Sub-signature Operators and the Kastler-Kalau-Walze type theorem for manifolds with boundary
aa r X i v : . [ m a t h . DG ] F e b Sub-signature Operators and the Kastler-Kalau-Walze type theorem formanifolds with boundary
Tong Wu, Sining Wei, Yong Wang ∗ School of Mathematics and Statistics, Northeast Normal University, Changchun, 130024, China
Abstract
In this paper, we obtain two Lichnerowicz type formulas for sub-signature operators. And we give the proof ofKastler-Kalau-Walze type theorems for sub-signature operators on 4-dimensional and 6-dimensional compactmanifolds with (resp.without) boundary.
Keywords:
Sub-signature operator; Lichnerowicz type formulas; Noncommutative residue;Kastler-Kalau-Walze type theorems.
1. Introduction
Until now, many geometers have studied non-commutative residues. In [8, 24], authors found noncommutative residues are of great importance to the study of noncommutative geometry. In [3], Connesused the noncommutative residue to derive a conformal 4-dimensional Polyakov action analogy. Connesshowed us that the noncommutative residue on a compact manifold M coincided with the Dixmier’s traceon pseudodifferential operators of order − dim M in [4]. And Connes clamid the noncommutative residue ofthe square of the inverse of the Dirac operator was proportioned to the Einstein-Hilbert action. Kastler [12]gave a brute-force proof of this theorem. Kalau and Walze proved this theorem in the normal coordinatessystem simultaneously in [11] . Ackermann proved that the Wodzicki residue of the square of the inverse ofthe Dirac operator Wres( D − ) in turn is essentially the second coefficient of the heat kernel expansion of D in [1].On the other hand, Wang generalized the Connes’ results to the case of manifolds with boundary in[19, 20], and proved the Kastler-Kalau-Walze type theorem for the Dirac operator and the signature operatoron lower-dimensional manifolds with boundary [21]. In [21, 22], Wang computed ] Wres[ π + D − ◦ π + D − ]and ] Wres[ π + D − ◦ π + D − ], where the two operators are symmetric, in these cases the boundary termvanished. But for ] Wres[ π + D − ◦ π + D − ], Wang got a nonvanishing boundary term [18], and give a theoreticalexplanation for gravitational action on boundary. In others words, Wang provides a kind of method to studythe Kastler-Kalau-Walze type theorem for manifolds with boundary. In [13], L´opez and his collaboratorsintroduced an elliptic differential operator which is called the Novikov operator. In [23], Wei and Wangproved Kastler-Kalau-Walze type theorem for modified Novikov operators on compact manifolds. In [27]and [28], Zhang introduced the sub-signature operators and proved a local index formula for these operators.In [2], Bao, Wang and Wang proved a local equivariant index theorem for sub-signature operators whichgeneralized the Zhang’s index theorem for sub-signature operators. In [5] and [14], by computing theadiabatic limit of eta-invariants associated to the so-called sub-signature operators, a new proof of theRiemann-Roch-Grothendieck type formula of Bismut-Lott was given. The motivation of this paper is to ∗ Corresponding author.
Email addresses: [email protected] (Tong Wu), [email protected] (Sining Wei), [email protected] (YongWang)
Preprint submitted to Elsevier February 9, 2021 rove the Kastler-Kalau-Walze type theorem for the sub-signature operators.The paper is organized in the following way. In Section 2, by using the definition of sub-signatureoperators, we compute the Lichnerowicz formulas for the sub-signature operators. In Section 3 and inSection 4, we prove the Kastler-Kalau-Walze type theorem for 4-dimensional and 6-dimensional manifoldswith boundary for the sub-signature operators respectively.
2. Sub-signature Operators and their Lichnerowicz formulas
Firstly we introduce some notations about sub-signature Operators. Let M be a n -dimensional ( n ≥ g M . And let F be a subbundle of T M , F ⊥ be the subbundle of T M orthogonal to F . Then we have the following orthogonal decomposition: T M = F M F ⊥ ,g T M = g F M g F ⊥ , (2.1)where g F and g F ⊥ be the induced metric on F and F ⊥ .Let ∇ L be the Levi-Civita connection about g M . In the local coordinates { x i ; 1 ≤ i ≤ n } and the fixedorthonormal frame { e , · · · , e n } , the connection matrix ( ω s,t ) is defined by ∇ L ( e , · · · , e n ) = ( e , · · · , e n )( ω s,t ) . (2.2)Let ǫ ( e j ∗ ), ι ( e j ∗ ) be the exterior and interior multiplications respectively and c ( e j ) be the Clifford action.Suppose that ∂ i is a natural local frame on T M and ( g ij ) ≤ i,j ≤ n is the inverse matrix associated to themetric matrix ( g ij ) ≤ i,j ≤ n on M . Write b c ( e j ) = ǫ ( e j ∗ ) + ι ( e j ∗ ); c ( e j ) = ǫ ( e j ∗ ) − ι ( e j ∗ ) . (2.3)which satisfies b c ( e i ) b c ( e j ) + b c ( e j ) b c ( e i ) = 2 g T M ( e i , e j ); c ( e i ) c ( e j ) + c ( e j ) c ( e i ) = − g T M ( e i , e j ); c ( e i ) b c ( e j ) + b c ( e j ) c ( e i ) = 0 . (2.4)By [25], we have D = d + δ = n X i =1 c ( e i ) (cid:20) e i + 14 X s,t ω s,t ( e i )[ b c ( e s ) b c ( e t ) − c ( e s ) c ( e t )] (cid:21) (2.5)Let π F (resp. π F ⊥ ) be the orthogonal projection from T M to F (resp. F ⊥ ). And let ∇ T M be theLevi-Civita connection of g T M . Set ∇ F = π F ∇ L π F , ∇ F ⊥ = π F ⊥ ∇ L π F ⊥ , (2.6)then ∇ F (resp. ∇ F ⊥ ) is a Euclidean connection on F (resp. F ⊥ ), let S be the tensor defined by ∇ L = ∇ F + ∇ F ⊥ + S. (2.7)2et e , e · ·· , e n be the orthonormal basis of F and f , · · · , f k be the orthonormal basis of F ⊥ . Thesub-signature Operators D t and D t ∗ acting on V ∗ T ∗ M N C are defined by D t = d + δ + t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α ) (2.8)= n X i =1 c ( e i ) (cid:20) e i + 14 X s,t ω s,t ( e i )[ b c ( e s ) b c ( e t ) − c ( e s ) c ( e t )] (cid:21) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ); D t ∗ = d + δ + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) (2.9)= n X i =1 c ( e i ) (cid:20) e i + 14 X s,t ω s,t ( e i )[ b c ( e s ) b c ( e t ) − c ( e s ) c ( e t )] (cid:21) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) . where b c ( e i ) = ( e i ) ∗ ∧ + ι ( e i ), c ( e i ) = ( e i ) ∗ ∧ − ι ( e i ), where e ∗ i = g ( e i , · ) and t is a complex number.Then when t = − ( − k , D t = ( √− − k ( k +1)2 ( − k ( k − b c ( f ) · · · b c ( f k ) e D F . where e D F is the sub-signature operator defined in Proposition 2 . D t is the sub-signature operator.Next by computation, we get Lichnerowicz formulas, Theorem 2.1.
The following equalities hold: D t ∗ D t = − h g ij ( ∇ ∂ i ∇ ∂ j − ∇ ∇ L∂i ∂ j ) i − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) (2.10)+ 14 X j [ tc ( e j ) n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j )] + 14 K + 12 [ tc ( e j ) ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) − t ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( e j )]+ tt [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] D t = − h g ij ( ∇ ∂ i ∇ ∂ j − ∇ ∇ L∂i ∂ j ) i − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) (2.11)+ 14 X j [ tc ( e j ) n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j )] + 14 K −
12 [ tc ( e j ) ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) − t ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( e j )]+ t [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] where K is the scalar curvature. M be a smooth compact oriented Riemannian n -dimensional manifolds without boundary and N bea vector bundle on M . If P is a differential operator of Laplace type, then it has locally the form P = − ( g ij ∂ i ∂ j + A i ∂ i + B ) , (2.12)where ∂ i is a natural local frame on T M and ( g ij ) ≤ i,j ≤ n is the inverse matrix associated to the metric matrix( g ij ) ≤ i,j ≤ n on M , and A i and B are smooth sections of End( N ) on M (endomorphism). If a Laplace typeoperator P satisfies (2.12), then there is a unique connection ∇ on N and a unique endomorphism E suchthat P = − h g ij ( ∇ ∂ i ∇ ∂ j − ∇ ∇ L∂i ∂ j ) + E i , (2.13)where ∇ L is the Levi-Civita connection on M . Moreover (with local frames of T ∗ M and N ), ∇ ∂ i = ∂ i + ω i and E are related to g ij , A i and B through ω i = 12 g ij (cid:0) A i + g kl Γ jkl id (cid:1) , (2.14) E = B − g ij (cid:0) ∂ i ( ω j ) + ω i ω j − ω k Γ kij (cid:1) , (2.15)where Γ jkl is the Christoffel coefficient of ∇ L .Let g ij = g ( dx i , dx j ), ξ = P k ξ j dx j and ∇ L∂ i ∂ j = P k Γ kij ∂ k , we denote that σ i = − X s,t ω s,t ( e i ) c ( e s ) c ( e t ); a i = 14 X s,t ω s,t ( e i ) b c ( e s ) b c ( e t ); ξ j = g ij ξ i ; Γ k = g ij Γ kij ; σ j = g ij σ i ; a j = g ij a i . (2.16)Then the sub-signature Operators D t and D t ∗ can be written as D t = n X i =1 c ( e i )[ e i + a i + σ i ] + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ); (2.17) D t ∗ = n X i =1 c ( e i )[ e i + a i + σ i ] + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) . (2.18)By [25] and [1], we have( d + δ ) = −△ − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) + 14 K. (2.19) − △ = ∆ = − g ij ( ∇ Li ∇ Lj − Γ kij ∇ Lk ) . (2.20)By (2.8) and (2.9), we have D t ∗ D t = ( d + δ ) + ( d + δ )[ t n X i =1 κ X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]+ [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( d + δ ) + tt [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] , (2.21)4 d + δ )[ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] + [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( d + δ )= X i,j g i,j h c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) i ∂ j + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + c ( ∂ i ) ∂ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) − c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i , (2.22)then we obtain D t ∗ D t = − X i,j g i,j h ∂ i ∂ j + 2 σ i ∂ j + 2 a i ∂ j − Γ ki,j ∂ k + ( ∂ i σ j ) + ( ∂ i a j ) + σ i σ j + σ i a j + a i σ j + a i a j − Γ ki,j σ k − Γ ki,j a k i + X i,j g i,j h c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) i ∂ j + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + c ( ∂ i ) ∂ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) − c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) + 14 K + tt [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] . (2.23)5imilarly, we have D t = − X i,j g i,j h ∂ i ∂ j + 2 σ i ∂ j + 2 a i ∂ j − Γ ki,j ∂ k + ( ∂ i σ j ) + ( ∂ i a j ) + σ i σ j + σ i a j + a i σ j + a i a j − Γ ki,j σ k − Γ ki,j a k i + X i,j g i,j h c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) i ∂ j + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + c ( ∂ i ) ∂ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) + 14 K + t [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] . (2.24)By (2.12), (2.13), (2.14) and (2.24), we have( ω i ) D t ∗ D t = σ i + a i − h c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) i , (2.25)6 D t ∗ D t = − c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) − c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) − K + 18 X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i − t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + 12 ∂ j [ c ( ∂ j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ j )] −
12 [ c ( ∂ j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ j )]( σ j + a j ) − g ij c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i )] · [ c ( ∂ j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ j )] −
12 Γ k [ c ( ∂ k ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ k )] −
12 ( σ j + a j )[ c ( ∂ j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ j )] − tt [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] − c ( ∂ i ) ∂ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) . (2.26)Since E is globally defined on M , taking normal coordinates at x , we have σ i ( x ) = 0, a i ( x ) = 0, ∂ j [ c ( ∂ j )]( x ) = 0, Γ k ( x ) = 0, g ij ( x ) = δ ji , then E D t ∗ D t ( x ) = 18 X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − K − tt [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] − X j [ c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j )] −
12 [ tc ( e j ) ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) − t ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( e j )] . (2.27)7imilarly, we have E D t ( x ) = 18 X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − K − t [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] − X j [ c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j )] + 12 [ t ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( e j ) − tc ( e j ) ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))] , (2.28)by (2.7),we get Theorem 2.1.From [1], we konw that the non-commutative residue of a generalized laplacian ∆ is expressed as( n − (∆) = (4 π ) − n Γ( n f res (∆ − n +1 ) , (2.29)where Φ (∆) denotes the integral over the diagonal part of the second coefficient of the heat kernel expansionof ∆. Now let ∆ = D t ∗ D t . Since D t ∗ D t is a generalized laplacian , we can suppose D t ∗ D t = ∆ − E , then,we have Wres( D t ∗ D t ) − n − = ( n − π ) n ( n − Z M tr( 16 K + E D t ∗ D t ) d Vol M , (2.30)Wres( D t ) − n − = ( n − π ) n ( n − Z M tr( 16 K + E D t ) d Vol M , (2.31)where Wres denote the noncommutative residue. By computation, tr (cid:18) [ n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] (cid:19) = n X i =1 n X j =1 k X α =1 k X β =1 tr (cid:18) c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) b c ( S ( e j ) f β ) b c ( f β ) (cid:19) = n X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )tr[ id ] . (2.32) tr (cid:18) X j [ c ( e j ) n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] (cid:19) = ( n − n X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )tr[ id ] . (2.33) tr (cid:18) c ( e j ) ∇ V ∗ T ∗ Me j ( n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) (cid:19) = n X i =1 k X α =1 tr (cid:18) c ( e j ) ∇ V ∗ T ∗ Me j ( c ( e i )) b c ( S ( e i ) f α ) b c ( f α ) (cid:19) + n X i =1 k X α =1 tr (cid:18) c ( e j ) c ( e i ) ∇ V ∗ T ∗ Me j ( b c ( S ( e i ) f α )) b c ( f α ) (cid:19) + n X i =1 k X α =1 tr (cid:18) c ( e j ) c ( e i ) b c ( S ( e i ) f α ) ∇ V ∗ T ∗ Me j ( b c ( f α )) (cid:19) = − [ n X i =1 k X α =1 g ( f α , ∇ T Me i S ( e i ) f α ) + n X i =1 k X α =1 g ( ∇ T Me i f α , S ( e i ) f α )]tr[ id ] . (2.34)8hen by (2.32), (2.33) and (2.34), we get tr ( E D t ∗ D t ) = (cid:20) − K − [ 14 ( t + t − tt )( n −
2) + tt ] n X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )+ 12 ( t − t )[ n X i =1 k X α =1 g ( f α , ∇ T Me i S ( e i ) f α ) + n X i =1 k X α =1 g ( ∇ T Me i f α , S ( e i ) f α )] (cid:21) tr[ id ] , (2.35) tr ( E D t ) = [ − K − t n X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )]tr[ id ] . (2.36) Theorem 2.2. If M is a n -dimensional compact oriented manifolds without boundary, and n is even, thenwe get the following equalities : Wres( D t ∗ D t ) − n − = ( n − π ) n ( n − Z M n (cid:18) − K − [ 14 ( t + t − tt )( n −
2) + tt ] n X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )+ 12 ( t − t )[ k X α =1 g ( f α , ∇ T Me j S ( e j ) f α ) + k X α =1 g ( ∇ T Me j f α , S ( e j ) f α )] (cid:19) d Vol M . (2.37)Wres( D t ) − n − = ( n − π ) n ( n − Z M n (cid:18) − K − t n X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α ) (cid:19) d Vol M . (2.38) where K is the scalar curvature.
3. A Kastler-Kalau-Walze type theorem for 4-dimensional manifolds with boundary
We firstly recall that some basic facts and formulas about Boutet de Monvel’s calculus and the definitionof the noncommutative residue for manifolds with boundary which will be used in the following. For moredetails, (see in Section 2 in [21]).Let U ⊂ M be a collar neighborhood of ∂M which is diffeomorphic with ∂M × [0 , h ( x n ) ∈ C ∞ ([0 , h ( x n ) >
0, there exists b h ∈ C ∞ (( − ε, b h | [0 , = h and b h > ε >
0. Then there exists a metric g ′ on f M = M S ∂M ∂M × ( − ε,
0] which has the form on U S ∂M ∂M × ( − ε, g ′ = 1 b h ( x n ) g ∂M + dx n , (3.1)such that g ′ | M = g . We fix a metric g ′ on the f M such that g ′ | M = g .Let Fourier transformation F ′ be F ′ : L ( R t ) → L ( R v ); F ′ ( u )( v ) = Z e − ivt u ( t ) dt and let r + : C ∞ ( R ) → C ∞ ( g R + ); f → f | g R + ; g R + = { x ≥ x ∈ R } . where Φ( R ) denotes the Schwartz space and Φ( g R + ) = r + Φ( R ), Φ( g R − ) = r − Φ( R ).We define H + = F ′ (Φ( g R + )); H − = F ′ (Φ( g R − )) which satisfies H + ⊥ H − . We have the following prop-erty: h ∈ H + ( H − ) if and only if h ∈ C ∞ ( R ) which has an analytic extension to the lower (upper) complexhalf-plane { Im ξ < } ( { Im ξ > } ) such that for all nonnegative integer l , d l hdξ l ( ξ ) ∼ ∞ X k =1 d l dξ l ( c k ξ k ) , | ξ | → + ∞ , Im ξ ≤ ξ ≥ H ′ be the space of all polynomials and H − = H − L H ′ ; H = H + L H − . Denote by π + ( π − )respectively the projection on H + ( H − ). For calculations, we take H = e H = { rational functions having nopoles on the real axis } ( ˜ H is a dense set in the topology of H ). Then on ˜ H , π + h ( ξ ) = 12 πi lim u → − Z Γ + h ( ξ ) ξ + iu − ξ dξ, (3.2)where Γ + is a Jordan close curve included Im( ξ ) > h in the upperhalf-plane and ξ ∈ R . Similarly, define π ′ on ˜ H , π ′ h = 12 π Z Γ + h ( ξ ) dξ. (3.3)So, π ′ ( H − ) = 0. For h ∈ H T L ( R ), π ′ h = π R R h ( v ) dv and for h ∈ H + T L ( R ), π ′ h = 0.Let M be a n -dimensional compact oriented manifold with boundary ∂M . Denote by B Boutet deMonvel’s algebra, we recall the main theorem in [6, 21].
