Higher order Dirichlet-to-Neumann maps on graphs and their eigenvalues
aa r X i v : . [ m a t h . DG ] A p r HIGHER ORDER DIRICHLET-TO-NEUMANN MAPSON GRAPHS AND THEIR EIGENVALUES
YONGJIE SHI AND CHENGJIE YU Abstract.
In this paper, we first introduce higher order Dirichlet-to-Neumann maps on graphs which can be viewed as a discrete ana-logue of the corresponding Dirichlet-to-Neumann maps on compactRiemannian manifolds with boundary and a higher order general-ization of the Dirichlet-to-Neumann map on graphs introduced byHua-Huang-Wang[10] and Hassannezhad-Miclo [8]. Then, someRaulot-Savo-type estimates on the eigenvalues of the DtN mapsintroduced are derived. Introduction
Let ( M n , g ) be a compact Riemannian manifold with boundary and u be a harmonic function on M . Then the classical Steklov operatormaps u | ∂M to ∂u∂n . The eigenvalues of this operator are called the Stekloveigenvalues of ( M n , g ). These kinds of notions were first introduced bySteklov [16, 22] when studying the frequency of liquid sloshing. Theywere also found to have deep relationship with the Calder´on inverseproblem in applied mathematics ([3, 23]). There has been many workson the Steklov eigenvalues. For more details of the topic, one may referto the survey article [6]. Steklov operators are also called Dirichlet-to-Neumann maps and usually abbreviated as DtN maps because theytransform Dirchlet boundary data of a harmonic function to its Neu-mann boundary data.Recently, a discrete version of Steklov’s operator on graphs was intro-duced by Hua-Huang-Wang [10] and Hassannezhad-Miclo [8] indepen-dently. In [10], the authors also obtained some Cheeger-type inequal-ities for the first eigenvalue of DtN maps on graphs that are discreteanalogue of previous results of Escobar [5] and Jammes [12]. In [8], theauthors also obtained some Cheeger-type inequalities for higher order Mathematics Subject Classification.
Primary 05C50; Secondary 39A12.
Key words and phrases.
Hodge Theory, Dirichlet-to-Neumann map, Stekloveigenvalue, graph. Research partially supported by NSF of China with contract no. 11701355. Research partially supported by NSF of China with contract no. 11571215. eigenvalues. In [18], the author obtained some interesting lower boundsfor Steklov eigenvalues of graphs. In [11], the authors introduced theSteklov eigenvalues for infinite subgraphs and obtained some Cheeger-type inequalities for infinite subgraphs. Recently, in [7], the authorsderived a discrete version of Brock’s estimate (see [2]).Note that the classical Steklov operators and eigenvalues for func-tions were extended to differential forms by Raulot-Savo [20]. In fact,different definition of DtN maps for differential forms were introducedin [1] and [14] before Raulot-Savo [20]. However, the definitions ofDtN maps in [1] and [14] were motivated by inverse problems in ap-plied mathematics which are not suitable for spectral analysis. Re-cently, Karpukhin [15] slightly modified the DtN maps in [1] so thatit is suitable for spectral analysis and extended the higher dimensionalgeneralization of Hersch-Payne-Schiffer inequality (see [9]) in [25] tomore general cases. According to all of these works, it is a naturalproblem to develop a discrete version of higher order DtN maps forgraphs. This is the motivation of the paper.It seems that a discrete version of Hodge theory on simplicial com-plexes was known to experts and goes back to the pioneer work [4]of Dodziuk. For its analogue on graphs, although looks very similarwith the Hodge theory on simplicial complexes, the authors find it onlyappeared in some recent works in applied mathematics. For example,in [13] (see also [17]), the authors developed a very interesting rankingmethod which is called the
HodgeRank by using the Hodge decom-position on graphs.In this paper, motivated by the works of Hua-Huang-Wang [10] andRaulot-Savo [20], we first introduce higher order DtN maps on graphs.For a simple graph G , we denote by V ( G ) and E ( G ) the set of verticesand the set of edges for G respectively. We also adopt the notion ofgraph with boundary in [18]. A pair ( G, B ) is said to be a graph withboundary B if G is a simple graph and B ⊂ V ( G ), such that any twovertices of B are not adjacent to each other, and each vertex in B isadjacent to some vertcies in Ω := B c which is called the interior of( G, B ).A k -form f of a graph G is a skew symmetric function with k + 1variables on V ( G ) such that f ( v , v , · · · , v k ) = 0 if { v , v , · · · , v k } isnot a clique in G . The collection of all k -forms on G is denoted as A k ( G ). The exterior differential df of a k -form f is a ( k + 1)-form with(1.1) df ( v , v , · · · , v k +1 ) = k +1 X i =0 ( − i f ( v , · · · , ˆ v i , · · · , v k +1 ) igher order DtN maps 3 for any ( k + 2)-clique { v , v , · · · , v k +1 } of G .For a weight w on a graph G , we mean a positive function definedon the collection of cliques for G . Define the inner product on A k ( G )as h α, β i = 1( k + 1)! X v ,v , ··· ,v k ∈ G α ( v , v , · · · , v k ) β ( v , v , · · · , v k ) w ( v , v , · · · , v k ) . (1.2)Then, for a finite graph G with weight w , define δ to be the adjointoperator of d according to the inner product just defined. The HodgeLaplacian operator is defined to be ∆ = δd + dδ as usual.For a finite graph ( G, B ) with boundary B , a clique with one vertexin B is called a boundary clique. The restriction of a k -form of G tothe collection of boundary cliques is called a boundary k -form. Thecollection of boundary k -form is denoted as A k ( ∂ Ω). Let w be a weighton G . Define the inner product on A k ( ∂ Ω) as h α, β i ∂ Ω = 1 k ! X v ∈ δ Ω X v ,v , ··· ,v k ∈ Ω α ( v, v , v , · · · , v k ) β ( v, v , v , · · · , v k ) w ( v, v , v , · · · , v k ) . (1.3)Then, we have the following two Green’s formulas for finite weightedgraphs with boundary:(1.4) h δα, β i Ω = h α, dβ i G − h N α, β i ∂ Ω and(1.5) h dα, β i Ω = h α, δβ i Ω − h D α, β i ∂ Ω . For the expressions of the operators N and D , see (3.9) and (3.10).Motivated by the Green’s formulas above, we introduced three kindsof DtN maps T ( k ) δd , T ( k ) dδ and T ( k ) for k -forms on graphs.For the definition of T ( k ) δd , for each ϕ ∈ A k ( ∂ Ω), let E ( k ) δd ( ϕ ) be asolution of the boundary value problem:(1.6) (cid:26) δdω = 0 in Ω ω | ∂ Ω = ϕ and define(1.7) T ( k ) δd ( ϕ ) = N dE ( k ) δd ( ϕ ) . The existence of E ( k ) δd ( ϕ ) is guaranteed by Theorem 3.1. Note that thesolution of the boundary value problem may not be unique. So, to Shi & Yu make sure that T ( k ) δd ( ϕ ) is well defined, we must verify that N dE ( k ) δd ( ϕ )is independent of the choice of solution to the boundary value problem.This is done in Theorem 3.2.The definitions for T ( k ) dδ and T ( k ) are similar. Let E ( k ) dδ ( ϕ ) and E ( k ) ( ϕ )be solutions of the boundary value problems:(1.8) (cid:26) dδω = 0 in Ω ω | ∂ Ω = ϕ and(1.9) (cid:26) ∆ ω = 0 in Ω ω | ∂ Ω = ϕ respectively. Then, we define(1.10) T ( k ) dδ ( ϕ ) = D δE ( k ) dδ ( ϕ )and(1.11) T ( k ) ( ϕ ) = D δE ( k ) ( ϕ ) + N dE ( k ) ( ϕ ) . By the Green’s formulas above, we have(1.12) D T ( k ) δd ϕ, ψ E ∂ Ω = D dE ( k ) δd ( ϕ ) , dE ( k ) δd ( ψ ) E G , (1.13) D T ( k ) dδ ϕ, ψ E ∂ Ω = D δE ( k ) dδ ( ϕ ) , δE ( k ) dδ ( ψ ) E Ω and(1.14) (cid:10) T ( k ) ϕ, ψ (cid:11) ∂ Ω = (cid:10) δE ( k ) ( ϕ ) , δE ( k ) ( ψ ) (cid:11) Ω + (cid:10) dE ( k ) ( ϕ ) , dE ( k ) ( ψ ) (cid:11) G . So, T ( k ) δd , T ( k ) dδ and T ( k ) are all nonnegative self-adjoint operators. Infact, the three identities above are the motivations of the definitions ofthe three kinds of DtN maps. T ( k ) can be viewed as a discrete analogueof Raulot-Savo’s DtN maps for differential forms. It seems that T ( k ) δd corresponds to the modification of Belishev-Sharafutdino’s DtN mapsin [1] by Karpukhin in [15]. Moreover, it is clear that T (0) δd = T (0) is thesame as the DtN maps introduced in [10] and [8].The second part of this paper is to derive some discrete version ofRaulot-Savo-type estimates in [19, 21, 24] for subgraphs of integer lat-tices or standard tessellation of R n . For a nonnegative self-adjointoperator T on a finite dimensional vector space with inner product, wedenote by λ i ( T ) the i -th positive eigenvalue of T (in ascending orderand counting multiplicities). Moreover, for a graph G and a subsetΩ ⊂ V ( G ), as in [10], we denote by ˜Ω the subgraph of G with the setof vertices Ω = Ω ∩ δ Ω and set of edges E (Ω , Ω) where δ Ω means theset of vertices that are not in Ω but adjacent to some vertices in Ω and igher order DtN maps 5 E (Ω , Ω) means the set of edges with one end in Ω and the other in Ω.It is clear that ( ˜Ω , δ
Ω) is a graph with boundary δ Ω.Let Ω be a nonempty finite subset of the integer lattice in R n andlet T (0) be the DtN maps defined above for ( ˜Ω , δ Ω) with normalizedweight. Then, using similar arguments as in [19, 21, 24], we have thefollowing estimate:(1.15) λ ( T (0) ) + λ ( T (0) ) + · · · + λ n ( T (0) ) ≤ | δ Ω | min ≤ i ≤ n {| E i ( ˜Ω) |} . where E i ( ˜Ω) means the set of edges in ˜Ω that are parallel to e i . Here { e , e , · · · , e n } is the standard basis of R n .Furthermore, for a nonempty finite subset Ω of the integer latticeswhich is also the set of vertices of the graph of standard tessellation of R n , consider ˜Ω as the subgraph of the graph of standard tessellation of R n with unit weight. For each 0 ≤ k ≤ n −
1, we have(1.16) C k +1 n X i =1 λ i ( T ( k ) δd ) ≤ nC kn − n − k − | C k +2 ( ∂ Ω) | min ≤ i
In this section, we recall some preliminaries for Hodge theory ongraphs. In this paper, we assume all graphs G be simple graphs (with-out loops and multiple edges) with V ( G ) the set of vertices and E ( G )the set of edges.We first introduce forms and exterior differentials on graphs. Recallthe notion of cliques on a graph. Definition 2.1. A k -clique of a graph G is a subset { v , v , · · · , v k } of V ( G ) with k distinct elements such that { v i , v j } ∈ E ( G ) for any1 ≤ i < j ≤ k . The collection of all k -cliques of G is denoted as C k ( G ).It is clear that C ( G ) = V ( G ) and C ( G ) = E ( G ).Similar to the smooth case, we define tensors, weights, which playthe role of metrics in the smooth case, and forms on graphs. Definition 2.2.
