Lagrangian of Self-dual Gauge Fields in Various Formulations

The Lagrangian of self-dual gauge theory in various formulations are reviewed. From these results we see a simple rule and use it to present some new non-covariant Lagrangian based on the decomposition of spacetime into $D=D_1+D_2+D_3$. Our prescription could be easily extended to more complex decomposition of spacetime and some more examples are presented therefore. The self-dual property of the new Lagrangian is proved in detail. We also show that the new non-covariant actions give field equations with 6d Lorentz invariance.


Introduction
Chiral p-forms, i.e. antisymmetric boson fields with self-dual (p+1)-form field strengths play a central role in supergravity and in string theory, such as D = 6 and type IIB D = 10 supergravity, heterotic strings [1] and M-theory five-branes [2]. In particular, they contribute to the "miraculous" cancelation of the gravitational anomaly in type-IIB supergravity or superstring theory. The first calculation of the gravitational anomaly for chiral p-forms was performed in [3] without using a Lagrangian but just guessing suitable Feynman rules that incorporate the chirality condition.
It is well known that there is a problem in Lagrangian description of chiral bosons, since manifest duality and spacetime covariance do not like to live in harmony with each other in one action, as first seen by Marcus and Schwarz [4]. Historically, the nonmanifestly spacetime covariant action for self-dual 0-form was proposed by Floreanini and Jackiw [5], which is then generalized to p-form by Henneaux and Teitelboim [6]. In general the field strength of chiral p-form A 1···p is split into electric density E i 1 ···i p+1 and magnetic density B i 1 ···i p+1 : (1.1) in whichF is the dual form of F . The Lagrangian is described by Note that in order for self-dual fields to exist, i.e.F = F , the field strength F and dual field strengthF should have the same number of component. As the double dual on field strength shall give the original field strength the spacetime dimension have to be 2 modulo 4. Above actions, however, lead to second class constraints and complicates the quantization procedure. Siegel in [7] proposed a manifestly spacetime covariant action of chiral p-form models by squaring the second-class constraints and introducing Lagrange multipliers λ ab into the action. The Lagrangian of chiral 2 form is described by (1.4) in which we define It is easy to see that the field equation 0 = δS δλ ab implies F = 0 and we get the selfdual property. Using this property the other field equation 0 = δS δA ab is automatically satisfied. Siegel action, however, does not have enough local symmetry to completely gauge the Lagrange multipliers away and suffers from anomaly of gauge symmetry.
Note that, the self-dual relationF = F is a first-order differential equation which defines the dynamics of the chiral boson, contrast to other bosonic fields whose equations of motion are usually second-order differential equations. This lead McClain, Wu and Yu to construct chiral field action in a first order form [8]. In this case, for the Lagrange multiplier itself not to carry propagating degrees of freedom one has to introduce an infinite number of auxiliary fields "compensating" the dynamics of each other. However, this infinite set corresponds to the infinite number of local symmetries which cause problems in choosing the right regularization procedure during the quantization.
Pasti, Sorokin and Tonin in 1995 constructed a Lorentz covariant formulation of chiral p-forms in D = 2(p+1) dimensions that contains a finite number of auxiliary fields in a non-polynomial way [9]. For example, 6D PST Lagrangian is in which a(x) is the auxiliary field. In the gauge ∂ r a = δ 1 r the PST formulation reduces to the non-manifestly covariant formulation [5,6]. On the other hand, Perry and Schwarz [10] had shown that the non-covariant action (1.3) gives field equations with 6d Lorentz invariance.
Recently, a new non-covariant Lagrangian formulation of a chiral 2-form gauge field in 6D, called as (3+3) decomposition, was derived in [11] from the Bagger-Lambert-Gustavsson (BLG) model [12]. The covariant formulation of the associated Lagrangian is constructed in [13], with the use of a triplet of auxiliary scalar fields. Later, a general non-covariant Lagrangian formulation of self-dual gauge theories in diverse dimensions was constructed [14]. In this general formulation the (2+4) decomposition of Lagrangian was found.
In section 2 we review above formulations of self-dual 2-form in the decomposition of D = D 1 + D 2 and find a simple rule. In section 3 we use the rule to construct new non-covariant actions of self-dual 2-form gauge theory in the decomposition of D = D 1 + D 2 + D 3 . We present a detailed proof about the self-dual property in the new Lagrangian. We also show in detail that the new non-covariant action gives field equations with 6d Lorentz invariance. In section 4 we generalize our prescription to construct a non-covariant action in the decomposition of D = D 1 + D 2 + D 3 + D 4 . Last section is devoted to a short conclusion.