Theorem 3.1. [6] (Fedosov-Golse-Leichtnam-Schrohe)
Let X and ∂X be connected, dim X = n ≥ , A = (cid:18) π + P + G KT S (cid:19) ∈ B , and denote by p , b and s the local symbols of P, G and S respectively. Define: ] Wres( A ) = Z X Z S tr E [ p − n ( x, ξ )] σ ( ξ ) dx + 2 π Z ∂X Z S ′ { tr E [(tr b − n )( x ′ , ξ ′ )] + tr F [ s − n ( x ′ , ξ ′ )] } σ ( ξ ′ ) dx ′ , (3.4) Then a) ] Wres([
A, B ]) = 0 , for any
A, B ∈ B ; b) It is a unique continuous trace on B / B −∞ . Definition 3.2. [21]
Lower dimensional volumes of spin manifolds with boundary are defined by
Vol ( p ,p ) n M := ] Wres[ π + D − p ◦ π + D − p ] , (3.5)By [21], we get ] Wres[ π + D − p ◦ π + D − p ] = Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − n ( D − p − p )] σ ( ξ ) dx + Z ∂M Φ , (3.6)and Φ = R | ξ ′ | =1 R + ∞−∞ P ∞ j,k =0 P ( − i ) | α | + j + k +1 α !( j + k +1)! × trace ∧ ∗ T ∗ M N C [ ∂ jx n ∂ αξ ′ ∂ kξ n σ + r ( D − p )( x ′ , , ξ ′ , ξ n ) × ∂ αx ′ ∂ j +1 ξ n ∂ kx n σ l ( D − p )( x ′ , , ξ ′ , ξ n )] dξ n σ ( ξ ′ ) dx ′ , (3.7)where the sum is taken over r + l − k − | α | − j − − n, r ≤ − p , l ≤ − p .Since [ σ − n ( D − p − p )] | M has the same expression as σ − n ( D − p − p ) in the case of manifolds withoutboundary, so locally we can compute the first term by [12], [11], [21], [15].For any fixed point x ∈ ∂M , we choose the normal coordinates U of x in ∂M (not in M ) and computeΦ( x ) in the coordinates e U = U × [0 , ⊂ M and the metric h ( x n ) g ∂M + dx n . The dual metric of g M on e U is h ( x n ) g ∂M + dx n . Write g Mij = g M ( ∂∂x i , ∂∂x j ); g ijM = g M ( dx i , dx j ), then[ g Mi,j ] = (cid:20) h ( x n ) [ g ∂Mi,j ] 00 1 (cid:21) ; [ g i,jM ] = (cid:20) h ( x n )[ g i,j∂M ] 00 1 (cid:21) , (3.8)and ∂ x s g ∂Mij ( x ) = 0 , ≤ i, j ≤ n − g Mij ( x ) = δ ij . (3.9)From [21], we can get three lemmas. 10 emma 3.3. [21] With the metric g M on M near the boundary ∂ x j ( | ξ | g M )( x ) = ( , if j < n,h ′ (0) | ξ ′ | g ∂M , if j = n, (3.10) ∂ x j [ c ( ξ )]( x ) = (cid:26) , if j < n,∂x n ( c ( ξ ′ ))( x ) , if j = n, (3.11) where ξ = ξ ′ + ξ n dx n . Lemma 3.4. [21]
With the metric g M on M near the boundary ω s,t ( e i )( x ) = ω n,i ( e i )( x ) = h ′ (0) , if s = n, t = i, i < n,ω i,n ( e i )( x ) = − h ′ (0) , if s = i, t = n, i < n,ω s,t ( e i )( x ) = 0 , other cases, (3.12) where ( ω s,t ) denotes the connection matrix of Levi-Civita connection ∇ L . Lemma 3.5. [21]Γ kst ( x ) = Γ nii ( x ) = h ′ (0) if s = t = i, k = n, i < n, Γ ini ( x ) = − h ′ (0) , if s = n, t = i, k = i, i < n, Γ iin ( x ) = − h ′ (0) , if s = i, t = n, k = i, i < n, Γ ist ( x ) = 0 , other cases. (3.13)By (3.6) and (3.7), we firstly compute ] Wres[ π + D t − ◦ π + ( D t ∗ ) − ] = Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − (( D t ∗ D t ) − )] σ ( ξ ) dx + Z ∂M Φ , (3.14)where Φ = Z | ξ ′ | =1 Z + ∞−∞ ∞ X j,k =0 X ( − i ) | α | + j + k +1 α !( j + k + 1)! × trace ∧ ∗ T ∗ M N C [ ∂ jx n ∂ αξ ′ ∂ kξ n σ + r ( D t − )( x ′ , , ξ ′ , ξ n ) × ∂ αx ′ ∂ j +1 ξ n ∂ kx n σ l (( D t ∗ ) − )( x ′ , , ξ ′ , ξ n )] dξ n σ ( ξ ′ ) dx ′ , (3.15)and the sum is taken over r + l − k − j − | α | = − , r ≤ − , l ≤ − ] Wres[ π + D t − ◦ π + ( D t ∗ ) − ], so Z M Z | ξ | =1 trace ∧ ∗ T ∗ M [ σ − (( D t ∗ D t ) − )] σ ( ξ ) dx = 32 π Z M (cid:18) − K − t + t ) X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )+ 8( t − t )[ n X i =1 k X α =1 g ( f α , ∇ T Me i S ( e i ) f α ) + n X i =1 k X α =1 g ( ∇ T Me i f α , S ( e i ) f α )] (cid:19) d Vol M . (3.16)Now we need to compute R ∂M Φ. Since, some operators have the following symbols.11 emma 3.6.
The following identities hold: σ ( D t ) = σ ( D t ∗ ) = ic ( ξ ); σ ( D t ) = 14 X i,s,t ω s,t ( e i ) c ( e i ) b c ( e s ) b c ( e t ) − X i,s,t ω s,t ( e i ) c ( e i ) b c ( e s ) b c ( e t ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ); σ ( D t ∗ ) = 14 X i,s,t ω s,t ( e i ) c ( e i ) b c ( e s ) b c ( e t ) − X i,s,t ω s,t ( e i ) c ( e i ) b c ( e s ) b c ( e t ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) . (3.17)Write D αx = ( − i ) | α | ∂ αx ; σ ( D t ) = p + p ; ( σ ( D t ) − ) = P ∞ j =1 q − j . (3.18)By the composition formula of pseudodifferential operators, we have1 = σ ( D t ◦ D t − ) = X α α ! ∂ αξ [ σ ( D t )] D tαx [ σ ( D t − )]= ( p + p )( q − + q − + q − + · · · )+ X j ( ∂ ξ j p + ∂ ξ j p )( D x j q − + D x j q − + D x j q − + · · · )= p q − + ( p q − + p q − + X j ∂ ξ j p D x j q − ) + · · · , (3.19)so q − = p − ; q − = − p − [ p p − + X j ∂ ξ j p D x j ( p − )] . (3.20) Lemma 3.7.
The following identities hold: σ − ( D t − ) = σ − (( D t ∗ ) − ) = ic ( ξ ) | ξ | ; σ − ( D t − ) = c ( ξ ) σ ( D t − ) c ( ξ ) | ξ | + c ( ξ ) | ξ | X j c ( dx j ) h ∂ x j ( c ( ξ )) | ξ | − c ( ξ ) ∂ x j ( | ξ | ) i ; σ − (( D t ∗ ) − ) = c ( ξ ) σ (( D t ∗ ) − ) c ( ξ ) | ξ | + c ( ξ ) | ξ | X j c ( dx j ) h ∂ x j ( c ( ξ )) | ξ | − c ( ξ ) ∂ x j ( | ξ | ) i . (3.21)When n = 4, then tr ∧ ∗ T ∗ M [ id ] = dim( ∧ ∗ (4)) = 16, where tr as shorthand of trace, the sum is taken over r + l − k − j − | α | = − , r ≤ − , l ≤ − , then we have the following five cases: case a) I) r = − , l = − , k = j = 0 , | α | = 1By (3.15), we getΦ = − Z | ξ ′ | =1 Z + ∞−∞ X | α | =1 tr[ ∂ αξ ′ π + ξ n σ − ( D t − ) × ∂ αx ′ ∂ ξ n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.22)By Lemma 3.3, for i < n , then ∂ x i (cid:18) ic ( ξ ) | ξ | (cid:19) ( x ) = i∂ x i [ c ( ξ )]( x ) | ξ | − ic ( ξ ) ∂ x i ( | ξ | )( x ) | ξ | = 0 , (3.23)12o Φ = 0. case a) II) r = − , l = − , k = | α | = 0 , j = 1By (3.15), we getΦ = − Z | ξ ′ | =1 Z + ∞−∞ trace[ ∂ x n π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.24)By Lemma 3.7, we have ∂ ξ n σ − (( D t ∗ ) − )( x ) = i (cid:18) − ξ n c ( dx n ) + 2 c ( ξ ′ ) | ξ | + 8 ξ n c ( ξ ) | ξ | (cid:19) ; (3.25) ∂ x n σ − ( D t − )( x ) = i∂ x n c ( ξ ′ )( x ) | ξ | − ic ( ξ ) | ξ ′ | h ′ (0) | ξ | . (3.26)By (3.2), (3.3) and the Cauchy integral formula we have π + ξ n (cid:20) c ( ξ ) | ξ | (cid:21) ( x ) | | ξ ′ | =1 = π + ξ n (cid:20) c ( ξ ′ ) + ξ n c ( dx n )(1 + ξ n ) (cid:21) = 12 πi lim u → − Z Γ + c ( ξ ′ )+ η n c ( dx n )( η n + i ) ( ξ n + iu − η n ) ( η n − i ) dη n = − ( iξ n + 2) c ( ξ ′ ) + ic ( dx n )4( ξ n − i ) . (3.27)Similarly we have, π + ξ n (cid:20) i∂ x n c ( ξ ′ ) | ξ | (cid:21) ( x ) | | ξ ′ | =1 = ∂ x n [ c ( ξ ′ )]( x )2( ξ n − i ) . (3.28)By (3.25), then π + ξ n ∂ x n σ − ( D t − ) | | ξ ′ | =1 = ∂ x n [ c ( ξ ′ )]( x )2( ξ n − i ) + ih ′ (0) (cid:20) ( iξ n + 2) c ( ξ ′ ) + ic ( dx n )4( ξ n − i ) (cid:21) . (3.29)By the relation of the Clifford action and tr AB = tr BA , we have the equalities:tr[ c ( ξ ′ ) c ( dx n )] = 0; tr[ c ( dx n ) ] = −
16; tr[ c ( ξ ′ ) ]( x ) | | ξ ′ | =1 = − ∂ x n c ( ξ ′ ) c ( dx n )] = 0; tr[ ∂ x n c ( ξ ′ ) c ( ξ ′ )]( x ) | | ξ ′ | =1 = − h ′ (0);tr[ b c ( e i ) b c ( e j ) c ( e k ) c ( e l )] = 0( i = j ) . (3.30)By (3.27), we have h ′ (0)tr (cid:20) ( iξ n +2) c ( ξ ′ )+ ic ( dx n )4( ξ n − i ) × (cid:18) ξ n c ( dx n )+2 c ( ξ ′ )(1+ ξ n ) − ξ n [ c ( ξ ′ )+ ξ n c ( dx n )](1+ ξ n ) (cid:19)(cid:21) ( x ) | | ξ ′ | =1 = − h ′ (0) − iξ n − ξ n + i ( ξ n − i ) ( ξ n + i ) . (3.31)13imilarly, we have − i tr (cid:20)(cid:18) ∂ x n [ c ( ξ ′ )]( x )2( ξ n − i ) (cid:19) × (cid:18) ξ n c ( dx n ) + 2 c ( ξ ′ )(1 + ξ n ) − ξ n [ c ( ξ ′ ) + ξ n c ( dx n )](1 + ξ n ) (cid:19)(cid:21) ( x ) | | ξ ′ | =1 = − ih ′ (0) 3 ξ n − ξ n − i ) ( ξ n + i ) . (3.32)Then Φ = − Z | ξ ′ | =1 Z + ∞−∞ ih ′ (0)( ξ n − i ) ( ξ n − i ) ( ξ n + i ) dξ n σ ( ξ ′ ) dx ′ (3.33)= − ih ′ (0)Ω Z Γ + ξ n − i ) ( ξ n + i ) dξ n dx ′ (3.34)= − ih ′ (0)Ω πi [ 1( ξ n + i ) ] (1) | ξ n = i dx ′ (3.35)= − πh ′ (0)Ω dx ′ , (3.36)where Ω is the canonical volume of S . case a) III) r = − , l = − , j = | α | = 0 , k = 1By (3.15), we getΦ = − Z | ξ ′ | =1 Z + ∞−∞ trace[ ∂ ξ n π + ξ n σ − ( D t − ) × ∂ ξ n ∂ x n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.37)By Lemma 3.7, we have ∂ ξ n ∂ x n σ − (( D t ∗ ) − )( x ) | | ξ ′ | =1 = − ih ′ (0) (cid:20) c ( dx n ) | ξ | − ξ n c ( ξ ′ ) + ξ n c ( dx n ) | ξ | (cid:21) − ξ n i∂ x n c ( ξ ′ )( x ) | ξ | ; (3.38) ∂ ξ n π + ξ n σ − ( D t − )( x ) | | ξ ′ | =1 = − c ( ξ ′ ) + ic ( dx n )2( ξ n − i ) . (3.39)Similar to case a) II), we havetr (cid:26) c ( ξ ′ ) + ic ( dx n )2( ξ n − i ) × ih ′ (0) (cid:20) c ( dx n ) | ξ | − ξ n c ( ξ ′ ) + ξ n c ( dx n ) | ξ | (cid:21)(cid:27) = 8 h ′ (0) i − ξ n ( ξ n − i ) ( ξ n + i ) (3.40)and tr (cid:20) c ( ξ ′ ) + ic ( dx n )2( ξ n − i ) × ξ n i∂ x n c ( ξ ′ )( x ) | ξ | (cid:21) = − ih ′ (0) ξ n ( ξ n − i ) ( ξ n + i ) . (3.41)So we haveΦ = − Z | ξ ′ | =1 Z + ∞−∞ h ′ (0)4( i − ξ n )( ξ n − i ) ( ξ n + i ) dξ n σ ( ξ ′ ) dx ′ − Z | ξ ′ | =1 Z + ∞−∞ h ′ (0)4 iξ n ( ξ n − i ) ( ξ n + i ) dξ n σ ( ξ ′ ) dx ′ = − h ′ (0)Ω πi
