1 11D spinor conventions

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1 This note spells out some details in the on-shell superfield formulation of D supergravity in the 980 paper of Brink and Howe. D spinor conventions. Gamma matrices Our convention for -dimensional Gamma matrices is as follows. The spinor indices are raised and lowered by contracting with the anti-symmetric invariant tensor C αβ or its inverse C αβ on the left. Cαβ and C αβ by convention obey For instance, Note that while The Gamma matrices obey C αβ C βγ = δα γ = C αβ C γβ. (.) (Γ m ) αβ = C αγ C βδ Γ m γδ. (.2) M α = C αβ M β = C αβ C βγ M γ = δ α γ M γ, (.3) M α N α = C αβ M β N α = M α N α. (.4) Γ (m αβ (Γn) ) βγ = η mn δ γ α = (Γ (m) ) γβ Γ n) βα. (.5) Note that Γ m αβ and Γαβ m are symmetric in (αβ). The anti-symmetrized product of gamma matrices are defined by products where an upper left index is contracted with a lower right index, for instance (Γ mn ) αβ = C αα (Γ [m ) α γ Γ n] γβ = C ββ (Γ [n ) β γ (Γ m] ) γα = (Γ mn ) βα. (.6) In summary, Γ m αβ, (Γmn ) αβ and (Γ m m 5 ) αβ are all symmetric in α, β, whereas C αβ, (Γ mnp ) αβ and (Γ mnpq ) αβ are anti-symmetric in α, β. The trace of a matrix M α β is trm = M α α = M α α. In connection to 0D spinors and 0D gamma matrices, we can take an D spinor U α and decompose it into a pair of chiral and anti-chiral 0D spinors, The D gamma matrices decompose according to U α (u a, ũ a ). (.7) Γ i αβ (Γ i ab, (Γ i ) ab ), i =,, 0, Γ αβ (δ b a, δ a b ), C αβ (δb a, δ a b ). (.8)

2 .2 Pure spinors The dimensional pure spinor λ α is defined analogously to a 0d pure spinor, in that it obeys the constraint λγ m λ λ α Γ m αβλ β = 0. (.9) While the 0d pure spinor has (complex) degrees of freedom, the d pure spinor has 23 degrees of freedom. To see this, let us write the -dimensional pure spinor in terms of a pair of spin(0) spinors (λ L, λ R ), that obey λ α LΓ i αβλ β L + (λ R) α (Γ i ) αβ (λ R ) β = 0, λ α L(λ R ) α = 0. (.0) Here i is an 0d vector index, and Γ i αβ and (Γi ) αβ are 0d gamma matrices with chiral or anti-chiral spinor indices. Take an arbitrary spin() chiral spinor λ L, and form the complex 0d vector v i = λ α LΓ i αβλ β L. (.) v i is null due to the relation Γ i (αβ (Γ i) γ)δ = 0. Now the condition (λ R ) α (Γ i ) αβ (λ R ) β = v i (.2) may be viewed as 9 independent equations, since the LHS is automatically null as well. Here we are assuming v i is nonzero, otherwise there would be more than 6 9 = 7 degrees of freedom in the solution to (.2) (namely, as we knew). Next, we go back to d and use the -dimensional Gamma matrix relation which implies (Γ m ) (αβ Γ mn γδ) = 0, (.3) (λγ m λ)(λγ mn λ) = (λγ i λ)(λγ in λ) + (λγ λ)(λγ,n λ) = 0. (.4) Once we have λγ i λ = 0, by our construction of λ L, λ R, with a total of = 23 degrees of freedom, we know that (λγ λ)(λγ,n λ) = 0. (.5) This relation is nontrivial of course only for n = j an so(0) index, and we have λγ,j λ = λ L Γ,j λ L + λ R Γ,j λ R = λ L Γ j λ L λ R Γ j λ R = 2v j, (.6) which is assumed to be nonzero. Therefore, we must have λγ λ = 2(λ L λ R ) = 0. (.7) Namely, the last equation of (.0) is automatically obeyed. 2

