consequences in Section 4 and a conclusion in Section 5.

The Grassmann Scheme
We summarize some needed notation and results for association schemes in the following.Delsarte's PhD thesis [4] provides a deeper introduction into the theory of combinatorial applications of association schemes.Definition 9. Let X be a finite set.A k-class association scheme is a pair (X, R), where R = {R 0 , . . .R k } is a set of symmetric binary relations on X with the following properties: (c) There are constants p ij such that for x, y ∈ X with (x, y) ∈ R there are exactly p ij elements z with (x, z) ∈ R i and (z, y) ∈ R j .
Clearly, (X, R i ) is a p ii 0 -regular graph.For convenience, we write v = |X|.The relation R i can be described by its adjacency matrix A i , so a (v × v)-0-1-matrix which is the indicator function of R i .As the A i s are Hermitian and commute, we can diagonalize them simultaneously, that is their eigenvectors are the same.It is well-known that there are k +1 common eigenspaces V 0 , V 1 , . . ., V k of the A i s.As the A i s are regular, the all-ones vector j is an eigenvector and we can assume that V 0 = j .Let E i denote the orthogonal projection onto the ith eigenspace.We can express the A i s as Note that P 0i = p ii 0 .The following stability version of Hoffman's bound for independent sets is surely known for a long time.Its first application, at least in the context of intersecting families, which the author is aware of, is due to Ellis [7].We include a proof, limited to the setting of association schemes, to keep this paper mostly self-contained.
Lemma 10.Let χ be the characteristic vector of an independent set of (X, R i ).Assume that P 1i is the smallest eigenvalue of A i and that P − is the second smallest eigenvalue of A i .Let E r be the orthogonal projection matrix onto the eigenspaces orthogonal to j and the eigenspace of P 1i .Then Proof.As χ is a 0-1-vector, we have Hence, Rearranging shows the claim.
The following is surely folklore; see for example the proof of Theorem 2 in [5] for a mostly identical statement.
Lemma 11.Let χ be the characteristic vector of a non-empty subset Y of X, where y = |Y |.Let E r be the orthogonal projection matrix onto the eigenspaces orthogonal to j + V 1 .Let P − be the smallest eigenvalue of A i .Then there exists a T ∈ Y such that at least Proof.As in the proof of Lemma 10 we obtain Now averaging shows the claim.
the electronic journal of combinatorics 25 (2018), #P00 In the Grassmann scheme J q (n, k) the set of all k-spaces of F n q is X and two subspaces x and y are in relation R i if their intersection is a subspace of dimension k − i.Clearly, R k corresponds to adjacency in qK 2k:k .The eigenvalues P ij of the Grassmann scheme are well-known.There are two useful formulas, one due to Delsarte [4] and one due to Eisfeld [6]: 3 The Weak Hilton-Milner Theorem We rely on the following result by Tokushige.Here a pair (Y, Z), Y, Z ⊆ X is a crossintersecting family if all elements in Y intersect all elements of Z non-trivially.Similarly, throughout this section we call an independent set of qK 2k:k an intersecting family.
Theorem 12 (Tokushige (2013)).Let (Y, Z) be a cross-intersecting family of qK n:k .Then For the rest of the section, set We also assume that k > 3 as the case k = 3 was taken care of in [2], and that Y is not a dictator or the dual of a dictator.

