The minimum number of nonnegative edges in hypergraphs
aa r X i v : . [ m a t h . C O ] M a y The minimum number of nonnegative edges in hypergraphs
Hao Huang ∗ Benny Sudakov † Abstract An r -unform n -vertex hypergraph H is said to have the Manickam-Mikl´os-Singhi (MMS)property if for every assignment of weights to its vertices with nonnegative sum, the number ofedges whose total weight is nonnegative is at least the minimum degree of H . In this paper weshow that for n > r , every r -uniform n -vertex hypergraph with equal codegrees has the MMSproperty, and the bound on n is essentially tight up to a constant factor. This result has twoimmediate corollaries. First it shows that every set of n > k real numbers with nonnegativesum has at least (cid:0) n − k − (cid:1) nonnegative k -sums, verifying the Manickam-Mikl´os-Singhi conjecture forthis range. More importantly, it implies the vector space Manickam-Mikl´os-Singhi conjecturewhich states that for n ≥ k and any weighting on the 1-dimensional subspaces of F nq withnonnegative sum, the number of nonnegative k -dimensional subspaces is at least (cid:2) n − k − (cid:3) q . We alsodiscuss two additional generalizations, which can be regarded as analogues of the Erd˝os-Ko-Radotheorem on k -intersecting families. Given an r -uniform n -vertex hypergraph H with minimum degree δ ( H ), suppose every vertex has aweight w i such that w + · · · + w n ≥
0. How many nonnegative edges must H have? An edge of H isnonnegative if the sum of the weights on its vertices is ≥
0. Let e + ( H ) be the number of such edges.By assigning weight n − − δ ( H ). It is a very natural questionto determine when this easy upper bound is tight, which leads us to the following definition. Definition 1.1.
A hypergraph H with minimum degree δ ( H ) has the MMS property if for everyweighting w : V ( H ) → R satisfying P x ∈ v ( H ) w ( x ) ≥ , the number of nonnegative edges is at least δ ( H ) . The question, which hypergraphs have MMS property, was motivated by two old conjecturesof Manickam, Mikl´os, and Singhi [8, 9], both of which were raised in their study of so-called firstdistribution invariant of certain association schemes.
Conjecture 1.2.
Suppose n ≥ k , and we have n real numbers w , · · · , w n such that w + · · · + w n ≥ ,then there are at least (cid:0) n − k − (cid:1) subsets A of size k satisfying P w i ∈ A w i ≥ . The second conjecture is an analogue of Conjecture 1.2 for vector spaces. Let V be a n -dimensionalvector space over a finite field F q . Denote by (cid:2) Vk (cid:3) the family of k -dimensional subspaces of V , andthe q -Gaussian binomial coefficient (cid:2) nk (cid:3) q is defined as Q ≤ i Research sup-ported in part by SNSF grant 200021-149111 and by a USA-Israel BSF grant. onjecture 1.3. Suppose n ≥ k , and let w : (cid:2) V (cid:3) → R be a weighting on the one-dimensionalsubspaces of V such that P v ∈ [ V ] w ( v ) = 0 , then the number of k -dimensional subspaces S with P v ∈ [ V ] ,v ⊂ S w ( v ) ≥ is at least (cid:2) n − k − (cid:3) q . Conjecture 1.2 can be regarded as an analogue of the famous Erd˝os-Ko-Rado theorem [4]. Thelatter says that for n ≥ k , a family of k -subsets of [ n ] with the property that every two subsets havea nonempty intersection has size at most (cid:0) n − k − (cid:1) . In both problems, the extremal examples correspondto a star, which consists of subsets containing a particular element in [ n ]. The Manickam-Mikl´os-Singhi conjecture has been open for more than two decades, and various partial results were proven.There are several works verifying the conjecture for small k [6, 7, 10]. But most of the research focuson proving the conjecture for every n greater than a given function f ( k ). Manickam and Mikl´os[8] verified the conjecture for n ≥ ( k − k k + k ) + k . Later Tyomkyn [13] improved this boundto n ≥ e ck log log k . Alon, Huang, and Sudakov [1] obtained the first polynomial bound n > k .Later, Frankl [5] gave a shorter proof for a cubic range n ≥ k . A linear bound n ≥ k wasobtained by Pokrovskiy [11]. He reduced the conjecture to finding a k -uniform hypergraph on n vertices satisfying the MMS property (similar techniques were also employed earlier in [8]). Thesecond conjecture, Conjecture 1.3, was very recently proved by Chowdhury, Sarkis, and Shahriari [3]simultaneously with our work. They also proved a quadratic bound n ≥ k for sets.We observe that both conjectures can be reduced to proving that certain hypergraph has theMMS property. For the first conjecture, simply let the hypergraph H be the complete k -uniformhypergraph on n vertices. For the second conjecture, one can take the (cid:2) k (cid:3) q -uniform hypergraph H with vertex set (cid:2) V (cid:3) and let edges correspond to k -dimensional subspaces. Both hypergraphs areregular, and moreover the codegree of every pair of vertices is the same. It is tempting to conjecturethat all such hypergraphs satisfy the MMS property. The requirement that all the codegrees areequal may not be dropped. For instance, the tight Hamiltonian cycle (the edges are consecutive r -tuples modulo n ) when n ≡ r ) is not MMS. This can be seen by choosing the weights w ( xr + 1) = n for x = 0 , · · · , n/r and all the other weights to be − n + rr − , which results in only r − r . Our main theorem indeed confirmsthat equal codegrees imply the MMS property. Theorem 1.4. Let H be an r -uniform n -vertex hypergraph with n > r and all the codegrees equalto λ . Then for every weighting w : V ( H ) → R with P v w v ≥ , we have e + ( H ) ≥ δ ( H ) . Moreoverin the case of equality, all nonnegative edges form a star, i.e., contain a fixed vertex of H . The lower bound on n in this theorem is tight up to a constant factor. Our result immediatelyimplies two corollaries. First it verifies Conjecture 1.2 for a weaker range n ≥ Ω( k ). Moreover italso provides a proof of Conjecture 1.3.As mentioned earlier, there are some subtle connections between Manickam-Mikl´os-Singhi con-jecture and the Erd˝os-Ko-Rado theorem on intersecting families. In [4], Erd˝os, Ko and Rado alsoinitiated the study of k -intersecting families (any two subsets have at least k common elements).They show that for k < t , there exists an integer n ( k, t ) such that for all n ≥ n ( k, t ) the largest k -intersecting family of t -sets are the k -stars, which are of