3! [ 4( i − ξ n )( ξ n + i ) ] (3) | ξ n = i dx ′ + h ′ (0)Ω πi
3! [ 4 iξ n ( ξ n + i ) ] (3) | ξ n = i dx ′ = 32 πh ′ (0)Ω dx ′ . (3.42)14 ase b) r = − , l = − , k = j = | α | = 0By (3.15), we getΦ = − i Z | ξ ′ | =1 Z + ∞−∞ trace[ π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.43)By Lemma 3.7 we have σ − ( D t − )( x ) = c ( ξ ) σ ( D t )( x ) c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0) | ξ | ∂M ] , (3.44)where σ ( D t )( x ) = 14 X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t ) − X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t )+ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) . (3.45)We denote Q ( x ) = 14 X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t ); Q ( x ) = − X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t ) . (3.46)Then π + ξ n σ − ( D t − ( x )) | | ξ ′ | =1 = π + ξ n h c ( ξ ) Q ( x ) c ( ξ )(1 + ξ n ) i + π + ξ n h c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( ξ )(1 + ξ n ) i + π + ξ n h c ( ξ ) Q ( x ) c ( ξ ) + c ( ξ ) c ( dx n ) ∂ x n [ c ( ξ ′ )]( x )(1 + ξ n ) − h ′ (0) c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) i . (3.47)By computation, we have π + ξ n h c ( ξ ) Q ( x ) c ( ξ )(1 + ξ n ) i = π + ξ n h c ( ξ ′ ) Q ( x ) c ( ξ ′ )(1 + ξ n ) i + π + ξ n h ξ n c ( ξ ′ ) Q ( x ) c ( dx n )(1 + ξ n ) i + π + ξ n h ξ n c ( dx n ) Q ( x ) c ( ξ ′ )(1 + ξ n ) i + π + ξ n h ξ n c ( dx n ) Q ( x ) c ( dx n )(1 + ξ n ) i = − c ( ξ ′ ) Q ( x ) c ( ξ ′ )(2 + iξ n )4( ξ n − i ) + ic ( ξ ′ ) Q ( x ) c ( dx n )4( ξ n − i ) + ic ( dx n ) Q ( x ) c ( ξ ′ )4( ξ n − i ) + − iξ n c ( dx n ) Q ( x ) c ( dx n )4( ξ n − i ) . (3.48)Since c ( dx n ) Q ( x ) = − h ′ (0) n − X i =1 c ( e i ) b c ( e i ) c ( e n ) b c ( e n ) , (3.49)15hen by the relation of the Clifford action and tr AB = tr BA , we have the equalities:tr[ c ( e i ) b c ( e i ) c ( e n ) b c ( e n )] = 0 ( i < n ); tr[ Q c ( dx n )] = 0;tr[ t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α ) c ( dx n )] = 0; tr[ b c ( ξ ′ ) b c ( dx n )] = 0 . (3.50)Since ∂ tξ n σ − (( D t ∗ ) − ) = ∂ ξ n q − ( x ) | | ξ ′ | =1 = i (cid:20) c ( dx n )1 + ξ n − ξ n c ( ξ ′ ) + 2 ξ n c ( dx n )(1 + ξ n ) (cid:21) . (3.51)By (3.47) and (3.50), we havetr[ π + ξ n h c ( ξ ) Q ( x ) c ( ξ )(1 + ξ n ) i × ∂ ξ n σ − (( D t ∗ ) − )( x )] | | ξ ′ | =1 = 12(1 + ξ n ) tr[ c ( ξ ′ ) Q ( x )] + i ξ n ) tr[ c ( dx n ) Q ( x )]= 12(1 + ξ n ) tr[ c ( ξ ′ ) Q ( x )] . (3.52)We note that i < n, R | ξ ′ | =1 { ξ i ξ i · · · ξ i d +1 } σ ( ξ ′ ) = 0, so tr[ c ( ξ ′ ) Q ( x )] has no contribution for computingcase b).By computation, we have π + ξ n h c ( ξ ) Q ( x ) c ( ξ ) + c ( ξ ) c ( dx n ) ∂ x n [ c ( ξ ′ )]( x )(1 + ξ n ) i − h ′ (0) π + ξ n h c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) i := C − C , (3.53)where C = − ξ n − i ) [(2 + iξ n ) c ( ξ ′ ) Q ( x ) c ( ξ ′ ) + iξ n c ( dx n ) Q ( x ) c ( dx n )+ (2 + iξ n ) c ( ξ ′ ) c ( dx n ) ∂ x n c ( ξ ′ ) + ic ( dx n ) Q ( x ) c ( ξ ′ ) + ic ( ξ ′ ) Q ( x ) c ( dx n ) − i∂ x n c ( ξ ′ )] (3.54)and C = h ′ (0)2 (cid:20) c ( dx n )4 i ( ξ n − i ) + c ( dx n ) − ic ( ξ ′ )8( ξ n − i ) + 3 ξ n − i ξ n − i ) [ ic ( ξ ′ ) − c ( dx n )] (cid:21) . (3.55)By (3.50) and (3.54), we havetr[ C × ∂ ξ n σ − (( D t ∗ ) − )] | | ξ ′ | =1 = i h ′ (0) − iξ n − ξ n +4 i ξ n − i ) ( ξ n + i ) tr[ id ]= 8 ih ′ (0) − iξ n − ξ n +4 i ξ n − i ) ( ξ n + i ) . (3.56)By (3.50) and (3.53), we havetr[ C × ∂ ξ n σ − (( D t ∗ ) − )] | | ξ ′ | =1 = − ic (1 + ξ n ) + 2 h ′ (0) ξ n − iξ n − ξ n − i )(1 + ξ n ) , (3.57)where Q = c c ( dx n ) and c = − h ′ (0). 16y (3.55) and (3.56), we have − i Z | ξ ′ | =1 Z + ∞−∞ trace[( C − C ) × ∂ ξ n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ = − Ω Z Γ + c ( ξ n − i ) + ih ′ (0)( ξ n − i ) ( ξ n + i ) dξ n dx ′ = 92 πh ′ (0)Ω dx ′ . (3.58)Similar to (3.53), we havetr[ π + ξ n h c ( ξ ) t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )( x ) c ( ξ )(1 + ξ n ) i × ∂ ξ n σ − (( D t ∗ ) − )( x )] | | ξ ′ | =1 = i ξ n ) tr[ c ( dx n ) t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )( x )] . (3.59)By (3.58), we have − i Z | ξ ′ | =1 Z + ∞−∞ trace[ π + ξ n h c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )) c ( ξ )(1 + ξ n ) i × ∂ ξ n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ = π c ( dx n ) t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]Ω dx ′ = 0 . (3.60)Then, we have Φ = 92 πh ′ (0)Ω dx ′ . (3.61) case c) r = − , l = − , k = j = | α | = 0By (3.15), we getΦ = − i Z | ξ ′ | =1 Z + ∞−∞ trace[ π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.62)By (3.2) and (3.3), Lemma 3.7, we have π + ξ n σ − ( D t − ) | | ξ ′ | =1 = c ( ξ ′ ) + ic ( dx n )2( ξ n − i ) . (3.63)Since σ − (( D t ∗ ) − )( x ) = c ( ξ ) σ ( D t ∗ )( x ) c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n ) (cid:20) ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0) | ξ | ∂ M (cid:21) , (3.64)where σ ( D t ∗ )( x ) = 14 X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t ) − X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t )+ ( t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α ))( x )= Q ( x ) + Q ( x ) + ( t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α ))( x ) , (3.65)17hen ∂ ξ n σ − (( D t ∗ ) − )( x ) | | ξ ′ | =1 = ∂ ξ n (cid:26) c ( ξ )[ Q ( x ) + Q ( x ) + ( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x )] c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0)] (cid:27) = ∂ ξ n (cid:26) [ c ( ξ ) Q ( x )] c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0)] (cid:27) + ∂ ξ n c ( ξ ) Q ( x ) c ( ξ ) | ξ | + ∂ ξ n c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( ξ ) | ξ | . (3.66)By computation, we have ∂ ξ n c ( ξ ) Q ( x ) c ( ξ ) | ξ | = c ( dx n ) Q ( x ) c ( ξ ) | ξ | + c ( ξ ) Q ( x ) c ( dx n ) | ξ | − ξ n c ( ξ ) Q ( x ) c ( ξ ) | ξ | ; (3.67) ∂ ξ n c ( ξ )( t P ni =1 P κα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( ξ ) | ξ | = c ( dx n )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( ξ ) | ξ | + c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( dx n ) | ξ | − ξ n c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( ξ ) | ξ | . (3.68)We denote q − = c ( ξ ) Q ( x ) c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0)] , then ∂ ξ n ( q − ) = 1(1 + ξ n ) (cid:20) (2 ξ n − ξ n ) c ( dx n ) Q c ( dx n ) + (1 − ξ n ) c ( dx n ) Q c ( ξ ′ )+ (1 − ξ n ) c ( ξ ′ ) Q c ( dx n ) − ξ n c ( ξ ′ ) Q c ( ξ ′ ) + (3 ξ n − ∂ tx n c ( ξ ′ ) − ξ n c ( ξ ′ ) c ( dx n ) ∂ tx n c ( ξ ′ ) + 2 h ′ (0) c ( ξ ′ ) + 2 h ′ (0) ξ n c ( dx n ) (cid:21) + 6 ξ n h ′ (0) c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) ; (3.69)By (3.62) and (3.66), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ ) Q c ( ξ ) | ξ | ]( x ) | | ξ ′ | =1 = − ξ − i )( ξ + i ) tr[ c ( ξ ′ ) Q ( x )] + i ( ξ − i )( ξ + i ) tr[ c ( dx n ) Q ( x )] . (3.70)By (3.49), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ ) Q c ( ξ ) | ξ | ]( x ) | | ξ ′ | =1 = − ξ − i )( ξ + i ) tr[ c ( ξ ′ ) Q ( x )] . (3.71)We note that i < n, R | ξ ′ | =1 { ξ i ξ i · · · ξ i d +1 } σ ( ξ ′ ) = 0, so tr[ c ( ξ ′ ) Q ( x )] has no contribution for computingcase c). 18y (3.61) and (3.68), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n ( q − )]( x ) | | ξ ′ | =1 = 12 h ′ (0)( iξ n + ξ n − i )( ξ − i ) ( ξ + i ) + 48 h ′ (0) iξ n ( ξ − i ) ( ξ + i ) , (3.72)then − i Ω Z Γ + [ 12 h ′ (0)( iξ n + ξ n − i )( ξ n − i ) ( ξ n + i ) + 48 h ′ (0) iξ n ( ξ n − i ) ( ξ n + i ) ] dξ n dx ′ = − πh ′ (0)Ω dx ′ . (3.73)By (3.61) and (3.66), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( ξ ) | ξ | ]( x ) | | ξ ′ | =1 = − ξ − i )( ξ + i ) tr[ c ( ξ ′ )( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x )]+ i ( ξ − i )( ξ + i ) tr[ c ( dx n )( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x )] . (3.74)By R | ξ ′ | =1 { ξ · · · ξ d +1 } σ ( ξ ′ ) = 0 and (3.49), we have − i Z | ξ ′ | =1 Z + ∞−∞ tr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( ξ ) | ξ | ]( x ) dξ n σ ( ξ ′ ) dx ′ = − i Z | ξ ′ | =1 Z + ∞−∞ i ( ξ − i )( ξ + i ) tr[ c ( dx n )( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))]( x ) dξ n σ ( ξ ′ ) dx ′ = − π c ( dx n )( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))]Ω dx ′ = 0 . (3.75)Then, Φ = − πh ′ (0)Ω dx ′ . (3.76)So Φ = Φ + Φ + Φ + Φ + Φ = 0. Theorem 3.8.
