General Lagrangian of Non-Covariant Self-dual Gauge Field
aa r X i v : . [ h e p - t h ] O c t General Lagrangian of Non-CovariantSelf-dual Gauge Field
Wung-Hong HuangDepartment of PhysicsNational Cheng Kung UniversityTainan, Taiwan
ABSTRACT
We present the general formulation of non-covariant Lagrangian of self-dual gaugetheory. After specifying the parameters therein the previous Lagrangian in the decom-position of spacetime into 6 = D + D and 6 = D + D + D can be obtained. Theself-dual property of the general Lagrangian is proved in detail. We furthermore showthat the new non-covariant actions give field equations with 6d Lorentz invariance. Themethod can be straightforward extended to any dimension and we also give a shortdiscussion about the 10D self-dual gauge theory.*E-mail: [email protected] 1 ontents x with x . . . . . . . . . . . . . . . . . . . . . . . . . . 134.2.2 Mixing x with x a . . . . . . . . . . . . . . . . . . . . . . . . . . 144.2.3 Mixing x a with x b . . . . . . . . . . . . . . . . . . . . . . . . . . 15 The problem in Lagrangian description of chiral p-forms, i.e. antisymmetric bosonfields with self-dual had been known before thirty years ago. As first noted by Marcusand Schwarz [1], the manifest duality and spacetime covariance do not like to live inharmony with each other in one action.Historically, the non-manifestly spacetime covariant action for self-dual 0-form wasproposed by Floreanini and Jackiw [2], which is then generalized to p-form by Henneaux2nd Teitelboim [3]. In general the field strength of chiral p-form A ··· p is split intoelectric density E i ··· i p +1 and magnetic density B i ··· i p +1 : E i ··· i p +1 ≡ F i ··· i p +1 ≡ ∂ [ i A i ··· i p +1 ] (1.1) B i ··· i p +1 ≡ p + 1)! ǫ i ··· i p +2 F i p +2 ··· i p +2 ≡ ˜ F i ··· i p +1 (1.2)in which ˜ F is the dual form of F . The Lagrangian is described by L = − p ! ~ B · ( ~ E − ~ B ) = 1 p ! ˜ F i ··· i p +1 F i ··· i p +1 (1.3)in which we define F i ··· i p +1 ≡ ˜ F i ··· i p +1 − F i ··· i p +1 (1.4)Note that in order for self-dual fields to exist, i.e. ˜ F = F , the field strength F anddual field strength ˜ F should have the same number of component. As the double dualon field strength shall give the original field strength the spacetime dimension have tobe 2 modulo 4.Four years ago, a new non-covariant Lagrangian formulation of a chiral 2-form gaugefield in 6D, called as 6 = 3 + 3 decomposition, was derived in [4] from the Bagger-Lambert-Gustavsson (BLG) model [5]. Later, a general non-covariant Lagrangian for-mulation of self-dual gauge theories in diverse dimensions was constructed [6]. In thisgeneral formulation the 6 = 2 + 4 decomposition of Lagrangian was found.In the last year we have constructed a new kind of non-covariant actions of self-dual2-form gauge theory in the decomposition of 6 = D + D + D [7]. In this paper we willpresent the general formulation of non-covariant Lagrangian of self-dual gauge theory.In section II we first present the general non-covariant Lagrangian of self-dual gaugetheory. We see that after specifying the parameters therein the all known Lagrangianin the decomposition of 6 = D + D [6] and 6 = D + D + D [7] can be obtained.In section III We discuss some properties which are crucial to formulate the generalLagrangian of non-covariant forms of self-dual gauge theory. We then prove in detailthe self-dual property of the general Lagrangian. In section IV we follow Perry andSchwarz [8] to show that the general non-covariant Lagrangian gives field equationswith 6d Lorentz invariance. Our prescription can be straightforward extended to anydimension and we also give a brief description about the 10D self-dual gauge theory insection V. Last section is devoted to a short conclusion.3 Lagrangian of Self-dual Gauge Fields in SimpleDecomposition
In this section we first present the general non-covariant Lagrangian L G of self-dualgauge theory in (2.2) and table 1. Then we collect all know non-covariant Lagrangianof self-dual gauge theory in six dimension [6,7] and compare them with L G . Table 2and table 3 are just those in our previous paper [7], while for convenience we reproducethem in this paper. We will see that, after specifying the parameters in L G the previousLagrangian in the decomposition of spacetime into 6 = D + D and 6 = D + D + D can be obtained.To begin with, let us first define a function L ijk : L ijk ≡ ˜ F ijk × ( F ijk − ˜ F ijk ) , without summation over indices i, j, k (2.1)which is useful in the following formulations.In terms of L ijk the most general non-covariant Lagrangian of self-dual gauge theorywe found is L G ( α i ) = X a L a + ( 12 + α L + ( 12 − α L + ( 12 + α L + ( 12 − α L ( 12 + α L + ( 12 − α L + ( 12 + α L + ( 12 − α L ( 12 + α L + ( 12 − α L + ( 12 + α L + ( 12 − α L (2.2)Let us make three comments about above Lagrangian.First, From table 1 we see that L G does not picks up L , L , L nor L , whichis denoted as ✓✓ L abc . This can ensure to the existence of gauge symmetry δA = Φ ,which is crucial to prove the self-duality of L G , as shown in next section.Next, We have chosen the coefficient before L a to be one. This is because thatthe overall constant of L G does not affect the self-duality.Third, we choose coefficient ( + α ) before L while choose coefficient ( − α ) be-fore L . This can ensure that adding the two coefficient ( + α ) + ( − α ) = 1, whichgive a proper normalization. This property is also crucial to prove the self-duality of L G , as shown in next section. 4able 1: Lagrangian in the general decompositions: D = 6. D=6123 456124 L a ✓✓ L abc
125 346126 345134 ( + α ) L
256 ( − α ) L
135 ( + α ) L
246 ( − α ) L
136 ( + α ) L
245 ( − α ) L
145 ( + α ) L
236 ( − α ) L
146 ( + α ) L
235 ( − α ) L
156 ( + α ) L
234 ( − α ) L In the (1+5) decomposition the spacetime index A = (1 , · · · ,
6) is decomposed as A = (1 , ˙ a ), with ˙ a = (2 , · · · , L ABC = ( L a ˙ b , L ˙ a ˙ b ˙ c ). In terms of L ABC , theLagrangian is expressed as [6] L = − X L a ˙ b = −
14 ˜ F a ˙ b ( F a ˙ b − ˜ F a ˙ b ) , has summation over ˙ a ˙ b (2.3)From table 2 we see that L picks up only L a ˙ b . Note that that L G ( α i = 1) = L .Table 2: Lagrangian in various decompositions: D = D + D . D=1+5123 456124 356125 346126 345134 L a ˙ b ✟✟✟ L ˙ a ˙ b ˙ c
135 246136 245145 236146 235156 234 D=3+3123 L abc ✓✓ L ˙ a ˙ b ˙ b
124 356125 346126 L ab ˙ a ✓✓ L a ˙ a ˙ b
134 256135 246136 245145 236146 ✓✓ L a ˙ a ˙ b L ab ˙ a
156 234 D=2+4123 456124 L ab ˙ a ✓✓ L ˙ a ˙ b ˙ c
125 346126 345134 256135 246136 L a ˙ a ˙ b L a ˙ a ˙ b
145 236146 235156 234 .2 Lagrangian in Decomposition: In the (2+4) decomposition [6] the spacetime index A is decomposed as A = ( a, ˙ a ),with a = (1 ,