size (cid:0) n − kt − k (cid:1) . This result is equivalent tosaying that in the (cid:0) tk (cid:1) -uniform hypergraph H whose vertices are k -subsets of [ n ] and edges corre-spond to t -subsets of [ n ], the maximum intersecting sub-hypergraph has size (cid:0) n − kt − k (cid:1) . The followingtheorem says that for large n , this hypergraph has the MMS property. Note that this is not implied2y Theorem 1.4, because the codegree of two vertices (as k -subsets) depends on the size of theirintersection. Theorem 1.5. Let k, t be positive integers with t > k , n > Ct k +3 for sufficiently large C and let { w X } X ∈ ( [ n ] k ) be a weight assignment with P X ∈ ( [ n ] k ) w X ≥ . Then there are always at least (cid:0) n − kt − k (cid:1) subsets T of size t such that P X ⊂ T w X ≥ . This result can be regarded as an analogue of the k -intersecting version of the Erd˝os-Ko-Radotheorem. Moreover, the Manickam-Mikl´os-Singhi conjecture is a special case of this theorem corre-sponding to k = 1 and t = r . Using a similar proof one can also obtain a generalization of the vectorspace version of Manickam-Mikl´os-Singhi conjecture. Theorem 1.6. Let k, t be positive integers with t > k , n > Ck ( t − k ) for sufficiently large C , V be the n -dimensional vector space over F q and let { w X } X ∈ [ Vk ] be a weight assignment with P X ∈ [ Vk ] w X ≥ .Then there are always at least (cid:2) n − kt − k (cid:3) q t -dimensional subspaces T such that P X ∈ [ Vk ] ,X ⊂ T w X ≥ . The rest of the paper is organized as follows. In Section 2 we prove Theorem 1.4 and deduceConjecture 1.3 as a corollary. In Section 3 we will have two constructions, showing that the n > Ω( k )bound for Theorem 1.4 is essentially tight. The ideas presented in Section 2 are not enough to proveTheorem 1.5. Hence, in Section 4 we develop more sophisticated techniques to prove this theorem.We also sketch the lemmas needed to obtain Theorem 1.6 and leave the proof details to the appendix.The final section contains some open problems and further research directions. In this section we prove Theorem 1.4. Without loss of generality, we may assume that V ( H ) = [ n ],and the weights are 1 = w ≥ w ≥ · · · ≥ w n , such that P ni =1 w i = 0. Suppose the number of edges in H is e . By double counting, we have that H is d -regular with d = n − r − λ and that dn = re . Considerthe 2 r -th largest weight w r , we will verify Theorem 1.4 for the following three cases respectively:(i) w r ≤ r ; (ii) w r ≥ r ; and (iii) r ≤ w r ≤ r . Lemma 2.1. If w r ≤ r , then e + ( H ) ≥ d .Proof. First we show that among the d edges containing w , the number of negative edges is at most d r . Denote the negative edges by e , · · · , e m and the nonnegative edges by e m +1 , · · · , e d . By thedefinition of a negative edge, for every 1 ≤ i ≤ m we have X j ∈ e i \{ } w j < − w = − . Summing these inequalities, we get m X i =1 X j ∈ e i \{ } w j < − m. Now we consider the sum P di = m +1 P j ∈ e i \{ } w j and rewrite it as P j α j w j . The sum of coefficients α j ’s is equal to ( d − m )( r − w , · · · , w r each appears at most λ times3their codegree with { } ) and they are bounded by 1, so in total they contribute no more than 2 rλ .The remaining variables w r +1 , · · · , w n contribute less than ( d − m )( r − w r < d − m r . Combiningthese three estimates, we obtain that X ∈ e X j ∈ e \{ } w j < − m + 2 rλ + d − m r . By double counting, the left hand side is equal to λ ( w + · · · + w n ) = − λ . Comparing these twoquantities and doing simple calculations we get m < rλ + d r + 1 < d r + d r + 1 < d r . Here we used that n > r and λ = r − n − d < d/ (10 r ). Therefore we may assume there are at least(1 − r ) d nonnegative edges through w .If every edge through w is positive, then we are already done; so we assume that there exists anegative edge e through w . Suppose w u is the largest positive weight of vertex not contained in e .Such u exists since otherwise P ni =1 w i < 0, so we may assume that u ≥ { , · · · , u − } ⊂ e .We claim that in this case there are many nonnegative edges through w u which are disjoint from e .Consider the set S of r -tuples consisting of all the edges through w u which are disjoint from e . Sinceeach of the r vertices of e has at most λ common neighbors with w u , we have | S | ≥ d − rλ. Denoteby S − the set of negative edges in S and consider the sum P f ∈ S − P j ∈ f \{ u } w j . Obviously it is atmost − w u | S − | . Rewrite this sum as λ · ( P j e ∪{ u } α j w j ). Since all codegrees are λ , α j ∈ [0 , 1] and P j e ∪{ u } α j = ( r − | S − | /λ , which implies that P j e ∪{ u } (1 − α j ) = ( n − r − − ( r − | S − | /λ .Therefore − w u < X j e ∪{ u } w j = X j e ∪{ u } α j w j + X j e ∪{ u } (1 − α j ) w j < − | S − | λ · w u + ( n − r − − ( r − | S − | /λ ) · w u , The first inequality uses that the sum of all the weights is zero and that e is a negative edge, so P j ∈ e w j < 0. To see the second inequality, just observe that w j ≤ w u for every j e ∪ { u } .By simplifying the last inequality we get | S − | λ < n − rr . Therefore the number of nonnegative edgescontaining w that are disjoint from e is at least | S | − | S − | > d − rλ − n − rr λ = n − ( r − r + 2 r ) r ( n − d, which is greater than d r if n > r . These nonnegative edges, together with the (1 − r ) d nonnegativeedges through w , already give more than d nonnegative edges. Lemma 2.2. If w r ≥ r , then e + ( H ) ≥ d .Proof. First we claim for any 1 ≤ i ≤ r , there are at least r d nonnegative edges containing w i . Let S i be the set of negative edges containing w i , then for any edge e ∈ S i , P j ∈ e \{ i } w j < − w i . Summingup these inequalities, we have X e ∈ S i X j ∈ e \{ i } w j < −| S i | w i . λ · P j = i α j w j , then α j ∈ [0 , P j = i α j = ( r − | S i | /λ , which implies P j = i (1 − α j ) = n − − ( r − | S i | /λ . Since P j = i w j = − w i ,we have − w i = X j = i α j w j + X j = i (1 − α j ) w j < − | S i | λ w i + X ≤ j ≤ r w j + w i · X j> r (1 − α j ) ≤ − | S i | λ w i + 2 rw + (cid:16) n − − ( r − | S i | λ (cid:17) w i Substituting λ = r − n − d and w i ≥ w r ≥ r , gives | S i | ≤ ( n + 4 r ) λr = ( r − n + 4 r ) r ( n − d, which is less than (1 − r ) d when n > r . So there are at least r d nonnegative edges containing w i . Note that for 1 ≤ i < j ≤ r , w i and w j are simultaneously