Let M be a -dimensional oriented compact manifolds with the boundary ∂M and the metric g M as above, D t and D t ∗ be sub-signature operators on f M , then ] Wres[ π + D t − ◦ π + ( D t ∗ ) − ] = 32 π Z M (cid:18) − K − t + t )( n − X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )+ 8( t − t )[ X i =1 k X α =1 g ( f α , ∇ T Me i S ( e i ) f α ) + X i =1 k X α =1 g ( ∇ T Me i f α , S ( e i ) f α )] (cid:19) d Vol M . (3.77) where K is the scalar curvature. Next, we also prove the Kastler-Kalau-Walze type theorem for 4-dimensional manifolds with boundaryassociated to D t . By (3.6) and (3.7), we will compute ] Wres[ π + D t − ◦ π + D t − ] = Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − ( D t − )] σ ( ξ ) dx + Z ∂M Φ , (3.78)19here Φ = Z | ξ ′ | =1 Z + ∞−∞ ∞ X j,k =0 X ( − i ) | α | + j + k +1 α !( j + k + 1)! × trace ∧ ∗ T ∗ M N C [ ∂ jx n ∂ αξ ′ ∂ kξ n σ + r ( D t − )( x ′ , , ξ ′ , ξ n ) × ∂ αx ′ ∂ j +1 ξ n ∂ kx n σ l ( D t − )( x ′ , , ξ ′ , ξ n )] dξ n σ ( ξ ′ ) dx ′ , (3.79)and the sum is taken over r + l − k − j − | α | = − , r ≤ − , l ≤ − ] Wres[ π + D t − ◦ π + D t − ], then Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − ( D t − )] σ ( ξ ) dx = 32 π Z M (cid:18) − K − t X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α ) (cid:19) d Vol M . (3.80)When n = 4, then tr ∧ ∗ T ∗ M [ id ] = dim( ∧ ∗ (4)) = 16, where tr as shorthand of trace, the sum is taken over r + l − k − j − | α | = − , r ≤ − , l ≤ − , then we have the following five cases: case a) I) r = − , l = − , k = j = 0 , | α | = 1By (3.78), we getΦ = − Z | ξ ′ | =1 Z + ∞−∞ X | α | =1 tr[ ∂ αξ ′ π + ξ n σ − ( D t − ) × ∂ αx ′ ∂ ξ n σ − ( D t − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.81) case a) II) r = − , l = − , k = | α | = 0 , j = 1By (3.78), we getΦ = − Z | ξ ′ | =1 Z + ∞−∞ trace[ ∂ x n π + ξ n σ − ( D t − ) × ∂ ξ n σ − ( D t − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.82) case a) III) r = − , l = − , j = | α | = 0 , k = 1By (3.78), we getΦ = − Z | ξ ′ | =1 Z + ∞−∞ trace[ ∂ ξ n π + ξ n σ − ( D t − ) × ∂ ξ n ∂ x n σ − ( D t − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.83)By Lemma 3.7, we have σ − ( D t − ) = σ − (( D t ∗ ) − ). Similarly, Φ + Φ + Φ = 0 . case b) r = − , l = − , k = j = | α | = 0By (3.78), we getΦ = − i Z | ξ ′ | =1 Z + ∞−∞ trace[ π + ξ n σ − ( D t − ) × ∂ ξ n σ − ( D t − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.84)By Lemma 3.7, we have σ − ( D t − ) = σ − (( D t ∗ ) − ). By (3.57)-(3.70), we haveΦ = 92 πh ′ (0)Ω dx ′ , (3.85)20here Ω is the canonical volume of S . case c) r = − , l = − , k = j = | α | = 0By (3.79), we getΦ = − i Z | ξ ′ | =1 Z + ∞−∞ trace[ π + ξ n σ − ( D − ) × ∂ ξ n σ − ( D t − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (3.86)By (3.2) and (3.3), Lemma 3.7, we have π + ξ n σ − ( D t − ) | | ξ ′ | =1 = c ( ξ ′ ) + ic ( dx n )2( ξ n − i ) . (3.87)Since σ − ( D t − )( x ) = c ( ξ ) σ ( D t )( x ) c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0) | ξ | ∂M ] , (3.88)where σ ( D t )( x ) = 14 X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t ) − X s,t,i ω s,t ( e i )( x ) c ( e i ) b c ( e s ) b c ( e t )+ [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x )= Q ( x ) + Q ( x ) + [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x ) , (3.89)then ∂ ξ n σ − ( D t − )( x ) | | ξ ′ | =1 = ∂ ξ n (cid:26) c ( ξ )[ Q ( x ) + Q ( x ) + t [ P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x )] c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0)] (cid:27) = ∂ ξ n (cid:26) c ( ξ ) Q ( x )] c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ tx n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0)] (cid:27) + ∂ ξ n c ( ξ ) Q ( x ) c ( ξ ) | ξ | + ∂ ξ n c ( ξ ) t [ P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x ) c ( ξ ) | ξ | . (3.90)By computation, we have ∂ ξ n c ( ξ ) Q ( x ) c ( ξ ) | ξ | = c ( dx n ) Q ( x ) c ( ξ ) | ξ | + c ( ξ ) Q ( x ) c ( dx n ) | ξ | − ξ n c ( ξ ) Q ( x ) c ( ξ ) | ξ | ; (3.91) ∂ ξ n c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x ) c ( ξ ) | ξ | = c ( dx n )[ t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x ) c ( ξ ) | ξ | + c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x ) c ( dx n ) | ξ | − ξ n c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ ) | ξ | . (3.92)21e denote q − = c ( ξ ) Q ( x ) c ( ξ ) | ξ | + c ( ξ ) | ξ | c ( dx n )[ ∂ x n [ c ( ξ ′ )]( x ) | ξ | − c ( ξ ) h ′ (0)] , then ∂ ξ n ( q − ) = 1(1 + ξ n ) (cid:20) (2 ξ n − ξ n ) c ( dx n ) Q c ( dx n ) + (1 − ξ n ) c ( dx n ) Q c ( ξ ′ )+ (1 − ξ n ) c ( ξ ′ ) Q c ( dx n ) − ξ n c ( ξ ′ ) Q c ( ξ ′ ) + (3 ξ n − ∂ x n c ( ξ ′ ) − ξ n c ( ξ ′ ) c ( dx n ) ∂ x n c ( ξ ′ ) + 2 h ′ (0) c ( ξ ′ ) + 2 h ′ (0) ξ n c ( dx n ) (cid:21) + 6 ξ n h ′ (0) c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) . (3.93)By (3.86) and (3.90), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ ) Q c ( ξ ) | ξ | ]( x ) | | ξ ′ | =1 = − ξ − i )( ξ + i ) tr[ c ( ξ ′ ) Q ( x )] + i ( ξ − i )( ξ + i ) tr[ c ( dx n ) Q ( x )] . (3.94)By (3.47), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ ) Q c ( ξ ) | ξ | ]( x ) | | ξ ′ | =1 = − ξ − i )( ξ + i ) tr[ c ( ξ ′ ) Q ( x )] . (3.95)We note that i < n, R | ξ ′ | =1 { ξ i ξ i · · · ξ i d +1 } σ ( ξ ′ ) = 0, so tr[ c ( ξ ′ ) Q ( x )] has no contribution for computingcase c).By (3.85) and (3.92), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n ( q − )]( x ) | | ξ ′ | =1 = 12 h ′ (0)( iξ n + ξ n − i )( ξ − i ) ( ξ + i ) + 48 h ′ (0) iξ n ( ξ − i ) ( ξ + i ) , (3.96)then − i Ω Z Γ + [ 12 h ′ (0)( iξ n + ξ n − i )( ξ n − i ) ( ξ n + i ) + 48 h ′ (0) iξ n ( ξ n − i ) ( ξ n + i ) ] dξ n dx ′ = − πh ′ (0)Ω dx ′ . (3.97)By (3.85) and (3.90), we havetr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]) c ( ξ ) | ξ | ]( x ) | | ξ ′ | =1 = − ξ − i )( ξ + i ) tr[ c ( ξ ′ )[ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x )]+ i ( ξ − i )( ξ + i ) tr[ c ( dx n )[ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )]( x )] . (3.98)22y R | ξ ′ | =1 { ξ · · · ξ d +1 } σ ( ξ ′ ) = 0 and (3.45), we have − i Z | ξ ′ | =1 Z + ∞−∞ tr[ π + ξ n σ − ( D t − ) × ∂ ξ n c ( ξ )( t P ni =1 P kα =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) c ( ξ ) | ξ | ]( x ) dξ n σ ( ξ ′ ) dx ′ = − i Z | ξ ′ | =1 Z + ∞−∞ i ( ξ − i )( ξ + i ) tr[ c ( dx n )( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))]( x ) dξ n σ ( ξ ′ ) dx ′ = − π c ( dx n )( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))]Ω dx ′ = 0 . (3.99)Then, Φ = − πh ′ (0)Ω dx ′ . (3.100)So Φ = Φ + Φ + Φ + Φ + Φ = 0. Theorem 3.9.