3 2 On-shell superspace formulation of -dimensional supergravity The -dimensional supergravity can be formulated in ( 32)-superspace based on spin() connections. The superspace coordinates are denoted Z M (x m, θ α ) as usual. We denote by E α and E m the super-vierbeins, which are -forms on the superspace, of the form (adopting Brink-Howe convention) E α = dz M E M α, E m = dz M E M m. (2.) The spin()-connection -form consists of Ω (αβ) or Ω [mn], related by Ω αβ = 4 (Γmn ) αβ Ω mn. (2.2) We have Ω M N dz P Ω P M N. Denote by t mn the spin() generators, with commutators (Euclidean convention) [t mn, t pq ] = δ np t mq + δ mq t np δ nq t mp δ mp t nq. (2.3) It will be convenient to define the spinorial Lorentz generators t αβ Γ mn αβ t mn, so that Ω α β t β α = Ω m n t n m, or equivalently Ω αβ t αβ = Ω mn t mn. (2.4) We see that our normalization convention is such that The torsion 2-form is defined by The curvature 2-form is defined by t mn = 4 Γmn αβ t αβ, t αβ = 6 (Γ mn) αβ t mn. (2.5) T M = 2 EP E N T NP M = de M + E N Ω N M. (2.6) R M N = 2 EQ E P R P Q,M N = dω M N + Ω M R Ω R N. (2.7) Note that the only nontrivial components of R M N are R α β and R m n. The latter are related by R αβ = R (αβ) = dω αβ + Ω α γ Ω γβ = 4 (Γmn ) αβ Ω mn + 6 (Γmn Γ pq ) (αβ) Ω mn Ω pq = 4 (Γmn ) αβ Ω mn + 4 (Γmq ) αβ Ω m n Ω nq (2.8) = 4 (Γmn ) αβ R mn. 3

4 In components, the relations are R P Q,αβ = 4 (Γmn ) αβ R P Q,mn. Note that Grimm-Wess-Zumino and Brink-Howe used a strange convention of taking d on a differential form from the right, and changes sign each time it passes through a -form towards the left. This is important in verifying the Bianchi identity with their sign convention. One may define the covariant exteria derivative as DF M N df M N Ω M P F P N + F M Q Ω Q N +. (2.9) There are Bianchi identities DT M dt M + T N Ω N M = E N R N M, DR M N dr M N Ω M P R P N + R M P Ω P N = 0. (2.0) Let us verify these (here we adopt the bizarre sign convention of Grimm-Wess-Zumino and Brink-Howe): DT M = E N dω N M de N Ω N M + (de N + E P Ω P N ) Ω N M = E N R N M, (2.) and (suppressing indices which are contracted upper left to lower right) DR = ΩdΩ dωω Ω(dΩ + Ω 2 ) + (dω + Ω 2 )Ω = 0. (2.2) Let us write out these Bianchi identities in components of the torsion and curvature. First, E N E Q E P R M P Q,N = d(e P E N T M NP ) + E Q E P T N M P Q Ω N = E P E N (dt M NP + T R NP Ω M R ) + E P de N T M NP de P E N M T NP = E P E N (E Q D Q T M NP + Ω Q N T M QP + Ω Q P T M NQ ) + E P (T N E Q Ω N Q )T M NP (T P E Q Ω P Q )E N M T NP = E P E N E Q D Q T M NP + 2 EP E R E Q T N QR T M NP 2 ER E Q T P QR E N M T NP R M P Q,N D P T M QN T R P Q T M RN = 0. P QN cyclic (2.3) 4