contained in at most λ edges.Therefore the total number of nonnegative edges is at least2 r · r d − (cid:18) r (cid:19) · λ ≥ d − r ( r − n − d. When n > r , this gives more than d nonnegative edges. Lemma 2.3. If r ≤ w r ≤ r , then e + ( H ) ≥ d .Proof. Let t be the index such that w t ≥ rw r and w t +1 < rw r . Since w = 1 ≥ rw r such t exists and is between 1 and 2 r . For arbitrary 1 ≤ i ≤ t , let T i be the set of negative edges containing w i . Similarly as before, we assume X e ∈ T i X j ∈ e \{ i } w j = λ · X j = i α j w j . Then P j = i α j w j < − w i | T i | /λ , and P j = i (1 − α j ) = n − − ( r − | T i | /λ. We also have − w i = X j = i w j = X j = i α j w j + X j = i (1 − α j ) w j < − | T i | λ w i + X ≤ j ≤ t (1 − α j ) w j + X t 1. Since w i ≥ w t , we then have (cid:18) | T i | λ − − (2 r − t ) − n − − ( r − | T i | /λ r (cid:19) w t ≤ t ≤ t · r w r ≤ rtw t . n > r , t ≤ r and λ = r − n − d and rearranging the last inequality gives | T i | ≤ r r − (cid:18) n − r + ( r − t + 2 r + 1 (cid:19) λ ≤ r r − (cid:18) n − r + 2 r + 1 (cid:19) r − n − d < d. Since λ < d/ (10 r ), the above inequality also holds when | T i | < λ . Therefore there are at least d nonnegative edges through w i . This completes the proof for t ≥ 2, as the number of nonnegativeedges through w and w is already at least d − λ > d (when n > r ).If t = 1, it means that w < rw r . For 2 ≤ i ≤ r , as before denote by U i the set of all thenegative edges through w i . Then similarly we define X f ∈ U i X j ∈ f \{ i } w j = λ X j = i α j w j . Note that α i ∈ [0 , P j = i α j w j < − w i | U i | /λ and P j = i (1 − α j ) = n − − ( r − | U i | /λ . So − w i = X j = i w j = X j = i α j w j + X j = i (1 − α j ) w j ≤ − | U i | λ w i + r X i =1 w i + X j> r (1 − α j ) w j ≤ − | U i | λ w i + 1 + 2 rw + (( n − − ( r − | U i | ) /λ ) w i We have (cid:18) r | U i | λ − n (cid:19) w i ≤ rw ≤ r w r + 2 r · rw r = 6 r w r ≤ r w i . Therefore when n > r , | U i | ≤ r + nr · r − n − · d ≤ (cid:16) − r (cid:17) d. Hence there are at least r d nonnegative edges through every w i when 2 ≤ i ≤ r , together with the d nonnegative edges through w . Thus the total number of nonnegative edges is at least815 d + 25 r d (2 r − − (cid:18) r (cid:19) λ = (cid:18) − r − r − r + rn − (cid:19) d > d, where we used that λ = r − n − d and n > r .Combining Lemma 2.1, Lemma 2.2 and Lemma 2.3 we show that e + ( H ) ≥ d . From the proofs itis not hard to see that when n > r the only way to achieve the inequality is when the nonnegativeedges form a star, i.e. contain a fixed vertex of H . This concludes the proof of Theorem 1.4. (cid:3) Next we use Theorem 1.4 to prove the vector space analogue of Manickam-Mikl´os-Singhi conjec-ture. Proof of Conjecture 1.3 : Let H be the hypergraph such that the vertex set V ( H ) consists of allthe 1-dimensional subspaces of V = F nq . Obviously the number of vertices is equal to (cid:2) n (cid:3) q . Every k -dimensional subspace of V defines an edge of H which contains exactly (cid:2) k (cid:3) q vertices. Therefore H is an r -uniform hypergraph on n ′ vertices with r = (cid:2) k (cid:3) q and n ′ = (cid:2) n (cid:3) q . Since every two 1-dimensional6ubspaces span a unique 2-dimensional subspace, so the codegree of any two vertices in H is equalto (cid:2) n − k − (cid:3) q . Applying Theorem 1.4, as long as n ′ > r , the minimum number of nonnegative edgesin H is at least equal to its degree, which is equal to (cid:2) n − k − (cid:3) q . Actually the condition that n ′ > r is equivalent to q n − q − > (cid:18) q k − q − (cid:19) . Since n ≥ k , we have q n − ≥ q k − 1. Moreover ( q k − / ( q k − = ( q k + q k + q k + 1) / ( q k − q k + 1) ≥ q k . From k ≥ 2, we also have ( q − q k ≥ q ( q − > 10 if q ≥ 3. Therefore q n − q − ≥ ( q k − q k q − > (cid:18) q k − q − (cid:19) . For q = 2, it is not hard to verify that the inequality is still satisfied when k ≥ 3. The only remainingcase is when ( q, k ) = (2 , n ≥ 9. The case n = 8 was resolved by Manickam and Singhi [9], who proved their conjecture when k divides n . Remark. The statement of Conjecture 1.3 is known to be false only for n < k . Hence, it wouldbe interesting to determine the minimal n = n ( k ) which implies this conjecture. Note that Theorem1.4 can be used to prove the assertion of Conjecture 1.3 also for n < k . For example it shows thatthis conjecture holds for n ≥ k and q ≥ n ≥ k + 1 and q ≥ n ≥ k + 2 and all q . For large q the proof will work already starting with n = 3 k − In the previous section, we show that for every r -uniform n -vertex hypergraph with equal codegreesand n > r , the minimum number of nonnegative edges is always achieved by the stars. Here wediscuss the tightness of this result. As a warm-up example, recall that a finite projective plane has N + N + 1 points and N + N + 1 lines such that every line contains N + 1 points. Moreover everytwo points determine a unique line, and every two lines intersect at a unique point. If we regardpoints as vertices and lines as edges, this naturally corresponds to a ( N + 1)-uniform ( N + 1)-regularhypergraph with all codegrees equal to 1. Let us assign weights 1 to the N + 1 points on a fixedline l , and weights − N +1 N to the other points. Obviously the sum is nonnegative. On the other handevery line other than l contains at most one point with positive weight, thus its sum of weight isat most 1 − N · N +1 N < 0. Therefore there is only one nonnegative edge. This already gives us ahypergraph with n ∼ r that is not MMS.The next theorem provides an example of a hypergraph with n ∼ r for which there is a con-figuration of edges, different from a star, that also achieves the minimum number of nonnegativeedges. Theorem 3.1. For infinitely many r there is an r -uniform hypergraph H with equal codegrees on ( r − r + 2 r ) vertices, and a weighting w : V ( H ) → R with nonnegative sum, such that there are δ ( H ) nonnegative edges that do not form a star.Proof. Let r = q + 1, where q is a prime power. Denote by F q the finite field with q elements. Definea hypergraph H with the vertex set V ( H ) consisting of points from the 3-dimensional projective7pace P G (3 , F q ). Here P G ( n, F q ) = ( F n +1 q \{ } ) / ∼ , with the equivalence relation ( x , · · · , x n ) ∼ ( σx , · · · , σx n ), where σ is an arbitrary number from F q . It is easy to see that