Let M be a -dimensional oriented compact manifold with the boundary ∂M and the metric g M as above, D t be sub-signature operator on f M , then ] Wres[ π + D t − ◦ π + D t − ] = 32 π Z M (cid:18) − K − t X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α ) (cid:19) d Vol M . (3.101) where K is the scalar curvature.
4. A Kastler-Kalau-Walze type theorem for 6-dimensional manifolds with boundary
Firstly, we prove the Kastler-Kalau-Walze type theorems for 6-dimensional manifolds with boundary.From [18], we know that ] Wres[ π + D t − ◦ π + ( D t ∗ D t D t ∗ ) − ] = Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − (( D t ∗ D t ) − )] σ ( ξ ) dx + Z ∂ t M Ψ , (4.1)where Ψ = Z | ξ ′ | =1 Z + ∞−∞ ∞ X j,k =0 X ( − i ) | α | + j + k +1 α !( j + k + 1)! × trace ∧ ∗ T ∗ M N C [ ∂ jx n ∂ αξ ′ ∂ tkξ n σ + r ( D t − )( x ′ , , ξ ′ , ξ n ) × ∂ αx ′ ∂ j +1 ξ n ∂ kx n σ l (( D t ∗ D t D t ∗ ) − )( x ′ , , ξ ′ , ξ n )] dξ n σ ( ξ ′ ) dx ′ , (4.2)and the sum is taken over r + ℓ − k − j − | α | − − , r ≤ − , ℓ ≤ − Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − (( D t ∗ D t ) − )] σ ( ξ ) dx = 128 π Z M (cid:18) − K − t + t − tt ) X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )+ 32( t − t )[ X i =1 k X α =1 g ( f α , ∇ T Me i S ( e i ) f α ) + X i =1 k X α =1 g ( ∇ T Me i f α , S ( e i ) f α )] (cid:19) d Vol M . (4.3)23ext, we compute R ∂M Ψ. By computation, we get D t ∗ D t D t ∗ = n X i =1 c ( e i ) h e i , dx l i ( − g ij ∂ l ∂ i ∂ j ) + n X i =1 c ( e i ) h e i , dx l i (cid:26) − ( ∂ l g ij ) ∂ i ∂ j − g ij (cid:18) σ i + a i ) ∂ j − kij ∂ k (cid:19) ∂ l (cid:27) + n X i =1 c ( e i ) h e i , dx l i (cid:26) − ∂ l g ij )( σ i + a i ) ∂ j + g ij ( ∂ l Γ kij ) ∂ k − g ij [( ∂ l σ i ) + ( ∂ l a i )] ∂ j + ( ∂ l g ij )Γ kij ∂ k + X j,k h ∂ l (cid:16) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) − c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) (cid:17)i h e j , dx k i ∂ k + X j,k (cid:16) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) − c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) (cid:17)h ∂ l h e j , dx k i i ∂ k (cid:27) + n X i =1 c ( e i ) h e i , dx l i ∂ l (cid:26) − g ij h ( ∂ i σ j ) + ( ∂ i a j ) + σ i σ j + σ i a j + a i σ j + a i a j − Γ ki,j σ k − Γ ki,j a k + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + c ( ∂ i ) ∂ i ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) + c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i + 14 K − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] (cid:27) + h ( σ i + a i ) + ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) i ( − g ij ∂ ti ∂ tj )+ n X i =1 c ( e i ) h e i , dx l i (cid:26) X j,k h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) + c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i × h e i , dx k i (cid:27) l ∂ k + h ( σ i + a i ) + ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) i(cid:26) − X i,j g i,j h σ i ∂ j + 2 a i ∂ j − Γ ki,j ∂ k + ( ∂ i σ j ) + ( ∂ i a j ) + σ i σ j + σ i a j + a i σ j + a i a j − Γ ki,j σ k − Γ ki,j a k i − X i,j g i,j h c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ ti ) i ∂ tj + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ ti ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + c ( ∂ i ) ∂ i ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) − c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) ∂ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i + 14 K − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] (cid:27) . (4.4)24hen, we obtain Lemma 4.1.
The following identities hold: σ ( D t ∗ D t D t ∗ ) = X i,j,l c ( dx l ) ∂ l ( g i,j ) ξ i ξ j + c ( ξ )(4 σ k + 4 a k − k ) ξ k + 2[ | ξ | t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) − c ( ξ ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ξ )] + 14 | ξ | X s,t,l ω s,t ( e l )[ c ( e l ) b c ( e s ) b c ( e t ) − c ( e l ) c ( e s ) c ( e t )] + | ξ | ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) ; σ ( D t ∗ D t D t ∗ ) = ic ( ξ ) | ξ | . (4.5)Write σ ( D t ∗ D t D t ∗ ) = p + p + p + p ; σ (( D t ∗ D t D t ∗ ) − ) = ∞ X j =3 q − j . (4.6)By the composition formula of pseudodifferential operators, we have1 = σ (( D t ∗ D t D t ∗ ) ◦ ( D t ∗ D t D t ∗ ) − ) = X α α ! ∂ αξ [ σ ( D t ∗ D t D t ∗ )] D αx [( D t ∗ D t D t ∗ ) − ]= ( p + p + p + p )( q − + q − + q − + · · · )+ X j ( ∂ ξ j p + ∂ ξ j p + ∂ ξ j p + ∂ ξ j p )( D x j q − + D x j q − + D x j q − + · · · )= p q − + ( p q − + p q − + X j ∂ ξ j p D x j q − ) + · · · , (4.7)by (4.7), we have q − = p − ; q − = − p − [ p p − + X j ∂ ξ j p D x j ( p − )] . (4.8)By Lemma 4.1, we have some symbols of operators. Lemma 4.2.
The following identities hold: σ − (( D t ∗ D t D t ∗ ) − ) = ic ( ξ ) | ξ | ; σ − (( D t ∗ D t D t ∗ ) − ) = c ( ξ ) σ ( D t ∗ D t D t ∗ ) c ( ξ ) | ξ | + ic ( ξ ) | ξ | (cid:16) | ξ | c ( dx n ) ∂ x n c ( ξ ′ ) − h ′ (0) c ( dx n ) c ( ξ )+ 2 ξ n c ( ξ ) ∂ x n c ( ξ ′ ) + 4 ξ n h ′ (0) (cid:17) . (4.9)When n = 6, then tr ∧ ∗ T ∗ M [ id ] = 64, where tr as shorthand of trace. Since the sum is taken over r + ℓ − k − j − | α | − − , r ≤ − , ℓ ≤ −
3, then we have the R ∂ M Ψ is the sum of the following five cases: case (a) (I) r = − , l = − , j = k = 0 , | α | = 1.By (4.2), we getΨ = − Z | ξ ′ | =1 Z + ∞−∞ X | α | =1 trace h ∂ αξ ′ π + ξ n σ − ( D t − ) × ∂ αx ′ ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.10)25y Lemma 4.2, for i < n , we have ∂ x i σ − (( D t ∗ D t D t ∗ ) − )( x ) = ∂ x i h ic ( ξ ) | ξ | i ( x ) = i∂ x i [ c ( ξ )] | ξ | − ( x ) − ic ( ξ ) ∂ x i [ | ξ | ] | ξ | − ( x ) = 0 , (4.11)so Ψ = 0. case (a) (II) r = − , l = − , | α | = k = 0 , j = 1.By (4.2), we haveΨ = − Z | ξ ′ | =1 Z + ∞−∞ trace h ∂ x n π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.12)By computation, we have ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) = i (cid:20) (20 ξ n − c ( ξ ′ ) + 12( ξ n − ξ n ) c ( dx n )(1 + ξ n ) (cid:21) . (4.13)Since n = 6, tr[ − id ] = −
64. By the relation of the Clifford action and tr AB = tr BA , thentr[ c ( ξ ′ ) c ( dx n )] = 0; tr[ c ( dx n ) ] = −
64; tr[ c ( ξ ′ ) ]( x ) | | ξ ′ | =1 = − ∂ x n [ c ( ξ ′ )] c ( d x n )] = 0; tr[ ∂ x n c ( ξ ′ ) c ( ξ ′ )]( x ) | | ξ ′ | =1 = − h ′ (0) . (4.14)By (3.28), (4.13) and (4.14), we gettrace h ∂ x n π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) = 64 h ′ (0) − − ξ n i + 5 ξ n + 3 iξ n ( ξ n − i ) ( ξ n + i ) . (4.15)Then we obtain Ψ = − Z | ξ ′ | =1 Z + ∞−∞ h ′ (0) dimF − − ξ n i + 40 ξ n + 24 iξ n ( ξ n − i ) ( ξ n + i ) dξ n σ ( ξ ′ ) dx ′ = 8 h ′ (0)Ω Z Γ + ξ n i − ξ n − iξ n ( ξ n − i ) ( ξ n + i ) dξ n dx ′ = h ′ (0)Ω πi h ξ n i − ξ n − iξ n ( ξ n + i ) i (5) | ξ n = i dx ′ = − πh ′ (0)Ω dx ′ , (4.16)where Ω is the canonical volume of S . case (a) (III) r = − , l = − , | α | = j = 0 , k = 1.By (4.2), we haveΨ = − Z | ξ ′ | =1 Z + ∞−∞ trace h ∂ ξ n π + ξ n σ − ( D t − ) × ∂ ξ n ∂ x n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.17)By computation, we have ∂ ξ n ∂ x n σ − (( D t ∗ D t D t ∗ ) − ) = − iξ n ∂ x n c ( ξ ′ )( x )(1 + ξ n ) + i h ′ (0) ξ n c ( ξ ′ )(1 + ξ n ) − i (2 − ξ n ) h ′ (0) c ( dx n )(1 + ξ n ) . (4.18)Combining (3.34) and (4.18), we havetrace h ∂ ξ n π + ξ n σ − ( D t − ) × ∂ ξ n ∂ x n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) | | ξ ′ | =1 = 8 h ′ (0) 8 i − ξ n − iξ n ( ξ n − i ) ( ξ + i ) . (4.19)26hen Ψ = − Z | ξ ′ | =1 Z + ∞−∞ h ′ (0) 8 i − ξ n − iξ n ( ξ n − i ) ( ξ + i ) dξ n σ ( ξ ′ ) dx ′ = − h ′ (0)8Ω Z Γ + i − ξ n − iξ n ( ξ n − i ) ( ξ + i ) dξ n dx ′ = − h ′ (0)Ω πi h i − ξ n − iξ n ( ξ + i ) i (4) | ξ n = i dx ′ = 252 πh ′ (0)Ω dx ′ . (4.20) case (b) r = − , l = − , | α | = j = k = 0.By (4.2), we haveΨ = − i Z | ξ ′ | =1 Z + ∞−∞ trace h π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ = i Z | ξ ′ | =1 Z + ∞−∞ trace[ ∂ ξ n π + ξ n σ − ( D t − ) × σ − (( D t ∗ D t D t ∗ ) − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (4.21)In the normal coordinate, g ij ( x ) = δ ji and ∂ x j ( g αβ )( x ) = 0, if j < n ; ∂ x j ( g αβ )( x ) = h ′ (0) δ αβ , if j = n .So by [21], when k < n , we have Γ n ( x ) = h ′ (0), Γ k ( x ) = 0, δ n ( x ) = 0 and δ k = h ′ (0) c ( e k ) c ( e n ). Then,we obtain σ − (( D t ∗ D t D t ∗ ) − )( x ) | | ξ ′ | =1 = c ( ξ ) σ (( D t ∗ D t D t ∗ ) − )( x ) | | ξ ′ | =1 c ( ξ ) | ξ | − c ( ξ ) | ξ | X j ∂ ξ j (cid:0) c ( ξ ) | ξ | (cid:1) D tx j (cid:0) ic ( ξ ) | ξ | (cid:1) = 1 | ξ | c ( ξ ) (cid:16) h ′ (0) c ( ξ ) X k 758 ) πh ′ (0)Ω dx ′ = ( − i − πh ′ (0)Ω dx ′ . (4.26) case (c) r = − , l = − , | α | = j = k = 0.By (4.2), we haveΨ = − i Z | ξ ′ | =1 Z + ∞−∞ trace h π + ξ n σ − ( D t − ) × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.27)By Lemma 4.1 and Lemma 4.2, we have σ − ( D t − )( x ) = c ( ξ ) σ ( D t ) c ( ξ ) | ξ | ( x ) + c ( ξ ) | ξ | X j c ( dx j ) h ∂ x j ( c ( ξ )) | ξ | − c ( ξ ) ∂ x j ( | ξ | ) i ( x ) , (4.28)28here σ ( D t ) = 14 X i,s,t ω s,t ( e i ) c ( e i ) b c ( e s ) b c ( e t ) − X i,s,t ω s,t ( e i ) c ( e i ) b c ( e s ) b c ( e t ) + t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α ) . (4.29)On the other hand, ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − ) = − iξ n c ( ξ ′ )(1 + ξ n ) + i (1 − ξ n ) c ( d x n )(1 + ξ n ) . (4.30)By (4.28), (3.2) and (3.3), we have π + ξ n (cid:16) σ − ( D t − ) (cid:17) ( x ) | | ξ ′ | =1 = π + ξ n h c ( ξ ) σ ( D t )( x ) c ( ξ ) + c ( ξ ) c ( dx n ) ∂ x n [ c ( ξ ′ )]( x )(1 + ξ n ) i − h ′ (0) π + ξ n h c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) i . (4.31)We denote σ ( D t )( x ) | ξ n = i = Q ( x ) = Q ( x ) + Q ( x ) + t n X i =1 k X α =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α ) . (4.32)Then, we obtain π + ξ n (cid:16) σ − ( D t − ) (cid:17) ( x ) | | ξ ′ | =1 = π + ξ n h c ( ξ ) Q ( x ) c ( ξ ) + c ( ξ ) c ( dx n ) ∂ x n [ c ( ξ ′ )]( x )(1 + ξ n ) − h ′ (0) c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) i + π + ξ n h c ( ξ )[ Q ( x )] c ( ξ )( x )(1 + ξ n ) i + π + ξ n h c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )] c ( ξ )( x )(1 + ξ n ) i . (4.33)Furthermore, π + ξ n h c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ )(1 + ξ n ) i = π + ξ n h c ( ξ ′ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ ′ )(1 + ξ n ) i + π + ξ n h ξ n c ( ξ ′ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( dx n )(1 + ξ n ) i + π + ξ n h ξ n c ( dx n )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ ′ )(1 + ξ n ) i + π + ξ n h ξ n c ( dx n )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( dx n )(1 + ξ n ) i = − c ( ξ ′ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ ′ )(2 + iξ n )4( ξ n − i ) + ic ( ξ ′ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( dx n )4( ξ n − i ) + ic ( dx n )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ ′ )4( ξ n − i ) + − iξ n c ( dx n )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( dx n )4( ξ n − i ) , (4.34) π + ξ n h c ( ξ ) Q ( x ) c ( ξ )(1 + ξ n ) i = π + ξ n h c ( ξ ′ ) Q ( x ) c ( ξ ′ )(1 + ξ n ) i + π + ξ n h ξ n c ( ξ ′ ) Q ( x ) c ( dx n )(1 + ξ n ) i + π + ξ n h ξ n c ( dx n ) Q ( x ) c ( ξ ′ )(1 + ξ n ) i + π + ξ n h ξ n c ( dx n ) Q ( x ) c ( dx n )(1 + ξ n ) i = − c ( ξ ′ ) Q ( x ) c ( ξ ′ )(2 + iξ n )4( ξ n − i ) + ic ( ξ ′ ) Q ( x ) c ( dx n )4( ξ n − i ) + ic ( dx n ) Q ( x ) c ( ξ ′ )4( ξ n − i ) + − iξ n c ( dx n ) Q ( x ) c ( dx n )4( ξ n − i ) . (4.35)29y the relation of the Clifford action and tr AB = tr BA , then we have equalities:tr[ Q c ( dx n )] = 0; tr[ b c ( ξ ′ ) b c ( dx n )] = 0 . (4.36)Then we havetr (cid:20) π + ξ n (cid:16) c ( ξ ) Q ( x ) c ( ξ )(1 + ξ n ) (cid:17) × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − )( x ) (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) | ξ ′ | =1 = 2 − iξ n − ξ n ξ n − i ) (1 + ξ n ) tr[ Q ( x ) c ( ξ ′ )] , (4.37)By computation, we have π + ξ n h c ( ξ ) Q ( x ) c ( ξ ) + c ( ξ ) c ( dx n ) ∂ x n ( c ( ξ ′ ))( x )(1 + ξ n ) i − h ′ (0) π + ξ n h c ( ξ ) c ( dx n ) c ( ξ )(1 + ξ n ) i := C − C , (4.38)where C = − ξ n − i ) (cid:2) (2 + iξ n ) c ( ξ ′ ) Q c ( ξ ′ ) + iξ n c ( dx n ) Q c ( dx n )+ (2 + iξ n ) c ( ξ ′ ) c ( dx n ) ∂ x n c ( ξ ′ ) + ic ( dx n ) Q c ( ξ ′ ) + ic ( ξ ′ ) Q c ( dx n ) − i∂ x n c ( ξ ′ ) (cid:3) = 14( ξ n − i ) h h ′ (0) c ( dx n ) − i h ′ (0) c ( ξ ′ ) − (2 + iξ n ) c ( ξ ′ ) c ( dx n ) ∂ ξ n c ( ξ ′ ) + i∂ ξ n c ( ξ ′ ) i ; (4.39) C = h ′ (0)2 h c ( dx n )4 i ( ξ n − i ) + c ( dx n ) − ic ( ξ ′ )8( ξ n − i ) + 3 ξ n − i ξ n − i ) (cid:0) ic ( ξ ′ ) − c ( dx n ) (cid:1)i . (4.40)By (4.30) and (4.40), we havetr[ C × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − )( x )] | | ξ ′ | =1 = tr n h ′ (0)2 h c ( dx n )4 i ( ξ n − i ) + c ( dx n ) − ic ( ξ ′ )8( ξ n − i ) + 3 ξ n − i ξ n − i ) [ ic ( ξ ′ ) − c ( dx n )] i × − iξ n c ( ξ ′ ) + ( i − iξ n ) c ( dx n )(1 + ξ n ) o = 8 h ′ (0) 4 i − ξ n − iξ n + 3 ξ n ( ξ n − i ) ( ξ n + i ) . (4.41)Similarly, we havetr[ C × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − )( x )] | | ξ ′ | =1 = tr n ξ n − i ) h h ′ (0) c ( dx n ) − i h ′ (0) c ( ξ ′ ) − (2 + iξ n ) c ( ξ ′ ) c ( dx n ) ∂ ξ n c ( ξ ′ ) + i∂ ξ n c ( ξ ′ ) i × − iξ n c ( ξ ′ ) + ( i − iξ n ) c ( dx n )(1 + ξ n ) o = 8 h ′ (0) 3 + 12 iξ n + 3 ξ n ( ξ n − i ) ( ξ n + i ) ; (4.42)tr (cid:20) π + ξ n (cid:16) c ( ξ )[ t P ni =1 P kα =1 c ( e i ) b c ( s ( e i ) f α ) b c ( f α )]( x ) c ( ξ )(1 + ξ n ) (cid:17) × ∂ ξ n σ − (( D t ∗ D t D t ∗ ) − )( x ) (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) | ξ ′ | =1 = 2 − iξ n − ξ n ξ n − i ) (1 + ξ n ) tr[( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ))( x ) c ( ξ ′ )]= 2 − iξ n − ξ n ξ n − i ) (1 + ξ n ) tr[ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )( x ) c ( ξ ′ )]= 0 . (4.43)30y R | ξ ′ | =1 ξ · · · ξ q +1 σ ( ξ ′ ) = 0 , we haveΨ = − ih ′ (0) Z | ξ ′ | =1 Z + ∞−∞ × − i + 26 ξ n + 15 iξ n ( ξ n − i ) ( ξ n + i ) dξ n σ ( ξ ′ ) dx ′ − i Z | ξ ′ | =1 Z + ∞−∞ (cid:20) − iξ n − ξ n ξ n − i ) (1 + ξ n ) [ − g ( θ ′ , ξ ′ )]tr[ id ] (cid:21)(cid:12)(cid:12)(cid:12)(cid:12) | ξ ′ | =1 dξ n σ ( ξ ′ ) dx ′ = − ih ′ (0) × πi h − i + 26 ξ n + 15 iξ n ( ξ n + i ) i (5) | ξ n = i Ω dx ′ = 552 πh ′ (0)Ω dx ′ . (4.44)Now Ψ is the sum of the cases (a), (b) and (c), thenΨ = ( 658 − i ) πh ′ (0)Ω dx ′ . (4.45) Theorem 4.3. Let M be a -dimensional compact oriented manifold with the boundary ∂M and the metric g M as above, D t and D t ∗ be sub-signature operators on f M , then ] Wres[ π + D t − ◦ π + ( D t ∗ D t D t ∗ ) − ]= 128 π Z M (cid:18) − K − t + t − tt ) X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α )+ 32( t − t )[ X i =1 k X α =1 g ( f α , ∇ T Me i S ( e i ) f α ) + X i =1 k X α =1 g ( ∇ T Me i f α , S ( e i ) f α )] (cid:19) d Vol M + Z ∂M (cid:20) ( 658 − i ) πh ′ (0) (cid:21) Ω dx ′ . (4.46) where K is the scalar curvature. Next, we prove the Kastler-Kalau-Walze type theorem for 6-dimensional manifold with boundary asso-ciated to D t . From [18], we know that ] Wres[ π + D t − ◦ π + D t − ] = Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − ( D t − )] σ ( ξ ) dx + Z ∂M Ψ , (4.47)where ] Wres denote noncommutative residue on minifolds with boundary,Ψ = Z | ξ ′ | =1 Z + ∞−∞ ∞ X j,k =0 X ( − i ) | α | + j + k +1 α !