Let G be a graph.(1) A function f : V ( G ) × V ( G ) × · · · × V ( G ) | {z } k +1 → R is called a k -tensor of G if f ( v , v , · · · , v k ) = 0 whenever { v , v , · · · , v k } 6∈ C k +1 ( G ). The collection of all k -tensors on G is denoted as T k ( G ). Moreover, let T ∗ ( G ) = L ∞ k =0 T k ( G ).(2) A k -tensor that is symmetric with respect to all of its variablesis called a k -weight. The collection of all k -weights on G isdenoted as W k ( G ). Moreover, let W ∗ ( G ) = L ∞ k =0 W k ( G )(3) A k -weight w with w ( v , v , · · · , v k ) > { v , v , · · · , v k } ∈ C k +1 ( G ) is called a positive k -weight.(4) A k -tensor that is skew symmetric with respect to all of itsvariables is called a k -form. The collection of all k -forms on G is denoted as A k ( G ). Moreover, let A ∗ ( G ) = L ∞ k =0 A k ( G ).As in the smooth case, we define the skew-symmetrization operatoras follows. Definition 2.3.
The skew-symmetrization operator A : T k ( G ) → A k ( G ) is defined as(2.1) A ( f )( v , v , · · · , v k ) = 1( k + 1)! X σ ∈ S k +1 sgn( σ ) · f ( v σ (0) , v σ (1) , · · · , v σ ( k ) )where S k +1 is the group of permutations on { , , , · · · , k } .On any graph, a natural weight with every clique of weight 1 is calledthe unit weight. igher order DtN maps 7 Definition 2.4.
We define the unit weight G ∈ W ∗ ( G ) on the graph G as(2.2) G ( v , v , · · · , v k ) = 1if and only if { v , v , · · · , v k } ∈ C k +1 ( G ) for k = 0 , , , · · · .Similar to cup product in algebraic topology, we define tensor prod-uct on graphs as follows. Definition 2.5.
Let G be a graph, f ∈ T r ( G ) and g ∈ T s ( G ). Wedefine the tensor product of f and g as f ⊗ g ( v , v , · · · , v r + s )= G ( v , v , · · · , v r + s ) f ( v , v , · · · , v r ) g ( v r , v r +1 , · · · , v r + s )(2.3)for any v , v , · · · , v r + s ∈ V ( G ).Similar to the smooth case, we define wedge product of forms ongraphs as skew-symmetrization of tensor product. Definition 2.6.
Let G be a graph, α ∈ A r ( G ) and β ∈ A s ( G ). Then,define the wedge product of α and β as(2.4) α ∧ β = A ( α ⊗ β ) . Similar to the boundary operators in algebraic topology, we definethe exterior differential operators on graphs as follows.
Definition 2.7.
Let G be a graph and α ∈ A k ( G ). Define the exteriordifferential of α as dα ( v , v , · · · , v k +1 )= G ( v , v , · · · , v k +1 ) k +1 X j =0 ( − j α ( v , v , · · · , ˆ v j , · · · , v k +1 ) . (2.5)for any v , v , · · · , v k +1 ∈ V ( G ).Similarly as the smooth case, we have the following conclusions forwedge product and exterior differentials on graphs. For their proofs,see the Appendix. Proposition 2.1.
Let G be a graph. Then, (1) d = 0 ; (2) for any α ∈ A r ( G ) and β ∈ A s ( G ) , α ∧ β = ( − rs β ∧ α ;(3) for any α ∈ A r ( G ) and β ∈ A s ( G ) , d ( α ∧ β ) = dα ∧ β + ( − r α ∧ dβ ; Shi & Yu (4) for any α ∈ A p ( G ) , β ∈ A q ( G ) and γ ∈ A r ( G ) with dα = dβ = dγ = 0 , ( α ∧ β ) ∧ γ = α ∧ ( β ∧ γ ) . Because of d = 0, we have the following chain complex which is theanalogue of the de Rham complex of differential forms:(2.6) 0 → A ( G ) → A ( G ) → · · · → A r ( G ) → A r +1 ( G ) → · · · . We call this the de Rham complex of the graph G . Let Z r ( G ) bethe kernel of d : A r ( G ) → A r +1 ( G ) and B r ( G ) = d ( A r − ( G )). Thecohomology groups:(2.7) H rdR ( G ) = Z r ( G ) B r ( G ) ( r = 0 , , , · · · )are called the de Rham cohomology groups of G . The dimension of H rdR ( G ) denoted as b r ( G ) is called the r -th Betti number of G . It isclear that b ( G ) is the number of connected components of G when G is a finite graph. In fact, if we associate a finite graph G with asimplical complex K ( G ) with each ( k + 1)-clique in G corresponding toa k -simplex of K ( G ) such that a k -simplex is a face of a l -simplex if andonly if the ( k + 1)-clique corresponding to the k -simplex is a subset ofthe ( l + 1)-clique corresponding the l -simplex, then b i ( G ) = b i ( K ( G )).Next, we introduce Hodge theory on graphs. Definition 2.8.
Let G be a graph and w ∈ W ∗ ( G ) be a positive weighton G . We call ( G, w ) a weighted graph.A weight on the edges of a graph induces weights on any clique inthe following way which is sometimes call the normalized weight.
Definition 2.9.