n = | V ( H ) | = q + q + q + 1 = r − r + 2 r . Every 1-dimensional subspace of P G (3 , F q ) defines an edge of H with q + 1 = r elements. It is not hard to check that H is d -regular for d = q + q + 1, and everypair of vertices has codegree 1.Now we assign the weights to V ( H ) in the following way. Let S be the set of points of a 2-dimensional projective subspace of V ( H ), then | S | = q + q + 1. Every vertex from S receives weight1, and every vertex outside S has weight − q + q +1 q , such that the total weight is zero. Note thatevery edge has size q + 1, so if it contains at most one vertex from S , its total weight is at most1 − q · q + q +1 q < 0. Therefore every nonnegative edge must contain at least two vertices from S .Since S is a subspace, the lines containing 2 points from S are completely contained in S . There areprecisely q + q + 1 = d lines in S (these are all the nonnegative edges in H ) and they do not forma star.Finally, we give an example which shows that one might find hypergraphs with n ∼ r andweights such that the number of nonnegative edges is strictly smaller than the vertex degree. Theorem 3.2. If q and q + 1 are both prime powers, then there exists a ( q + 1) -uniform ( q + 1) -regular hypergraph H on ( q + 2 q + q + 1) vertices with all codegrees equal to , and an assignmentof weights with nonnegative sum such that there are strictly less than ( q + 1) nonnegative edges in H . In particular if there are infinitely many Mersenne primes, then we obtain infinitely many suchhypergraphs.Proof. Let V ( H ) = V ∪ V , such that | V | = q + q + 1, and | V | = q ( q + 1). We first take H tobe the projective plane P G (2 , F q ) on V with edges corresponding to the projective lines. In otherwords H is a ( q + 1)-uniform hypergraph with degree q + 1 and codegree 1. The hypergraph H consists of some ( q + 1)-tuples that intersect V in exactly one vertex and intersect V in q vertices,such that e ( H ) = ( q + q + 1)( q + q ). H is a ( q + 1)-uniform hypergraph on V with q edges. Wewill carefully define the edges of H and H soon.We hope H and H to satisfy the following properties: (i) for every pair of vertices u ∈ V and v ∈ V , their codegree in H is equal to 1; (ii) note that every edge in H naturally induces a cliqueof size q in (cid:0) V (cid:1) ; while every edge in H induces a clique of size q + 1 in (cid:0) V (cid:1) . We hope these cliquesform an edge partition of the complete graph K | V | = K q ( q +1) . It is not hard to see that if (i), (ii)are both satisfied, then the hypergraph H = H ∪ H ∪ H has codegree δ = 1. And H is a regularhypergraph with degrees equal to δ · n − r − q + 2 q + q + 1) − q + 1) − q + 1) . Now we assign weights to V ( H ), such that every vertex in V receives a weight − 1, while everyvertex in V receives a weight q ( q +1) q + q +1 , so the total weight is zero. If an edge is nonnegative, it mustcontain at least two vertices from V , since q ( q +1) q + q +1 + ( − · q < 0. Such edge can only come from H .However we have e ( H ) = q + q + 1, which is strictly smaller than the degree ( q + 1) . Thereforewhat remains is to show the existence of H and H satisfying (i), (ii). In other words, we need tofind a clique partition in (ii) with K q ( q +1) = q · K q +1 ∪ ( q + q + 1)( q + q ) · K q . K q ’s can be partitioned into K q -factors.A natural idea is to partition [ q ( q + 1)] = S ∪ · · · ∪ S q with | S i | = q ( q + 1). Observe that theprojective plane P G (2 , F q ) defines a clique partition K q + q +1 = ( q + q + 1) K q +1 . By removing onevertex from it, we obtain a partition K q + q = ( q + 1) · K q ∪ q · K q +1 . By doing this for every S i , weget the q copies of K q +1 we want, and ( q + 1) q copies of K q , which clearly form a K q -factor, sincethe ( q + 1) copies of K q from each S i are pairwise disjoint. We still need to find an edge partitionof the balanced complete q -partite graph K q + q, ··· ,q + q into ( q + q ) copies of K q , so that they alsocan be grouped into q + q disjoint K q -factors.Suppose we know that q + 1 is also a prime power. Label the vertices in K q + q, ··· ,q + q by ( x, y, z )where x ∈ F q , y ∈ F q , and z ∈ F q +1 . Two vertices ( x, y, z ) and ( x ′ , y ′ , z ′ ) are adjacent iff x = x ′ .Now we define ( q + q ) cliques C i,j,k,l ’s for i, k ∈ F q and j, l ∈ F q +1 . The clique C i,j,k,l consists ofvertices in the form of ( x, i + kx, j + lf ( x )) for all x ∈ F q , where f is a fixed injective map from F q to F q +1 . Suppose ( x, y, z ) and ( x ′ , y ′ , z ′ ) with x = x ′ are both contained in the clique C i,j,k,l , thenwe have i + kx = y i + kx ′ = y ′ j + lf ( x ) = z j + lf ( x ′ ) = z ′ . Since x = x ′ , the first two equations uniquely determine i and k . Moreover, f ( x ) and f ( x ′ ) aredifferent elements of F q +1 since f is injective, thus j, l are also uniquely determined. Therefore { C i,j,k,l } forms a K q -partition of the edges of K q + q, ··· ,q + q , and it is not hard to see that they canbe partitioned into K q -factors by fixing k and l .By the above discussions, if we have both of q and q + 1 to be powers of prime, in particularwhen q = 2 n − In the next two subsections we discuss generalizations of the two Manickam-Mikl´os-Singhi conjecturesand prove Theorem 1.5 and Theorem 1.6. For k = 1 they follow from Theorem 1.4, thus we canassume that k ≥ 2. In that case, as we mentioned earlier in the introduction, these two theoremsare not direct consequences of Theorem 1.4 because the codegrees in the corresponding hypergraphsare not equal. In this subsection we will prove Theorem 1.5. This requires some new ideas and techniques sincedirect adaptation of the proof of Theorem 1.4 does not work. Indeed, it is easy to construct aweighting such that there is no nonnegative edge through the vertex (a k -set) of maximal weight.For example say k = 2, one can take w { , } = 1, the weights of all the 2 n − − n − , and the rest to have weights roughly . For sufficiently large n , no t -set containing { , } has nonnegative total weights.First we prove a simple lemma from linear algebra. Lemma 4.1. Suppose the s × s lower triangular matrix β = { β i,j } satisfies that β i,i > and forevery j < k ≤ i , ≤ β i,j ≤ β i,k . Then for a given vector ~b = ( b , · · · , b s ) such that b ≥ · · · ≥ b s ≥ ,the equation ~b = ~γ · β has a unique solution ~γ = ( γ , · · · , γ s ) and moreover ≤ γ i ≤ b i /β i,i . roof. The existence and uniqueness of ~γ follow from the fact that β is invertible. Next we inductivelyprove 0 ≤ γ i ≤ b i /β i,i . We start from γ s , from the equation we know b s = γ s β s,s . So γ s = b s /β s,s and the inductive hypothesis is true. Suppose 0 ≤ γ i ≤ b i /β i,i for every i > k . Now from the linearequation, we have b k = β k,k γ k + β k +1 ,k γ k +1 · · · + β s,k γ s . Since γ i , i > k and β i,j are nonnegative, wehave γ k ≤ b k /β k,k . Note that β i,j is increasing in j , so for every k +1 ≤ i ≤ s , β i,k ≤ β i,k +1 . Therefore b k ≤ β k,k γ k + P si = k +1 β i,k +1 γ i = β k,k γ k + b k +1 . Since 0 ≤ b k +1 ≤ b k , we know that γ k ≥ P X ⊂ ( [ n ] k ) w X = 0; and w { , ··· ,k } , or alternativelywritten as w [ k ] , has the largest positive weight. We may let w [ k ] = 1, then w X ≤ k -set X .Throughout this section, we also assume that n > Ct k +3 , here C is some sufficiently large constant.The next lemma shows that if the sum of weights of certain edges is very negative, then we alreadyhave enough nonnegative edges. Lemma 4.2. If for some subset | L | = k , X L ⊂ Y, | Y | = t X L = X ⊂ Y w X ≤ − t k (cid:18) n − kt − k (cid:19) , and P X = L w X ≥ − , then there are more than (cid:0) n − kt − k (cid:1) nonnegative edges in H .Proof. We may rewrite the left hand side of the inequality as (cid:18) n − kt − k (cid:19) X | X ∩ L | =0 w X + (cid:18) n − k + 1 t − k + 1 (cid:19) X | X ∩ L | =1 w X + · · · + (cid:18) n − k − t − k − (cid:19) X | X ∩ L | = k − w X = (cid:18) n − k − t − k − (cid:19) X | X ∩ L |≤ k − w X − k − X j =0 (cid:16) b j · X | X ∩ L | = j w X (cid:17) . Here we let b j = (cid:0) n − k +1 t − k +1 (cid:1) − (cid:0) n − k + jt − k + j (cid:1) . Note that P | X ∩ L |≤ k − w X = P X = L w X ≥ − 1. Since n >Ct k +3 , this implies k − X j =0 (cid:16) b j · X | X ∩ L | = j w X (cid:17) ≥ t k (cid:18) n − kt − k (cid:19) − (cid:18) n − k − t − k − (cid:19) ≥ t k (cid:18) n − kt − k (cid:19) . (1)For a fixed integer 0 ≤ y ≤ k − 1, denote by D y the number of nonnegative t -sets Z with | Z ∩ L | = y . If D y > (cid:0) n − kt − k (cid:1) then we are done. Otherwise assume D y ≤ (cid:0) n − kt − k (cid:1) for every y . Weestimate the following sum: X | Z ∩ L | = y, | Z | = t X X ⊂ Z w X . Since every nonnegative t -set contributes to the sum at most (cid:0) tk (cid:1) , it is at most (cid:0) tk (cid:1) D y ≤ (cid:0) tk (cid:1)(cid:0) n − kt − k (cid:1) . Bydouble counting, the above sum also equals P yj =0 ( β y,j · P | X ∩ L | = j w X ) , where β y,j = (cid:0) k − jy − j (cid:1)(cid:0) n − k + jt − k − y + j (cid:1) ,note that β y,j = 0 when j < k + y − t . When j ≥ k + y − t , since n ≫ t , for fixed y , β y,j is increasingin j . Also note that b j is decreasing in j . Let ~γ = ( γ , · · · , γ k − ) be the unique solution of the system10f equations ~b = ~γ · β , then by Lemma 4.1 k − X j =0 (cid:16) b j · X | X ∩ L | = j w X (cid:17) = k − X j =0 k − X y = j β y,j γ y X | X ∩ L | = j w X = k − X y =0 γ y · y X j =0 (cid:16) β y,j · X | X ∩ L | = j w X (cid:17) ≤ (cid:18) tk (cid:19)(cid:18) n − kt − k (cid:19) k − X y =0 γ y ≤ (cid:18) tk (cid:19)(cid:18) n − kt − k (cid:19) k − X y =0 b y β y,y . Since b y /β y,y = (cid:16)(cid:0) n − k − t − k − (cid:1) − (cid:0) n − k + yt − k + y (cid:1)(cid:17) / (cid:0) n − k + yt − k (cid:1) ≤ (cid:0) n − k − t − k − (cid:1) / (cid:0) n − kt − k (cid:1) . We have k − X j =0 (cid:16) b j · X | X ∩ L | = j w X (cid:17) ≤ (cid:18) tk (cid:19)(cid:18) n − kt − k (cid:19) · ( k − · (cid:0) n − k − t − k − (cid:1)(cid:0) n − kt − k (cid:1) ≤ t k +1 n (cid:18) n − kt − k (cid:19) . For n > Ct k +3 this contradicts (1).We now assume that the t k -th largest weight in H is w P and consider several cases. Lemma 4.3. If w P > t k , there are more than (cid:0) n − kt − k (cid:1) nonnegative edges in the hypergraph H .Proof. We will show that every vertex whose weight is larger than w P is contained in at least t k (cid:0) n − kt − k (cid:1) nonnegative edges, otherwise there are already > (cid:0) n − kt − k (cid:1) nonnegative edges. For simplicity we justneed to prove this statement for w P itself. Suppose there are S negative edges containing w P , whichare denoted by e , · · · , e S (as t -subsets). And e S +1 , · · · , e ( n − kt − k ) are the other (thus nonnnegative)edges containing w P . We have( n − kt − k ) X i =1 X P = X ⊂ e i w X = S X i =1 X P = X ⊂ e i w X + ( n − kt − k ) X i = S +1 X P = X ⊂ e i w X ≤ − w P · S + w P · (cid:18)(cid:18) n − kt − k (cid:19) − S (cid:19) · (cid:18)(cid:18) tk (cid:19) − (cid:19) + t k · (cid:18) n − k − t − k − (cid:19) (2)Here we used that there are at most t k sets X whose weight is larger than w P (but always ≤ (cid:0) n − k − t − k − (cid:1) . If S ≥ (1 − t k ) (cid:0) n − kt − k (cid:1) ,then the above expression is at most − (cid:18) n − kt − k (cid:19) (cid:18)(cid:16) − t k (cid:17) w P − t k · (cid:18) tk (cid:19) · w P − t k +1 n (cid:19) , which can be further bounded by − (cid:18) n − kt − k (cid:19) (cid:18)(cid:16) − t k − · k ! (cid:17) w P − t k +1 n (cid:19) < − (cid:18) n − kt − k (cid:19) · w P ≤ − (cid:18) n − kt − k (cid:19) · t k . The last inequality uses that n > Ct k +3 , t > k ≥ − t k − · k ! ≥ − · − · = .Since we also have P X = P w X = − w P ≥ − 1, Lemma 4.2 for L = P immediately gives > (cid:0) n − kt − k (cid:1) nonnegative edges. 