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The Hessian matrix of Lagrange function
Linear Algebra and its Applications 531 (2017) 537–546
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
The Hessian matrix of Lagrange function Pingge Chen, Yuejian Peng 1 , Suijie Wang ∗,2 Institute of Mathematics, Hunan University, Changsha, China
a r t i c l e
i n f o
Article history: Received 6 January 2016 Accepted 8 June 2017 Available online 13 June 2017 Submitted by R. Brualdi MSC: 05C50 15A18
a b s t r a c t We study the Hessian matrix of the Lagrange function of a dense 3-uniform hypergraph and show that the Lagrange function of a dense 3-uniform hypergraph has a unique optimal weight. We also give a characterization to the optimal weight of a dense 3-uniform hypergraph by the Hessian matrix of the Lagrange function and a simple application on the regular 3-uniform hypergraphs. © 2017 Elsevier Inc. All rights reserved.
Keywords: Lagrange function Hessian matrix Dense graph Regular graph
1. Introduction We start with some definitions and notations. For a positive integer r, an r-uniform hypergraph or r-graph G consists of a vertex set V (G) and an edge set E(G) ⊆ V (G) , r V (G) where r denotes the family of all r-subsets of V (G). A subgraph H of G is an r-graph (r) with V (H) ⊆ V (G) and E(H) ⊆ E(G), denoted H ⊆ G. Let Kt denote the complete * Corresponding author. E-mail addresses: [email protected] (P. Chen), [email protected] (Y. Peng), [email protected] (S. Wang). 1 Supported in part by National Natural Science Foundation of China (No. 11271116). 2 Supported in part by National Natural Science Foundation of China (No. 11401196, 11571097). http://dx.doi.org/10.1016/j.laa.2017.06.012 0024-3795/© 2017 Elsevier Inc. All rights reserved.
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r-graph on t vertices, that is the r-graph on t vertices containing all r-subsets of the vertex set as edges. We assume that all hypergraphs have the vertex set [n] = {1, 2, . . . , n} throughout the paper if it is not specified. For an r-graph G with the edge set E ⊆ [n] r , define the Lagrange function of G to be LG (x) = xi , where x = (x1 , . . . , xn ) ∈ Rn . e∈E i∈e
The Hessian matrix HG (x) of LG (x) is an n × n square matrix defined as follows, ⎡ ∂ 2 LG (x) HG (x) =
∂ 2 LG (x) ∂xi ∂xj
n×n
⎢ ⎢ ⎢ =⎢ ⎢ ⎣
∂x21 ∂ 2 LG (x) ∂x2 ∂x1
.. . ∂ 2 LG (x) ∂xn ∂x1
∂ 2 LG (x) ∂x1 ∂x2 ∂ 2 LG (x) ∂x22
···
∂ 2 LG (x) ∂xn ∂x2
···
···
.. .
∂ 2 LG (x) ∂x1 ∂xn ∂ 2 LG (x) ∂x2 ∂xn
.. . ∂ 2 LG (x) ∂x2n
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
Let Δn be the standard (n − 1)-dimensional closed simplex, i.e., Δn = x = (x1 , . . . , xn ) ∈ Rn | xi ≥ 0, xet = 1 , where e denotes the vector of all entries 1 and et the transpose of e. The Lagrangian λ(G) of a graph G is the supremum of the Lagrange function LG (x) in Δn , i.e., λ(G) = sup LG (x). x∈Δn
It is obvious from the compactness of Δn that the supremum can be attained at some vector in Δn . A vector y ∈ Δn is called an optimal weight of G if λ(G) = LG (y) = maxx∈Δn LG (x). An r-graph G is said to be dense if λ(H) < λ(G) for any proper subgraph H of G. This is equivalent to that all optimal weights of G are attained in the interior of Δn (denoted Int(Δn )), i.e., no optimal weight has coordinates zero. In 1965, Motzkin and Straus [4] showed that any graph has the same Lagrangian as its maximum cliques, since that only complete graphs are dense when r = 2, and gave a new proof of the Turán’s classical result on Turán densities of complete graphs. Sidorenko [5] showed that the Turán density of an r-uniform hypergraph F is equal to the supremum of r!λ(G) over all dense F -hom-free r-uniform hypergraphs G. More studies on developing the Lagrange method for hypergraph Turán problems can be found in [1,3,5,6]. In this paper, we will study the Hessian matrix HG (x) of the Lagrange function LG (x) and obtain that the optimal weight y of a dense 3-graph can be characterized by yHG (y) = 6LG (y)e and the negativity of the second largest eigenvalue of HG (y). Finally, we give a simple application on the structure of the regular dense 3-graphs.
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2. Hessian matrix In this section, our arguments and proofs are practiced on 3-graphs without multiple edges. However, when multiple edges are allowed, all results still hold automatically after some simple amendments of notations. Let G be a 3-graph with edge set E ⊆ [n] 3 . An edge {i, j, k} will be simply denoted by (ijk) throughout the paper. For i ∈ [n], denote by Ei (G) the link of vertex i, i.e.,
(jk) ∈
Ei (G) =
[n] 2
| (ijk) ∈ E .