Let G be a graph and w ∈ W ( G ) be a positive 1-weight on G . Then, w induces a positive weight w ∗ ∈ W ∗ ( G ) on G by(2.8) w ∗ ( v ) = X v ∈ V w ( v , v )for any v ∈ V ( G ) and(2.9) w ∗ ( v , v , · · · , v k ) = G ( v , v , · · · , v k ) X ≤ i G, w ) an edge-weightedgraph.Similar to the smooth case, we define inner products for k -forms ongraphs as follows. igher order DtN maps 9 Definition 2.10. Let ( G, w ) be a finite weighted graph. Define theinner product on A k ( G ) as h α, β i = 1( k + 1)! X v ,v , ··· ,v k ∈ G α ( v , v , · · · , v k ) β ( v , v , · · · , v k ) w ( v , v , · · · , v k ) . (2.10)Let δ : A k +1 ( G ) → A k ( G ) be the adjoint operator of d . By directcomputation, one has the following identity. Proposition 2.2 (see [17]) . Let ( G, w ) be a finite weighted graph. Then δα ( v , v , · · · , v k )= 1 w ( v , v , · · · , v k ) X v ∈ V ( G ) α ( v, v , v , · · · , v k ) w ( v, v , v , · · · , v k )(2.11) for any α ∈ A k +1 ( G ) . Let ∆ = dδ + δd : A k ( G ) → A k ( G ) which is called the Hodge Lapla-cian operator on G . Forms in the kernel of ∆ are call harmonic forms.The collection of harmonic k -forms on G is denoted as H k ( G ). Byusing simple linear algebra, it is not hard to see that a discrete versionof Hodge decomposition theorem holds for finite weighted graphs. Theorem 2.1. Let ( G, w ) be a finite weighted graph. Then (2.12) A r ( G ) = d ( A r − ( G )) ⊕ H r ( G ) ⊕ δ ( A r +1 ( G )) which is an orthogonal decomposition of A r ( G ) , for r = 0 , , , · · · . Asa consequence, (2.13) H rdR ( G ) ∼ = H r ( G ) for r = 0 , , , · · · . Higher order Dirichlet-to-Neumann maps In this section, we define higher order DtN maps for graphs withboundary. First, we adopt the notations in [10]. Let G be a graph, andΩ be a subset of V ( G ). Let(3.1) δ Ω = { x ∈ V ( G ) \ Ω | x is adjacent to some vertices in Ω . } which is called the vertex boundary of Ω. Let Ω = Ω ∪ δ Ω. For any twosubsets A and B of V ( G ), we denote by E ( A, B ) the set of edges in G with one end in A and the other in B . Denote by ˜Ω the subgraph of G with Ω the set of vertices and E (Ω , Ω) the set of edges. We will also use Ω to denote the subgraph of G spanned by Ω. Denote E (Ω , δ Ω) by ∂ Ω, which is called edge boundary of Ω.Moreover, let(3.2) C k ( ∂ Ω) = {{ v , v , · · · , v k } ∈ C k ( ˜Ω) : v ∈ δ Ω , v , · · · , v k ∈ Ω } which is called the k -clique boundary of Ω. It is clear that C ( ∂ Ω) = δ Ωand C ( ∂ Ω) = ∂ Ω. Let A k ( ∂ Ω) = { f : δ Ω × Ω × · · · × Ω | {z } k → R : f ( v , v , v , · · · , v k ) = 0when { v , v , · · · , v k } 6∈ C k +1 ( ∂ Ω) and f is skew symmetrywith respect to the last k variables. } (3.3)An elements in A k ( ∂ Ω) is called a boundary k -form of Ω. For a weightedgraph, we define the inner product on A k ( ∂ Ω) as the follows: h α, β i ∂ Ω = 1 k ! X v ∈ δ Ω X v ,v , ··· ,v k ∈ Ω α ( v, v , v , · · · , v k ) β ( v, v , v , · · · , v k ) w ( v, v , v , · · · , v k ) . (3.4)More generally, for S ⊂ C k +1 ( ˜Ω), we define(3.5) h α, β i S = X { v ,v , ··· ,v k }⊂ S α ( v , v , · · · , v k ) β ( v , v , · · · , v k ) w ( v , v , · · · , v k )for any α, β ∈ A k ( ˜Ω). It is clear that(3.6) h α, β i ˜Ω = h α, β i Ω + h α, β i ∂ Ω for any α, β ∈ A k ( ˜Ω).We also adopt the notion of graphs with boundary in [18]. A pair( G, B ) is said to be a graph with boundary if(1) G is a simple graph;(2) B ⊂ V ( G ) with δ Ω = B and E ( B, B ) = ∅ where Ω = V ( G ) \ B is called the interior of G and B is called the boundary of G .It is clear that ( ˜Ω , δ Ω) in the notations above adopted from [10] is agraph with boundary.For weighted graphs with boundary, we have the following Green’sformulas. igher order DtN maps 11 Proposition 3.1. Let ( G, B, w ) be a finite weighted graph with bound-ary and with Ω its interior. Then, for any α ∈ A k +1 ( G ) and β ∈ A k ( G ) , (3.7) h δα, β i Ω = h α, dβ i G − h N α, β i ∂ Ω and for any α ∈ A k − ( G ) and β ∈ A k ( G )(3.8) h dα, β i Ω = h α, δβ i Ω − h D α, β i ∂ Ω where N α, D α ∈ A k ( ∂ Ω) are given by (3.9) N α ( v, v , v , · · · , v k ) = P u ∈ Ω α ( u, v, v , · · · , v k ) w ( u, v, v , · · · , v k ) w ( v, v , v , · · · , v k ) and (3.10) D α ( v, v , v , · · · , v k ) = α ( v , v , · · · , v k ) respectively, when { v, v , v , · · · , v k } ∈ C k +1 ( ∂ Ω) .Proof. Note that h α, dβ i G = h δα, β i G = h δα, β i Ω + h δα, β i ∂ Ω . (3.11)Then, by Proposition 2.2 and that E ( B, B ) = ∅ , we get the first iden-tity.Moreover, by that h dα, β i Ω + h dα, β i ∂ Ω = h dα, β i G = h α, δβ i G = h α, δβ i Ω + h α, δβ i ∂ Ω , (3.12)(3.13) h dα, β i Ω = h α, δβ i Ω + h α, δβ i ∂ Ω − h dα, β i ∂ Ω . Note that h α, δβ i ∂ Ω = 1( k − X v ∈ B,v , ··· ,v k − ∈ Ω α ( v, v , · · · , v k − ) δβ ( v, v , · · · , v k − ) w ( v, v , · · · , v k − )= ( − k ( k − X v ∈ B,v , ··· ,v k − ,v k ∈ Ω α ( v, v , · · · , v k − ) β ( v, v , · · · , v k ) w ( v, v , · · · , v k )(3.14) by Proposition 2.2 and E ( B, B ) = ∅ , and h dα, β i ∂ Ω = 1 k ! X v ∈ B,v ,v , ··· ,v k ∈ Ω dα ( v, v , · · · , v k ) β ( v, v , · · · , v k ) w ( v, v , · · · , v k )= 1 k ! X v ∈ B,v ,v , ··· ,v k ∈ Ω α ( v , · · · , v k ) β ( v, v , · · · , v k ) w ( v, v , · · · , v k )+1 k ! k X j =1 X v ∈ B,v ,v , ··· ,v k ∈ Ω ( − j α ( v, v , · · · , ˆ v j , · · · , v k ) β ( v, v , · · · , v k ) w ( v, v , · · · , v k )= h D α, β i ∂ Ω + h α, δβ i ∂ Ω . (3.15)From these, we get the second identity. (cid:3) For the definition of DtN maps, we need to solve some boundaryvalue problems. We are interested in the following three kinds of bound-ary value problems:(3.16) (cid:26) δdω = 0 in Ω ω = ϕ in C k +1 ( ∂ Ω) , (3.17) (cid:26) dδω = 0 in Ω ω = ϕ in C k +1 ( ∂ Ω) , and(3.18) (cid:26) ∆ ω = 0 in Ω ω = ϕ in C k +1 ( ∂ Ω) , where ω ∈ A k ( G ) is the unknown k -form and ϕ ∈ A k ( ∂ Ω) is the givenboundary k -form. We will show the solvability of the boundary valueproblems (3.16), (3.17) and (3.18) by a variation argument. We firstneed the following simple lemma in linear algebra. Lemma 3.1. Let V be a finite dimensional vector space over R and Q be a nonnegative quadratic form on V . Then, for any x ∈ V and anysubspace W ⊂ V , the function f ( x ) = Q ( x + x, x + x ) with x ∈ W achieves its minimum in W .Proof. Let ξ , · · · , ξ r be a basis of W such that(3.19) Q ( λ ξ + λ ξ + · · · + λ r ξ r , λ ξ + λ ξ + · · · + λ r ξ r ) = r X i =1 Q ( ξ i , ξ i ) λ i . igher order DtN maps 13 Then, for any x = λ ξ + λ ξ + · · · + λ r ξ r ,(3.20) f ( x ) = r X i =1 (cid:0) Q ( ξ i , ξ i ) λ i + 2 Q ( x , ξ i ) λ i (cid:1) + Q ( x , x ) . By that Q is nonnegative, we have Q ( ξ i , ξ i ) ≥ Q ( x , ξ i ) = 0whenever Q ( ξ i , ξ i ) = 0. This gives us the conclusion. (cid:3) We are now ready to show that the boundary value problems (3.16),(3.17) and (3.18) are