Uniform hypergraphs with the first two smallest spectral radii
UUniform hypergraphs with the first twosmallest spectral radii
Jianbin Zhang a ∗ , Jianping Li b † , Haiyan Guo a ‡ a School of Mathematical Sciences, South China Normal University,Guangzhou 510631, P. R. China b Faculty of Applied Mathematics, Guangdong University of Technology,Guangzhou 510090, P. R. China
Abstract
The spectral radius of a uniform hypergraph G is the the maxi-mum modulus of the eigenvalues of the adjacency tensor of G . For k ≥
2, among connected k -uniform hypergraphs with m ≥ k -uniform loose path with m edges is the unique onewith minimum spectral radius, and we also determine the unique oneswith second minimum spectral radius when m ≥ KEY WORDS: spectral radius, adjacency tensor, uniform hyper-graph edge.
Let G be a hypergraph with vertex set V ( G ) and edge set E ( G ), where E ( G )is a set whose elements are subsets of V ( G ). For an integer k ≥
2, if each edgeof G contains exactly k distinct vertices, then G is a k -uniform hypergraph.Two vertices u and v are adjacent if u and v are contained in some edge.An e is incident with vertex v if v ∈ e . An alternating sequence of verticesand edges is called a path if all vertices and edges are distinct, and a cycleif the first and last vertices are the same, the other vertices and all edgesare distinct. If there exists a path between any two vertices of G , then G ∗ Corresponding author. E-mail:[email protected] † E-mail:[email protected] ‡ E-mail:[email protected] a r X i v : . [ m a t h . SP ] A ug s connected. A hypertree is a connected acyclic hypergraph. A vertex ofdegree one is called a pendant vertex.For a k -uniform hypergraph G with vertex set V ( G ) = { , . . . , n } , itsadjacency tensor is the tensor A ( G ) = ( a i ...i k ) of order k and dimension n with a i ...i k = k − if { i . . . i k } ∈ E ( G ), 0 otherwise, where i j ∈ { , . . . , n } and j ∈ { , . . . , k } .For some complex λ , if there exists a nonzero vector x = ( x , . . . , x n ) T such that A ( G ) x = λx k − , then λ is called an eigenvalue of G and x is calledthe eigenvector of G corresponding to λ , where A ( G ) x is a n -dimensionalvector whose i -th component is( A ( G ) x ) i = n (cid:88) i ,...,i k =1 a ii ...i k x i · · · x i k and x k − = ( x k − , . . . , x k − n ) T , Moreover, if λ and x are both real, then wecall λ an H -eigenvalue of G . Note that x T ( A ( G ) x ) = k (cid:80) e ∈ E ( G ) (cid:81) v ∈ e x v .The spectral radius of a k -uniform hypergraph G , denoted by ρ ( G ), isdefined as the maximum modulus of eigenvalues of A ( G ). Since A ( G ) issymmetric and thus ρ ( G ) is the largest H -eigenvalue of A ( G ), see [13]. Itis proved in [4, 12] that for a connected k -uniform hypergraph G , A ( G ) hasa unique positive eigenvector x with (cid:80) v ∈ V ( G ) x kv = 1 corresponding to ρ ( G ),which is called the principal eigenvector of G .The problem to determine the hypergraphs in some given classes of hyper-graphs with maximum spectral radius received much attention. Li et al. [5]determined the uniform hypertree with maximum spectral radius. Yuan etal. [17] extended to this to determine the first eight uniform hypertrees