Let Ai (G) be the n × n matrix whose (j, k)-entry is 1 if (jk) ∈ Ei (G), and 0 otherwise. Note that all entries of the i-th column and i-th row of Ai (G) are zero. Denote by A(Gi ) the (n − 1) × (n − 1) matrix obtained by deleting the i-th column and i-th row of Ai (G), which is the adjacent matrix of the graph Gi with the vertex set [n] \ {i} and edge set n Ei (G). For any x, y ∈ Rn , we have HG (x) = xi Ai (G) and i=1
yHG (x) = (yA1 (G)xt , . . . , yAn (G)xt ) = (xA1 (G)y t , . . . , xAn (G)y t ) = xHG (y). In addition, we have xHG (x)xt =
i=j
xi xj
(ijk)∈E
xk =
xi xj xk = 6LG (x),
(ijk)∈E
and 1 (y + αx)HG (y + αx)(y + αx)t 6 1 1 = LG (y) + αxHG (y)y t + α2 xHG (y)xt + α3 LG (x) 2 2 1 1 = LG (y) + αyHG (x)y t + α2 xHG (y)xt + α3 LG (x). 2 2
LG (y + αx) =
(1) (2)
Lemma 2.1. Let G be a 3-graph and y ∈ Δn . The following three statements are equivalent, (i). yHG (y) = 6LG (y)e, ∂LG ∂LG (ii). (y) = (y), ∀ i, j ∈ [n], ∂yi ∂yj (iii). xHG (y)y t = yHG (x)y t = 0 for all x ∈ Rn satisfying xet = 0.
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∂LG ∂LG ∂LG (y), (y), . . . , (y) and yHG (y)y t = ∂y1 ∂y2 ∂yn 6LG (y), then (i) and (ii) are equivalent. It is obvious that (i) implies (iii). Recall that the orthogonal complement of the subspace {x ∈ Rn | xet = 0} is the line Re. If xHG (y)y t = yHG (x)y t = 0 for all x ∈ Rn with xet = 0, then yHG (y) ∈ Re, i.e., yHG (y) = ce for some constant c ∈ R. Thus we have c = 6LG (y) since yHG (y)y t = 6LG (y) and yet = 1. 2 Proof. Note that yHG (y) = 2
Next we shall prove the following result for dense 3-graphs which is in fact a partial result of [2, Theorem 2.1]. Lemma 2.2. If y ∈ Int(Δn ) is an optimal weight of a dense 3-graph G, then for any x satisfying xet = 0, we have xHG (y)y t = yHG (x)y t = 0, and yHG (y) = 6LG (y)e. Proof. Suppose the vector x ∈ Rn satisfies xet = 0 and xHG (y)y t = 0, then we can choose α small enough such that y + αx ∈ Int(Δn ) and 1 1 αxHG (y)y t + α2 xHG (y)xt + α3 LG (x) > 0. 2 2 From (1), we have LG (y + αx) > LG (y), which is a contradiction to the optimality of y. Thus xHG (y)y t = yHG (x)y t = 0 for all x in the hyperplane xet = 0. Hence, yHG (y) = 6LG (y)e by Lemma 2.1. 2 Theorem 2.3. The optimal weight of a dense 3-graph is unique. Proof. Suppose that y, y ∈ Int(Δn ) are two optimal weights of a dense 3-graph G and set z = y −y. Then we have LG (y) = LG (y ) and zet = 0. Applying (1) and Lemma 2.2, we have 1 LG (y + αz) = LG (y) + α2 zHG (y)z t + α3 LG (z). 2 Taking α = 1, we obtain
1 zHG (y)z t + LG (z) = 0. Thus 2
1 LG (y + αz) = LG (y) + α2 (1 − α)zHG (y)z t . 2 Note that for 0 ≤ α ≤ 1, we have y + αz ∈ Int(Δn ) since Δn is convex and y, y ∈ Int(Δn ). It follows that zHG (y)z t ≤ 0, otherwise, LG (y + αz) > LG (y) for 0 ≤ α ≤ 1,
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which is a contradiction. Suppose z = 0, it is easily seen that there exists an α0 ∈ R with α0 > 1 such that y + α0 z ∈ ∂Δn . Then 1 LG (y + α0 z) = LG (y) + α02 (1 − α0 )zHG (y)z t ≥ LG (y). 2 Which is impossible since G is dense. Hence, z = y − y = 0.
2
Theorem 2.4. Let G be a dense 3-graph, then y is the optimal weight of G if and only if xHG (y)xt < 0 and yHG (x)y t = 0 for all x = 0 and xet = 0. Proof. To prove the necessity, it is enough by Lemma 2.2 to show that HG (y) is negative definite on the hyperplane xet = 0, i.e., xHG (y)xt < 0 for x = 0 and xet = 0. For any x = 0 and xet = 0, by (2) and Lemma 2.2, we have 1 LG (y + αx) = LG (y) + α2 xHG (y)xt + α3 LG (x). 2 Suppose xHG (y)xt > 0. We can choose α small enough such that y + αx ∈ Int(Δn ) and 1 2 α xHG (y)xt +α3 LG (x) > 0, which implies LG (y +αx) > LG (y), contradicting the as2 sumption that y is optimal. Thus xHG (y)xt ≤ 0. Indeed, we shall prove xHG (y)xt < 0. Suppose xHG (y)xt = 0. Then we have LG (y + αx) = LG (y) + α3 LG (x). It follows that LG (x) = 0, otherwise, we choose α small enough such that α3 LG (x) > 0 and y + αx ∈ Int(Δn ) which implies LG (y + αx) > LG (y), a contradiction. However, when xHG (y)xt = 0 and LG (x) = 0, we have LG (y +αx) = LG (y), which is a contradiction to the uniqueness of the optimal weight y by Theorem 2.3. Hence, we obtain xHG (y)xt < 0 for all x = 0 with xet = 0. To prove the sufficiency, suppose y ∈ Int(Δn ) is the unique optimal weight of G and y = y. Let z = y − y, then zet = 0 and 1 LG (y + αz) = LG (y) + α2 zHG (y)z t + α3 LG (z). 2 1 zHG (y)z t < 0, b = LG (z), and f (t) = at2 + bt3 . Since a < 0 and 2 f (1) > 0, then f (1) = 3(a + b) − a = 3f (1) − a > 0, which implies that there exists α > 1 satisfying f (α) > f (1) and y + (α − 1)z = y + αz ∈ Int(Δn ). Then we have LG (y + αz) = LG (y) + f (α) > LG (y) + f (1) = LG (y ), which contradicts the optimality of y . 2 Denote by a =
Corollary 2.5. [2] If G is a dense 3-graph, then for any pair {i, j} ∈ edge e ∈ E such that {i, j} ⊆ e.