all solvable. Theorem 3.1. Let ( G, B, w ) be a finite weighted graph with bound-ary and with Ω its interior. Then, for any given boundary data ϕ ∈ A k ( ∂ Ω) , the boundary values problems (3.16) , (3.17) and (3.18) are allsolvable.Proof. We only show the solvability of (3.16). The proofs for the othertwo boundary value problems are similar.Consider the quadratic form Q ( α, α ) = h dα, dα i G on V = A k ( G ).Let(3.21) W = { α ∈ A k ( G ) | α = 0 on C k +1 ( ∂ Ω) } and x ∈ A k ( G ) be the zero-extension of ϕ . By Lemma 3.1, there is aminimizer of Q when restricting Q on(3.22) x + W = { α ∈ A k ( G ) | α = ϕ on C k +1 ( ∂ Ω) } . Let ω be such a minimizer. We claim that ω is a solution to theboundary value problem (3.16). Indeed, by that ω is a minimizer, weknow that(3.23) h dω, dξ i G = 0for any ξ ∈ W . By (3.7),(3.24) h δdω, ξ i Ω = 0for any ξ ∈ W . So, δdω = 0 in Ω. (cid:3) Moreover, the Dirichlet principle holds for the boundary value prob-lems (3.16), (3.17) and (3.18). Proposition 3.2. (1) The solutions of the boundary value problem (3.16) are minimizers of the quadratic form h dα, dα i G with α ∈ A k ( G ) and α = ϕ on C k +1 ( ∂ Ω) . (2) The solutions of the boundary value problem (3.17) are min-imizers of the quadratic form h δα, δα i Ω with α ∈ A k ( G ) and α = ϕ on C k +1 ( ∂ Ω) . (3) The solutions of the boundary value problem (3.18) are minimiz-ers of the quadratic form h dα, dα i G + h δα, δα i Ω with α ∈ A k ( G ) and α = ϕ on C k +1 ( ∂ Ω) .Proof. We only need to show (1). The proofs of (2) and (3) are similar.Let ω ϕ be a solution of the boundary value problem (3.16) and ω ∈ A k ( G ) with ω = ϕ on C k +1 ( ∂ Ω). By (3.7),(3.25) h dω ϕ , d ( ω − ω ϕ ) i G = 0since ω − ω ϕ = 0 on C k +1 ( ∂ Ω). So,(3.26) h dω, dω i G = h dω ϕ , dω ϕ i G + h d ( ω − ω ϕ ) , d ( ω − ω ϕ ) i G ≥ h dω ϕ , dω ϕ i G . This completes the proof. (cid:3) We are now ready to introduce DtN maps for forms on graphs. Foreach ϕ ∈ A k ( ∂ Ω), let E ( k ) δd ( ϕ ), E ( k ) dδ ( ϕ ) and E ( k ) ( ϕ ) be solutions of theboundary value problems (3.16), (3.17) and (3.18) respectively. Wedefine three kinds of DtN maps for forms on graphs corresponding tothe three kinds of boundary value problems:(3.27) T ( k ) δd ( ϕ ) = N ( dE ( k ) δd ( ϕ )) , (3.28) T ( k ) dδ ( ϕ ) = D ( δE ( k ) dδ ( ϕ ))and(3.29) T ( k ) ( ϕ ) = N ( dE ( k ) ( ϕ )) + D ( δE ( k ) ( ϕ )) . However, because the solutions of the boundary value problems (3.16),(3.17) and (3.18) may not be unique, we should check that the defini-tions of the DtN maps T ( k ) δd , T ( k ) dδ and T k are independent of the choiceof solutions of the boundary value problems (3.16), (3.17) and (3.18)respectively. Theorem 3.2. The operators T ( k ) δd , T ( k ) dδ and T ( k ) are well defined.Proof. We only need to show the conclusion for T ( k ) δd . The proofs forthe other two operators are similar.For ϕ ∈ A k ( G ), let ω and ω be two solutions of the boundary valueproblem (3.16). Then, by (3.7), we have(3.30) 0 = h δd ( ω − ω ) , ω − ω i Ω = h d ( ω − ω ) , d ( ω − ω ) i G since ω − ω = 0 on C k +1 ( ∂ Ω). So(3.31) d ( ω − ω ) = 0 igher order DtN maps 15 on G . Moreover, for any ξ ∈ A k ( G ), by (3.7) again,(3.32)0 = h δd ( ω − ω ) , ξ i Ω = h d ( ω − ω ) , dξ ) i G − h N d ( ω − ω ) , ξ i ∂ Ω . So(3.33) h N d ( ω − ω ) , ξ i ∂ Ω = 0for any ξ ∈ A k ( G ). This means that(3.34) N dω = N dω . (cid:3) Moreover, as usual, the DtN maps just defined are all nonnegativeand self-adjoint operators. Proposition 3.3. The DtN maps T ( k ) δd , T ( k ) dδ and T ( k ) are all nonnega-tive self-adjoint operators on A k ( ∂ Ω) . Moreover, (3.35) ker T ( k ) δd = { ω | C k +1 ( ∂ Ω) : ω ∈ A k ( G ) with dω = 0 on G. } , (3.36) ker T ( k ) dδ = { ω | C k +1 ( ∂ Ω) : ω ∈ A k ( G ) with δω = 0 on Ω . } , and (3.37)ker T ( k ) = { ω | C k +1 ( ∂ Ω) : ω ∈ A k ( G ) with dω = 0 on G and δω = 0 on Ω . } . Proof. We only need to show the conclusion for T ( k ) δd . The proofs forthe other two operators are similar. By (3.7),(3.38) D T ( k ) δd ( ϕ ) , ψ E ∂ Ω = h dω ϕ , dω ψ i G for any ϕ, ψ ∈ A k ( ∂ Ω) where ω ϕ and ω ψ mean solutions to the boundaryvalue problem (3.16) with boundary data ϕ and ψ respectively. Fromthis we get the conclusion. (cid:3) It is clear that T (0) δd = T (0) is the same as the DtN maps for functionsintroduced in [10] and [8] for edge -weighted graphs. Moreover, we havethe following upper bounds for the norms of the DtN maps that aregeneralizations of the similar conclusion in [10]. Proposition 3.4. Let ( G, B, w ) be a finite weighted graph with bound-ary and with Ω its interior. Then k T ( k ) δd k ≤ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) , k T ( k ) dδ k ≤ max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k ) for k ≥ , and k T ( k ) k ≤ max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k )+ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) for k ≥ .Proof. Note that for each ϕ ∈ A k +1 ( ∂ Ω) h N ϕ, N ϕ i ∂ Ω = 1 k ! X v ∈ B X u ,u , ··· ,u k ∈ Ω ( N ϕ ( v, u , · · · , u k )) w ( v, u , · · · , u k )= 1 k ! X v ∈ B w ( v, u , u , · · · , u k ) X u ,u , ··· ,u k ∈ Ω X u ∈ Ω ϕ ( u, v, u , · · · , u k ) w ( u, v, u , · · · , u k ) ! ≤ k ! X v ∈ B X u , ··· ,u k ∈ Ω X u ∈ Ω ϕ ( v, u, u , · · · , u k ) w ( v, u, u , · · · , u k ) ! × X u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) ! ! ≤ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) h ϕ, ϕ i ∂ Ω . So, for each ϕ ∈ A k ( ∂ Ω), D T ( k ) δd ( ϕ ) , T ( k ) δd ( ϕ ) E ∂ Ω = D N dE ( k ) δd ( ϕ ) , N dE ( k ) δd ( ϕ ) E ∂ Ω ≤ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) D dE ( k ) δd ( ϕ ) , dE ( k ) δd ( ϕ ) E ∂ Ω ≤ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) D dE ( k ) δd ( ϕ ) , dE ( k ) δd ( ϕ ) E G ≤ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) h dϕ, dϕ i G =( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) h dϕ, dϕ i ∂ Ω ≤ ( k + 1) (cid:18) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) (cid:19) h ϕ, ϕ i ∂ Ω igher order DtN maps 17 where ϕ means the trivial extension of ϕ and we have used the Dirich-let principle for the boundary value problem (3.16) and the Cauchy-Schwartz inequality. This give us the first inequality.Moreover, for any ϕ ∈ A k (Ω), h D ϕ, D ϕ i ∂ Ω = 1( k + 1)! X v ∈ B X u ,u , ··· ,u k +1 ∈ Ω ( D ϕ ( v, u , · · · , u k +1 )) w ( v, u , · · · , u k +1 )= 1( k + 1)! X u ,u , ··· ,u k +1 ∈ Ω ϕ ( u , · · · , u k +1 ) X v ∈ B w ( v, u , · · · , u k +1 ) ≤ max { u ,u , ··· ,u k +1 }∈ C k +1 (Ω) P v ∈ B w ( v, u , · · · , u k +1 ) w ( u , u , · · · , u k +1 ) h ϕ, ϕ i Ω . So, for any ϕ ∈ A k ( ∂ Ω), D T ( k ) dδ ( ϕ ) , T ( k ) dδ ( ϕ ) E ∂ Ω = D D δE ( k ) dδ ( ϕ ) , D δE ( k ) dδ ( ϕ ) E ∂ Ω ≤ max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k ) D δE ( k ) dδ ( ϕ ) , δE ( k ) dδ ( ϕ ) E Ω ≤ max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k ) h δϕ, δϕ i Ω ≤ (cid:18) max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k ) (cid:19) h ϕ, ϕ i ∂ Ω where we have used the Dirichlet principle for the boundary value prob-lem (3.17) and the Cauchy-Schwartz inequality. This completes theproof of the second inequality.A combination of the proofs of the first two inequalities will give usthe last inequality. (cid:3) In the rest of the paper, for a nonnegative self-adjoint liner operator T : V → V defined on a finite dimensional vector space V with innerproduct. We will denote by λ i ( T ) the i -th positive eigenvalue of T (inascending order and counting multiplicities).As a direct corollary of Proposition 3.4, we have the following upperbounds of the eigenvalues of T ( k ) δd , T ( k ) dδ and T ( k ) . Corollary 3.1. Let ( G, B, w ) be a finite weighted graph with boundaryand with Ω its interior. Then, λ i ( T ( k ) δd ) ≤ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) ,λ i ( T ( k ) dδ ) ≤ max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k ) for k ≥ , and λ i ( T ( k ) ) ≤ max { u ,u , ··· ,u k }∈ C k (Ω) P v ∈ B w ( v, u , · · · , u k ) w ( u , u , · · · , u k )+ ( k + 1) max { v,u ,u , ··· ,u k }∈ C k +1 ( ∂ Ω) P u ∈ Ω w ( u, v, u , · · · , u k ) w ( v, u , · · · , u k ) for k ≥ . A Raulot-Savo-type estimate for subgraphs of integerlattices In this section, we derive a discrete version of Raulot-Savo-type es-timates in [19, 21, 24] for finite subgraphs of integer lattices. We firstneed the following discrete version of Lemma 2.1 in [24]. Proposition 4.1. Let ( G, B, w ) be a finite weighted graph with bound-ary and with Ω its interior. Let V be a subspace of (4.1) { ξ ∈ A k +1 ( G ) : ξ = dη for some η ∈ A k ( G ) and δξ = 0 on Ω . } . Suppose dim V = n . Let A : V → V be the linear transformation on V such that (4.2) h Aξ, η i G = h N ξ, N η i ∂ Ω for any ξ, η ∈ V . Then, A is positive and self-adjoint. Moreover (4.3) λ i ( T ( k ) δd ) ≤ λ i ( A ) and (4.4) λ i ( T ( k ) ) ≤ λ i ( A ) for i = 1 , , · · · , n .Proof. By definition of A , it is clear that A is self-adjoint. For positivityof A , let ξ ∈ V be such that N ξ = 0. Suppose that ξ = dη . Then, byGreen’s formula (3.7),(4.5) h ξ, ξ i G = h dη, ξ i G = h η, δξ i Ω + h N ξ, η i ∂ Ω = 0 . So, ξ = 0 on G and hence A is positive. igher order DtN maps 19 Let ξ , ξ , · · · , ξ n be an orthonormal frame of V such that Aξ i = λ i ( A ) ξ i . Suppose that ξ i = dη i . By (3.35), without loss of generality,we can assume that η i | C k +1 ( ∂ Ω) ⊥ ker T ( k ) δd .We first claim that η | C k +1 ( ∂ Ω) , η | C k +1 ( ∂ Ω) , · · · , η n | C k +1 ( ∂ Ω) are linearindependent. Indeed, suppose that(4.6) c η | C k +1 ( ∂ Ω) + c η | C k +1 ( ∂ Ω) + · · · + c n η n | C k +1 ( ∂ Ω) = 0 . Let η = c η + c η + · · · + c n η n . Then, by (3.7),(4.7) h dη, dη i G = h δdη, η i Ω + h N dη, η i ∂ Ω = 0 . So,(4.8) n X i =1 c i ξ i = dη = 0on G which implies that c = c = · · · = c n = 0.Let ϕ , ϕ , · · · , ϕ N ∈ (ker T ( k ) δd ) ⊥ be an orthonormal basis of (ker T ( k ) δd ) ⊥ such that T ( k ) δd ( ϕ i ) = λ i ( T ( k ) δd ) ϕ i for i = 1 , , · · · , N . Let U = span { η | C k +1 ( ∂ Ω) , η | C k +1 ( ∂ Ω) , · · · , η i | C k +1 ( ∂ Ω) } ⊂ (ker T ( k ) δd ) ⊥ . Then dim U = i . Let W = span { ϕ i , ϕ i +1 , · · · , ϕ N } . Then dim W = N − i + 1. So, U ∩ W = 0. Let ϕ = 0 be in U ∩ W . Because ϕ ∈ W ,suppose that ϕ = P Nj = i c j ϕ j . Then, D N dE ( k ) δd ( ϕ ) , N dE ( k ) δd ( ϕ ) E ∂ Ω D dE ( k ) δd ( ϕ ) , dE ( k ) δd ( ϕ ) E G = D T ( k ) δd ( ϕ ) , T ( k ) δd ( ϕ ) E ∂ Ω D T ( k ) δd ( ϕ ) , ϕ E ∂ Ω = P Nj = i c j λ j ( T ( k ) δd ) P Nj = i c j λ j ( T ( k ) δd ) ≥ λ i ( T ( k ) δd ) . (4.9) On the other hand, because ϕ ∈ U , suppose that ϕ = P ij =1 c j η j | C k +1 ( ∂ Ω) .Then, D N dE ( k ) δd ( ϕ ) , N dE ( k ) δd ( ϕ ) E ∂ Ω D dE ( k ) δd ( ϕ ) , dE ( k ) δd ( ϕ ) E G = D N d P ij =1 c j η j , N d P ij =1 c j η j E ∂ Ω D d P ij =1 c j η j , d P ij =1 c j η j E G = DP ij =1 c j N ξ j , P ij =1 c j N ξ j E ∂ Ω DP ij =1 c j ξ j , P ij =1 c j ξ j E G = P ij =1 c j λ j ( A ) P ij =1 c j ≤ λ i ( A ) . (4.10)So, we get (4.3).Furthermore, let ξ i be the same as before and suppose that ξ i = dη i .Without loss of generality, we assume that η i ⊥ B k ( G ). Then, for any α ∈ A k − ( G ) with α | C k ( ∂ Ω) = 0, by (3.7),(4.11) h δη i , α i Ω = h η i , dα i G − h N η i , α i ∂ Ω = 0 . So, δη i = 0 on Ω. Let η i | C k +1 ( ∂ Ω) ∈ ker T ( k ) be the orthogonal projec-tion of η i | C k +1 ( ∂ Ω) into ker T ( k ) and let ζ i = η i − η i . Then, by (3.37), dζ i = ξ i , δζ i = 0 and ζ i ⊥ ker T ( k ) . By the same argument as before, ζ | C k +1 ( ∂ Ω) , ζ | C k +1 ( ∂ Ω) , · · · , ζ n | C k +1 ( ∂ Ω) are linearly independent. Fromthis, a similar argument as before will give us (4.4). (cid:3) We are now ready to prove the main result of this section. Theorem 4.1. Let G be the graph with V ( G ) = Z n and (4.12) E ( G ) = ( { x, y } : x, y ∈ Z n , n X i =1 | x i − y i | = 1 ) . Let Ω be a nonempty finite subset of Z n . Consider the graph ˜Ω as anedge-weighted graph with each edge of weight . Then, (4.13) λ ( T (0) ) + λ ( T (0) ) + · · · + λ n ( T (0) ) ≤ X v ∈ δ Ω v n X i =1 deg i v | E i ( ˜Ω) | where deg v means the number of edges in ˜Ω adjacent to v , deg i v meansthe number of edges in ˜Ω adjacent to v that are parallel to e i , and E i ( ˜Ω) igher order DtN maps 21 means the set of edges in ˜Ω that are parallel to e i . Here { e , e , · · · , e n } is the standard basis of R n . As a consequence, we have (4.14) λ ( T (0) ) + λ ( T (0) ) + · · · + λ n ( T (0) ) ≤ | δ Ω | min ≤ i ≤ n {| E i ( ˜Ω) |} . Proof. Let x , x , · · · , x n be the coordinate functions of R n . Note that dx , dx , · · · , dx n ∈ A ( ˜Ω) are linearly independent. Moreover, for each x ∈ Ω,(4.15) δdx i ( x ) = 12 n n X j =1 (2 x i ( x ) − x i ( x + e j ) − x i ( x − e j )) = 0 . Let V = span { dx , dx , · · · , dx n } and let A be the linear transforma-tion defined in (4.2). Then, by Proposition 4.1, we know that(4.16) n X i =1 λ i ( T (0) ) ≤ tr A. Moreover, note that(4.17) h dx i , dx j i ˜Ω = (cid:26) i = j | E i ( ˜Ω) | i = j and h N dx i , N dx i i ∂ Ω = X v ∈ δ Ω v X u ∈ Ω ,u ∼ v ( x i ( v ) − x i ( u )) ! ≤ X v ∈ δ Ω deg i v deg v . (4.18)So,(4.19) tr A = n X i =1 h N dx i , N dx i i ∂ Ω h dx i , dx i i ˜Ω ≤ X v ∈ δ Ω v n X i =1 deg i v | E i ( ˜Ω) | . This completes the proof of Theorem 4.1 (cid:3) Raulot-Savo-type estimates for subgraphs of thestandard tessellation of R n In this section, we want to make some applications of Proposition4.1 to higher order forms. Because integer lattices do not contain anytriangle, we consider the graph G of standard