2) and ˙ a = (3 , · · · , L ABC = ( L ab ˙ a , L a ˙ a ˙ b , L ˙ a ˙ b ˙ c ). From table 2 it iseasy to see that in terms of L ABC the Lagrangian can be expressed as [7] L = − (cid:16) X L ab ˙ a + 12 X L a ˙ a ˙ b (cid:17) (2.4)The factor before L a ˙ a ˙ b arising from the property that L a ˙ a ˙ b contains both of left-handside element and right-hand side element (for example, it includes L and L ), thusthere is double counting. Note that that L G ( α i = 0) = L . In the (3+3) decomposition [6] the spacetime index A is decomposed as A = ( a, ˙ a ),with a = (1 , ,
3) and ˙ a = (4 , , L ABC = ( L abc , L ab ˙ a , L a ˙ a ˙ b , L ˙ a ˙ b ˙ c ). Using table 2it is easy to see that in terms of L ABC the Lagrangian can be expressed as L = − (cid:16) X L abc + 3 X L ab ˙ a (cid:17) (2.5)The “3” factor before L ab ˙ a arising from the property that we have to include threekinds of L ijk : L ab ˙ a , L a ˙ ab and L ˙ aab . Note that L G ( α = α = α = 1; α = α = α = −
1) = L .Self-dual property of above decomposition had been proved in [6]. In the (1+1+4) decomposition the spacetime index A is decomposed as A = (1 , , ˙ a ),with ˙ a = (3 , , , L ABC = ( L
12 ˙ a , L ˙ a ˙ b ˙ c , L a ˙ b , L a ˙ b ). From table 3 it is easy to seethat, in terms of L ABC , the Lagrangian can be expressed as [7] L = 6 X L
12 ˙ a + 3(1 + α )2 X L a ˙ b + 3(1 − α )2 X L a ˙ b (2.6)We neglect overall constant in Lagrangian, which is irrelevant to the self-duality.It is easy to see that L in the case of α = 0 is just L , in the case of α = 1is just L , and in the case of α = − L while exchanging indices 1 and 2,as can be seen from table 2. Note that L G ( α i = α ) = L .6able 3: Lagrangian in various decompositions: D = D + D + D D=1+1+4123 456124 L
12 ˙ a ✓✓ L ˙ a ˙ b ˙ c
125 346126 345134 256135 (1 + α ) 246 (1 − α )136 L a ˙ b L a ˙ b
145 236146 235156 234 D=1+2+3123 ✓✓ L ab L ˙ a ˙ b ˙ c
124 356125 346126 L a ˙ a ✓✓ L a ˙ a ˙ b
134 256135 246136 245145 236146 L a ˙ b ✓✓ L ab ˙ a
156 234 D=2+2+2123 L ab ˙ a ✓✓ L ˙ a ¨ a ¨ b
124 356125 L ab ¨ a ✓✓ L ˙ a ˙ b ¨ a
126 345134 L a ˙ a ˙ b ✓✓ L a ¨ a ¨ b
135 246136 L a ˙ a ¨ a L a ˙ a ¨ a
145 236146 235156 ✓✓ L a ¨ a ¨ b L a ˙ a ˙ b In the (1+2+3) decomposition the spacetime index A is decomposed as A = (1 , a, ˙ a ),with a = (2 , a = (4 , , L ABC = ( L ab , L a ˙ a , L a ˙ b , L ˙ a ˙ b ˙ c , L a ˙ a ˙ b , L ab ˙ a ). Fromtable 3 it is easy to see that, in terms of L ABC , the Lagrangian can be expressed as [7] L = X L ˙ a ˙ b ˙ c + 6 X L a ˙ a + 3 X L a ˙ b (2.7)Choosing L ab + L a ˙ a + L a ˙ b is just L , and choosing L ab + L a ˙ a + L ab ˙ a is just L ,as can be seen from table 2. Note that, exchanging indices 2 with 5 and 3 with 6 then L G ( α = α = α = α = α = 1 , α = −
1) = L . In the (2+2+2) decomposition the spacetime index A is decomposed as A = ( a, ˙ a, ¨ a ),with a = (1 , a = (3 ,
4) and ¨ a = (5 , L ABC = ( L a ˙ a ˙ b , L ˙ a ¨ a ¨ b , L ab ¨ a , L ˙ a ˙ b ¨ a , L a ˙ a ˙ b , L a ¨ a ¨ b , L a ˙ a ¨ a ). Then, in terms of L ABC the Lagrangian can beexpressed as [7] L = X L ab ˙ a + X L ab ¨ a + X L a ˙ a ˙ b + X L a ˙ a ¨ a (2.8)Note that L G ( α = 1 , α = α = α = α = 0 , α = −
1) = L .Self-dual property of above decompositions had been proved in [6, 7].7 General Lagrangian of Self-dual Gauge Fields
In order to understand how to find the general formulation we need to find someconstrains in formulating the non-covariant Lagrangian of self-dual gauge theory [6,7].