Proof of Theorem 5
Lemma 13.Let be a line in F n q .Let Z be a set of k-spaces which meet in a fixed point For z ∈ Z we have that z, meets C in a (k − 1)-space z as dim(C) = n − 2 and dim( z, ) = k + 1.There are at most k 1 k-spaces z through p in z , .Hence, at most k 1 k-spaces in Z correspond to the same (k − 1)-space in Z .Lemma 14.Let k 5 and q 3. Let Y be an intersecting family of qK 2k:k of size at least y, then no point lies in more than (q Proof.First we show that no point p 1 lies on more than k We continue by showing that if one point p 1 lies in at least (q −α k 1 + 1) 2k−2 k−2 elements of Y , then all other points lie in at most (1 + q −1 )q α+1 2k−3 k−2 elements of Y which are not on p 1 .
Suppose to the contrary that there are two points p 1 and p 2 such that p 1 lies in at least (q −1 k 1 + 1) 2k−2 k−2 elements of Y .Let Z 1 , respectively, Z 2 denote the elements of Y on p 1 , respectively, p 2 .Let be p 1 , p 2 .Let Z i = { , z / : z ∈ Z i and not ⊆ z} for i ∈ {1, 2}.Let x be the number of elements of Y containing .By Lemma 13, we conclude that By Theorem 12, we obtain For q 3, this simplifies to Let R ∈ Y .As no point on R except for p 1 lies in more than b elements of Y , R has k 1 points and all elements of Y meet R in at least one point, we have Suppose that no point p 1 lies in at least Recall that we can assume that α 1 as p 1 lies in at most k 1 2k−2 k−2 elements of Y .For k 5 and α 1, this is less than 1 + 3q −1 , a contradiction as long as 3 + α − k 0. Hence, we can choose α = k − 3 which shows the assertion.Lemma 15.Let either k 5 and q 3, or k 6 and q 5. Let Y be an intersecting family of qK 2k:k of size at least y.Then there are no points p 1 and p 2 such that the number of elements of Y on p 1 and the number of elements of Y on p 2 is at least the electronic journal of combinatorics 25 (2018), #P00 As Y is an intersecting family, (Z 1 , Z 2 ) is a cross-intersecting family of (k − 1)-sets in . By the bound in Theorem 12, taking the square root and rearranging, we obtain By using y = (1 + 3q −1 ) k 1 2k−2 k−2 and rearranging, we obtain This is easily verified to be a contradiction under the conditions on k and q.
Lemma 16.Let either k 5 and q 3, or k 6 and q 5. Let Y be an intersecting family of qK 2k:k of size at least y.Let s be the dimension of a smallest subspace meeting all elements of Y .Then s ∈ {1, k}.
Proof.Let S be a subspace meeting all elements of Y .We suppose that 1 < dim(S) < k and will arrive at a contradiction, so suppose that dim(S) = k − 1 from now on.Let p 1 and p 2 the points in S which lie on the most elements of Y .Let Z 1 , respectively, Z 2 denote the elements of Y on p 1 , respectively, p 2 .Set . By Lemma 15, this is a contradiction.Hence, s ∈ {1, k}.
By duality, we obtain the following: Corollary 17.Let either k 5 and q 3, or k 6 and q 5. Let Y be an intersecting family of qK 2k:k of size at least y.Let s be the dimension of a largest subspace S such that the hyperplanes through S contain all elements of Y .Then s ∈ {2k − 1, k}.
Hence, in the notation of Lemma 16 and Corollary 17, (s, s ) ∈ {(1, 2k−1), (1, k), (k, 2k− 1), (k, k)}.If s = 1, then all elements of Y lies on a fixed point, so Y is a subset of a dictator.Similarly, if s = 2k − 1, then all elements of Y lie in a fixed, so is a subset of the dual of a dictator.Hence, we only need to rule out the case (s, s ) = (k, k).
Lemma 18.Let Y be an intersecting family of qK 2k:k of size at least y.Then an element in Y meets more than k Proof.We assume without loss of generality that y = |Y |.By Equation (1), we have P 0k = q k 2 , P 1k = −q k(k−1) as the smallest eigenvalue and P 3k = −q k(k−3)+3 as the second smallest eigenvalue of A k .By Lemma 10, (P 3k − P 1k )χ T E r χ y(−P 1k − P 0k − P 1k v y).
the electronic journal of combinatorics 25 (2018), #P00 elements of Z and a point p 2 in at least elements of Z.By Lemma 15, this is a contradiction.Hence, (s, s ) = (k, k) does not occur and the proof is complete.Proof.Our setup is as in the proof of Lemma 15, just that this time the resulting inequality is By using y = 3 k 1 2k−2 k−2 and rearranging, we obtain This is a contradiction.The assertion follows.
From here on it is easy to copy the steps which we took for the proof of Theorem 5, replacing Lemma 14 and Lemma 15 accordingly.

The Chromatic Number
In [2, p. 192] it was established that if q k + q k−1 is not the chromatic number and f is the size of the largest independent set which is not contained in a dictator or its dual, then the electronic journal of combinatorics 25 (2018), #P00 for some > 0. By Theorem 5, For q 5 this is easily verified to be less than 1 and therefore the right hand side of Equation ( 3) is negative.Similarly, (q k − q k−1 ) 2k−1 k−1 q k−1 q k−1 − 1 q 2k−1 − 1 • (q k − 1) 2 (q k+1 − 1) (q 2k−1 − q k−1 )(q − 1) 3 < 1, so the left hand side of Equation ( 3) is positive.As this is a contradiction, we have shown Theorem 3. Note that [2, Proposition 5.1] gives a characterization of the case of equality.
Theorem 8 is a simple consequence of replacing Theorem 4 with Theorem 5 in the proof of Theorem 7. See [16] for details.

Future Work
Clearly, the most urgent open cases are the determination of the chromatic number qK 2k:k for q = 2, 3, 4. For q = 3, 4 it is sufficient to obtain slightly better stability type results which show f as then 2f < 2k−1 k−1 .For q = 2 the current approach of determining the chromatic number cannot work as there are examples very close in size to a dictator and its dual.
Classical polar spaces are the geometries induced by non-degenerate sesquilinear forms onto F n q .There are currently barely any stability results known for intersecting families of maximals of finite classical polar spaces in literature and there is an interesting diversity of largest families [19], similar to qK 2k:k .Results for cross-intersecting families are known for finite classical polar spaces [13], so it might be feasible to determine their chromatic number in a similar fashion.

1 2k− 2 k− 2 1 2k− 2 k− 2
elements of Y .This is clear as otherwise there is a T ∈ Y with p 1 / ∈ T .We want to bound the number of R ∈ Y with p 1 ∈ R. We have k 1 choices for one point p in R ∩ T and then 2k−2 k−2 choices for choosing the k-space R through p 1 , p .Hence, there are at most k elements of Y on p 1 .theelectronic journal of combinatorics 25 (2018), #P00

3. 2 6 Now y = 3 k 1 2k− 2 k− 2 . 1 2k− 2 k− 2 Lemma 19 .
Proof of Theorem For this case the proof is nearly identical to the proof of Theorem 5. Instead of Lemma 14, we just use the crude bound that no point lies on more than k elements of Y .The key difference is that we can replace Lemma 15 with the following.Let k 4. Let Y be an intersecting family of qK 2k:k of size at least y.Then there are no points p 1 and p 2 such that the number of elements of Y on p 1 or the number of elements of Y on p 2 is more than