tessellation of R n with V ( G ) = Z n and(5.1) E ( G ) = (cid:26) { x, y } : x = y ∈ Z n , ≤ min ≤ i ≤ n ( y i − x i ) ≤ max ≤ i ≤ n ( y i − x i ) = 1 (cid:27) . For simplicity, we first consider estimates for eigenvalues of DtNmaps for functions. Theorem 5.1. Let G be the standard tessellation of R n and Ω be anonempty finite subset of Z n . Consider ˜Ω as a weighted graph withunit weight. Then n X i =1 λ i ( T (0) ) ≤ n − n X i =1 | E i ( ˜Ω) | n X k =1 X ≤ i < i < · · · < i k ≤ ni ∈ { i , i , · · · , i k } | E i i ··· i k ( ∂ Ω) | (5.2) where E i i ··· i k ( ∂ Ω) means the number of edges in ∂ Ω that are parallelto e i + e i + · · · + e i k . As a consequence, (5.3) n X i =1 λ i ( T (0) ) ≤ n n − | ∂ Ω | min ≤ i ≤ n {| E i ( ˜Ω) |} . Proof. Consider V = span { dx , dx , · · · , dx n } , it is not hard to see that dx , dx , · · · , dx n are linearly independent and we still have(5.4) δdx i ( u ) = 0for any u ∈ Ω. Moreover,(5.5) h dx i , dx j i ˜Ω = n X k =1 X ≤ i < · · · < i k ≤ n { i, j } ⊂ { i , i , · · · , i k } | E i i ··· i k ( ˜Ω) | , and h N dx i , N dx i i ∂ Ω = X v ∈ δ Ω X u ∈ Ω ,u ∼ v ( v i − u i ) ! ≤ n − X v ∈ δ Ω n X k =1 X ≤ i < i < · · · < i k ≤ ni ∈ { i , i , · · · , i k } deg i i ··· i k v =2 n − n X k =1 X ≤ i < i < · · · < i k ≤ ni ∈ { i , i , · · · , i k } | E i i ··· i k ( ∂ Ω) | (5.6)for i = 1 , , · · · , n , where deg i i ··· i k v means the number of edges adja-cent to v that are parallel to e i + e i + · · · + e i k . The factor 2 n − in thelast inequality comes from the fact that there are at most 2 n − vertices u adjacent to v in ˜Ω such that v i − u i all equal to 1 or all equal to − igher order DtN maps 23 Let B = (cid:0) h dx i , dx j i ˜Ω (cid:1) i,j =1 , , ··· ,n and C = (cid:0) h N dx i , N dx j i ∂ Ω (cid:1) i,j =1 , , ··· ,n .We claim that(5.7) B ≥ D := diag {| E ( ˜Ω) | , | E ( ˜Ω) | , · · · , | E n ( ˜Ω) |} . In fact, for any ( ξ , ξ , · · · , ξ n ) ∈ R n . n X i,j =1 ( B − D ) ij ξ i ξ j = n X k =2 n X i,j =1 X ≤ i < · · · < i k ≤ n { i, j } ⊂ { i , i , · · · , i k } | E i i ··· i k ( ˜Ω) | ξ i ξ j = n X k =2 X ≤ i
More generally, we have the following estimate. Theorem 5.2. Let G be the standard tessellation of R n and Ω be anonempty finite subset of Z n and ≤ k ≤ n − . Consider ˜Ω as a weighted graph with unit weight and suppose that C i i ··· i k +1 ( ˜Ω) = ∅ forany ≤ i < i < · · · < i k +1 ≤ n . Here C i i ··· i k +1 ( ˜Ω) means theset of ( k + 2) -cliques in ˜Ω such that the ( k + 1) -simplex formed by the ( k + 2) vertices of the ( k + 2) -clique is parallel to the ( k + 1) -simplexformed by o, e j , e j + e j , · · · , e j + · · · + e j k +1 with ( j , j , · · · , j k +1 ) somepermutation of ( i , i , · · · , i k +1 ) . Then, (5.9) C k +1 n X i =1 λ i ( T ( k ) δd ) ≤ ( k + 1)2 n − k − X ≤ i
Let(5.13) V = span { dx i ∧ dx i ∧ · · · ∧ dx i k +1 : 1 ≤ i < i < · · · < i k +1 ≤ n } . Let(5.14) C ∗ k +2 ( ˜Ω) = ∪ ≤ i
In this section, we give the proof of Proposition 2.1 and some detailsof computation in the proof of Theorem 5.2.Although the definitions in Section 2 and the facts in Proposition 2.1are very similar to that of exterior calculus for simplicial complexes.In the first part of this section, we present the proof of Proposition 2.1because our notations and definitions are slightly different with thatfor simplicial complexes. Proof of Proposition 2.1. (1) It is just by simple direct computation.(2) For any { v , v , · · · , v r + s } ∈ C r + s +1 ( G ),( r + s + 1)! α ∧ β ( v , v , · · · , v r + s )= X σ ∈ S r + s +1 sgn( σ ) α ( v σ (0) , v σ (1) , · · · , v σ ( r ) ) β ( v σ ( r ) , v σ ( r +2) , · · · , v σ ( r + s ) )= X σ ∈ S r + s +1 sgn( σ )( − r r + s s β ( v σ ( r + s ) , v σ ( r + s − , · · · , v σ ( r ) ) α ( v σ ( r ) , v σ ( r − , · · · , v σ (0) )= X σ ∈ S r + s +1 sgn( σ )( − r r + s s β ( v σ ( τ (0)) , v σ ( τ (1)) , · · · , v σ ( τ ( s )) ) × α ( v σ ( τ ( s )) , v σ ( τ ( s +1)) , · · · , v σ ( τ ( r + s )) )= X σ ∈ S r + s +1 sgn( σ · τ − )( − r r + s s β ( v σ (0) , v σ (1) , · · · , v σ ( s ) ) α ( v σ ( s ) , v σ ( s +1) , · · · , v σ ( r + s ) )= X σ ∈ S r + s +1 sgn( σ )( − r r + s s +( r + s )2+( r + s )2 β ( v σ (0) , v σ (1) , · · · , v σ ( s ) ) × α ( v σ ( s ) , v σ ( s +1) , · · · , v σ ( r + s ) )=( − rs ( r + s + 1)! β ∧ α ( v , v , · · · , v r + s )where τ = (cid:18) · · · r + s − r + sr + s r + s − r + s − · · · (cid:19) . (3) For any { v , v , · · · , v r + s +1 } ∈ C r + s +2 ( G ), d ( α ∧ β )( v , v , · · · , v r + s +1 )= r + s +1 X j =0 ( − j α ∧ β ( v , v , · · · , ˆ v j , · · · , v r + s +1 )= 1( r + s + 1)! r + s +1 X j =0 ( − j X σ ∈ S jr + s +2 sgn( σ ) α ⊗ β ( v σ (0) , v σ (1) , · · · , ˆ v j , · · · , v σ ( r + s +1) )where S jr + s +2 = { σ ∈ S r + s +2 | σ ( j ) = j } . Moreover, dα ∧ β + ( − r α ∧ dβ ( v , v , · · · , v r + s +1 )= 1( r + s + 2)! X σ ∈ S r + s +2 sgn σdα ( v σ (0) , v σ (1) , · · · , v σ ( r +1) ) β ( v σ ( r +1) , v σ ( r +2) , · · · , v σ ( r + s +1) )+( − r ( r + s + 2)! X σ ∈ S r + s +2 sgn σα ( v σ (0) , v σ (1) , · · · , v σ ( r ) ) dβ ( v σ ( r ) , v σ ( r +1) , v σ ( r +2) , · · · , v σ ( r + s +1) )= I + I + I + I igher order DtN maps 27 where I = 1( r + s + 2)! X σ ∈ S r + s +2 sgn σ r X j =0 ( − j α ( v σ (0) , · · · , ˆ v σ ( j ) , · · · , v σ ( r +1) ) × β ( v σ ( r +1) , v σ ( r +2) , · · · , v σ ( r + s +1) )= 1( r + s + 2)! X σ ∈ S r + s +2 sgn σ r X j =0 ( − j α ⊗ β ( v σ (0) , · · · , ˆ v σ ( j ) , · · · , v σ ( r + s +1) )= 1( r + s + 2)! r X j =0 ( − j r + s +1 X i =0 X σ ∈ S j,ir + s +2 sgn σα ⊗ β ( v σ (0) , · · · , ˆ v σ ( j ) , · · · , v σ ( r + s +1) )= 1( r + s + 2)! r X j =0 ( − j r + s +1 X i =0 X τ ∈ S ir + s +2 ( − j − i sgn τ α ⊗ β ( v τ (0) , · · · , ˆ v τ ( i ) , · · · , v τ ( r + s +1) )= r + 1( r + s + 2)! r + s +1 X i =0 ( − i X τ ∈ S ir + s +2 sgn τ α ⊗ β ( v τ (0) , · · · , ˆ v τ ( i ) , · · · , v τ ( r + s +1) )(6.1) I = ( − r +1 ( r + s + 2)! X σ ∈ S r + s +2 sgn σα ( v σ (0) , · · · , v σ ( r ) ) β ( v σ ( r +1) , v σ ( r +2) , · · · , v σ ( r + s +1) ) , (6.2) I = ( − r ( r + s + 2)! X σ ∈ S r + s +2 sgn σα ( v σ (0) , · · · , v σ ( r ) ) β ( v σ ( r +1) , v σ ( r +2) , · · · , v σ ( r + s +1) )= − I , (6.3)and similarly as in (6.1), I = 1( r + s + 2)! X σ ∈ S r + s +2 sgn σ r + s +1 X j = r +1 ( − j α ( v σ (0) , v σ (1) , · · · , v σ ( r ) ) × β ( v σ ( r ) , · · · , ˆ v σ ( j ) , · · · , v σ ( r + s +1) )= s + 1( r + s + 2)! r + s +1 X i =0 ( − i X τ ∈ S ir + s +2 sgn τ α ⊗ β ( v τ (0) , · · · , ˆ v τ ( i ) , · · · , v τ ( r + s +1) )(6.4)with S j,ir + s +2 = { σ ∈ S r + s +2 | σ ( j ) = i } . So,(6.5) d ( α ∧ β ) = dα ∧ β + ( − r α ∧ dβ. (4) For any { u , u , · · · , u p + q + r } ∈ C p + q + r +1 ( G ),( α ∧ β ) ∧ γ ( u , u , · · · , u p + q + r )= 1( p + q + r + 1)! X σ ∈ S p + q + r +1 sgn σα ∧ β ( u σ (0) , · · · , u σ ( p + q ) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= 1( p + q + r + 1)!