withmaximum spectral radius. Fan et al. [3] determined the hypergraphs withmaximum spectral radius among uniform hypergraphs with few edges. Xiaoet al. [15] determined the hypertree with maximum spectral radius amonguniform hypertrees with a given degree sequence. See [] for more work onhypergraphs with maximum spectral radius in someclasses of uniform hy-pergraphs. However, there is much less work on the problem to determinethe hypergraphs in some given classes of hypergraphs with minimum spec-tral radius. Li et al. [5] determined the unique one with minimum spec-tral radius among k -uniform power hypertrees with fixed number of edges.Here a k -uniform power hypertree is a k -uniform hypertree in which everyedge contains at least k − k -uniform hypergraphs with spectral radius at most k √
4. Thereseems no more result in this line. In this note, for k ≥
2, we show that the k -uniform loose path with m edges is the unique one with minimum spectralradius among connected k -uniform hypergraphs with m ≥ k -uniform hypergraphs with m ≥ Let r ≥ G be a hypergraph with u ∈ V ( G ) and e , . . . , e r ∈ E ( G ). Supposethat v i ∈ e i and u (cid:54)∈ e i for i = 1 , , . . . , r . Let e (cid:48) i = ( e i \{ v i } ) ∪ { u } for i =1 , . . . , r . Suppose that e (cid:48) i (cid:54)∈ E ( G ) for i = 1 , . . . , r . Let G (cid:48) be the hypergraphobtained from G by deleting e , e , . . . , e r and adding e (cid:48) , e (cid:48) , . . . , e (cid:48) r . Then wesay that G (cid:48) is obtained from G by moving ( e , e , . . . , e r ) from ( v , v , . . . , v r )to u . Lemma 1 [5] Let r ≥ , G be a hypergraph with u ∈ V ( G ) and e , . . . , e r ∈ E ( G ) . If G (cid:48) is obtained from G by moving ( e , e , . . . , e r ) from ( v , v , . . . , v r ) to u , and x u ≥ max ≤ i ≤ r x v i , then ρ ( G (cid:48) ) > ρ ( G ) . Lemma 2 [15] Let G be a connected k -uniform hypergraph, and e = U ∪ U , f = V ∪ V be two edges of G , where e ∩ f = ∅ , and ≤ | U | = | V | < k .Let e (cid:48) = U ∪ V and f (cid:48) = V ∪ U . Suppose that e (cid:48) , f (cid:48) (cid:54)∈ E ( G ) . Let G (cid:48) bethe hypergraph obtained from G by deleting edges e and f and adding edges e (cid:48) and f (cid:48) . Let x be the principle eigenvector of G . If x U ≥ x V , x U ≤ x V ,and one is strict, then ρ ( G ) < ρ ( G (cid:48) ) . A path ( u , e , u , . . . , e p , u p ) in a k -unoform hypergraph G is called apendant path at u if d ( u ) ≥ , d ( u i ) = 2 for i = 1 , , . . . , p − d ( u p ) = 1and d ( u ) = 1 for any u ∈ e i \ { u i − , u i } with i = 1 , , . . . , p . If p = 1, then itis a pendant edge at u .For a k -uniform hypergraph G with a pendant path P at u , we say that G is obtained from H by attaching a pendant path P at u , where H = G [ V ( G ) \ ( V ( P ) \ { u } )]. We write G = H ( u, p ) if the length of P is p . Let H ( u,
0) = H .For a k -unifoorm hypergraph G with u ∈ V ( G ), and p ≥ q ≥
0, let G u ( p, q ) = ( G u ( p )) u ( q ). Lemma 3 [16] Let u be a vertex of a connected k -uniform hypergraph G with | E ( G ) | ≥ . If p ≥ q ≥ , then ρ ( G u ( p, q )) > ρ ( G u ( p + 1 , q − . Let G be a connected k -uniform hypergraph with u, v ∈ V ( G ), and p ≥ q ≥
0, let G u,v ( p, q ) = ( G u ( p )) v ( q ). Lemma 4