5 Similarly, we have 0 = d(e Q E P R N P Q,M ) Ω P M E R E Q R N QR,P + E R E Q R P N QR,M Ω P = E Q E P (dr N P Q,M Ω R M R N P Q,R + R S P Q,M Ω N S ) + E Q de P R N P Q,M de Q E P N R P Q,M = E Q E P (E S D S R N P Q,M + Ω R P R N RQ,M + Ω R Q R N P R,M ) + E Q (T P E R Ω P R )R N P Q,M (T Q E R Ω Q R )E P N R P Q,M = E Q E P E S D S R N P Q,M + 2 EQ E S E R T P RS R N P Q,M 2 ES E R T Q RS E P N R P Q,M D S R N P Q,M + T R SP R N RQ,M = 0. SP Q cyclic (2.4) The covariant derivatives also obey a commutation relation, which can be seen from D 2 F M = d(df M ) + DF N M Ω N = d(f N Ω M N ) + DF N M Ω N = F N dω M N (DF N F P Ω N P )Ω M N + DF N M Ω N = F N M R N (2.5) = 2 F N E Q E P R P Q,N M. The LHS can be written as (the awkward ordering convention strikes again) D 2 F M = D N (D P F M E P )E N = 2 D P F M E R E Q T QR P + 2( [DN, D P ]F M) E P E N. (2.6) From this we derive and more generally [D N, D P ]F M = T NP Q D Q F M + F Q R NP,Q M, (2.7) [D M, D N ] = T MN P D P + R MN (2.8) acting on any tensor. Note that unlike in gauge theories, here the Lorentz generators in R MN will act on superspace indices of covariant derivatives as well, which would be crucial if one tries to derive Bianchi identities from this commutation relation. The next piece of ingredient is a closed super 4-form H = 4! EQ E P E N E M H MNP Q. (2.9) 5

6 The condition dh = 0 can be written in components as 0 = E Q E P E N E M dh MNP Q + 4E Q E P E N de M H MNP Q = E Q E P E N E M (E R D R H MNP Q + 4Ω R M H RNP Q ) + 4E Q E P E N (T M E R Ω M R )H MNP Q = E Q E P E N E M E R D R H MNP Q + 2E Q E P E N E S E R T M RS H MNP Q D [R H MNP Q] + 2T [RM S H SNP Q] = 0. (2.20) So far, all we did is that we have defined the torsion and curvature in terms of the frame and spin connection, and introduced the closed 4-form H. Now we will introduce the dynamical equations. These equations can be entirely phrased in terms of simple constraints on components of the torsion and of H. They are T αβ m = iγ m αβ, T np m = 0, T αn m = 0, T αβ γ = 0, (2.2) and H αβmn = i(γ mn ) αβ, H αβγδ = H αβγm = H αmnp = 0. (2.22) The dynamical components of T and H are T αm β, T mn α, and H mnpq. (2.23) Note that H mnpq obeys D [m H npqr] = 0 since T mn p vanishes. Note also that the (αβγδm) component of the Bianchi identity for H is solved by the torsion constraints because of the d gamma matrix identity Γ m (αβ (Γ nm) γδ) = 0. Note that the super-covariant derivatives of T αβ m and H αβmn vanish as well: DT αβ m = T αβ n Ω n m Ω α γ T γβ m Ω β γ T αγ m = iγ n αβω n m + 2iΓ m γ(αω β) γ = iγ n αβω n m + i 2 Γm γ(α(γ pq ) β) γ Ω pq (2.24) = iγ n αβω n m + i 2 (Γpq Γ m ) (αβ) Ω pq = iγ n αβω n m + iγ p αβ Ω p m = 0, and DH αβmn = 2Ω (α γ H β)γmn 2Ω m p H αβpn = 2i(Γ mn ) γ(α Ω β) γ + 2i(Γ pm ) αβ Ω n p = i 2 (Γpq Γ mn ) (αβ) Ω pq + 2i(Γ pm ) αβ Ω n p (2.25) = 2i(Γ p n) αβ Ω pm + 2i(Γ pm ) αβ Ω n p = 0. 6