( j + k + 1)! × trace ∧ ∗ T ∗ M N C [ ∂ jx n ∂ αξ ′ ∂ kξ n σ + r ( D t − )( x ′ , , ξ ′ , ξ n ) × ∂ αx ′ ∂ j +1 ξ n ∂ kx n σ l ( D t − )( x ′ , , ξ ′ , ξ n )] dξ n σ ( ξ ′ ) dx ′ , (4.48)and the sum is taken over r + ℓ − k − j − | α | − − , r ≤ − , ℓ ≤ − Z M Z | ξ | =1 trace ∧ ∗ T ∗ M N C [ σ − ( D t − )] σ ( ξ ) dx = 128 π Z M (cid:20) − K − t X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α ) (cid:21) d Vol M . (4.49)31o we only need to compute R ∂M Ψ. Let us now turn to compute the specification of D t . D t = n X i =1 c ( e i ) h e i , dx l i ( − g ij ∂ l ∂ i ∂ j ) + n X i =1 c ( e i ) h e i , dx l i (cid:26) − ( ∂ l g ij ) ∂ i ∂ j − g ij (cid:18) σ i + a i ) ∂ j − kij ∂ k (cid:19) ∂ l (cid:27) + n X i =1 c ( e i ) h e i , dx l i (cid:26) − ∂ l g ij )( σ i + a i ) ∂ j + g ij ( ∂ tl Γ kij ) ∂ k − g ij [( ∂ l σ i ) + ( ∂ l a i ) ∂ j + ( ∂ l g ij )Γ kij ∂ k + X j,k h ∂ l (cid:16) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) + c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) (cid:17)i h e j , dx k i ∂ k + X j,k (cid:16) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) + c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) (cid:17)h ∂ l h e j , dx k i i ∂ k (cid:27) + n X i =1 c ( e i ) h e i , dx l i ∂ l (cid:26) − g ij h ( ∂ i σ j ) + ( ∂ ti a j ) + σ i σ j + σ i a j + a i σ j + a i a j − Γ ki,j σ k − Γ ki,j a k + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ i ) a i + c ( ∂ i ) ∂ i ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) + c ( ∂ i ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i + 14 K − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] (cid:27) + h ( σ i + a i ) + ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) i ( − g ij ∂ i ∂ j )+ n X i =1 c ( e i ) h e i , dx l i (cid:26) X j,k h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( e j ) + c ( e j ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i × h e i , dx k i (cid:27) l ∂ k + h ( σ i + a i ) + ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) i(cid:26) − X i,j g i,j h σ i ∂ j + 2 a i ∂ j − Γ ki,j ∂ k + ( ∂ i σ j ) + ( ∂ i a j ) + σ i σ j + σ i a j + a i σ j + a i a j − Γ ki,j σ k − Γ ki,j a k i + X i,j g i,j h c ( ∂ i ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ ti ) i ∂ tj + X i,j g i,j h t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ ti ) σ i + t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ∂ ti ) a i + c ( ∂ i ) ∂ i ( t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )) + c ( ∂ ti ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) ∂ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) + c ( ∂ ti ) σ i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )+ c ( ∂ i ) a i t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) i + 14 K − X ijkl R ijkl b c ( e i ) b c ( e j ) c ( e k ) c ( e l ) − [ t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] (cid:27) . (4.50)32hen, we obtain Lemma 4.4. The following identities hold: σ ( D t ) = X i,j,l c ( dx l ) ∂ l ( g i,j ) ξ i ξ j + c ( ξ )(4 σ k + 4 a k − k ) ξ k − c ( ξ ) t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α ) c ( ξ ) − | ξ | t n X i =1 k X α =1 c ( e i ) b c ( S ( e i ) f α ) b c ( f α )] + 14 | ξ | X s,t,l ω s,t ( e l )[ c ( e l ) b c ( e s ) b c ( e t ) − c ( e l ) c ( e s ) c ( e t )]+ | ξ | (( e l )[ c ( e l ) b c ( e s ) b c ( e t )); σ ( D t ) = ic ( ξ ) | ξ | . (4.51)Write σ ( D t ) = p + p + p + p ; σ ( D t − ) = ∞ X j =3 q − j . (4.52)By the composition formula of pseudodifferential operators, we have1 = σ ( D t ◦ D t − ) = X α α ! ∂ αξ [ σ ( D t )] D tαx [ σ ( D t − )]= ( p + p + p + p )( q − + q − + q − + · · · )+ X j ( ∂ ξ j p + ∂ ξ j p + ∂ ξ j p + ∂ ξ j p )( D x j q − + D x j q − + D x j q − + · · · )= p q − + ( p q − + p q − + X j ∂ ξ j p D x j q − ) + · · · , (4.53)by (4.53), we have q − = p − ; q − = − p − [ p p − + X j ∂ ξ j p D x j ( p − )] . (4.54)By (4.50)-(4.54), we have some symbols of operators. Lemma 4.5. The following identities hold: σ − ( D t − ) = ic ( ξ ) | ξ | ; σ − ( D t − ) = c ( ξ ) σ ( D t ) c ( ξ ) | ξ | + ic ( ξ ) | ξ | (cid:16) | ξ | c ( dx n ) ∂ x n c ( ξ ′ ) − h ′ (0) c ( dx n ) c ( ξ )+ 2 ξ n c ( ξ ) ∂ x n c ( ξ ′ ) + 4 ξ n h ′ (0) (cid:17) . (4.55)When n = 6, then tr ∧ ∗ T ∗ M [ id ] = 64, where tr as shorthand of trace. Since the sum is taken over r + ℓ − k − j − | α | − − , r ≤ − , ℓ ≤ − 3, then we have the R ∂ t M Ψ is the sum of the following five cases: case (a) (I) r = − , l = − , j = k = 0 , | α | = 1.By (4.48), we getΨ = − Z | ξ ′ | =1 Z + ∞−∞ X | α | =1 trace h ∂ αξ ′ π + ξ n σ − ( D t − ) × ∂ αx ′ ∂ ξ n σ − ( D t − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.56)33 ase (a) (II) r = − , l = − , | α | = k = 0 , j = 1.By (4.48), we haveΨ = − Z | ξ ′ | =1 Z + ∞−∞ trace h ∂ x n π + ξ n σ − ( D t − ) × ∂ ξ n σ − ( D t − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.57) case (a) (III) r = − , l = − , | α | = j = 0 , k = 1.By (4.48), we haveΨ = − Z | ξ ′ | =1 Z + ∞−∞ trace h ∂ ξ n π + ξ n σ − ( D t − ) × ∂ ξ n ∂ x n σ − ( D t − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ . (4.58)By Lemma 4.2 and Lemma 4.5, we have σ − (( D t ∗ D t D t ∗ ) − ) = σ − ( D t − ) , by (4.10)-(4.20), we obtainΨ + Ψ + Ψ = 5 πh ′ (0)Ω dx ′ , where Ω is the canonical volume of S . case (b) r = − , l = − , | α | = j = k = 0.By (4.48), we haveΨ = − i Z | ξ ′ | =1 Z + ∞−∞ trace h π + ξ n σ − ( D t − ) × ∂ ξ n σ − ( D t − ) i ( x ) dξ n σ ( ξ ′ ) dx ′ = i Z | ξ ′ | =1 Z + ∞−∞ trace[ ∂ ξ n π + ξ n σ − ( D t − ) × σ − ( D t − )]( x ) dξ n σ ( ξ ′ ) dx ′ . (4.59)In the normal coordinate, g ij ( x ) = δ ji and ∂ x j ( g αβ )( x ) = 0, if j < n ; ∂ x j ( g αβ )( x ) = h ′ (0) δ αβ , if j = n .So by [21], when k < n , we have Γ n ( x ) = h ′ (0), Γ k ( x ) = 0, δ n ( x ) = 0 and δ k = h ′ (0) c ( e k ) c ( e n ). Then,we obtain σ − ( D t − )( x ) | | ξ ′ | =1 = c ( ξ ) σ ( D t − )( x ) | | ξ ′ | =1 c ( ξ ) | ξ | − c ( ξ ) | ξ | X j ∂ tξ j (cid:0) c ( ξ ) | ξ | (cid:1) D tx j (cid:0) ic ( ξ ) | ξ | (cid:1) = 1 | ξ | c ( ξ ) (cid:16) h ′ (0) c ( ξ ) X k Let M be a -dimensional compact oriented manifold with the boundary ∂M and the metric M as above, D t be a sub-signature operator on f M , then ] Wres[ π + D t − ◦ π + ( D t − )]= 128 π Z M (cid:20) − K − t X i =1 k X α =1 g ( S ( e i ) f α , S ( e i ) f α ) (cid:21) d Vol M + Z ∂M (cid:20) ( 658 − i ) πh ′ (0) (cid:21) Ω dx ′ . (4.65) where K is the scalar curvature. Acknowledgements This work was supported by NSFC. 11771070 . The authors thank the referee for his (or her) carefulreading and helpful comments. ReferencesReferences [1] Ackermann T.: A note on the Wodzicki residue. J. 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