V (G) 2
, there is an
Proof. Let y be the optimal weight of G. Suppose that no edge of G contains {i, j}, then the (i, j) entry of HG (y) is 0. Let x be the vector whose i-th coordinate is 1,
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j-th coordinate −1, and 0 otherwise, we have x = 0 and xet = 0. We obtain that xHG (y)xt = 0, contradicting Theorem 2.4. 2 Corollary 2.6. If y is the optimal weight of a dense 3-graph G, then HG (y) is invertible. Proof. Applying the theory of linear algebra, for a square matrix M , the equation xM = 0 has a nonzero solution if and only if the matrix M is degenerate. Suppose x = 0 is a solution of the equation xHG (y) = 0, we have yHG (y) = ce for some constant c by Lemma 2.2. Then xHG (y)y t = cxet = 0. From (2), we have LG (y + αx) = LG (y) +α3 LG (x), which will lead to a contradiction. In fact, if LG (x) = 0, we can choose α small enough such that y+αx ∈ Int(Δn ) and α3 LG (x) > 0, then we have LG (y+αx) > LG (y), contradicting the optimality of y. If LG (x) = 0, we have LG (y + αx) = LG (y), contradicting the uniqueness of the optimal weight of G. 2 3. The second largest eigenvalue Let G be a dense 3-graph. In the following, we shall characterize the optimal weight of G via the second largest eigenvalue of its Hessian matrix. Given any real vector y ∈ Δn , the Hessian matrix HG (y) is a symmetric matrix whose entries are nonnegative and diagonal entries are all zero. Then all eigenvalues of HG (y) are real and can be linearly ordered as λ1 (y) ≥ λ2 (y) ≥ · · · ≥ λn (y). Let ΛG (y) = diag(λ1 (y), . . . , λn (y)) be the diagonal matrix with diagonal entries λi (y)(i ∈ [n]), then there is an orthogonal matrix PG (y), i.e., PG (y)PGt (y) = I, such that HG (y) = PG (y)ΛG (y)PGt (y). If we set v G (y) = yPG (y) = (v1 (y), . . . , vn (y)), aG (y) = ePG (y) = (a1 (y), . . . , an (y)), then x = 0 and xet = 0 are equivalent to u = 0 and uatG (y) = 0, where u = xPG (y) = (u1 , . . . , un ). Remark 3.1. Theorem 2.4 can be restated as follows, y is the optimal weight of a dense 3-graph G if and only if for all u = 0 satisfying uatG (y) = 0, we have uΛG (y)v tG (y) = 0, uΛG (y)ut < 0.
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Lemma 3.2. If y is the optimal weight of a dense 3-graph G, then the largest eigenvalue λ1 (y) of HG (y) is positive. n Proof. Since λ1 (y), . . . , λn (y) are eigenvalues of HG (y), then i=1 λi (y) is the trace n of HG (y) which is obviously zero, i.e., i=1 λi (y) = 0. It follows that λ1 (y) ≥ 0 since λ1 (y) is the maximal eigenvalue of HG (y). However if λ1 (y) = 0, then we have λi (y) = 0 for all i ∈ [n], namely HG (y) = 0, a contradiction. Hence we obtain λ1 (y) > 0. 2 Lemma 3.3. If y is the optimal weight of a dense 3-graph G, then a1 (y) = 0. Proof. Suppose a1 (y) = 0. Let u = (1, 0, . . . , 0), then uatG (y) = 0 and uΛG (y)ut = λ1 (y) > 0, contradicting Remark 3.1. 2 Lemma 3.4. If y is the optimal weight of a dense 3-graph G, then the second largest eigenvalue λ2 (y) of HG (y) is negative. Proof. Suppose λ2 (y) ≥ 0. Let u = (a2 (y), −a1 (y), 0, . . . , 0) = 0, then uatG (y) = 0 and uΛG (y)ut = λ1 (y)a22 (y) + λ2 (y)a21 (y) ≥ 0. Contradicting Remark 3.1.
2
Now we are ready to prove our main result in this section. Theorem 3.5. A dense 3-graph G has the optimal weight y if and only if the second largest eigenvalue λ2 (y) of HG (y) is negative and yHG (y) = 6LG (y)e. Proof. The necessity is given by Lemma 2.2 and Lemma 3.4. To prove the sufficiency, it is enough to show that uΛG (y)v tG (y) = 0 and uΛG (y)ut < 0 for all u = 0 and uatG (y) = 0 from Remark 3.1. If u = 0, uatG (y) = 0, then x = 0, xet = 0. By Lemma 2.1, we have uΛG (y)v tG (y) = xPG (y)ΛG (y)PGt (y)y t = xHG (y)y t = 0. It remains to prove uΛG (y)ut < 0 for all u = 0 and uatG (y) = 0. Since 6LG (y)e = yHG (y) = yPG (y)ΛG (y)PGt (y) = v G (y)ΛG (y)PGt (y), multiplying PG (y) on both sides, we obtain 6LG (y)aG (y) = v G (y)ΛG (y). On the other hand, we have v G (y)aG (y)t = yPG (y)PGt (y)et = yet = 1. It follows that 1=
t 6LG (y)aG (y)Λ−1 G (y)aG (y)
= 6LG (y)
n a2 (y) i
i=1
λi (y)
.
