Micro-Note on solving super p=2 form A MN

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1 Micro-Note on solving super p=2 form A MN Sunny Guha April 10, Introduction Superforms are extended versions of differential forms which include superspace coordinates. For example a one-form superform can be written as, A = dz M A M = dx m A m + dθ α A α + d θ α Ā α (1) The purpose of this note is to solve the constraints and Bianchi identities for a super 2-form Gauge field. An analogous calculation has been done in [1] for 1-form gauge field. We will proceed similarly. Variation of a p-form gauge field is given in terms of a (p-1)-form. Field strength of a gauge p-form is given as, Bianchi identities are given as, δa p = dk p 1 (2) F p+1 = da p (3) df p+1 = 0 (4) All the details of d operator in superspace are given in [1]. 2 Antisymmetric Gauge Field Second rank antisymmetric gauge field are contained in chiral spinor superfield. Chiral spinor superfield is given as [2], V α = η α + θ α (A(x) + ib(x)) + θ β F αβ + θ 2 (χ α + i α β η β) (5) Its gauge transformation is given as, The spinor superfield strength is given as, δψ α = i D 2 D α K (6) G = D α Ψ α + D α Ψ α (7) So from this we know that once we enter the superfield language every super gauge field component can be represented in terms of chiral spinor superfield. 1

2 3 Field Strengths Here we have listed field strengths of p=1,2,3 and 4 superforms. Following from (2-4) equations for p-form become Bianchi identities for (p-1)-form and hence those equations must be set to 0. Similarly variation of p-forms are given by (p-1)-form set of equations. 3.1 p=1 Field strength of the zero-form A is a one-form, 3.2 p=2 F α = D α A (8) F a = a A (9) F ab = a A b b A a (10) F αβ = D (α A β) (11) F αb = D α A b b A α (12) F α β = D α Ā β + D βa α i2σ c A (13) α β The gauge transformation of 1-form gauge field can be read from last section, δa α = D α B (14) here B is just a zero-form. δa a = a B (15) 3.3 p=3 F αβγ = D (α A βγ) (16) F αβ γ = D (α A β) γ + D γ A αβ + i2σ c (α γ A β)c (17) F αβc = D (α A β)c + c A αβ (18) F α βc = D α A βc + D βa αc + c A α β + iσ d α β A cd (19) F αbc = D α A bc b A αc c A αb (20) F abc = a A bc b A ca c A ab (21) 2

3 3.4 p=4 B αβνδ = D (α F βνδ) (22) B αβν δ = D (α F βν) δ + D δf αβν i2σ d (α δ F βν)d (23) B αβ ν δ = D (α Fβ) ν δ + D ( ν F αβ δ) i2σ d (α( ν F β) δ)d (24) B αβνd = D (α F βν)d d F αβν (25) B αβ νd = D (α F β) νd + D ν F αβd d F αβ ν i2σ d (α ν F β)de (26) B αβcd = D (α F β)cd + [c F αβd] (27) B α βcd = D α F βcd D βf αcd + [c F α βd] iσ e α β F cde (28) B αbcd = D α F bcd [b F αcd] (29) B abcd = [a F bcd] (30) These equations become the Bianchi identities for a 2-form gauge field and hence each of these should vanish. 4 Constraints To consistently solve Bianchi identities (22-30) for a 2-form we need certain constraints on the Field strengths. The reason for constraints is that when we went from an ordinary form to a superform we increased the degrees of freedom. To restore the actual number we need to impose certain constraints (for eg putting certain field strengths to 0) such that we get our original dofs back. The constraints should be consistent with Bianchi identities. To begin with we will impose the following constraints, F αβγ = 0 F αβ γ = 0 F αβc=0 (31) Let us first look at F αβγ = 0 constraint. This trivially satisfies the Bianchi identity (22). Now to solve it we look at field strength equation (16) D α A βγ + D β A γα + D γ A αβ = 0 (32) Introducing a new spinor superfield U β we see that the following form of A αβ solves the constraint, A αβ = D α U β + D β U α (33) Thus we have solved A αβ in terms of a spinor superfield U β. (31b) and (31c) again trivially satisfy the Bianchi identity [23]. Now plugging the constraints in [24] we get, F α βc = iσ cα βg (34) Here G is a Superfield. Thus overall we have the constraint (31) and (34). So now we have everything we need to solve for A MN. These constraints are sufficient to solve the 2-form. 3