From previous studies [6,7] we see that the proof of the self-dual property has used a
Gauge symmetry . The property tells us that terms involved A , for example, onlythrough total derivative terms and we have gauge symmetry δA = Φ (3.1)for arbitrary functions Φ . In order to have this property we must not choose the dualtransformationof L a in table 1, where a=(3,4,5,6). Thus the general non-covariantLagrangian of self-dual gauge theory is L = X a L a + C ( 12 + α L + C ( 12 − ˜ α L + C ( 12 + α L + C ( 12 − ˜ α L C ( 12 + α L + C ( 12 − ˜ α L + C ( 12 + α L + C ( 12 − ˜ α L C ( 12 + α L + C ( 12 − ˜ α L + C ( 12 + α L + C ( 12 − ˜ α L (3.2)in which we have let the coefficient before L a to be one as the overall constant of L does not affect the self-duality. We will now show that above Lagrangian has desiredgauge symmetry, and after proper choosing the parameters C i it becomes L G in (2.2)and the associated field strength has self-dualtiy property.First, the variation of the action gives δLδA = − ∂ ˜ F − ∂ ˜ F − ∂ ˜ F − ∂ ˜ F = ∂ a ˜ F a = 0 (3.3)which is identically zero. This means that terms involved A only through totalderivative terms and we have gauge symmetry δA = Φ (3.4)for arbitrary functions Φ . 8 .2 Self-duality Next, simply using above gauge symmetry does not guarantee that the Lagrangian hasself-dual property. Let us find the another constrain.The variation of the Lagrangian L gives0 = δLδA = − h C ∂ ˜ F + C ∂ ˜ F + ∂ ˜ F + ∂ ˜ F i + C (1 − α + ˜ α ∂ ˜ F − C (1 − ˜ α ) ∂ F + C (1 − α + ˜ α ∂ ˜ F − C (1 − ˜ α ) ∂ F +2( ∂ F + ∂ F ) (3.5)Now, if we require each L i,j,k has a same normalization then C i = 1. Under thisconstrain we find that0 = δLδA = (1 − α + ˜ α ∂ ˜ F − (1 − ˜ α ) ∂ F +(1 − α + ˜ α ∂ ˜ F − (1 − ˜ α ) ∂ F +2( ∂ F + ∂ F ) (3.6)where we have used the property ∂ a ˜ F a = 0.To procced, we can from table 2 and table 3 see that, for example, the differencebetween the Lagrangian in decomposition D = 2 + 4 and D = 1 + 5 is that we havechosen left-hand (electric) part and right-hand (magnetic) part in D = 2 + 4, while in D = 1 + 5 we choose only left-hand (electric) part. However, in the self-dual theorythe electric part is equal to magnetic part. Thus the Lagrangian choosing electricpart is equivalent to that choosing magnetic part. In the decomposition into differentdirect-product of spacetime one can choose different part of L ijk to mixing to eachother and we have many kind of decomposition. This observation lead us to find moredecomposition 6 = D + D + D in [7].This property can be applied to find the more general formulation of non-covariantLagrangian of self-dual gauge theory. Thus the another constrain is that (cid:16)