(3)
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n Note that i=1 λi (y) = 0. According to the assumption that the second largest eigenvalue of HG (y) is negative, it is easily seen that λ1 (y) > 0 > λ2 (y) ≥ · · · ≥ λn (y) ai (y) , then uatG (y) = 0 implies and a1 (y) = 0. For 2 ≤ i ≤ n, let bi (y) = a (y) 1 n u1 = − i=2 bi (y)ui . Let dn (y) = λ1 (y), we have t
uΛG (y)u =
n
λi (y)u2i
i=1
=
n
λi (y)u2i
i=2
+ dn (y)
n
2 bi (y)ui
(4)
i=2
2 n−1 dn (y)bn (y) un + bi (y)ui = λn (y) + λn (y) + dn (y)b2n (y) i=2 2 n−1 n−1 d2n (y)b2n (y) 2 + λi (y)ui + dn (y) − bi (y)ui . λn (y) + dn (y)b2n (y) i=2 i=2
dn (y)b2n (y)
Let dn−1 (y) = dn (y) −
λn (y)dn (y) d2n (y)b2n (y) = , we have 2 λn (y) + dn (y)bn (y) λn (y) + dn (y)b2n (y)
n−1 dn (y)bn (y) t 2 uΛG (y)u = λn (y) + dn (y)bn (y) un + bi (y)ui λn (y) + dn (y)b2n (y) i=2 n−1 2 n−1 2 + λi (y)ui + dn−1 (y) bi (y)ui .
i=2
2
i=2
Applying similar operations as (4) to the formula n−1
λi (y)u2i
+ dn−1 (y)
n−1
i=2
2 bi (y)ui
,
i=2
we can obtain ⎛ n uΛG (y)ut = λi (y) + di (y)b2i (y) ⎝ui + i=2
di (y)bi (y) λi (y) + di (y)b2i (y)
i−1
⎞2 bj (y)uj ⎠ , (5)
j=2
where the sequence di (y) is defined recursively from i = n to i = 1 as follows, dn (y) = λ1 (y), di−1 (y) =
di (y)λi (y) , for i = 2, . . . , n. λi (y) + di (y)b2i (y)
It is easy to check that di−1 (y) =
1+
λ1 (y) n λ1 (y)
2 j=i λj (y) bj (y)
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by induction on i. To prove uΛG (y)ut < 0 for all u = 0 and uatG (y) = 0, it suffices from (5) to prove λi (y) + di (y)b2i (y) < 0 for i = 2, . . . , n. Note that λi (y) are negative and λi (y) + di (y)b2i (y) =
di (y)λi (y) , it is enough to show di−1 (y)
that di (y) > 0 for i = 2, . . . , n. di−1 (y) n Recall that dn (y) = λ1 (y) = − i=2 λi (y) > 0, then we only need to prove di (y) > 0 for i = 2, . . . , n. Since λ1 (y) > 0 > λ2 (y) ≥ · · · ≥ λn (y), we have 1+
n λ1 (y) j=i
λj (y)
b2j (y) ≥ 1 +
n λ1 (y) i=2
λi (y)
b2i (y).
Since di−1 (y) =
1+
λ1 (y) n λ1 (y)
2 j=i λj (y) bj (y)
,
it is enough to prove 1+
n λ1 (y) i=2
λi (y)
λ1 (y) a2i (y) > 0. a21 (y) i=2 λi (y) n
b2i (y) = 1 +
Applying (3), it remains to prove λ1 (y) > 0, 6LG (y)a21 (y) which is obvious since λ1 (y) > 0, LG (y) > 0, and a1 (y) = 0. 2 4. Application on regular 3-graphs Given a 3-graph G with edge set E ⊆ i.e.,
[n] 3 , we denote by Eij (G) the link of {i, j} ⊆ [n],
Eij (G) = {k ∈ [n] | (ijk) ∈ E} . A 3-graph G is called s-regular if |Eij (G)| = s for all {i, j} ⊆ [n].
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Theorem 4.1. If G is an s-regular dense 3-graph, then y = ( n1 , n1 , · · · , n1 ) is the unique optimal weight. Proof. Since HG (y) =
1 s HG (e) = (J − I) and n n yHG (y) =
s s(n − 1) e(J − I) = e. n2 n2
Meanwhile, |E(G)| 1 n s s(n − 1) = = . n3 3 2 n3 6n2
LG (y) =
Then we have yHG (y) = 6LG (y)e. By Theorem 3.5, it remains to show that the second s largest eigenvalue of HG (y) = (J −I) is negative, where J is the matrix with all entries n 1 and I the identity matrix. Note that J − I has eigenvalues n − 1, −1, . . . , −1, which completes the proof. 2 Corollary 4.2. If G is an s-regular dense 3-graph, then s > 29 n. Proof. Since G is an s-regular dense 3-graph. By Theorem 4.1, we have LG (y) =
s(n − 1) ≥ LG (x) for all x ∈ Δn . 6n2
Taking x = ( 13 , 13 , 13 , 0, · · · , 0), we have 1 s(n − 1) . ≥ 2 6n 27 A simple calculation then yields s≥
2 2n2 > n. 2 9(n − 1) 9
References [1] P. Frankl, Z. Füredi, Extremal problems whose solutions are the blow-ups of the small Witt-designs, J. Combin. Theory Ser. A 52 (1989) 129–147. [2] P. Frankl, V. Rödl, Hypergraphs do not jump, Combinatorica 4 (2–3) (1984) 149–159. [3] P. Keevash, Hypergraph Turán problems, in: Surveys in Combinatorics, Cambridge University Press, 2011, pp. 83–140. [4] T. Motzkin, E. Straus, Maxima for graphs and a new proof of a theorem of Turán, Canad. J. Math. 17 (1965) 533–540. [5] A.F. Sidorenko, The maximal number of edges in a homogeneous hypergraph containing no prohibited subgraphs, Math. Notes 41 (1987) 247–259, translated from Mat. Zametki. [6] J. Talbot, Lagrangians of hypergraphs, Combin. Probab. Comput. 11 (2002) 199–216.