Let G be a k -uniform hypergraph with k ≥ . Let e be a pendantedge of G , and u and v be two pendant vertices in e . If p ≥ q ≥ , then ρ ( G u,v ( p, q )) > ρ ( G u,v ( p + 1 , q − . roof. Suppose that P = ( u , e , u , e , . . . , u p +1 , e p +1 , u p +2 ) and Q = ( v , f ,v , f , . . . , v q − , f q − , v q ) are two pendant paths of G u,v ( p + 1 , q −
1) at u oflength p + 1 and at v of length q −
1, respectively, where u = u and v = v .If q = 1, then let Q = ( v ).Suppose that ρ ( G u,v ( p, q )) ≤ ρ ( G u,v ( p + 1 , q − x be the principaleigenvector of G u,v ( p + 1 , q − x u p +1 ≤ x v q . Let G (cid:48) be the hypergraph obtained from G u,v ( p + 1 , q −
1) by moving edge e p +1 from u p +1 to v q . It is obvious that G (cid:48) ∼ = G u,v ( p, q ). By Lemma 1, we have ρ ( G u,v ( p, q )) = ρ ( G (cid:48) ) > ρ ( G u,v ( p +1 , q − x u p +1 > x v q .Suppose that i ≥ x u p +1 − i > x v q − i for i ≤ q −
2. We want to showthat x u p − i > x v q − i − .Suppose that x u p − i ≤ x v q − i − . If x e p − i \{ u p − i ,u p − i +1 } > x f q − i − \{ v q − i − ,v q − i } .Let U = e p − i \{ u p − i } and V = f q − i − \{ v q − i − } . Let G (cid:48) be the hypergraph G (cid:48) obtained from G u,v ( p + 1 , q −
1) by deleting edges e p − i and f q − i − and addingedges e (cid:48) p − i and f (cid:48) q − i − , where e (cid:48) p − i = U ∪ ( f q − i − \ V ) and f (cid:48) q − i − = V ∪ ( e p − i \ U ). Note that G (cid:48) ∼ = G u,v ( p, q ). We have by Lemma 2 that ρ ( G u,v ( p, q )) = ρ ( G (cid:48) ) > ρ ( G u,v ( p + 1 , q − x e p − i \{ u p − i ,u p − i +1 } ≤ x f q − i − \{ v q − i − ,v q − i } . Now let U = { u p − i +1 } and V = { v p − i } . Let G (cid:48) bethe hypergraph obtained from G u,v ( p + 1 , q −
1) by deleting edges e p − i and f q − i − and adding edges e (cid:48) p − i and f (cid:48) q − i − , where e (cid:48) p − i = V ∪ ( e p − i \ U ) and f (cid:48) q − i − = U ∪ ( f q − i − \ V ). Note that G (cid:48) ∼ = G u,v ( p, q ). By Lemma 2, we have ρ ( G (cid:48) ) = ρ ( G u,v ( p, q )) > ρ ( G u,v ( p + 1 , q − x u p − i > x v q − i − , as desired.Let e = { u , v , w , w , . . . , w k − } with d G ( w ) ≥ e p − q +1 = { u p − q +1 , u p − q +2 , w (cid:48) , w (cid:48) , . . . , w (cid:48) k − } . Clearly, x w = · · · = x w k − and x w (cid:48) = · · · = x w (cid:48) k − . If x w ≤ x w (cid:48) , then we can obtain a G (cid:48) from G u,v ( p + 1 , q −
1) bymoving all the edges except e incident with w from w to w (cid:48) . By Lemma 1and the fact that G (cid:48) ∼ = G u,v ( p, q ), we have ρ ( G u,v ( p, q )) > ρ ( G u,v ( p +1 , q − x w > x w (cid:48) .Suppose that x e \{ v ,w } ≥ x e p − q +1 \{ u p − q +2 ,w (cid:48) } . Let U = e \ { v } and V = e p − q +1 \ { u p − q +2 } . Thus we can form a hypergraph G (cid:48) from G u,v ( p + 1 , q − e and e p − q +1 and adding edges e (cid:48) and e (cid:48) p − q +1 , where e (cid:48) = U ∪ ( e p − q +1 \ V ) and e (cid:48) p − q +1 = V ∪ ( e \ U ). It is obvious that G (cid:48) ∼ = G u,v ( p, q ).By Lemma 2, we have ρ ( G u,v ( p, q )) > ρ ( G u,v ( p + 1 , q − x e \{ v ,w } < x e p − q +1 \{ u p − q +2 ,w (cid:48) } . Now let U = { w } and V = { w (cid:48) } . Let G (cid:48) be the hypergraph obtained from G u,v ( p + 1 , q −