7 3 Recovering the equations of motion from torsion and H constraints Now we will examine the consequences of these constraints on the Bianchi identities. This will allow us to determine T αm β in terms of H mnpq, and T mn α in terms of the supercovaraint derivative of H mnpq. In the end, H mnpq is not an independent superfield, but is rather determined in terms of the torsion components, which are defined in terms of the frame and Lorentz connection. We will inspect the Bianchi identity for H, R, and DR, respectively. The nontrivial components of the Bianchi identity for H are D [m H npqr] = 0, ) D α H mnpq + 2 (T β mn H βpqα + T β np H βqαm + T β pq H βαmn = 0 D α H mnpq + 6i(Γ mn ) αβ T β pq = 0, 3 T αβ q H qmnp T γ βm H γnpα + T γ pα H γβmn = 0 iγ q αβ H qmnp 6i(Γ mn ) γ(α T γ β)p = 0, T (αβ n H γδ)mn = 0 Γ m (αβ(γ mn ) γδ) = 0. (3.) Here we used the underline notation for complete anti-symmetrization for bosonic indices or complete symmetrization for fermionic indices. The last equation is automatically satisfied thanks to the d gamma matrix identity. The third equation relates T αm β to H mnpq, H mnpq = 3 6 Γαβ q (Γ mn ) γα T βp γ = 3 6 (Γ qγ mn ) β γt βp γ (3.2) The complete anti-symmetry on [mnpq] implies that T βp γ when contracted with (Γ k ) β γ vanishes, and that H mnpq = 3 6 (Γ mnp) β α T αq β 3 6 tr(γ [mnpt q] ). (3.3) To write T αm β in terms of H mnpq, note that the matrix (T m ) α β T αm β can be decomposed in terms of anti-symmetrized product of gamma matrices, in the form From T αm β = c (Γ npq ) α β H mnpq + c 2 (Γ mnpqr ) α β H npqr. (3.4) 6 Γq αβ H qmnp = (T [p Γ mn] ) (αβ) = c H pqrs (Γ qrs Γ mn ) (αβ) + c 2 H qrst (Γ pqrst Γ mn ) (αβ) = 6c H pqmn Γ q αβ + c H pqrs (Γ qrs mn) αβ + 8c 2 H qrs m(γ qrsnp ) αβ, (3.5) 7

8 we find and so c = 36, c 2 = c 8 = 288, (3.6) T αm β = 36 (Γnpq ) α β H mnpq 288 (Γ mnpqr) α β H npqr. (3.7) The second equation of (3.) implies D α H mnpq = 6i(Γ mn ) αβ T pq β 6i(Γ [mn T pq] ) α. (3.8) This is not the complete set of equations on D α H mnpq or T mn α, however. We will see that there are additional constraints on T mn α coming from the Bianchi identities for R. Let us turn to the Bianchi identity for R. The component equations are (recall that R P Q,m α and R P Q,α m vanish identically) R [pq,n] m = 0, D [p T qn] α T pq β T βn α = 0, 2R α[m,n] p T mn β T βα p = 0 R α[m,n] p = i 2 Γp αβ T mn β i 2 (T mnγ p ) α, R β mn,α D α T β mn 2D [m T β n]α 2T γ αm T β γn = 0, R n αβ,m 2T γ m(α T n β)γ = 0 R n αβ,m = 2iΓ n γ(αt γ β)m 2i(T m Γ n ) αβ, 2R γ m(α,β) 2D (α T γ β)m T n αβ T γ nm = 0 (3.9) R γ m(α,β) D (α T γ β)m i 2 Γn αβt γ mn = 0, R δ (αβ,γ) + T m (αβ T δ γ)m = 0 R δ (αβ,γ) iγ m (αβt δ γ)m = 0. Now we will use the relation R P Q,αβ = 4 (Γmn ) αβ R P Q,mn, (3.0) 8