Contents lists available at ScienceDirect
Linear Algebra and its Applications www.elsevier.com/locate/laa
The Hessian matrix of Lagrange function Pingge Chen, Yuejian Peng 1 , Suijie Wang ∗,2 Institute of Mathematics, Hunan University, Changsha, China
a r t i c l e
i n f o
Article history: Received 6 January 2016 Accepted 8 June 2017 Available online 13 June 2017 Submitted by R. Brualdi MSC: 05C50 15A18
a b s t r a c t We study the Hessian matrix of the Lagrange function of a dense 3-uniform hypergraph and show that the Lagrange function of a dense 3-uniform hypergraph has a unique optimal weight. We also give a characterization to the optimal weight of a dense 3-uniform hypergraph by the Hessian matrix of the Lagrange function and a simple application on the regular 3-uniform hypergraphs. © 2017 Elsevier Inc. All rights reserved.
Keywords: Lagrange function Hessian matrix Dense graph Regular graph
1. Introduction We start with some definitions and notations. For a positive integer r, an r-uniform hypergraph or r-graph G consists of a vertex set V (G) and an edge set E(G) ⊆ V (G) , r V (G) where r denotes the family of all r-subsets of V (G). A subgraph H of G is an r-graph (r) with V (H) ⊆ V (G) and E(H) ⊆ E(G), denoted H ⊆ G. Let Kt denote the complete * Corresponding author. E-mail addresses: [email protected] (P. Chen), [email protected] (Y. Peng), [email protected] (S. Wang). 1 Supported in part by National Natural Science Foundation of China (No. 11271116). 2 Supported in part by National Natural Science Foundation of China (No. 11401196, 11571097). http://dx.doi.org/10.1016/j.laa.2017.06.012 0024-3795/© 2017 Elsevier Inc. All rights reserved.
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r-graph on t vertices, that is the r-graph on t vertices containing all r-subsets of the vertex set as edges. We assume that all hypergraphs have the vertex set [n] = {1, 2, . . . , n} throughout the paper if it is not specified. For an r-graph G with the edge set E ⊆ [n] r , define the Lagrange function of G to be LG (x) = xi , where x = (x1 , . . . , xn ) ∈ Rn . e∈E i∈e
The Hessian matrix HG (x) of LG (x) is an n × n square matrix defined as follows, ⎡ ∂ 2 LG (x) HG (x) =
∂ 2 LG (x) ∂xi ∂xj
n×n
⎢ ⎢ ⎢ =⎢ ⎢ ⎣
∂x21 ∂ 2 LG (x) ∂x2 ∂x1
.. . ∂ 2 LG (x) ∂xn ∂x1
∂ 2 LG (x) ∂x1 ∂x2 ∂ 2 LG (x) ∂x22
···
∂ 2 LG (x) ∂xn ∂x2
···
···
.. .
∂ 2 LG (x) ∂x1 ∂xn ∂ 2 LG (x) ∂x2 ∂xn
.. . ∂ 2 LG (x) ∂x2n
⎤ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦
Let Δn be the standard (n − 1)-dimensional closed simplex, i.e., Δn = x = (x1 , . . . , xn ) ∈ Rn | xi ≥ 0, xet = 1 , where e denotes the vector of all entries 1 and et the transpose of e. The Lagrangian λ(G) of a graph G is the supremum of the Lagrange function LG (x) in Δn , i.e., λ(G) = sup LG (x). x∈Δn
It is obvious from the compactness of Δn that the supremum can be attained at some vector in Δn . A vector y ∈ Δn is called an optimal weight of G if λ(G) = LG (y) = maxx∈Δn LG (x). An r-graph G is said to be dense if λ(H) < λ(G) for any proper subgraph H of G. This is equivalent to that all optimal weights of G are attained in the interior of Δn (denoted Int(Δn )), i.e., no optimal weight has coordinates zero. In 1965, Motzkin and Straus [4] showed that any graph has the same Lagrangian as its maximum cliques, since that only complete graphs are dense when r = 2, and gave a new proof of the Turán’s classical result on Turán densities of complete graphs. Sidorenko [5] showed that the Turán density of an r-uniform hypergraph F is equal to the supremum of r!λ(G) over all dense F -hom-free r-uniform hypergraphs G. More studies on developing the Lagrange method for hypergraph Turán problems can be found in [1,3,5,6]. In this paper, we will study the Hessian matrix HG (x) of the Lagrange function LG (x) and obtain that the optimal weight y of a dense 3-graph can be characterized by yHG (y) = 6LG (y)e and the negativity of the second largest eigenvalue of HG (y). Finally, we give a simple application on the structure of the regular dense 3-graphs.
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2. Hessian matrix In this section, our arguments and proofs are practiced on 3-graphs without multiple edges. However, when multiple edges are allowed, all results still hold automatically after some simple amendments of notations. Let G be a 3-graph with edge set E ⊆ [n] 3 . An edge {i, j, k} will be simply denoted by (ijk) throughout the paper. For i ∈ [n], denote by Ei (G) the link of vertex i, i.e.,
(jk) ∈
Ei (G) =
[n] 2
| (ijk) ∈ E .