1) by deleting edges e and e p − q +1 and adding edges e (cid:48) and e (cid:48) p − q +1 , where e (cid:48) = V ∪ ( e \ U ) and e (cid:48) p − q +1 = U ∪ ( e p − q +1 \ V ). Note that G (cid:48) ∼ = G u,v ( p, q ). By Lemma 2, wehave ρ ( G (cid:48) ) = ρ ( G u,v ( p, q )) > ρ ( G u,v ( p + 1 , q − (cid:50) A hypergraph G is said to be reducible if every edge contains at least onependant vertex. For a reducible k -uniform hypergraph G with e ∈ E ( G ),4et v e be a pendant vertex in e , and let G (cid:48) be the hypergraph with V ( G (cid:48) ) = V ( G ) \ { v e : e ∈ E ( G ) } and E ( G (cid:48) ) = { e \ { v e } : e ∈ E ( G ) } . We say that G (cid:48) is reduced from G . Lemma 5 [7] Let G be a reducible k -uniform hypergraph. If G (cid:48) is reducedfrom G , then ρ k ( G ) = ρ k − ( G (cid:48) ) . For a k -uniform hypertree G with E ( G ) = { e , . . . , e m } , if V ( G ) = { v , . . . , v n } with n = ( k − m + 1, and e i = { v ( i − k − , . . . , v ( i − k − k } for i = 1 , . . . , m , then we call G a k -uniform loose path, denoted by P ( k ) m .For k ≥ m ≥
3, let D ( k ) m be the k -uniform hypertree obtained froma k -uniform loose path P ( k ) m − = ( u , e , u , . . . , e m − , u m − ) by attaching apendant edge at u .For k ≥ m ≥
3, let D (cid:48) ( k ) m be the k -uniform hypertree obtainedfrom a k -uniform loose path P ( k ) m − = ( u , e , u , . . . , e m − , u m − ) by attachinga pendant edge at a vertex of degree 1 in e . Lemma 6 [7] Let G be a k -uniform hypergraph with k ≥ . ( i ) If k = 3 and ρ ( G ) < k √ , then G is isomorphic to one of the followinghypergraphs: P (3) m for m ≥ , D (3) m for m ≥ , D (cid:48) (3) m for m ≥ , B (3) m for m ≥ , B (cid:48) (3) m for m ≥ , ¯ B (3) m for m ≥ , BD (3) m for m ≥ , and thirty-oneadditional hypergraphs: E (3)1 , , , E (3)1 , , , E (3)1 , , , F (3)2 , , , F (3)2 , ,l ( for ≤ l ≤ , F (3)1 , ,l ( for ≤ l ≤ , F (3)1 , ,l ( for ≤ l ≤ , F (3)1 , , , and G (3)1 , l :1 , ( for ≤ l ≤ see Figure 1 ) . ( ii ) If k = 4 , G is not reducible, and ρ ( G ) ≤ k √ , then G ∼ = H , , ,t with t = 1 , , , see Figure 2 ) . ( iii ) If k ≥ and ρ ( G ) ≤ k √ , then G is reducible. We remark that in Figure 1, E (3) i,j,l consists of three pendant paths of length i , j and l at a common vertex, F (3) i,j,l consists of three pendant paths of length i , j and l at three different vertices of a single edge, and G (3) i,j : l : p,q consists of a3-uniform loose path of length l + 2 with two pendant paths of length i and j at two pendant vertices in the first edge and two pendant paths of length p and q at two pendant vertices in the last edge. Now we are ready to show our main result.
Theorem 1
Let G be a connected k -uniform hypergraph with m ≥ edges,where k ≥ . Suppose that G (cid:54)∼ = P ( k ) m . Then ρ ( G ) ≥ ρ ( D (cid:48) ( k ) m ) with equality ifand only if G ∼ = D (cid:48) ( k ) m . (3) m ( m ≥ D (3) m ( m ≥ D (cid:48) (3) m ( m ≥ B (3) m ( m ≥ B (cid:48) (3) m ( m ≥
6) ¯ B (3) m ( m ≥ BD (3) m ( m ≥ E (3)1 , ,l F (3) i,j,l G (3)1 , l :1 , Figure 1: Hypergraphs in Lemma 6(i)6 , , , H , , , H , , , H , , , Figure 2: Hypergraphs H , , ,i for i = 1 , , , Proof.