9 and rewrite the above equations in terms of R P Q,mn only, R [pq,n] m = 0, D [p T qn] α + T pq β T βn α = 0, R α[m,n] p = i 2 Γp αβ T mn β i 2 (T mnγ p ) α, 4 (Γpq ) α β R mn,pq D α T mn β 2D [m T n]α β 2T αm γ T γn β = 0, R αβ,m n = 2iΓ n γ(αt β)m γ 2i(T m Γ n ) αβ, 4 (Γnp ) (α γ R mβ),np D (α T β)m γ i 2 Γn αβt mn γ = 0, 4 (Γmn ) (α δ R βγ),mn iγ m (αβt γ)m δ = 0. (3.) Let us first consider the constraints on R αβ,mn, coming from the fifth and last equations in (3.). We can eliminate R αβ,mn in the last equation, and write it as a constraint on T β αm, i 2 (Γmn ) δ (α (T m Γ n ) βγ) iγ m (αβt δ γ)m = 0. (3.2) Using T m = 36 H mnpqγ npq 288 Hnpqr Γ mnpqr, (3.3) we can write (3.2) as 2 (Γmn ) (α δ (H mpqr Γ pqr Γ n + 8 Hpqrs Γ mpqrs Γ n ) βγ) Γ m (αβ(h mnpq Γ npq + 8 Hnpqr Γ mnpqr ) γ) δ = 0 2 (Γmn ) δ (α (3H mnpq Γ pq + 8 Hpqrs Γ mnpqrs ) βγ) Γ m (αβ(h mnpq Γ npq + 8 Hnpqr Γ mnpqr ) δ γ) = 0 [ 3 H mnpq 2 (Γmn ) δ (α (Γ pq ) βγ) Γ m (αβ(γ npq ) δ γ) + 6 (Γ rs) δ (α (Γ mnpqrs ) βγ) ] 8 (Γ r) (αβ (Γ mnpqr δ ) γ) = 0. (3.4) One can verify using GAMMA.m package that the RHS when contracted with (Γ a ) βγ, (Γ ab ) βγ, and (Γ abcde ) βγ indeed vanishes, and hence this component of the Bianchi identity is automatically satisfied. Next let us consider the constraints on R αm,np, from the third and sixth equation of (3.). Using the third equation of (3.), we can write (Γ np ) α γ R mβ,np = 2(Γ np ) α γ R β[m,n]p + (Γ np ) α γ R β[n,p]m = i(γ np ) α γ (T mn Γ p ) β i 2 (Γnp ) α γ (T np Γ m ) β. (3.5) Plugging this into the sixth equation of (3.), we find i 4 (Γnp ) (α γ (T mn Γ p ) β) i 8 (Γnp ) (α γ (T np Γ m ) β) D (α T β)m γ i 2 Γn αβt mn γ = 0, (3.6) 9

10 Using the expression of T βm γ in terms of H mnpq, and D α H mnpq in terms of T mn α, we can write D α T βm γ = 36 (Γnpq ) β γ D α H mnpq 288 (Γ mnpqr) β γ D α H npqr = i 6 (Γnpq ) β γ (Γ [mn T pq] ) α i 48 (Γ mnpqr) β γ (Γ np T qr ) α. Plugging this into (3.6), we now obtain a linear constraint on T mn α entirely, i 4 (Γnp ) (α γ (T mn Γ p ) β) i 8 (Γnp ) (α γ (T np Γ m ) β) + i 6 (Γnpq ) (α γ (Γ [mn T pq] ) β) + i 48 (Γ mnpqr) (α γ (Γ np T qr ) β) i 2 Γn αβt mn γ = 0 2 (Γnp ) γ (α(γ p T mn ) β) + 4 (Γnp ) γ (α(γ m T np ) β) 3 (Γnpq ) γ (α(γ [mn T pq] ) β) + 24 (Γ mnpqr) γ (α(γ np T qr ) β) Γ n αβt mn γ = 0. (3.7) (3.8) We now contract this equation with (Γ a ) αβ, (Γ ab ) αβ, and (Γ abcde ) αβ respectively. The relation obtained by contraction with (Γ a ) αβ is equivalent to (Γ n T mn ) α = 0, (Γ mn T mn ) α = 0, (Γ mnp T mn ) α = 0. (3.9) Once this is obeyed, one can then verify that the contraction of (3.8) with (Γ ab ) αβ also vanishes. The contraction with (Γ abcde ) αβ should then vanish as well, but I haven t verified this by mathematica. It follows from (3.9) that we have the identity Now multiplying Γ pq on (3.8), we have (Γ pq Γ [mn T pq] ) α = 7(T mn ) α. (3.20) T mn α = i 42 (Γpq D) α H mnpq. (3.2) Plugging this back into (3.8), we obtain an equation for H, D α H mnpq = 7 (Γ [mnγ rs D) α H pq]rs. (3.22) Next we examine the fourth equation of (3.). Contracting it with (Γ pq ) β α, we find R mn,pq = 6 (Γ pqd) α T mn α 8 tr(γ pqd [m T n] ) 8 tr(γ pqt [m T n] ) = 6 Dα (Γ pq T mn ) α 8 tr(γ pqd [m T n] ) 8 tr(γ pqt [m T n] ) (3.23) = 6 Dα (Γ pq T mn ) α 8 tr(γ pqt [m T n] ). 0