11herefore we can assume that for the t k sets with largest weights, the number of nonnegativeedges containing each such set is at least t k (cid:0) n − kt − k (cid:1) . Using the union bound, the number of nonnegativeedges is at least t k · t k (cid:18) n − kt − k (cid:19) − (cid:18) t k (cid:19)(cid:18) n − k − t − k − (cid:19) , which is also larger than (cid:0) n − kt − k (cid:1) .The next lemma covers the case when w P is smaller than t k , and there are significant numberof negative edges containing { , · · · , k } . Lemma 4.4. If the t k -th largest weight w P is smaller than /t k , and there are less than (1 − t k ) (cid:0) n − kt − k (cid:1) nonnegative edges containing { , · · · , k } , then there are at least (cid:0) n − kt − k (cid:1) nonnegative edges in H .Proof. We consider all the t -tuples containing { , · · · , k } , similarly as before suppose there are S ≥ t k (cid:0) n − kt − k (cid:1) negative edges e , · · · , e S and nonnegative edges e S +1 , · · · , e ( n − kt − k ), we get X [ k ] ⊂ Z, | Z | = t X X ⊂ Z,X =[ k ] w X = S X i =1 X X ⊂ e i ,X =[ k ] w X + ( n − kt − k ) X i = S +1 X X ⊂ e i ,X =[ k ] w X ≤ − S + 1 t k · (cid:18)(cid:18) n − kt − k (cid:19) − S (cid:19) · (cid:18)(cid:18) tk (cid:19) − (cid:19) + t k · (cid:18) n − k − t − k − (cid:19) ≤ − (cid:18) S − k ! · t k (cid:18)(cid:18) n − kt − k (cid:19) − S (cid:19) − t k (cid:18) n − k − t − k − (cid:19)(cid:19) The first inequality is by bounding the t k largest weights in the second sum by 1 and the rest by t k . It also uses the fact that two sets are contained in at most (cid:0) n − k − t − k − (cid:1) edges. Since S ≥ t k (cid:0) n − kt − k (cid:1) ,we have X [ k ] ⊂ Z, | Z | = t X X ⊂ Z,X =[ k ] w X ≤ − (cid:18) t k (cid:18) n − kt − k (cid:19) − t k (cid:18) n − k − t − k − (cid:19)(cid:19) ≤ − (cid:18) t k − t k +1 n (cid:19) (cid:18) n − kt − k (cid:19) . For large n the right hand side is at most − t k (cid:0) n − kt − k (cid:1) . We also have P X =[ k ] w X = − w [ k ] = − 1. Nowwe once again can apply Lemma 4.2 for L = { , · · · , k } to show the existence of > (cid:0) n − kt − k (cid:1) nonnegativeedges.It remains to prove the case when { , · · · , k } is contained in at least (1 − t k ) (cid:0) n − kt − k (cid:1) nonnegativeedges. Lemma 4.5. If { , · · · , k } is contained in at least (1 − t k ) (cid:0) n − kt − k (cid:1) nonnegative edges, then there areat least (cid:0) n − kt − k (cid:1) nonnegative edges in H .Proof. Note that if every edge containing { , · · · , k } is nonnegative, this already gives (cid:0) n − kt − k (cid:1) nonneg-ative edges and the lemma is proved. So we may assume that there is a negative edge f (as t -subset)through { , · · · , k } with P X ⊂ f w X < 0. Suppose the largest weight outside the edge f is w Q , where | Q ∩ f | ≤ k − 1. Now we define new weights w ′ , such that w ′ X = ( − (cid:0) tk (cid:1) if X ⊂ fw X /w Q otherwise.12hen for every X f , w ′ X ≤ w ′ Q = 1. Now we consider all the (cid:0) n − kt − k (cid:1) t -tuples containing the k -set Q . As usual, assume that S of them has negative sum according to w ′ . If S ≥ (1 − t k ) (cid:0) n − kt − k (cid:1) ,we have the following estimate: X Q ⊂ Y, | Y | = t X X ⊂ Y,X = Q w ′ X ≤ − S + (cid:18)(cid:18) n − kt − k (cid:19) − S (cid:19) · (cid:18) tk (cid:19) ≤ − (cid:18) n − kt − k (cid:19) · (cid:18) − t k − t k (cid:18) tk (cid:19)(cid:19) ≤ − (cid:18) n − kt − k (cid:19) · (cid:18) − t k − k ! (cid:19) ≤ − (cid:18) n − kt − k (cid:19) . Note that since P X ⊂ f w X < 0, we have X X = Q w ′ X = X X = Q w X /w Q − X X ⊂ f (cid:18) w X /w Q + (cid:18) tk (cid:19)(cid:19) ≥ − − (cid:18) tk (cid:19) ≥ − t k . If we apply lemma 4.2 for L = Q and the weight w ′′ X = w ′ X /t k , we get > (cid:0) n − kt − k (cid:1) nonnegative edges forthe new weight function w ′ . Note that every such nonnegative edge can not share with f a common k -subset, otherwise its total weight is at most ( (cid:0) tk (cid:1) − − (cid:0) tk (cid:1) < 0. Hence these nonnegative edgesare also nonnegative edges for the original weight function w .By the above discussion, it remains to consider the case S < (1 − t k ) (cid:0) n − kt − k (cid:1) . Then there are atleast t k (cid:0) n − kt − k (cid:1) nonnegative edges containing Q , together with the (1 − t k ) (cid:0) n − kt − k (cid:1) nonnegative edgescontaining { , · · · , k } . Since { , · · · , k } and w Q have codegree at most (cid:0) n − k − t − k − (cid:1) < t k (cid:0) n − kt − k (cid:1) , we havein total more than (cid:0) n − kt − k (cid:1) nonnegative edges. Our techniques from the previous section also allow us to prove a generalization of the vector spaceversion of Manickam-Mikl´os-Singhi conjecture. Since the proof of this result is very similar to thatof Theorem 1.5 we only state the appropriate variants of the lemmas involved. The detailed proofsof these lemmas can be found in the appendix of this paper. The proof of Theorem 1.6 followsimmediately from combining these lemmas. As before, we define the hypergraph H to have thevertex set (cid:2) Vk (cid:3) and every edge corresponds to a t -dimensional subspace. It is easy to check that thehypergraph is (cid:2) tk (cid:3) q -uniform on (cid:2) nk (cid:3) q vertices. Like the previous section, we also assume that [ k ] is the k -dimensional subspace with w [ k ] = 1 and for every X , w X ≤ 1. All the following lemmas are provenunder the assumption that n > C ( t − k ) k for sufficiently large constant C . Lemma 4.6. If for some k -dimensional subspace L , X L ⊂ Y,Y ∈ [ Vt ] X L = X ⊂ Y w X ≤ − (cid:2) tk (cid:3) q (cid:20) n − kt − k (cid:21) q , and P X = L w X ≥ − , then there are more than (cid:2) n − kt − k (cid:3) q nonnegative edges in H . We now assume that the 3 (cid:2) tk (cid:3) q -th largest weight in H is w P , and consider the following severalcases. Lemma 4.7. If w P > / (4 (cid:2) tk (cid:3) q ) , there are more than (cid:2) n − kt − k (cid:3) q nonnegative edges in H . emma 4.8. If w P ≤ / (4 (cid:2) tk (cid:3) q ) , and there are less than (1 − [ tk ] q ) (cid:2) n − kt − k (cid:3) q nonnegative edges containing [ k ] , then there are at least (cid:2) n − kt − k (cid:3) q nonnegative edges in H . Lemma 4.9. If [ k ] is contained in at least (1 − [ tk ] q ) (cid:2) n − kt − k (cid:3) q nonnegative edges, then there are at least (cid:2) n − kt − k (cid:3) q nonnegative edges in H . A r –( n, t, λ ) block design is a collection of t -subsets of [ n ] such that every r elements are containedin exactly λ subsets. In [12], Rands proved the following generalization of Erd˝os-Ko-Rado theorem:given a r –( n, t, λ ) block design H and 0 < s < r , then there exists a function f ( t, r, s ) such that if H has an s -intersecting subhypergraph H ′ , then if n > f ( t, r, s ), the number of edges in H ′ is at most b s , which is the number of blocks through s vertices. Note that Erd˝os-Ko-Rado theorem correspondsto the very special case when H = (cid:0) [ n ] t (cid:1) and s = 1. Moreover, when ( s, r ) = (1 , r –( n, t, λ ) design H , for j = 1 , · · · , t , let d j be the number ofblocks containing a fixed set of j elements. Obviously d r = λ , and by double counting, d j = ( n − jr − j )( t − jr − j ) λ . Theorem 5.1. Let k, r, t be positive integers with t ≥ r ≥ k , n > Ct k +3 for sufficiently large C and