( p + q + 1)! X σ ∈ S p + q + r +1 sgn σ X τ ∈ S p + q +1 sgn τ α ( u σ ( τ (0)) , · · · , u σ ( τ ( p )) ) × β ( u σ ( τ ( p )) , · · · , u σ ( τ ( p + q )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= 1( p + q + r + 1)!( p + q + 1)! ( I + I + I ) . Here I = p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 sgn τ α ( u σ ( τ (0)) , · · · , u σ ( τ ( p )) ) β ( u σ ( τ ( p )) , · · · , u σ ( τ ( p + q )) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 sgn τ p X j =0 ( − p − j α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( j )) , · · · , u σ ( τ ( p +1)) ) × q X k =0 ( − q − k β ( u σ ( τ ( p )) , · · · , \ u σ ( τ ( p + k )) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( i )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= J + J + J where we have used the fact that dα = dβ = 0, and J = p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 ( − p + q − i sgn τ × α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( i )) , · · · , u σ ( τ ( p )) , u σ ( τ ( p +1)) ) × β ( u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( i )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= p − X i =0 X σ ∈ S p + q + r +1 sgn σ X µ ∈ S p + qp + q +1 sgn µα ( u σ ( µ (0)) , · · · , u σ ( µ ( p )) ) β ( u σ ( µ ( p )) , · · · , u σ ( µ ( p + q )) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) ) igher order DtN maps 29 with the relation of µ and τ given by µ (0) = τ (0) , · · · , µ ( i − 1) = τ ( i − , µ ( i ) = τ ( i + 1) , · · · , µ ( p + q − 1) = τ ( p + q ) , µ ( p + q ) = τ ( i ) = p + q , J = p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 ( − p + q +1 − i sgn τ × α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( i )) , · · · , u σ ( τ ( p )) , u σ ( τ ( p +1)) ) × β ( u σ ( τ ( p )) , \ u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( i )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 ( − p + q − i sgn τ × α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( i )) , · · · , u σ ( τ ( p +1)) , u σ ( τ ( p )) ) × β ( u σ ( τ ( p )) , \ u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( i )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= − p − X i =0 X σ ∈ S p + q + r +1 sgn σ X µ ∈ S p + qp + q +1 sgn µα ( u σ ( µ (0)) , · · · , u σ ( µ ( p )) ) β ( u σ ( µ ( p )) , · · · , u σ ( µ ( p + q )) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= − J , with the relation of µ and τ given by µ (0) = τ (0) , · · · , µ ( i − 1) = τ ( i − , µ ( i ) = τ ( i + 1) , · · · , µ ( p − 2) = τ ( p − , µ ( p − 1) = τ ( p + 1) , µ ( p ) = τ ( p ) , µ ( p + 1) = τ ( p + 2) , · · · , µ ( p + q − 1) = τ ( p + q ) , µ ( p + q ) = τ ( i ) = p + q , and J =( − q − p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 sgn τ α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( p )) , u σ ( τ ( p +1)) ) × β ( u σ ( τ ( p )) , \ u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( i )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= p − X i =0 X σ ∈ S p + q + r +1 sgn σ X τ ∈ S i,p + qp + q +1 ( − p +1 − i sgn τ × α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( i )) , · · · , \ u σ ( τ ( p )) , u σ ( τ ( p +1)) , u σ ( τ ( i )) ) × β ( u σ ( τ ( i )) , u σ ( τ ( p )) , \ u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= p X σ ∈ S p + q + r +1 sgn σ X µ ∈ S p,p + qp + q +1 sgn µα ( u σ ( µ (0)) , · · · , u σ ( µ ( p )) ) β ( u σ ( µ ( p )) , · · · , u σ ( µ ( p + q )) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= pI , with the relation of µ and τ given by µ (0) = τ (0) , · · · , µ ( i − 1) = τ ( i − , µ ( i ) = τ ( i + 1) , · · · , µ ( p − 2) = τ ( p − , µ ( p − 1) = τ ( p + 1) , µ ( p ) = τ ( i ) = p + q, µ ( p +1) = τ ( p ) , µ ( p +2) = τ ( p +2) , · · · , µ ( p + q ) = τ ( p + q ).The other terms of I vanish because there are the same pair of verticesin the arguments of α and β in those terms. I = X σ ∈ S p + q + r +1 sgn σ X τ ∈ S p,p + qp + q +1 sgn τ α ( u σ ( τ (0)) , · · · , u σ ( τ ( p )) ) β ( u σ ( τ ( p )) , · · · , u σ ( τ ( p + q )) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= X σ ∈ S p + q + r +1 sgn σ X τ ∈ S p,p + qp + q +1 sgn τ p X i =0 ( − p + q − i α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( i )) , · · · , u σ ( τ ( p +1)) ) × β ( u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( p )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= X σ ∈ S p + q + r +1 sgn σ X τ ∈ S p,p + qp + q +1 ( − q sgn τ α ( u σ ( τ (0)) , · · · , \ u σ ( τ ( p )) , u σ ( τ ( p +1)) ) × β ( u σ ( τ ( p +1)) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( p )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= X σ ∈ S p + q + r +1 sgn σ X µ ∈ S p + q,p + qp + q +1 sgn µα ( u σ ( µ (0)) , · · · , u σ ( µ ( p )) ) β ( u σ ( µ ( p )) , · · · , u σ ( µ ( p + q )) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )=( p + q )! X σ ∈ S p + q + r +1 sgn σα ( u σ (0) , · · · , u σ ( p ) ) β ( u σ ( p ) , · · · , u σ ( p + q ) ) × γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )=( p + q + r + 1)!( p + q )! A ( α ⊗ β ⊗ γ )( u , u , · · · , u p + q + r +1 )with the relation of τ and µ given by µ (0) = τ (0) , · · · , µ ( p − 1) = τ ( p − , µ ( p ) = τ ( p + 1) , · · · , µ ( p + q − 1) = τ ( p + q ) , µ ( p + q ) = τ ( p ) = p + q . igher order DtN maps 31 Finally, I = X σ ∈ S p + q + r +1 sgn σ p + q X i = p +1 X τ ∈ S i,p + qp + q +1 sgn τ α ( u σ ( τ (0)) , · · · , u σ ( τ ( p )) ) × β ( u σ ( τ ( p )) , · · · , u σ ( τ ( p + q )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= X σ ∈ S p + q + r +1 sgn σ p + q X i = p +1 X τ ∈ S i,p + qp + q +1 ( − p + q − i sgn τ α ( u σ ( τ (0)) , · · · , u σ ( τ ( p )) ) × β ( u σ ( τ ( p )) , · · · , \ u σ ( τ ( i )) , · · · , u σ ( τ ( p + q )) , u σ ( τ ( i )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= p + q X i = p +1 X σ ∈ S p + q + r +1 sgn σ X µ ∈ S p + q,p + qp + q +1 sgn µα ( u σ ( µ (0)) , · · · , u σ ( µ ( p )) ) × β ( u σ ( µ ( p )) , · · · , u σ ( µ ( p + q )) ) γ ( u σ ( p + q ) , · · · , u σ ( p + q + r ) )= q ( p + q )!( p + q + r + 1)! A ( α ⊗ β ⊗ γ )( u , u , · · · , u p + q + r ) , with the relation of µ and τ given by µ (0) = τ (0) , · · · , µ ( i − 1) = τ ( i − , µ ( i ) = τ ( i + 1) , · · · , µ ( p + q − 1) = τ ( p + q ) , µ ( p + q ) = τ ( i ) = p + q .Combining the expressions above together, we have(6.6) ( α ∧ β ) ∧ γ = A ( α ⊗ β ⊗ γ ) . Similarly, we have(6.7) α ∧ ( β ∧ γ ) = A ( α ⊗ β ⊗ γ ) . This completes the proof of (4). (cid:3) In the second part of this section, we present some details of compu-tation in the proof of Theorem 5.2. We first need the following relationof parallel differential forms on R n and forms for graphs embedded into R n . Proposition 6.1. Let G be a graph embedded in R n . Let Φ : V k ( R n ) ∗ → A k ( G ) be defined as follows: (6.8) (Φ ω )( u , u , · · · , u k ) = ω ( u − u , u − u , · · · , u k − u ) when { u , u , · · · , u k } ∈ C k +1 ( G ) . Then, (6.9) Φ α ∧ Φ β = k ! l !