Lagrangian in Decomposition:
To begin with, let us first define a useful function L ijk : which is useful in the following formulations.

D=3+3
In the (3+3) decomposition [14] the spacetime index A is decomposed as A = (a,ȧ), with a = (1, 2, 3) andȧ = (4,5,6). Then L ABC = (L abc , L abȧ , L aȧḃ , L˙a˙b˙c). Using table 1 it is easy to see that in terms of L ABC the Lagrangian can be expressed as [14] L 3+3 = − 1 12 Let us make following interesting comments: 1. Why there is the "3" factor before L abȧ in above equation ? This is because that we have to include three kinds of L ijk : L abȧ , L aȧb and Lȧ ab .
2. Note that choosing L 3+3 ∼ L abc + 3 L aȧḃ will spoil the gauge symmetry δA ab = Φ ab which is crucial in proving the self-dual property of the Lagrangian. A simple rule to have this symmetry is that the choosing Lagrangian L 3+3 shall contain all possible index "ab" in L ABC . More precisely, as L ABC = (L abc , L abȧ , L aȧḃ , L˙a˙b˙c) the all possible term with index "ab" in L ABC is L abc , L abȧ . As both terms have been included in L 3+3 , the Lagrangian thus has the crucial gauge symmetry. Self-dual property of L 3+3 had been proved in [13,14].

D=2+4
In the (2+4) decomposition the spacetime index A is decomposed as A = (a,ȧ), with a = (1, 2) andȧ = (3, · · ·, 6). Then L ABC = (L abȧ , L aȧḃ , L˙a˙b˙c). From table 1 it is easy to see that in terms of L ABC the Lagrangian can be expressed as [14] Self-dual property of L 2+4 had been proved in [14]. Let us make following interesting comments: 1. Why there is the 1 2 factor before L aȧḃ in above equation ? This is because that in table 1 L aȧḃ contains both of left-line element and right-line element (for example, it includes L 134 and L 256 ), thus there is double counting.
2. From table 1 we see that the difference between the Lagrangian in decomposition D = 2 + 4 and D = 1 + 5 is that we have chosen 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. In the self-dual theory the electric part is equal to magnetic part. Thus the Lagrangian choosing electric part is equivalent to that choosing magnetic part. However, in the decomposition into different direct-product of spacetime one shall choose different part of L ijk to mixing to each other. This renders the results to be different and we have many kinds of formulation, as shown in the next section.

Lagrangian in Decomposition:
We now consider another decomposition of Lagrangian by: D = D 1 + D 2 + D 3

D=1+1+4
In the (1+1+4) decomposition the spacetime index A is decomposed as A = (1, 2,ȧ), withȧ = (3,4,5,6), and L ABC = (L 12ȧ , L˙a˙b˙c, L 1ȧḃ , L 2ȧḃ ).  From table 2 it is easy to see that, in terms of L ABC , the Lagrangian can be expressed as We neglect overall constant in Lagrangian, which is irrelevant to the following proof. Note that the case of α = 0 is just L 2+4 , the case of α = −1 is just L 1+5 , and the case of α = 1 is just L 1+5 while exchanging indices 1 and 2, as can be seen from table 1.
We now follow the method in [14] to prove the self-dual property of L 1+1+4 and follow the method in [10] to prove that the new non-covariant action gives field equations with 6d Lorentz invariance.

Self-duality in D=1+1+4
First, we rewrite the Lagrangian as The variation of the action S 1+1+4 gives which is identically zero. This means that terms involved A 12 only through total derivative terms and we have gauge symmetry for arbitrary functions Φ 12 . The Gauge symmetry is crucial to prove the self-duality in following. Next, the field equations for arbitrary functions Φ 2ḋ and Ψ 1ḋ .
We can now follow [14] to find a self-dual relation. First, taking the Hodge-dual of F 1ȧḃ in above solution and identifying it to the solution F 2ȧḃ in above equation we find that Acting ∂˙a on both sides gives ∂˙a∂ȧΦ 2ḃ = 0 (3.10) Following [14], imposing the boundary condition that the regular field Φ 2ḃ be vanished at infinities will lead to the unique solution Φ 2ḃ = 0 and we arrive at the self-duality conditions Taking the Hodge-dual of F 1ȧḃ we also obtain Using above result the another field equation becomes which has solution F˙a˙b˙c = ǫ 12ȧḃċḋ ∂ḋΦ 12 (3.14) We can now use the gauge symmetry of δA 12 = Φ 12 to totally remove Φ 12 in F˙a˙b˙c and we find a self-dual relation These complete the proof.