By Lemma 6(i) , ρ ( D (cid:48) (3) m ) < √
4. Then by Lemma 5, we have ρ k ( D (cid:48) ( k ) m ) = ρ ( D (cid:48) (3) m ) <
4, and thus ρ ( D (cid:48) ( k ) m ) < k √ G be a connected k -uniform hypergraph with m ≥ G (cid:54)∼ = P ( k ) m having minimum spectral radius. We need only to show that G ∼ = D (cid:48) ( k ) m .Since ρ ( D (cid:48) ( k ) m ) < k √
4, we have ρ ( G ) < k √ Case 1. k = 3.By Lemma 6(i) , G is isomorphic to one of the following hypergraphs: P (3) m for m ≥ D (3) m for m ≥ D (cid:48) (3) m for m ≥ B (3) m for m ≥ B (cid:48) (3) m for m ≥ B (3) m for m ≥ BD (3) m for m ≥
5, and thirty-one additional hypergraphs: E (3)1 , , , E (3)1 , , , E (3)1 , , , F (3)2 , , , F (3)2 , ,l (for 2 ≤ l ≤ F (3)1 , ,l (for 3 ≤ l ≤ F (3)1 , ,l (for 4 ≤ l ≤ F (3)1 , , , and G (3)1 , l :1 , (for 0 ≤ l ≤ ρ ( D (3) m ) > ρ ( D (cid:48) (3) m ).By Lemma 3, we have ρ ( BD (3) m ) > ρ ( D (cid:48) (3) m ), and for E (3)1 , ,l with l = 2 , , m = l + 3, and thus ρ ( E (3)1 , ,l ) > ρ ( D (3) l +3 ) > ρ ( D (cid:48) (3) m ).By Lemma 4, we have ρ ( B (3) m ) > ρ ( D (cid:48) (3) m ), ρ ( B (cid:48) (3) m ) > ρ ( D (cid:48) (3) m ), and ρ ( ¯ B (3) m ) >ρ ( D (cid:48) (3) m ).For F (3) i,j,l , we have m = i + j + l + 1. By Lemma 4, we have ρ ( F (3) i,j,l ) >ρ ( F (3) i, ,m − i − ) > ρ ( F (3)1 , ,m − ) = ρ ( D (cid:48) (3) m ). Thus ρ ( F (3)2 , , ) > ρ ( D (cid:48) (3)9 ), ρ ( F (3)2 , ,m − ) >ρ ( D (cid:48) (3) m ) for (7 ≤ m ≤ ρ ( F (3)1 , ,m − ) > ρ ( D (cid:48) (3) m ) (for 8 ≤ m ≤ ρ ( F (3)1 , ,m − ) > ρ ( D (cid:48) (3) m ) (for 10 ≤ m ≤ ρ ( F (3)1 , , ) > ρ ( D (cid:48) (3)12 ).By Lemma 4, we have ρ ( G (3)1 , m − , ) > ρ ( F (3)1 , ,m − ) = ρ ( D (cid:48) (3) m ) for 8 ≤ m ≤ G (cid:54)∼ = D (cid:48) ( k ) m , then ρ ( G ) > ρ ( D (cid:48) (3) m ). It follows that G ∼ = D (cid:48) (3) m . Case 2. k = 4. 7f G is reducible, then by Lemma 5, for the hypergraph G reduced from G , we have ρ ( G ) < √
4, and by the proof in Case 1, we have G ∼ = D (cid:48) (3) m ,implying that G ∼ = D (cid:48) (4) m .Next suppose that G is not reducible. Then by Lemma 6(ii), G ∼ = H , , ,i with i = 1 , , ,
4. We will show that these are impossible. Since ρ ( D (3) m ) >ρ ( D (cid:48) (3) m ), by Lemma 3 we have ρ ( D (4) m ) > ρ ( D (cid:48) (4) m ). Thus it suffices to showthat ρ ( G ) > ρ ( D (4) m ).Suppose that G ∼ = H , , , . Then m = 8. From the table of [2] we have ρ ( D (2)8 ) = 1 . ρ ( D (4)8 ) = ( ρ ( D (2)8 ) , implying that( ρ ( D (4)8 )) = 3 . x be the principal eigenvector of G , and let u i , w i for i = 1 , , , , v , v be the vertices of H , , , as labeled in Figure 2). Let ρ ( G ) = ρ and x u i = x i for i = 1 , , , ,
5. Then x v = x v = x u = x . We have ρx w = x w x , and then x w = x ρ . Similarly, we have that x w = x ρ , x w i = (cid:113) x i x i +1 ρ for i = 2 , ,
4. Thus ρx = x ρ x + x x ,ρx = x x + x x ρ ,ρx = x x ρ + x x ρ ,ρx = x x ρ + x x ρ ,ρx = x x ρ + x ρ x . From the first two equations, we have ( ρ − ρ ( ρ − ) x = x , and from thethe other three equations, we have x = ρ x − x = ( ρ − ρ + ρ ) x and x = ρ x − x = ( ρ − x . Thus( ρ ) − ρ ) + 21( ρ ) − ρ ) + 13( ρ ) − . Since P (4)6 is a subhypergraph of G , we have ρ > ρ ( P (4)6 ) = ( (cid:112) π ) =2 + √ f ( t ) = t − t + 21 t − t + 13 t −
3. Note that f ( ) = − < , f (2 − √
2) = − √ > , f (1) = 1 > , f (2 + √
2) = − − √ < , f (3 .