12 + α i (cid:17) + (cid:16) − ˜ α i (cid:17) = 1 ⇒ α i = ˜ α i (3.7)From now on we will use this property and Lagrangian L becomes L G in (2.2).9hus 0 = δLδA = (1 − α ) ∂ F + (1 + α ) ∂ F + 2 ∂ F + 2 ∂ F = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.8)in which we have normalized each F by the associated factor (1 − α i ) or 2 for conve-nience.Similary, we have the relations0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.9)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.10)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.11)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.12)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.13)Above six equations can be expressed as ∂ a ¯ F abc = 0 , a, b, c = 1 , F abc = ǫ abcd ∂ d Φ (3.15)for arbitrary functions Φ . As the gauge symmetry of δA = Φ can totally removeΦ in F abc we have found a self-dual relation F abc = 0 , a, b, c = 1 , δLδA = (1 − α ) ∂ F + (1 − α ) ∂ F + (1 − α ) ∂ F = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.17)where we have normalized each F by the associated factor (1 − α ). In the same waywe have the relations 0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.18)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.19)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.20)10bove 4 equations give the solution of F ab ( a, b = 2) F ab = ǫ abcd ∂ c Φ d (3.21)In the same way, we can find0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.22)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.23)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.24)0 = δLδA = ∂ ¯ F + ∂ ¯ F + ∂ ¯ F (3.25)Above 4 equations give the solution of F ab ( a, b = 1) F ab = ǫ abcd ∂ c Φ d (3.26)We can now follow [6,7] to find another self-dual relation. First, taking the Hodge-dualof both sides in above equation we find that F ab = ∂ [ a Φ b , ( a, b = 2) (3.27)Identifying this solution with previou found solution of F ab , then ∂ [ a Φ b = ǫ abcd ∂ c Φ d (3.28)Acting ∂ a on both sides gives ∂ a ∂ a Φ b = 0 (3.29)under the Lorentz gauge ∂ a Φ a = 0. Now, following [6,7], imposing the boundarycondition that the field Φ b be vanished at infinities will lead to the unique solutionΦ b = 0 and we arrive at the self-duality conditions F ab = 0 , a, b = 1 (3.30)In the same way, we can find another self-duality conditions F ab = 0 , a, b = 2 (3.31)These complete the proof. 11 Lorentz Invariance of Self-dual Field Equation
In this we follow the method of Perry and Schwarz [8] to show that the above generalnon-covariant actions give field equations with 6d Lorentz invariance. Note that sec.4.1, 4.2.1 and 4.2.2 are just those in our previous paper [7], while for completeness wereproduce them in below.