Lorentz Invariance in D=1+1+4
As covariant symmetry on 4D coordinates xȧ is manifest we only need to examine transformations (I) mixing x 1 with xȧ, (II) mixing x 2 with xȧ and (II) mixing x 1 with x 2 .
(I) For the mixing x 1 with xȧ we shall consider the transformation Use above transformation we can find which are zero for self-dual theory and the non-covariant action gives field equations with 6d Lorentz transformation mixing x 1 with xȧ.
(II) With exchange the index 1 ↔ 2 above result also shows that the non-covariant action gives field equations with 6d Lorentz transformation mixing x 2 with xȧ.
(III) Finally, we consider the mixing x 1 with x 2 . The transformation is Use above transformation we can calculate the transformations ofF 12ã andF 1ȧḃ . Then we see that which are zero for self-dual theory and the non-covariant action gives field equations with 6d Lorentz transformation mixing x 1 with x 2 .
In summary, we have found the non-covariant action of self-dual 2-form in decomposition D = 1 + 1 + 4 and have checked that the non-covariant action gives field equations with 6d Lorentz transformation.
We now follow the method in [14] to prove the self-dual property of L 1+2+3 and follow the method in [10] to prove that the new non-covariant action gives field equations with 6d Lorentz invariance.

Self-duality in D=1+2+3
First, we rewrite the Lagrangian as The variation of the action S 1+2+3 gives which is identically zero. This means that terms involved A 1ȧ only through total derivative terms and we have a gauge symmetry for arbitrary functions Φ 1ȧ . Next, the field equation has solution for arbitrary functions Φ 1ċ . Using the above gauge symmetry to completely remove Φ 1ċ in F aȧḃ we obtain a self-dual relation To proceed we need to find more gauge symmetry. First, as term Φ 1ċ is shown as in which W 1ȧ (x 1 , xȧ) is an arbitrary function independing on the coordinate x a . In short, after using the gauge symmetry to find a self-dual relation we still have above "residual gauge symmetries". Next, as field A˙a˙b only appears as ∂ a A˙a˙b in F aȧḃ , therefore the previous results does not be modified under the variation in which W˙a˙b(x 1 , xȧ) is an arbitrary function independing on the coordinate x a . We will use the two residual gauge symmetries to find other self-duality relations. Note that these residual gauge symmetries do not spoil any of the self-duality conditions already satisfied. Now, the field equation 0 = δS 1+2+3 δA aȧ = −6(∂ 1F 1aȧ + ∂˙bF˙b aȧ + ∂ bF baȧ ) + 12(∂˙bF˙b aȧ + ∂ b F baȧ ) tells us that F baȧ is independent of the coordinate"x a " . In the same way, the field equation tells us that F 1ab is independent of the coordinate "x a " . Finally, using above properties the field equation As W˙b˙c and W 1ċ are arbitrary functions independing on the coordinates "x a " we can use the residual gauge symmetries to completely remove W˙b˙c and W 1ċ . Thus we obtain the self-dual relations These complete the proof.

Lorentz Invariance in D=1+2+3
As covariant symmetry on 2D coordinates x a and 3D coordinates xȧ are manifest we only need to examine transformations (I) mixing x 1 with x a , (II) mixing x 1 with xȧ and (III) mixing x a with xȧ.
(I) For the mixing x 1 with xȧ we shall consider the transformation then we see that which are zero for self-dual theory and the non-covariant action gives field equation with 6d Lorentz transformation mixing x 1 with xȧ.
(II) For the mixing x 1 with x a we shall consider the transformation then we see that which are zero for self-dual theory and the non-covariant action gives field equation with 6d Lorentz transformation mixing x 1 with x a . (III) Finally, we consider the mixing x a with xȧ. In this case the transformation is then we see that which are zero for self-dual theory and the non-covariant action gives field equation with 6d Lorentz transformation mixing x a with xȧ.
In summary, we have found the non-covariant action of self-dual 2-form in decomposition D = 1 + 2 + 3 and have checked that the non-covariant action gives field equation with 6d Lorentz transformation.