4 5 Solution We will solve for A MN using equations(16-21) and (31,34). In the last section we have already found A αβ. Using constraint (31b) in field equation (17) we get A βc in terms of A β ν and U α, i2σ c (α ν A β)c = D (α A β) ν D ν A αβ (35) Now we plug this in the field strength equation (18) and also use constraint (31c). Finally we get the solution as, A α β = D α Ū β + D βu α i2σ c α β U c (36) The U c superfield was added as it would vanish in a symmetric combination. Now we will solve for A αb Using constraint (31) in Field equation (18) we get, D α A β ν + D β A α ν + D ν A αβ + i2σ c (α ν A β)c = 0 (37) We have put equation (36) to use here, D α D ν U β + D β D ν U α + D ν D α U β + D ν D β U α 2iσ d (α ν D β)u d + i2σ c (α ν A β)c = 0 (38) Simplifying this we get, i2σ c (α ν A β)c = 2iσ d (α ν D β)u d + i2σ c (α ν A β)c (39) From this we can read of the expression A βc. A βc = D α U b b U α + iσ bα ν V ν (40) The additional chiral superfield V α was added since it does not appear in a symmetric combination of σ c (α ν A β)c. In a similar manner A ab can be found. To summarize we have found the following results for A MN, A αβ = D α U β + D β U α (41) A α β = D α Ū β + Ḋ βu α i2σ c α β U c (42) A αb = iσ b α ν T ν + D α U b b U α (43) ν δ A ab = i 4 [σνδ ab D ν T δ + σ D ab ν T δ] + a U b b U a (44) 6 Transformation D α Vβ = 0 (45) With this we have expressed all components of A MN in terms of U a,u α and V α. Now we know that variation of a p-form is given in terms of (p-1)-form from equation (11). So we need to have, δa αβ = D α B β + D β B α (46) 4

5 Now from (33) we have, Thus we can infer from (46) and (47) that, δa αβ = D α δu β + D β δu α (47) δu α = B α + id α Λ (48) Here Λ is a real superfield. Following a similar procedure we find that, δt α = i D 2 D α Λ (49) δu α = B α + id α Λ (50) δu a = B a + σa β ν [D β, D ν ]Λ (51) B a and B α can be used to gauge away U α and U a. Thus everything an be written in terms of spinor superfield V α. From the Field Strength equation (19) we find the form of G, G = 1 2 (Dα V α + D α V α ) (52) This matches with the expected value and thus, we have fully solved the equations. 7 Conclusion We saw that to solve Bianchi identities we need the following constraints. F αβγ = 0 F αβ γ = 0 F αβc=0 (53) F α βc = iσ cα βg (54) This gave us each component of A MN in terms of a spinor superfield V α, just like we wanted as in [2]. We also saw that the field strength was given as G (52) whose form again matched with [2]. Hence we consistently solved the p=2 superform. The non-trivial aspect of this calculation is finding the constraints. In this short note our main concern was to solve the two form relations and hence not a lot of thought was put in the origin of these constraints. For a detailed account on how to find the constraints refer to [4]. 5

6 8 References 1. J.Wess and J.Bagger, Supersymmetry and Supergravity,Princeton Univ Press 2. Warren Siegel, Gauge Spinor Supefields as Scalar Multiplet, Physics Letter Vol 85B,number 4 3. S. James Gates, Super p-form Gauge Superfields, Nucl. Phys. B184 (1981) W.D. Linch III and S. Randall, Superspace de Rham Complex and Relative Cohomology, JHEP 09(2015)190,[hep-th/ ] 6