We first describe the Lorentz transformation of 2-form field strength. For the coordinatetransformation : x a → ¯ x a ≡ x a + δx a the tensor field H ¯ M ¯ N ¯ P ( x a ) will becomes H MNP ( x a ) → H ¯ M ¯ N ¯ P ( x a + δx a ) ≡ ∂x Q ∂ ¯ x M ∂x R ∂ ¯ x N ∂x S ∂ ¯ x P H QRS ( x a + δx a ) ≈ H MNP ( x a + δx a ) + ∂x Q ∂ ¯ x M ∂x R ∂ ¯ x N ∂x S ∂ ¯ x P H QRS ( x a ) (4.1)For the transformation mixing between x with x µ ( µ = 1) the relation δx a = ω ab x b leads to δx = − Λ µ x µ and δx µ = Λ µ x in which we define ω µ = − ω µ ≡ Λ µ .The orbital part of transformation [8] is defined by δ orb H MNP ≡ H MNP ( x a + ω ab x b ) − H MNP ( x a ) ≈ [ δx a ] · ∂ a H MNP = [Λ µ x µ ∂ ] H MNP − x [Λ µ ∂ µ ] H MNP = [(Λ · x ) ∂ − x (Λ · ∂ )] H MNP ≡ (Λ · L ) H MNP (4.2)Note that δ orb is independent of index M N P and is universal for any tensor.The spin part of transformation [8] becomes δ spin H µνλ ≡ ∂x Q ∂x µ ∂x R ∂x ν ∂x S ∂x λ H QRS ( x ) − H µνλ = ∂ ( δx ) ∂x µ ∂x R ∂x ν ∂x S ∂x λ H RS ( x ) + ∂x Q ∂x µ ( δx ) ∂x ν ∂x S ∂x λ H Q S ( x ) + ∂x Q ∂x µ ∂x R ∂x ν ∂ ( δx ) ∂x λ H QR ( x )= h − Λ µ H νλ i + h − Λ ν H µ λ i + h − Λ λ H µν i (4.3)and δH µνλ = δ orb H µνλ + δ spin H µνλ In a same way δ spin H µν ≡ ∂x Q ∂x µ ∂x R ∂x ν ∂x S ∂x H QRS ( x ) − H µν = ∂ ( δx ) ∂x µ ∂x R ∂x ν ∂x S ∂x H RS ( x ) + ∂x Q ∂x µ ( δx ) ∂x ν ∂x S ∂x H Q S ( x ) + ∂x Q ∂x µ ∂x R ∂x ν ∂ ( δx λ ) ∂x H QRλ ( x )= 0 + 0 + Λ λ H µνλ (4.4)and δH µν = δ orb H µν + δ spin H µν = (Λ · L ) H µν + Λ λ H µνλ .12 .2 Lorentz Invariance of Self-dual Field Equation We now use above Lorentz transformation need to examine transformations (I) mixing x with x , (II) mixing x with x a and (IV) mixing x a with x b , a, b = 3 , , , x with x (I) Consider first the mixing x with x . The transformation is δx = ω x ≡ Λ x , (4.5) δx = ω x = − Λ x (4.6)Define Λ · L ≡ (Λ x ) ∂ − x (Λ ∂ ) (4.7)then δF a = (Λ · L ) F a (4.8) δF abc = (Λ · L ) F abc (4.9) δF ab = (Λ · L ) F ab − Λ F ab (4.10) δF ab = (Λ · L ) F ab + Λ F ab (4.11)Using above transformation we can find δ ˜ F a = (Λ · L ) ˜ F a + 16 ǫ abcd ( δ spin F bcd ) = (Λ · L ) ˜ F a (4.12) δ ˜ F ab = (Λ · L ) ˜ F ab + 16 ǫ ab cd ( δ spin F cd · · L ) ˜ F ab + 12 ǫ ab cd [Λ F cd ]= (Λ · L ) ˜ F ab − Λ ˜ F ab (4.13)Thus δ ( F a − ˜ F a ) = (Λ · L )( F a − ˜ F a ) = 0 (4.14) δ ( F ab − ˜ F ab ) = (Λ · L )( F ab − ˜ F ab ) − Λ( F ab − ˜ F ab ) = 0 (4.15)which are zero for self-dual theory. Taking Hodge of above relations we also get δ ( F abc − ˜ F abc ) = 0 , δ ( F ab − ˜ F ab ) = 0 (4.16)and the non-covariant action gives field equations with 6d Lorentz invariance undertransformation mixing x with x . 13 .2.2 Mixing x with x a (II) For the mixing x with x a , a = 3 , , ,