Let Ai (G) be the n × n matrix whose (j, k)-entry is 1 if (jk) ∈ Ei (G), and 0 otherwise. Note that all entries of the i-th column and i-th row of Ai (G) are zero. Denote by A(Gi ) the (n − 1) × (n − 1) matrix obtained by deleting the i-th column and i-th row of Ai (G), which is the adjacent matrix of the graph Gi with the vertex set [n] \ {i} and edge set n Ei (G). For any x, y ∈ Rn , we have HG (x) = xi Ai (G) and i=1
yHG (x) = (yA1 (G)xt , . . . , yAn (G)xt ) = (xA1 (G)y t , . . . , xAn (G)y t ) = xHG (y). In addition, we have xHG (x)xt =
i=j
xi xj
(ijk)∈E
xk =
xi xj xk = 6LG (x),
(ijk)∈E
and 1 (y + αx)HG (y + αx)(y + αx)t 6 1 1 = LG (y) + αxHG (y)y t + α2 xHG (y)xt + α3 LG (x) 2 2 1 1 = LG (y) + αyHG (x)y t + α2 xHG (y)xt + α3 LG (x). 2 2
LG (y + αx) =
(1) (2)
Lemma 2.1. Let G be a 3-graph and y ∈ Δn . The following three statements are equivalent, (i). yHG (y) = 6LG (y)e, ∂LG ∂LG (ii). (y) = (y), ∀ i, j ∈ [n], ∂yi ∂yj (iii). xHG (y)y t = yHG (x)y t = 0 for all x ∈ Rn satisfying xet = 0.
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∂LG ∂LG ∂LG (y), (y), . . . , (y) and yHG (y)y t = ∂y1 ∂y2 ∂yn 6LG (y), then (i) and (ii) are equivalent. It is obvious that (i) implies (iii). Recall that the orthogonal complement of the subspace {x ∈ Rn | xet = 0} is the line Re. If xHG (y)y t = yHG (x)y t = 0 for all x ∈ Rn with xet = 0, then yHG (y) ∈ Re, i.e., yHG (y) = ce for some constant c ∈ R. Thus we have c = 6LG (y) since yHG (y)y t = 6LG (y) and yet = 1. 2 Proof. Note that yHG (y) = 2
Next we shall prove the following result for dense 3-graphs which is in fact a partial result of [2, Theorem 2.1]. Lemma 2.2. If y ∈ Int(Δn ) is an optimal weight of a dense 3-graph G, then for any x satisfying xet = 0, we have xHG (y)y t = yHG (x)y t = 0, and yHG (y) = 6LG (y)e. Proof. Suppose the vector x ∈ Rn satisfies xet = 0 and xHG (y)y t = 0, then we can choose α small enough such that y + αx ∈ Int(Δn ) and 1 1 αxHG (y)y t + α2 xHG (y)xt + α3 LG (x) > 0. 2 2 From (1), we have LG (y + αx) > LG (y), which is a contradiction to the optimality of y. Thus xHG (y)y t = yHG (x)y t = 0 for all x in the hyperplane xet = 0. Hence, yHG (y) = 6LG (y)e by Lemma 2.1. 2 Theorem 2.3. The optimal weight of a dense 3-graph is unique. Proof. Suppose that y, y ∈ Int(Δn ) are two optimal weights of a dense 3-graph G and set z = y −y. Then we have LG (y) = LG (y ) and zet = 0. Applying (1) and Lemma 2.2, we have 1 LG (y + αz) = LG (y) + α2 zHG (y)z t + α3 LG (z). 2 Taking α = 1, we obtain
1 zHG (y)z t + LG (z) = 0. Thus 2
1 LG (y + αz) = LG (y) + α2 (1 − α)zHG (y)z t . 2 Note that for 0 ≤ α ≤ 1, we have y + αz ∈ Int(Δn ) since Δn is convex and y, y ∈ Int(Δn ). It follows that zHG (y)z t ≤ 0, otherwise, LG (y + αz) > LG (y) for 0 ≤ α ≤ 1,
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which is a contradiction. Suppose z = 0, it is easily seen that there exists an α0 ∈ R with α0 > 1 such that y + α0 z ∈ ∂Δn . Then 1 LG (y + α0 z) = LG (y) + α02 (1 − α0 )zHG (y)z t ≥ LG (y). 2 Which is impossible since G is dense. Hence, z = y − y = 0.
2
Theorem 2.4. Let G be a dense 3-graph, then y is the optimal weight of G if and only if xHG (y)xt < 0 and yHG (x)y t = 0 for all x = 0 and xet = 0. Proof. To prove the necessity, it is enough by Lemma 2.2 to show that HG (y) is negative definite on the hyperplane xet = 0, i.e., xHG (y)xt < 0 for x = 0 and xet = 0. For any x = 0 and xet = 0, by (2) and Lemma 2.2, we have 1 LG (y + αx) = LG (y) + α2 xHG (y)xt + α3 LG (x). 2 Suppose xHG (y)xt > 0. We can choose α small enough such that y + αx ∈ Int(Δn ) and 1 2 α xHG (y)xt +α3 LG (x) > 0, which implies LG (y +αx) > LG (y), contradicting the as2 sumption that y is optimal. Thus xHG (y)xt ≤ 0. Indeed, we shall prove xHG (y)xt < 0. Suppose xHG (y)xt = 0. Then we have LG (y + αx) = LG (y) + α3 LG (x). It follows that LG (x) = 0, otherwise, we choose α small enough such that α3 LG (x) > 0 and y + αx ∈ Int(Δn ) which implies LG (y + αx) > LG (y), a contradiction. However, when xHG (y)xt = 0 and LG (x) = 0, we have LG (y +αx) = LG (y), which is a contradiction to the uniqueness of the optimal weight y by Theorem 2.3. Hence, we obtain xHG (y)xt < 0 for all x = 0 with xet = 0. To prove the sufficiency, suppose y ∈ Int(Δn ) is the unique optimal weight of G and y = y. Let z = y − y, then zet = 0 and 1 LG (y + αz) = LG (y) + α2 zHG (y)z t + α3 LG (z). 2 1 zHG (y)z t < 0, b = LG (z), and f (t) = at2 + bt3 . Since a < 0 and 2 f (1) > 0, then f (1) = 3(a + b) − a = 3f (1) − a > 0, which implies that there exists α > 1 satisfying f (α) > f (1) and y + (α − 1)z = y + αz ∈ Int(Δn ). Then we have LG (y + αz) = LG (y) + f (α) > LG (y) + f (1) = LG (y ), which contradicts the optimality of y . 2 Denote by a =
Corollary 2.5. [2] If G is a dense 3-graph, then for any pair {i, j} ∈ edge e ∈ E such that {i, j} ⊆ e.