Self-duality in D=2+2+2
First, we rewrite the Lagrangian as The variation of the action S 2+2+3 gives δS 2+2+2 δA ab = −(∂ȧF˙a ab + ∂äFä ab ) = 0 (3.72) which is identically zero and terms involved A ab only through total derivative terms. Thus, as before, we have a gauge symmetry for arbitrary functions Φ ab .
Next, the field equation for arbitrary functions Φ ab . As before, we can now use the gauge symmetry of A ab to reduce F˙a˙bä to be zero and find the first self-dual relation.
To proceed we need to find more gauge symmetry. First, as term Φ ab is shown as ∂bΦ ab in F˙a˙bä we have a furthermore symmetry in which W ab (x a , xȧ) is an arbitrary function independing on the coordinate xä. In short, after using the gauge symmetry to find a self-dual relation we still have above residual gauge symmetries. Next, as field A˙a˙b appears only as ∂äA˙a˙b in F˙a˙bä the above relations do not be modified under the variation δA˙a˙b = W˙a˙b(x a , xȧ) (3.78) in which W˙a˙b(x a , xȧ) is an arbitrary function independing on the coordinate x a . We need the above two residual gauge symmetries to find other self-duality relations in below. Note that these residual gauge symmetries do not spoil any of the self-duality conditions already satisfied. Now, consider the field equation which has solution F aȧä = ǫ abȧḃäb ∂bΦ bḃ (3.80) We can now follow [14] to find another self-dual relation. First, taking the Hodge-dual of both sides in above equation we find that F aȧä = ∂äΦ aȧ (3.81) Identifying above two solutions leads to ∂äΦ aȧ = ǫ abȧḃäb ∂bΦ bḃ (3.82) Acting ∂ä a on both sides gives ∂ä∂äΦ aȧ = 0 (3.83) Following [14], imposing the boundary condition that the regular field Φ aȧ be vanished at infinities will lead to the unique solution Φ aȧ = 0 and we arrive at the self-duality conditions To proceed we need to find one more gauge symmetry. As term Φ aȧ is shown as ∂äΦ aȧ in F aȧä we have a furthermore symmetry in which W aȧ (x a , xȧ) is an arbitrary function independing on the coordinate xä. In short, after using the gauge symmetry to find a self-dual relation we still have above residual gauge symmetries. We need the above residual gauge symmetries to find other self-duality relations in below. Note that these residual gauge symmetries do not spoil any of the self-duality conditions already satisfied. As W˙a˙b, W aȧ and W ab are arbitrary functions independing on the coordinates "xä" we can use the "residual gauge symmetries" to completely remove them. Thus we obtain the self-dual relations These complete the proof.

Lorentz Invariance in D=2+2+2
As covariant symmetry on 2D coordinates x a , 2D coordinates xȧ and 2D coordinates xä are manifest we only need to examine transformations (I) mixing x a with xȧ, (II) mixing xȧ with xä and (III) mixing x a with xä. We consider the mixing x a with xȧ. In this case the transformation is Use above transformation we can calculate the transformations ofF abȧ ,F abä ,F aȧḃ and F aȧä . Then we see that which are zero for self-dual theory and the non-covariant action gives field equations with 6d Lorentz transformation mixing x a with xȧ. The transformation mixing x a with xä and mixing xȧ with xä have the similar results.
In summary, we have found the non-covariant action of self-dual 2-form in decomposition D = 2 + 2 + 2 and checked that the non-covariant action gives field equations with 6d Lorentz transformation.

Other Decomposition : D=1+1+1+3
Besides the decomposition in section 2 and section 3 there are many other possible decompositions. We will in this subsection see that it is possible to found the Lagrangian of self-dual 2 form in decomposition : D = 1 + 1 + 1 + 3.
In the (1+1+1+3) decomposition the spacetime index A is decomposed as A = (1, 2, 3,ȧ), withȧ = (4, 5, 6),and L ABC = (L 123 , L 12ȧ , L 13ȧ , L 1ȧḃ , L˙a˙b˙c, L 2ȧḃ , L 3ȧḃ , L 23ȧ ). From table 3 it is easy to see that, in terms of L ABC , the Lagrangian can be expressed as Other choices will becomes the decompositions in section 2 or section 3, as can be seen from table 1 and table 2. Note that the factor 6 or 3 in (4.1) is to count the number of possible permutation and the factor 1 2 in (4.1) revels that fact we have counted both the left-hand side and right-hand side in table 3. We follow the method in [14] to prove the self-dual property of L 1+1+1+3 and follow the method in [10] to prove that the new non-covariant action gives field equations with 6d Lorentz invariance.