11 Contracting n with q, we find an expression for the Ricci tensor (Ric) mp = R mn,p n = 6 Dα (T mp ) α 8 tr(γ p n T [m T n] ). (3.24) To simplify this further, we make use of D α (T mp ) α = D α T α mp = i 42 (Γqr ) αβ D α D β H mpqr = i 84 (Γqr ) αβ {D α, D β }H mpqr = i ( ) 84 (Γqr ) αβ iγ s αβd s H mpqr 4R n αβ,m H npqr = 2 2 (Γqr ) αβ (T m Γ n ) αβ H npqr (3.25) Now using = 2 2 tr(γqrn T m )H npqr = 6 63 H [m qrs H p]qrs = 0. tr(γ p n T [m T n] ) = 4 9 H m qrs H pqrs H m qrs H pqrs 8 δ mnh pqrs H pqrs = 2 3 H m qrs H pqrs 8 δ mnh pqrs H pqrs, (3.26) we arrive at (Ric) mn = R mp,n p = 2 H m pqr H npqr + 44 δ mnh pqrs H pqrs. (3.27) There is another equation involving D m H mnpq, which can be obtained by contracting the fourth equation of (3.) with (Γ p ) β α. This gives (Γ p D) α T mn α 2tr(Γ p D [m T n] ) 2tr(Γ p T [m T n] ) = 0 (Γ p D) α T mn α + 2tr(Γ p T [m T n] ) = 0 i 42 (DΓ pγ qr D)H mnqr ɛ mnpa a 8 H a a 2 a 3 a 4 H a 5a 6 a 7 a 8 = 0. (3.28) To proceed, we act on (3.8) with Γ m, which together with the constraints on T mn α gives (Γ m D) α H mnpq = 8i(Γ [n T pq] ) α = 3 7 (Γ [nγ rs D) α H pq]rs. (3.29) Now anti-symmetrize [mnp] on the third line of (3.28), then using (3.29), we obtain i 8 (DΓq D)H mnpq ɛ mnpa a 8 H a a 2 a 3 a 4 H a 5a 6 a 7 a 8 = 0 D m H mnpq = 52 ɛ npqa a 8 H a a 2 a 3 a 4 H a 5a 6 a 7 a 8. (3.30)

12 This again differs from Brink-Howe by a factor of 2/3. Next, we want to reduce D α D β H mnpq to vector derivatives on H, H 2, or Riemann curvature terms. Firstly, we already know D (α D β) H mnpq = 2 {D α, D β }H mnpq = i 2 Γr αβd r H mnpq + 2R αβ,[m r H npq]r = i 2 Γr αβd r H mnpq i 3 (Γ st) αβ H [m rst H npq]r i 72 (Γ abcd[m r ) αβ H abcd H npq]r. If we anti-symmetrize [α, β], on the other hand, we have (3.3) and as well as D α D α H mnpq = 6i(Γ [mn ) α β β D α T pq] ( ) α = 6i(Γ [mn ) β 4 R pq],rs(γ rs ) β α 2D p T β q]α 2T γ β αp T γq] = 96iR [mn,pq] 2itr(Γ [mn D p T q] ) 2itr(Γ [mn T p T q] ) = 2itr(Γ [mn T p T q] ) = 32i 3 H [mn rs H pq]rs, D α (Γ abc ) β α D β H mnpq = 6i(Γ abc Γ [mn ) α β βd α T pq] ( ) = 2i(Γ abc Γ [mn ) α β D p T β q]α + T γ β αp T γq], (3.32) (3.33) D α (Γ abcd ) β α D β H mnpq = 6i(Γ abcd Γ [mn ) α β βd α T pq] ( ) = 6i(Γ abcd Γ [mn ) α β 4 R pq],rs(γ rs ) β α 2D p T β q]α 2T γ β αp T γq] = 3i ( ) 2 tr(γabcd Γ [mn Γ rs )R pq],rs 2i(Γ abcd Γ [mn ) α β D p T β q]α + T γ β αp T γq] ( ) = 52iδ [ab [mn R pq], cd] 2i(Γ abcd Γ [mn ) α β D p T β q]α + T γ β αp T γq]. (3.34) In the last expression, it appears that we are stuck with the Riemann (or Weyl) curvature and the expression cannot be reduced to vector derivatives on H. 2