( k + l )! Φ( α ∧ β ) . for any α ∈ V k ( R n ) ∗ and β ∈ V l ( R n ) ∗ . Proof. For any { u , u , · · · , u k , u k +1 , · · · , u k + l } ∈ C k + l ( G ),Φ α ∧ Φ β ( u , u , · · · , u k + l )= 1( k + l + 1)! X σ ∈ S k + l +1 sgn σ · Φ α ( u σ (0) , u σ (1) , · · · , u σ ( k ) )Φ β ( u σ ( k ) , · · · , u σ ( k + l ) )= 1( k + l + 1)! X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u σ (0) , · · · , u σ ( k ) − u σ (0) ) × β ( u σ ( k +1) − u σ ( k ) , · · · , u σ ( k + l ) − u σ ( k ) )= 1( k + l + 1)! X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u + u − u σ (0) , · · · , u σ ( k ) − u + u − u σ (0) ) × β ( u σ ( k +1) − u + u − u σ ( k ) , · · · , u σ ( k + l ) − u + u − u σ ( k ) )= 1( k + l + 1)! ( I + I + I + I + I )where I = X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u , · · · , u σ ( k ) − u ) β ( u σ ( k +1) − u , · · · , u σ ( k + l ) − u )= X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u , · · · , u σ ( k ) − u ) β ( u σ ( k +1) − u , · · · , u σ ( k + l ) − u ) , = k ! l !Φ( α ∧ β )( u , u , · · · , u k + l ) ,I = k X i =1 X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u , · · · , \ u σ ( i ) − u , u − u σ (0) , · · · , u σ ( k ) − u ) × β ( u σ ( k +1) − u , · · · , u σ ( k + l ) − u )= kI by swapping the values of σ (0) and σ ( i ), I = k + l X i = k +1 X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u , · · · , u σ ( k ) − u ) × β ( u σ ( k +1) − u , · · · , \ u σ ( i ) − u , u − u σ ( k ) , · · · , u σ ( k + l ) − u )=0 igher order DtN maps 33 by swapping the values of σ ( i ) and σ (0), I = k − X i =1 k + l X j = k +1 X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u , · · · , \ u σ ( i ) − u , u − u σ (0) , · · · , u σ ( k ) − u ) × β ( u σ ( k +1) − u , · · · , \ u σ ( j ) − u , u − u σ ( k ) , · · · , u σ ( k + l ) − u )= 0by swapping the value of σ ( i ) and σ ( j ), and I = k + l X j = k +1 X σ ∈ S k + l +1 sgn σ · α ( u σ (1) − u , · · · , u σ ( k − − u , u − u σ (0) ) × β ( u σ ( k +1) − u , · · · , \ u σ ( j ) − u , u − u σ ( k ) , · · · , u σ ( k + l ) − u )= k + l X j = k +1 X σ ∈ S k + l +1 sgn σ · α ( u τ (1) − u , · · · , u τ ( k − − u , u σ ( k ) − u ) × β ( u τ ( k +1) − u , · · · , u τ ( j ) − u , · · · , u τ ( k + l ) − u )= lI by re-ordering σ to τ with τ (0) = σ ( j ), τ ( k ) = σ (0) , τ ( j ) = σ ( k ) and τ ( i ) = σ ( i ) for i 6∈ { , k, j } .Therefore,(6.10) Φ α ∧ Φ β = k ! l !( k + l )! Φ( α ∧ β ) . (cid:3) With the help of Proposition 6.1, we give the details of computationin the proof of Theorem 5.2. Proposition 6.2. Let G be the graph of standard tessellation of R n , ≤ k ≤ n − . Then, (1) For each ≤ i < i < · · · < i k +1 ≤ n , dx i ∧ dx i ∧ · · · ∧ dx i k +1 ( u , u , · · · , u k +1 ) = ± k +1)! or . (2) For each ≤ i < i < · · · < i k +1 ≤ n and each ( k + 1) -clique { u , · · · , u k } in G , (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 1( k + 1)! (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = − k + 1)! (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) ≤ n − k − . (6.11) Proof. (1) Equip with Z n the partial order: x ≤ y if y i − x i ≥ i = 1 , , · · · , n . Then, a ( k + 2)-clique { u , u , · · · , u k +1 } of G is atotally ordered subset of Z n . Without loss of generality, suppose that u < u < · · · < u k +1 . Then, there is a flag ∅ $ I $ I · · · $ I k +1 of { , , · · · , n } , such that(6.12) u i − u = e I i for i = 1 , , · · · , k +1. Here, for a subset I of { , , · · · , n } , e I = P i ∈ I e i .Then, by Proposition 6.1, dx i ∧ · · · ∧ dx i k +1 ( u , u , · · · , u k +1 )= 1( k + 1)! dx i ∧ · · · ∧ dx i k +1 ( e I , e I , · · · , e I k +1 )= 1( k + 1)! dx i ∧ · · · ∧ dx i k +1 ( e J , e J , · · · , e J k +1 )= (cid:26) sgn σ ( k +1)! { i σ ( j ) } = J j for j = 1 , , · · · , k + 10 otherwise(6.13)where J i = ( I i \ I i − ) ∩ { i , i , · · · , i k +1 } for i = 1 , , · · · , k + 1 with I = ∅ .(2) By the same argument as in (1), we may assume that u = o, u = e I , · · · , u k = e I k where ∅ $ I $ I · · · $ I k +1 is a flag of { , , · · · , n } .Then, by Proposition 6.1, dx i ∧ · · · ∧ dx i k +1 ( u , u , · · · , u k , u )= 1( k + 1)! dx i ∧ · · · ∧ dx i k +1 ( e I , e I , · · · , e I k , u )= 1( k + 1)! dx i ∧ · · · ∧ dx i k +1 ( e J , e J , · · · , e J k , u )(6.14)where J i = ( I i \ I i − ) ∩ { i , i , · · · , i k +1 } for i = 1 , , · · · , k with I = ∅ .If there is some J i = ∅ , then dx i ∧ · · · ∧ dx i k +1 ( e J , e J , · · · , e J k , u ) = 0 igher order DtN maps 35 for all u . The conclusion is clearly true.If J i is not empty for any i = 1 , , · · · , k , only the following two caseshappen.(i) Each J i contains only one element in { i , i , · · · , i k +1 } . In thiscase, without loss of generality, we can assume that J j = i j for j = 1 , , · · · , k . Then, to make sure that dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 0 , we must have u < o or u > e I k . So, (cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 1( k + 1)! (cid:27) = { e I k + e i k +1 + e J : J ⊂ I ck \ { i k +1 }} . and (cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = − k + 1)! (cid:27) = {− e i k +1 − e J : J ⊂ I ck \ { i k +1 }} . Hence (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 1( k + 1)! (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = − k + 1)! (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) =2 n −| I k |− ≤ n − k − . (ii) Some J i contains two elements of { i , i , · · · , i k +1 } . Withoutloss of generality, we may assume that J = { i } , · · · , J k − = { i k − } and J k = { i k , i k +1 } . To make sure that dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 0 , we must have e I k − < u < e I k . So (cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 1( k + 1)! (cid:27) = { e I k − + e i k +1 + e J : J ⊂ I k \ ( I k − ∪ { i k , i k +1 } ) } . and (cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = − k + 1)! (cid:27) = { e I k − + e i k + e J : J ⊂ I k \ ( I k − ∪ { i k , i k +1 } ) } . Hence (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = 1( k + 1)! (cid:27)(cid:12)(cid:12)(cid:12)(cid:12) = (cid:12)(cid:12)(cid:12)(cid:12)(cid:26) u ∈ G : dx i ∧ · · · dx i k +1 ( u , u , · · · , u k , u ) = − k + 1)! 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Department of Mathematics, Shantou University, Shantou, Guang-dong, 515063, China E-mail address : [email protected] Department of Mathematics, Shantou University, Shantou, Guang-dong, 515063, China E-mail address ::