V (G) 2
, there is an
Proof. Let y be the optimal weight of G. Suppose that no edge of G contains {i, j}, then the (i, j) entry of HG (y) is 0. Let x be the vector whose i-th coordinate is 1,
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j-th coordinate −1, and 0 otherwise, we have x = 0 and xet = 0. We obtain that xHG (y)xt = 0, contradicting Theorem 2.4. 2 Corollary 2.6. If y is the optimal weight of a dense 3-graph G, then HG (y) is invertible. Proof. Applying the theory of linear algebra, for a square matrix M , the equation xM = 0 has a nonzero solution if and only if the matrix M is degenerate. Suppose x = 0 is a solution of the equation xHG (y) = 0, we have yHG (y) = ce for some constant c by Lemma 2.2. Then xHG (y)y t = cxet = 0. From (2), we have LG (y + αx) = LG (y) +α3 LG (x), which will lead to a contradiction. In fact, if LG (x) = 0, we can choose α small enough such that y+αx ∈ Int(Δn ) and α3 LG (x) > 0, then we have LG (y+αx) > LG (y), contradicting the optimality of y. If LG (x) = 0, we have LG (y + αx) = LG (y), contradicting the uniqueness of the optimal weight of G. 2 3. The second largest eigenvalue Let G be a dense 3-graph. In the following, we shall characterize the optimal weight of G via the second largest eigenvalue of its Hessian matrix. Given any real vector y ∈ Δn , the Hessian matrix HG (y) is a symmetric matrix whose entries are nonnegative and diagonal entries are all zero. Then all eigenvalues of HG (y) are real and can be linearly ordered as λ1 (y) ≥ λ2 (y) ≥ · · · ≥ λn (y). Let ΛG (y) = diag(λ1 (y), . . . , λn (y)) be the diagonal matrix with diagonal entries λi (y)(i ∈ [n]), then there is an orthogonal matrix PG (y), i.e., PG (y)PGt (y) = I, such that HG (y) = PG (y)ΛG (y)PGt (y). If we set v G (y) = yPG (y) = (v1 (y), . . . , vn (y)), aG (y) = ePG (y) = (a1 (y), . . . , an (y)), then x = 0 and xet = 0 are equivalent to u = 0 and uatG (y) = 0, where u = xPG (y) = (u1 , . . . , un ). Remark 3.1. Theorem 2.4 can be restated as follows, y is the optimal weight of a dense 3-graph G if and only if for all u = 0 satisfying uatG (y) = 0, we have uΛG (y)v tG (y) = 0, uΛG (y)ut < 0.
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Lemma 3.2. If y is the optimal weight of a dense 3-graph G, then the largest eigenvalue λ1 (y) of HG (y) is positive. n Proof. Since λ1 (y), . . . , λn (y) are eigenvalues of HG (y), then i=1 λi (y) is the trace n of HG (y) which is obviously zero, i.e., i=1 λi (y) = 0. It follows that λ1 (y) ≥ 0 since λ1 (y) is the maximal eigenvalue of HG (y). However if λ1 (y) = 0, then we have λi (y) = 0 for all i ∈ [n], namely HG (y) = 0, a contradiction. Hence we obtain λ1 (y) > 0. 2 Lemma 3.3. If y is the optimal weight of a dense 3-graph G, then a1 (y) = 0. Proof. Suppose a1 (y) = 0. Let u = (1, 0, . . . , 0), then uatG (y) = 0 and uΛG (y)ut = λ1 (y) > 0, contradicting Remark 3.1. 2 Lemma 3.4. If y is the optimal weight of a dense 3-graph G, then the second largest eigenvalue λ2 (y) of HG (y) is negative. Proof. Suppose λ2 (y) ≥ 0. Let u = (a2 (y), −a1 (y), 0, . . . , 0) = 0, then uatG (y) = 0 and uΛG (y)ut = λ1 (y)a22 (y) + λ2 (y)a21 (y) ≥ 0. Contradicting Remark 3.1.
2
Now we are ready to prove our main result in this section. Theorem 3.5. A dense 3-graph G has the optimal weight y if and only if the second largest eigenvalue λ2 (y) of HG (y) is negative and yHG (y) = 6LG (y)e. Proof. The necessity is given by Lemma 2.2 and Lemma 3.4. To prove the sufficiency, it is enough to show that uΛG (y)v tG (y) = 0 and uΛG (y)ut < 0 for all u = 0 and uatG (y) = 0 from Remark 3.1. If u = 0, uatG (y) = 0, then x = 0, xet = 0. By Lemma 2.1, we have uΛG (y)v tG (y) = xPG (y)ΛG (y)PGt (y)y t = xHG (y)y t = 0. It remains to prove uΛG (y)ut < 0 for all u = 0 and uatG (y) = 0. Since 6LG (y)e = yHG (y) = yPG (y)ΛG (y)PGt (y) = v G (y)ΛG (y)PGt (y), multiplying PG (y) on both sides, we obtain 6LG (y)aG (y) = v G (y)ΛG (y). On the other hand, we have v G (y)aG (y)t = yPG (y)PGt (y)et = yet = 1. It follows that 1=
t 6LG (y)aG (y)Λ−1 G (y)aG (y)
= 6LG (y)
n a2 (y) i
i=1
λi (y)
.