9) = − . <
0, and f (4) = 1 >
0. Thus f ( t ) = 0 has threereal roots t , t and t satisfying < t < − √
2, 1 < t < √
2, and3 . < t <
4. Let t and t be the remaining two roots of f ( t ) = 0. Then t t > t + t > − (2 −√ − (2+ √ − t + t < − − − . . < √
2. Note that ρ > √
2. So whether t and t are real or not,8hey can not be equal to ρ . Thus ρ = t > . > . ρ ( D (4)8 ) , i.e., ρ ( G ) > ρ ( D (4)8 ), as desired.In the following, we consider the cases when G ∼ = H , , ,i for i = 1 , , ρ ( D (2)5 ) = 1 . ρ ( D (2)6 ) = 1 . ρ ( D (2)7 ) =1 . ρ ( D (4)5 ) = 1 . ρ ( D (4)6 )) = 3 . ρ ( D (4)7 )) = 3 . ρ ( H , , , ), ρ ( H (4)1 , , , ), and ρ ( H , , , ) are roots of ρ − ρ − ρ ) − ρ ) + 10( ρ ) − ρ ) + 2 = 0, and ( ρ ) − ρ ) + 15( ρ ) − ρ ) +6( ρ ) − ρ ( G ) > . > . ρ ( D (4)5 ) if i = 1 ,ρ ( G ) > . > .
733 = ρ ( D (4)6 ) if i = 2 ,ρ ( G ) > . > . ρ ( D (4)7 ) if i = 3 , i.e., ρ ( G ) > ρ ( D (4) m ), as desired. Case 3. k ≥ G is reducible. By Lemma 5, for the hypergraph G reduced from G , we have ρ ( G )) < k − √
4. Repeating this process by usingLemmas 6(iii) and 5, we have a hypergraph sequence G , G , . . . , G k − with G = G , where G i +1 is reduced from G i for i = 0 , . . . , k −
5. It is easily seenthat G k − is 4-uniform and ρ k ( G ) = ρ ( G k − ) <
4. By the proof of Case 2, G k − ∼ = D (4) m , and thus G ∼ = D ( k ) m . (cid:50) Among connected 2-uniform hypergraphs with m edges, the ones withspectral radius less than 2 are determined in [8] to be the trees P (2) m , D (2) m ,and three additional trees with m = 5 , ,
7, obtained from D (2) m − by attachinga pendant edge at a pendant vertex that is adjacent to a vertex of degree3, and by Lemma 3, it is easy to see that P (2) m for m ≥ D (2) m for m ≥ G be a connected k -uniform hypergraph with 2 edges, where k ≥ a be the number of common vertices of the two edges. Obviously,1 ≤ a ≤ k −
1. By direct calculation, ρ ( G ) = 2 ak . Therefore P ( k )2 and thehypergraph in which two edges share two vertices in common are the uniquehypergraphs with minimum and second minimum spectral radii among con-nected k -uniform hypergraphs with exactly 2 edges.Let G be a connected k -uniform hypergraphs with 3 edges, where k ≥
3. If there is a subhypergraph consisting two edges containing at least twovertices in common, then ρ ( G ) ≥ k √
4. If any two edges of G contain at mostone common vertex, then G is a cycle of length 3, D ( k )3 or P ( k )3 . If G is acycle of length 3, then ρ ( G ) = k √
4. By Lemma 3, ρ ( D ( k )3 ) > ρ ( P ( k )3 ). ByLemmas 5 and 6(i) , ρ k ( D ( k )3 ) = ρ ( D (3)3 ) <
4. Therefore P ( k )3 and D ( k )3 are9he unique hypergraphs with minimum and second minimum spectral radiiamong connected k -uniform hypergraphs with exactly 3 edges.For m ≥ k ≥
3, by Lemma 4, we have ρ ( D (cid:48) ( k ) m ) > ρ ( P ( k ) m ).Combining the above facts and Theorem 1, we have Theorem 2
Among connected k -uniform hypergraphs with m edges, P ( k ) m for m ≥ is the unique one with minimum spectral radius, and the hypergraphin which two edges share two vertices in common for m = 2 and k ≥ , D ( k )3 for k = 2 or m = 3 , and D (cid:48) ( k ) m for m ≥ and k ≥ are the unique ones withsecond minimum spectral radius. Acknowledgements
This work was supported by National Natural Science Foundation ofChina (No. 11071089 and No. 11701102), Natural Science Foundation ofGuangdong Province (No. 2017A030313032 and No. 2017A030310441).
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