let { w X } X ∈ ( [ n ] k ) be a weight assignment with P X ∈ ( [ n ] k ) w X ≥ . Then for a given r – ( n, t, λ ) design H , the number of blocks B with P X ⊂ B,X ∈ ( [ n ] k ) w X ≥ is at least d k = ( n − kr − k )( t − kr − k ) λ . It would be interesting if one can remove the condition t ≥ r ≥ k in this statement. This willgive a general result unifying our Theorems 1.4 and 1.5. The only additional ingredient needed toprove the above theorem is the following fact. For two disjoint vertex subsets | A | = a and | B | = b of a r –( n, t, λ ) design, the number of edges containing every vertex from A while not containing anyvertex in B is equal to ( n − r − bt − r )( n − rt − r ) ( n − a − br − a )( t − ar − a ) · λ . We will omit any further details here and will return tothis problem in the future.In Section 3, we gave an example of infinitely many r -uniform n -vertex hypergraphs with equalcodegrees and n ∼ r not having the MMS property, based on the assumption that there are infinitelymany Mersenne primes. Since the largest known Mersenne number has more than ten million digits,our example already gives (unconditionally) a huge hypergraph with n cubic in r . Still it would beinteresting to construct infinitely many such hypergraphs directly, without relying on the existenceof Mersenne primes?In Section 4, we proved two additional generalizations of the Manickam-Mikl´os-Singhi conjecture.Both results can be regarded as the analogues of the Erd˝os-Ko-Rado theorem on the k -intersectingfamilies for sufficiently large n . It would be interesting to determining the exact range for whichthese theorems hold. For example when k = 1, Theorem 1.5 only gives n > t while we know from[11] that it is true already for n linear in t . Acknowledgment. We would like to thank Ameerah Chowdhury for bringing to our attention aManikam-Miklos-Sighi conjecture for vector spaces and for sharing with us her preprint on this topic.14 eferences [1] N. Alon, H. Huang, and B. Sudakov, Nonnegative k -sums, fractional covers, and probability ofsmall deviations, J. Combin. Theory Ser. B , (2012), 784–796.[2] A. Chowdhury, A note on the Manickam-Mikl´os-Singhi conjecture, European J. Combin. , C(2014), 145–154.[3] A. Chowdhury, G. Sarkis, and S. Shahriari, The Manickam-Miklos-Singhi Conjectures for Setsand Vector Spaces, preprint.[4] P. Erd˝os, C. Ko, and R. Rado, Intersection theorem for system of finite sets. Quart. J. Math.Oxford Ser. , (1961), 313–318.[5] P. Frankl, On the number of nonnegative sums. J. Combin. Theory Ser. B , to appear.[6] S. Hartke, D. Stolee, A branch-and-cut strategy for the Manickam-Mikl´os-Singhi conjecture.preprint available on arxiv: http://arxiv.org/abs/1302.3636.[7] N. Manickam, On the distribution invariants of association schemes. PhD thesis, Ohio StateUniversity, 1986.[8] N. Manickam and D. Mikl´os, On the number of non-negative partial sums of a nonnegativesum. Colloq. Math. Soc. J´anos Bolyai , (1987), 385–392.[9] N. Manickam and N. Singhi, First distribution invariants and EKR theorems. J. Combin.Theory Ser. A , (1988), 91–103.[10] G. Marino and G. Chiaselotti, A method to count the positive 3-subsets in a set of real numberswith non-negative sum. European J. Combin. , (2002), 619–629.[11] A. Pokrovskiy, A linear bound on the Manickam-Mikl´os-Singhi Conjecture, preprint availableon arxiv, http://arxiv.org/pdf/1308.2176.[12] B. Rands, An extension of the Erd˝os-Ko-Rado theorem to t -designs. J. Combin. Theory Ser.A. , (3) (1982), 391–395.[13] M. Tyomkyn, An improved bound for the Manickam-Mikl´os-Singhi conjecture. European J.Combin. , (1) (2012), 27–32. A Missing proofs from Section 4.2 Throughout this section we use that for a > b , q ( a − b ) b ≤ (cid:2) ab (cid:3) q ≤ q ( a − b ) b + b . Proof of Lemma 4.6 : We may rewrite the left hand side of the inequality as (cid:20) n − kt − k (cid:21) q X dim( X ∩ L )=0 w X + (cid:20) n − k + 1 t − k + 1 (cid:21) q X dim( X ∩ L )=1 w X + · · · + (cid:20) n − k − t − k − (cid:21) q X dim( X ∩ L )= k − w X = (cid:20) n − k − t − k − (cid:21) q X dim( X ∩ L ) ≤ k − w X − k − X j =0 (cid:16) b j · X dim( X ∩ L )= j w X (cid:17) . b j = (cid:2) n − k +1 t − k +1 (cid:3) q − (cid:2) n − k + jt − k + j (cid:3) q . Note that P dim( X ∩ L ) ≤ k − w X = P X = L w X ≥ − 1. Since n > Ck ( t − k ), this implies k − X j =0 (cid:16) b j · X | X ∩ L | = j w X (cid:17) ≥ (cid:2) tk (cid:3) q (cid:20) n − kt − k (cid:21) q − (cid:20) n − k − t − k − (cid:21) q ≥ (cid:2) tk (cid:3) q (cid:20) n − kt − k (cid:21) q . (3)For a fixed integer 0 ≤ y ≤ k − 1, denote by D y the number of nonnegative t -dimensionalsubspaces Z with dim( Z ∩ L ) = y . If D y > (cid:2) n − kt − k (cid:3) q then we are done. Otherwise assume D y ≤ (cid:2) n − kt − k (cid:3) q for every y . We estimate the following sum: X dim( Z ∩ L )= y, dim Z = t X X ⊂ Z w X . Since every nonnegative t -dimensional subspace contributes to the sum at most (cid:2) tk (cid:3) q , it is at most (cid:2) tk (cid:3) q D y ≤ (cid:2) tk (cid:3) q (cid:2) n − kt − k (cid:3) q . By double counting, the above sum also equals P yj =0 ( β y,j · P dim( X ∩ L )= j w X ) . Here for a k -dimensional subspace X with dim( X ∩ L ) = j , β y,j denotes the number of t -dimensionalsubspaces Z such that X ⊂ Z and dim( Z ∩ L ) = y . There are ( q k − q j ) ··· ( q k − q y − )( q y − q j ) ··· ( q y − q y − ) ways to extend X ∩ L to Z ∩ L . Let Q = span { X, Z ∩ L } , and R = span { X, L } . Then dim Q = k + y − j , dim R = 2 k − j ,and Q ⊂ R . The next step is to extend Q to Z such that Z ∩ R = Q . The number of ways is equalto ( q n − q k − j ) ··· ( q n − q t + k − y − )( q t − q k + y − j ) ··· ( q t − q t − ) . Note that this is only nonzero for j ≥ k + y − t , in this case β y,j is theproduct of these two expressions, which is roughly q ( k − y )( y − j )+( n − t )( t − k + j − y ) . Since t − k + j − y ≥ j for large n . Also note that b j is decreasing in j . Let ~γ = ( γ , · · · , γ k − ) be theunique solution of the system of equations ~b = ~γ · β , then by Lemma 4.1 k − X j =0 (cid:16) b j · X dim( X ∩ L )= j w X (cid:17) = k − X j =0 k − X y = j β y,j γ y X dim( X ∩ L )= j w X = k − X y =0 γ y · y X j =0 (cid:16) β y,j · X dim( X ∩ L )= j w X (cid:17) ≤ (cid:20) tk (cid:21) q (cid:20) n − kt − k (cid:21) q k − X y =0 γ y ≤ (cid:20) tk (cid:21) q (cid:20) n − kt − k (cid:21) q