Self-duality in D=1+1+1+3
First, we rewrite the Lagrangian as The variation of the action S 1+1+1+3 gives which are identically zero. This means that terms involved A 12 and A 13 only through total derivative terms and we have a gauge symmetry δA 12 = Φ 12 (4.5) for arbitrary functions Φ 12 and Φ 13 .
F 2ȧḃ = 0 (4.17) In the same way, the field equation Thus, use the gauge symmetry δA 12 = Φ 12 we can completely remove the function Φ 12 and find the self-dual relation.
Finally, through the calculation the field equation becomes in which we have used the found self-duality rlations. Above field equation has solution where Φ(x 1 , x 2 , x 3 ) is independent of the coordinates x˙a. As Φ(x 1 , x 2 , x 3 ) can be written as ∂ i f i (x 1 , x 2 , x 3 ), with i = 1, 2, 3, the function Φ(x 1 , x 2 , x 3 ) can be absorbed by a field redefinition A ij → A ij + ǫ ijk f k (x 1 , x 2 , x 3 ). Thus, we have another self-duality condition Together with other self-duality conditions we have found all self-duality conditions and complete the proof.

Lorentz Invariance in D=1+1+1+3
As covariant symmetry on 2D coordinates xȧ is manifest we only need to examine transformations (I) mixing x 1 with x 2 and (II) mixing x 1 with xȧ. Other transformations needed to examine are just the exchange of 1 ↔ 2, 1 ↔ 3 or 2 ↔ 3.
(I) For the mixing x 1 with xȧ we shall consider the transformation δx˙a = ω˙a 1 x 1 ≡ Λ˙a x 1 , (4.24) then we see that which are zero for self-dual theory and the non-covariant action gives field equations with 6d Lorentz transformation mixing x 1 with x a .
(II) For the mixing x 1 with x 2 we shall consider the transformation then we see that which are zero for self-dual theory and the non-covariant action gives field equations with 6d Lorentz transformation mixing x 1 with x 2 .
In summary, we have found the non-covariant action of self-dual 2-form in decomposition D = 1 + 1 + 1 + 3 and checked that the non-covariant action gives field equations with 6d Lorentz transformation.

Decomposition in Other Spacetime
We have seen that the Lagrangian of self-dual 2-form gauge field could be expressed as many different formulations. The extension to other form is straightforward.
The decomposition in D = D 1 + D 2 + D 3 can also be performed as before. In the case of D = 1 + 1 + 8 then, as that in section 3.1, the spacetime index A is decomposed as A = (1, 2,ȧ), withȧ = (3, ···, 10), and L ABCEF = (L 12ȧḃċ , L˙a˙b˙cḋ˙e, L 1ȧḃċḋ , L 2ȧḃċḋ ). From table 4 we see that Lagrangian can be expressed as Choosing L 12ȧḃċ + L 2ȧḃċḋ is just the decomposition D = 1 + 9 in [14]. The proof of self-dual relation is the same as those in section 3.1.
In conclusion, there are many different formulations of the self-dual gauge field. As the decomposition has many kind it seems not easy to provide a general proof of the self-dual relation in there and we have used some examples to illuminate the property.

Conclusion
In this paper we have first reviewed the Lagrangian of self-dual gauge theory in various non-covariant formulations. Then, we see a simple rule in there and use it to present some new Lagrangian of non-covariant forms of self-dual gauge theory. We see that the existence of gauge symmetry δA = Φ in the Lagrangian play important role of the self-dual relation. Using this as the guiding principle we have found many different formulations. It is interesting to see that in some cases it remains only one possible choice for the specified decomposition. Especially, we have followed the prescription in [14] to prove the self-dual property in the new Lagrangian in a detailed way. We also have followed the method of Perry and Schwarz in to show that these new noncovariant actions give field equations with 6d Lorentz invariance.
Finally, the covariant form in each decomposition may be found by following the method in [9,10,13] and we leave the study in future research. It also remains to see whether the non-abelin self-dual gauge theory in 1+5 dimension [15] could be decomposed in other way.
Acknowledgments : The author thanks Kuo-Wei Huang for interesting discussion in the initial stage of investigation. This work is supported in part by the Taiwan National Science Council.

APPENDIX A Coordinate Transformation
In this appendix we evaluate in detail the Lorentz transformation of 2-form field strength.
Under the coordinate transformation : x a → x a + δx a we consider the tensor field FMNP (x a ) defined by For the transformation mixing between x 1 with x µ (µ = 1) the relation δx a = ω ab x b leads to δx 1 = −Λ µ x µ and δx µ = Λ µ x 1 in which we define ω µ1 = −ω 1µ ≡ Λ µ . The orbital part of transformation [10] is defined by Note that δ orb is independent of index MNP and is universal for all type tensor.