6, we shall consider the transformation δx a = ω a x ≡ Λ a x , (4.17) δx = ω a x a = − Λ a x a = − Λ · x (4.18)Define Λ · L ≡ (Λ · x ) ∂ − x (Λ · ∂ ) (4.19)then δF a = (Λ · L ) F a + Λ b F b a (4.20) δF abc = (Λ · L ) F abc − Λ a F bc − Λ b F a c − Λ c F ab (4.21) δF ab = (Λ · L ) F ab + Λ c F cab (4.22) δF ab = (Λ · L ) F ab − Λ a F b − Λ b F a (4.23)Use above transformation we can find δ ˜ F a = (Λ · L ) ˜ F a + 16 ǫ abcd ( δ spin F bcd )= (Λ · L ) ˜ F a + 16 ǫ abcd [ − Λ b F cd − Λ c F b d − Λ d F bc ]= (Λ · L ) ˜ F a + Λ b ˜ F ab (4.24) δ ˜ F ab = (Λ · L ) ˜ F ab + 16 ǫ ab cd ( δ spin F cd · · L ) ˜ F ab + 12 ǫ ab cd [ − Λ c F d − Λ d F c ]= (Λ · L ) ˜ F ab + Λ c ˜ F cab (4.25)Thus δ ( F a − ˜ F a ) = (Λ · L )( F a − ˜ F a ) + Λ b ( F ab − ˜ F ab ) = 0 (4.26) δ ( F ab − ˜ F ab ) = (Λ · L )( F ab − ˜ F ab ) + Λ c ( F cab − ˜ F cab ) = 0 (4.27)which are zero for self-dual theory. Taking Hodge of above relations we also get δ ( F abc − ˜ F abc ) = 0 , δ ( F ab − ˜ F ab ) = 0 (4.28)and the non-covariant action gives field equations with 6d Lorentz invariance undertransformation mixing x with x a . 14 .2.3 Mixing x a with x b (II) For the mixing x a with x b , a = 3 , , ,
6, we shall consider the transformation δx a = ω ab x b ≡ Λ ab x b (4.29)Define Λ · L ≡ Λ ab ( x a ∂ b − x b ∂ a ) (4.30)then δF a = (Λ · L ) F a − Λ ea F e (4.31) δF abc = (Λ · L ) F abc − Λ ea F ebc − Λ eb F aec − Λ ec F abe − Λ c F ab (4.32) δF ab = (Λ · L ) F ab − Λ ea F eb − Λ eb F ae (4.33) δF ab = (Λ · L ) F ab − Λ ea F eb − Λ eb F ae (4.34)Using above transformations we can find δ ˜ F a = (Λ · L ) ˜ F a + 16 ǫ abcd ( δ spin F bcd )= (Λ · L ) ˜ F a − ǫ abcd (Λ be F ecd + Λ ce F bed + Λ de F bce )= (Λ · L ) ˜ F a − Λ ea ˜ F e (4.35) δ ˜ F ab = (Λ · L ) ˜ F ab + 16 ǫ abcd ( δ spin F cd · · L ) ˜ F ab − ǫ abcd (Λ ce F ed · de F ce · · L ) ˜ F ab − Λ ea ˜ F eb − Λ eb ˜ F ae (4.36)Therefore δ ( F a − ˜ F a ) = (Λ · L )( F a − ˜ F a ) − Λ ea ( F e − ˜ F e ) = 0 (4.37) δ ( F ab − ˜ F ab ) = (Λ · L )( F ab − ˜ F ab ) − Λ ea ( F eb − ˜ F eb ) − Λ eb ( F ae − ˜ F ae ) = 0(4.38)which are zero for self-dual theory. Taking Hodge of above relations we also get δ ( F abc − ˜ F abc ) = 0 , δ ( F ab − ˜ F ab ) = 0 (4.39)and the non-covariant action gives field equations with 6d Lorentz invariance undertransformation mixing x a with x b . 15
10D Self-dual Gauge Theory
The extension above prescription to 10D self-dual gauge theory is straightforward. Asbefore, let us first define a function L ijkℓmn : L ijkℓm ≡ ˜ F ijkℓm × ( F ijkℓm − ˜ F ijkℓm ) , without summation over indices (5.1)In terms of L ijkℓm the most general non-covariant Lagrangian of self-dual gauge theoryis L G ( α i ) = X a L a + ( 12 + α L + ( 12 − α L Q + ( 12 + α L + ( 12 − α L Q + · · · (5.2)in which Q denotes as tenth dimension hereafter.Table 4: Lagrangian in the general decompositions: D = 10. D=1012345 6789Q12346 5789Q12347 L a ✓✓ L abcde + α ) L − α ) L Q + α ) L − α ) L Q · · · ·· · · ·· · · · From table 4 we see that L G does not picks up L Q , · · · , L . This is a crucialproperty to have a gauge symmetry. We now summarize the proof of self-duality ofabove Lagrangian. 