(3)
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n Note that i=1 λi (y) = 0. According to the assumption that the second largest eigenvalue of HG (y) is negative, it is easily seen that λ1 (y) > 0 > λ2 (y) ≥ · · · ≥ λn (y) ai (y) , then uatG (y) = 0 implies and a1 (y) = 0. For 2 ≤ i ≤ n, let bi (y) = a (y) 1 n u1 = − i=2 bi (y)ui . Let dn (y) = λ1 (y), we have t
uΛG (y)u =
n
λi (y)u2i
i=1
=
n
λi (y)u2i
i=2
+ dn (y)
n
2 bi (y)ui
(4)
i=2
2 n−1 dn (y)bn (y) un + bi (y)ui = λn (y) + λn (y) + dn (y)b2n (y) i=2 2 n−1 n−1 d2n (y)b2n (y) 2 + λi (y)ui + dn (y) − bi (y)ui . λn (y) + dn (y)b2n (y) i=2 i=2
dn (y)b2n (y)
Let dn−1 (y) = dn (y) −
λn (y)dn (y) d2n (y)b2n (y) = , we have 2 λn (y) + dn (y)bn (y) λn (y) + dn (y)b2n (y)
n−1 dn (y)bn (y) t 2 uΛG (y)u = λn (y) + dn (y)bn (y) un + bi (y)ui λn (y) + dn (y)b2n (y) i=2 n−1 2 n−1 2 + λi (y)ui + dn−1 (y) bi (y)ui .
i=2
2
i=2
Applying similar operations as (4) to the formula n−1
λi (y)u2i
+ dn−1 (y)
n−1
i=2
2 bi (y)ui
,
i=2
we can obtain ⎛ n uΛG (y)ut = λi (y) + di (y)b2i (y) ⎝ui + i=2
di (y)bi (y) λi (y) + di (y)b2i (y)
i−1
⎞2 bj (y)uj ⎠ , (5)
j=2
where the sequence di (y) is defined recursively from i = n to i = 1 as follows, dn (y) = λ1 (y), di−1 (y) =
di (y)λi (y) , for i = 2, . . . , n. λi (y) + di (y)b2i (y)
It is easy to check that di−1 (y) =
1+
λ1 (y) n λ1 (y)
2 j=i λj (y) bj (y)
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by induction on i. To prove uΛG (y)ut < 0 for all u = 0 and uatG (y) = 0, it suffices from (5) to prove λi (y) + di (y)b2i (y) < 0 for i = 2, . . . , n. Note that λi (y) are negative and λi (y) + di (y)b2i (y) =
di (y)λi (y) , it is enough to show di−1 (y)
that di (y) > 0 for i = 2, . . . , n. di−1 (y) n Recall that dn (y) = λ1 (y) = − i=2 λi (y) > 0, then we only need to prove di (y) > 0 for i = 2, . . . , n. Since λ1 (y) > 0 > λ2 (y) ≥ · · · ≥ λn (y), we have 1+
n λ1 (y) j=i
λj (y)
b2j (y) ≥ 1 +
n λ1 (y) i=2
λi (y)
b2i (y).
Since di−1 (y) =
1+
λ1 (y) n λ1 (y)
2 j=i λj (y) bj (y)
,
it is enough to prove 1+
n λ1 (y) i=2
λi (y)
λ1 (y) a2i (y) > 0. a21 (y) i=2 λi (y) n
b2i (y) = 1 +
Applying (3), it remains to prove λ1 (y) > 0, 6LG (y)a21 (y) which is obvious since λ1 (y) > 0, LG (y) > 0, and a1 (y) = 0. 2 4. Application on regular 3-graphs Given a 3-graph G with edge set E ⊆ i.e.,
[n] 3 , we denote by Eij (G) the link of {i, j} ⊆ [n],
Eij (G) = {k ∈ [n] | (ijk) ∈ E} . A 3-graph G is called s-regular if |Eij (G)| = s for all {i, j} ⊆ [n].
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Theorem 4.1. If G is an s-regular dense 3-graph, then y = ( n1 , n1 , · · · , n1 ) is the unique optimal weight. Proof. Since HG (y) =
1 s HG (e) = (J − I) and n n yHG (y) =
s s(n − 1) e(J − I) = e. n2 n2
Meanwhile, |E(G)| 1 n s s(n − 1) = = . n3 3 2 n3 6n2
LG (y) =
Then we have yHG (y) = 6LG (y)e. By Theorem 3.5, it remains to show that the second s largest eigenvalue of HG (y) = (J −I) is negative, where J is the matrix with all entries n 1 and I the identity matrix. Note that J − I has eigenvalues n − 1, −1, . . . , −1, which completes the proof. 2 Corollary 4.2. If G is an s-regular dense 3-graph, then s > 29 n. Proof. Since G is an s-regular dense 3-graph. By Theorem 4.1, we have LG (y) =
s(n − 1) ≥ LG (x) for all x ∈ Δn . 6n2
Taking x = ( 13 , 13 , 13 , 0, · · · , 0), we have 1 s(n − 1) . ≥ 2 6n 27 A simple calculation then yields s≥
2 2n2 > n. 2 9(n − 1) 9
References [1] P. Frankl, Z. Füredi, Extremal problems whose solutions are the blow-ups of the small Witt-designs, J. Combin. Theory Ser. A 52 (1989) 129–147. [2] P. Frankl, V. Rödl, Hypergraphs do not jump, Combinatorica 4 (2–3) (1984) 149–159. [3] P. Keevash, Hypergraph Turán problems, in: Surveys in Combinatorics, Cambridge University Press, 2011, pp. 83–140. [4] T. Motzkin, E. Straus, Maxima for graphs and a new proof of a theorem of Turán, Canad. J. Math. 17 (1965) 533–540. [5] A.F. Sidorenko, The maximal number of edges in a homogeneous hypergraph containing no prohibited subgraphs, Math. Notes 41 (1987) 247–259, translated from Mat. Zametki. [6] J. Talbot, Lagrangians of hypergraphs, Combin. Probab. Comput. 11 (2002) 199–216.