k − X y =0 b y β y,y . It is easy to check that β y,y ≥ q ( n − t )( t − k ) and so b y /β y,y = ( (cid:2) n − k − t − k − (cid:3) q − (cid:2) n − k + yt − k + y (cid:3) q ) /q ( n − t )( t − k ) ≤ (cid:2) n − k − t − k − (cid:3) q /q ( n − t )( t − k ) . Therefore k − X j =0 (cid:16) b j · X | X ∩ L | = j w X (cid:17) ≤ (cid:20) tk (cid:21) q (cid:20) n − kt − k (cid:21) q · ( k − · (cid:2) n − k − t − k − (cid:3) q q ( n − t )( t − k ) ≤ ( k − (cid:20) tk (cid:21) q q t − k − ( n − t ) (cid:20) n − kt − k (cid:21) q , that for n > Ck ( t − k ) contradicts (3). Proof of Lemma 4.7 : We will show that every k -subspace whose weight is larger than w P iscontained in at least [ tk ] q (cid:2) n − kt − k (cid:3) q nonnegative edges, otherwise there are already more than (cid:2) n − kt − k (cid:3) q nonnegative edges. For simplicity we just need to prove this statement for w P itself. Suppose there16re S negative edges containing w P , which are denoted by e , · · · , e S (as t -dimensional subspaces).And e S +1 , · · · , e [ n − kt − k ] q are the other (thus nonnnegative) edges containing w P . We have[ n − kt − k ] q X i =1 X P = X ⊂ e i w X = S X i =1 X P = X ⊂ e i w X + [ n − kt − k ] q X i = S +1 X P = X ⊂ e i w X ≤ − w P · S + w P · (cid:20) n − kt − k (cid:21) q − S ! · (cid:20) tk (cid:21) q − ! + 3 (cid:20) tk (cid:21) q · (cid:20) n − k − t − k − (cid:21) q (4)Here we used that there are at most 3 (cid:2) tk (cid:3) q vertices X whose weight is larger than w P (but always ≤ (cid:2) n − k − t − k − (cid:3) q . If S ≥ (cid:16) − [ tk ] q (cid:17)(cid:2) n − kt − k (cid:3) q , then the above expression is at most − (cid:20) n − kt − k (cid:21) q (cid:16) − (cid:2) tk (cid:3) q (cid:17) w P − (cid:2) tk (cid:3) q · (cid:20) tk (cid:21) q · w P − (cid:20) tk (cid:21) q · q t − k − q n − k − ! , which can be further bounded by − (cid:20) n − kt − k (cid:21) q (cid:16) − (cid:2) tk (cid:3) q (cid:17) w P − (cid:20) tk (cid:21) q · q t − k − q n − k − ! < − (cid:20) n − kt − k (cid:21) q · w P ≤ − (cid:20) n − kt − k (cid:21) q · (cid:2) tk (cid:3) q . The first inequality is because t > k ≥ q ≥ 2, so (cid:2) tk (cid:3) q ≥ 7, and also because n > Ck ( t − k )for large C . Since we also have P X = P w X = − w P ≥ − . Lemma 4.6 for L = P immediately gives > (cid:2) n − kt − k (cid:3) q nonnegative edges.Therefore we can assume that for the 3 (cid:2) tk (cid:3) q vertices with largest weights, the number of nonneg-ative edges containing each such vertex is at least [ tk ] q (cid:2) n − kt − k (cid:3) q . Using the union bound, the numberof nonnegative edges is at least3 (cid:20) tk (cid:21) q · (cid:2) tk (cid:3) q (cid:20) n − kt − k (cid:21) q − (cid:18) (cid:2) tk (cid:3) q (cid:19)(cid:20) n − k − t − k − (cid:21) q ≥ − (cid:2) tk (cid:3) q q n − t (cid:20) n − kt − k (cid:21) q , which is also larger than (cid:2) n − kt − k (cid:3) q when n > Ck ( t − k ). Proof of Lemma 4.8 : We consider all the t -dimensional subspaces containing [ k ], similarly as beforesuppose there are S ≥ [ tk ] q (cid:2) n − kt − k (cid:3) q negative edges e , · · · , e S and nonnegative edges e S +1 , · · · , e [ n − kt − k ] q ,17e get X [ k ] ⊂ Z, dim Z = t X X ⊂ Z,X =[ k ] w X = S X i =1 X X ⊂ e i ,X =[ k ] w X + [ n − kt − k ] q X i = S +1 X X ⊂ e i ,X =[ k ] w X ≤ − S + 14 (cid:2) tk (cid:3) q · (cid:20) n − kt − k (cid:21) q − S ! · (cid:20) tk (cid:21) q − ! + 3 (cid:20) tk (cid:21) q · (cid:20) n − k − t − k − (cid:21) q ≤ − (cid:20) n − kt − k (cid:21) q S (cid:2) n − kt − k (cid:3) q − (cid:2) tk (cid:3) q − (cid:20) tk (cid:21) q · q t − k − q n − k − ≤ − (cid:20) n − kt − k (cid:21) q (cid:2) tk (cid:3) q − (cid:2) tk (cid:3) q − (cid:20) tk (cid:21) q q − ( n − t ) ! The first inequality is by bounding the 3 (cid:2) tk (cid:3) q largest weights in the second sum by 1 and the restby [ tk ] q . It also uses the fact that two k -dimensional subspaces are contained in at most (cid:2) n − k − t − k − (cid:3) q t -dimensional subspaces. For n > Ck ( t − k ), we have X [ k ] ⊂ Z, dim Z = t X X ⊂ Z,X =[ k ] w X ≤ − (cid:20) n − kt − k (cid:21) q · (cid:2) tk (cid:3) q . We also have P X =[ k ] w X = − w [ k ] = − . Now we once again can apply Lemma 4.6 for L = [ k ] toshow the existence of > (cid:2) n − kt − k (cid:3) q nonnegative edges. Proof of Lemma 4.9 : Note that if every t -dimensional subspaces containing [ k ] is nonnegative,this already gives (cid:2) n − kt − k (cid:3) q nonnegative edges and the lemma is proved. So we may assume that thereis a negative edge f (as t -dimensional subspace) containing [ k ] with P X ⊂ f w X < 0. Suppose thelargest weight outside the edge f is w Q , where dim( Q ∩ f ) ≤ k − 1. Now we define new weights w ′ ,such that w ′ X = ( − (cid:2) tk (cid:3) q if X ⊂ fw X /w Q otherwise.Then for every X f , w ′ X ≤ w ′ Q = 1. Now we consider all the (cid:2) n − kt − k (cid:3) q t -dimensionalsubspaces containing Q . As usual, assume that S of them has negative sum according to w ′ . If S ≥ (cid:16) − [ tk ] q (cid:17)(cid:2) n − kt − k (cid:3) q , we have the following estimate: X Q ⊂ Y, dim Y = t X X ⊂ Y,X = Q w ′ X ≤ − S + (cid:20) n − kt − k (cid:21) q − S ! · (cid:20) tk (cid:21) q ≤ − (cid:20) n − kt − k (cid:21) q · − (cid:2) tk (cid:3) q − ! ≤ − (cid:20) n − kt − k (cid:21) q . P X ⊂ f w X < 0, we have X X = Q w ′ X = X X = Q w X /w Q − X X ⊂ f w X /w Q + (cid:20) tk (cid:21) q ! ≥ − − (cid:20) tk (cid:21) q . If we apply Lemma 4.6 for L = Q and the new weighting w ′′ X = w ′ X / (2 (cid:2) tk (cid:3) q ), we get > (cid:2) n − kt − k (cid:3) q nonnegative edges for weight w ′ . Note that every such nonnegative edge cannot share with f acommon k -dimensional subspace, otherwise its total weight is at most ( (cid:2) tk (cid:3) q − − (cid:2) tk (cid:3) q < 0. Hencethese nonnegative edges are also nonnegative edges for the original weighting w .By the above discussion, it remains to consider the case S < (cid:16) − [ tk ] q (cid:17)(cid:2) n − kt − k (cid:3) q . Then there areat least [ tk ] q (cid:2) n − kt − k (cid:3) q nonnegative edges containing Q , together with the (cid:16) − [ tk ] q (cid:17)(cid:2) n − kt − k (cid:3) q nonnegativeedges containing [ k ]. Since [ k ] and w Q have codegree at most (cid:2) n − k − t − k − (cid:3) q ≤ [ tk ] q (cid:2) n − kt − k (cid:3) q , we have intotal more than (cid:2) n − kt − k (cid:3) qq