16 .1 Self-duality of 10D Self-dual Gauge Theory First, the variation of the action gives δL G ( α i ) δA = ∂ a ˜ F a = 0 (5.3)which is identically zero. This means that terms involved A only through totalderivative terms and we have gauge symmetry δA = Φ (5.4)for arbitrary functions Φ .Next, we can find that ∂ a ¯ F abcde = 0 , a, b, c, d, e = 1 , , , F abcde = ǫ abcdef ∂ f Φ (5.6)for arbitrary functions Φ . As the gauge symmetry of δA = Φ can totallyremove Φ in F abcde we have found a self-dual relation F abcde = 0 , a, b, c, d, e = 1 , , , ∂ a ¯ F abcd = 0 , a, b, c, d = 2 , , F abcd = ǫ abcdef ∂ f Φ e a, b, c, d = 2 , , e .We can also find that ∂ a ¯ F ab = 0 , a, b = 1 (5.10)which has solution F ab = ǫ abcdef ∂ c Φ def , a, b = 1 (5.11)for arbitrary functions Φ e . 17e can now follow [6,7] to find another self-dual relation. First, taking the Hodge-dual of both sides in above equation we find that F abcd = ∂ [ a Φ bcd ] , ( a, b, c, d = 1 , , ,
4) (5.12)Identifying this solution with previoue found solution of F abcd , then ∂ [ a Φ bcd ] = ǫ abcdef ∂ f Φ e , ( a, b, c, d = 1 , , ,
4) (5.13)Acting ∂ a on both sides gives ∂ a ∂ a Φ bcd = 0 , ( a, b, c, d = 1 , , ,
4) (5.14)under the Lorentz gauge ∂ a Φ abc = 0. Now, following [6,7], imposing the boundarycondition that the field Φ bcd be vanished at infinities will lead to the unique solutionΦ bcd = 0 and we arrive at the self-duality conditions F ab = 0 , a, b = 1 (5.15)In the same way, we can find all other self-duality conditions. Finally, the method in section IV can be easily applied to prove that general non-covariant actions give field equations with 10d Lorentz invariance. Essentially, wemerely add more index in field strength.For example, in considering mixing x with x we can find that δF abc = (Λ · L ) F abc (5.16) δF abcde = (Λ · L ) F abcde (5.17) δF abcd = (Λ · L ) F abcd − Λ F abcd (5.18) δF abcd = (Λ · L ) F abcd + Λ F abcd (5.19)Using above transformation we can find that δ ( F abc − ˜ F abc ) = (Λ · L )( F abc − ˜ F abc ) = 0 (5.20) δ ( F abcd − ˜ F abcd ) = (Λ · L )( F abcd − ˜ F abcd ) − Λ( F abcd − ˜ F abcd ) = 0 (5.21)which are zero for self-dual theory. Taking Hodge of above relations we also get δ ( F abcde − ˜ F abcde ) = 0 , δ ( F abde − ˜ F abde ) = 0 (5.22)and the non-covariant action gives field equations with 10d Lorentz invariance undertransformation mixing x with x . 18 Conclusion