Lorentz Covariance in Epstein-Glaser Renormalization

Home , Lorentz covariance

arXiv:hep-th/9904136v2 27 Apr 1999 nrdcincnb on n[r9] eeaiaint ma to generalization a [Pra99], Let in found be can introduction aeili nteapni.I h atscin5w ieacov a tran give fourier we by 5 momentum section spacelike) last (totally the arbitrary calculu In for group appendix. the permutation in some s is general material these symm in for the absent Since construction is inductive covariance. problem an Lorentz spinorial give and we tensorial 4 3, sections solu covariant In group a to leads that analysis for cohomological solution covariant a renormaliza construct invariant to Lorentz methods a same for the formula use We a gave a we There n prto ndsrbtosi endb h correspondi the by defined is distributions on operation any okspotdb h F rdirekle “Theoretische Graduiertenkolleg DFG the by supported Work erve h xeso rcdr nteE prah[EG73]. approach EG the in procedure extension the review We erve h Gsbrcinadgv oeueu omlsin formulas useful some give and subtraction EG the review We a is mainly paper this since brief rather is introduction Our OET OAINEI EPSTEIN-GLASER IN COVARIANCE LORENTZ D ω ain PZsbrcina oal pclk oetmuse momentum spacelike totally automat at This subtraction BPHZ approach. variant EG the in renormalization covariant A ( R BSTRACT d ) etesbpc fts ucin aihn oorder to vanishing functions test of subspace the be egv xlctadidciefrua o h constructi the for formulas inductive and explicit give We . I NTTTFRTERTSH PHYSIK THEORETISCHE FÜR INSTITUT II. RENORMALIZATION .T 1. -26 ABR GERMANY HAMBURG D-22761 EMAIL: UUE HUSE149 CHAUSSEE LURUPER HAMBURG UNIVERSITÄT IKPRANGE DIRK HE I NTRODUCTION EG SUBTRACTION 1 lmnaticepyi”Hamburg. Elementarteilchenphysik” sformation. u o asestheories. massless for ful ini eiwdi eto 2. section in reviewed is tion clypoie o co- a for provides ically gato ntestfunctions. on action ng er fteoevariable one the of metry iod n[F6 BF99]. [BF96, in nifolds otnaino [BPP99]. of continuation ratBH subtraction BPHZ ariant iletr Necessary enter. will s oevariables. more ltosi h aeof case the in olutions ini n coordinate. one in tion no Lorentz a of on ω oegeneral more A t0 susual, As 0. at eto .The 1. section 2 DIRK PRANGE Define Rd ω Rd ϕ ϕ W(ω;w) : D( ) D ( ), W(ω;w) , → α→ ϕ ϕ x ∂ ϕ 1 W(ω;w) (x) = (x) w(x) ∑ α α w− (0), (1) − α ω ! | |≤ with w D(Rd),w(0) = 0. If 0t is a distribution on D(Rd 0 ) with singular order ∈ 6 ω \{ } ω then 0t can be defined uniquely on D (Rd), and ϕ . 0 ϕ t(ω;w), = t,W(ω;w) (2) defines an extension – called renormalization – of 0t to the whole testfunction space, d t ω D (R ). With the Leibnitz rule ( ;w) ∈ ′ ∂ γ 1 ∂ ∂ γ( fg) = ! ∑ ν µ f νg, (3) µ+ν=γ µ! ! we find α ∂ x ∂ ϕ 1 ∂ ϕ γ w ∑ α α( w− )(0) (0) = γ (0), (4) α ω ! | |≤ for γ ω, verifying the projector properties of W. The W -operation (1) is simplified | | ≤ if we require w(0) = 1 and ∂αw(0) = 0, for 0 < α ω (this was our assumption in | | ≤ 1 1 [BPP99]). It can be achieved by the following V -operation (∂µw− means ∂µ(w− )): µ d d . x 1 Vω : D(R ) D(R ), (Vωw)(x) = w(x) ∑ ∂µw− (0), (5) 7→ µ ω µ! | |≤ where w(0) = 0 is still assumed. We can write W as 6 α ϕ ϕ x ∂ ϕ W(ω;w) (x) = (x) ∑ α Vω α w α (0). (6) − α ω ! −| | | |≤ The extension (2) is not unique. We can add any polynomial in derivatives of δ up to order ω: α ϕ ϕ a ∂ ϕ t(ω;w), = t(ω;w), + ∑ α α (0), (7) α ω ! | |≤ or rearranging the coefficients α ϕ c ∂ ϕ 1 = t(ω;w), + ∑ α α w− (0) (8) α ω ! | |≤ α α α ω Since W(ω;w)(wx ) = W(ω;w)(x Vω α w) = 0 for , c resp. a are given by −| | | | ≤ α α α α a = t(ω;w),x Vω α w , c = t(ω;w),x w . (9) −| |

LORENTZCOVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 3 They are related through:

µ µ α α c 1 α α a a = c ∑ ∂µw− (0), c = a ∑ ∂µw(0), 1 α ω, µ ω α µ! µ ω α µ! ≤| | ≤ | |≤ −| | | |≤ −| | µ µ 0 c 1 0 a a = ∑ ∂µw− (0), c = ∑ ∂µw(0). (10) µ ω µ! µ ω µ! | |≤ | |≤ The equation for a follows from the Leibnitz rule in (8), while the equation for c is derived from (9). In quantum field theory the coefficients a are called counter terms. They are not arbitrary. They have to be chosen in such a way that Lorentz covariance of 0t is pre- served in the extension. This follows in the next sections. The remaining freedom is further restricted by discrete symmetries like permutation symmetry or C,P and T symmetries. At the end gauge invariance or renormalization constraints will fix the extension uniquely. But up to now there is no local prescription of the latter.

2. THE G-COVARIANT EXTENSION We will ﬁrst deﬁne the notion of a G-covariant distribution. So let G be a linear transformation group on Rd i.e. x gx, g G. Then 7→ ∈ α α β α x g βx =(gx) (11) 7→ denotes the corresponding tensor representation. G acts on functions in the following way:

. 1 (gϕ)(x) = ϕ(g− x), (12) so that D is made a G-module. We further have

g(ϕψ)=(gϕ)(gψ), (13) α 1 α 1 x ∂α(g− ϕ)=(gx) g− (∂αϕ), (14) α 1 α x ∂α(g− ϕ)(0)=(gx) ∂αϕ(0). (15)

0 d Now assume we have a distribution t D′(R 0 ) that transforms covariantly under the Group G as a density, i.e. ∈ \{ }

0t(gx) detg = D(g)0t(x), (16) | | where D is the corresponding representation. That means: . 0t,gψ = D(g)0t,ψ = D(g) 0t,ψ . (17)

4 DIRK PRANGE We will now investigate what happens to the covariance in the extension process. We compute: 1 D(g) t ω ,g− ϕ t ω ,ϕ (18) ( ;w) − ( ;w) 0 1 0 = D(g) t,W ω g− ϕ t,W ω ϕ (19) ( ;w ) − ( ;w) α (6) 0 1ϕ x ∂ 1ϕ 0 ϕ = D(g) t,g− ∑ α Vω α w α(g− )(0) t,W(ω;w) (20) * − α ω ! −| | + − | |≤ α (13,15) 0 1 ϕ x ∂ ϕ 0 ϕ = D(g) t,g− ∑ α gVω α w α (0) t,W(ω;w) (21) * − α ω ! −| | + − | |≤ ∂ ϕ (17) 0 α I α (0) = ∑ t,x ( g)(Vω α w) α (22) α ω − −| | ! | |≤ . α ∂αϕ(0) = ∑ b (g) α . (23) α ω ! | |≤ Then (23) defines a map from G to a finite dimensional complex vectorspace. Now we follow [SP82], [Sch95][chapter 4.5]: Applying two transformations α 0 α b (g1g2) = t,x (I g1g2)(Vω α w) (24) − −| | 0 α = t,x ((I g1) + g1(I g2 ))(Vω α w) (25) − − −| | α 0 α = b (g1) + detg1 t(g1x),(g1x) (I g2 )(Vω α w) , (26) | | − −| | and omitting the indices we see b(g1g2) = b(g1)+D(g1)g1b(g2), which is a 1-cocycle for b(g). Its trivial solutions are the 1-coboundaries b(g)=(I D(g)g)a, (27) − and these are the only ones if the first cohomology group of G is zero. In that case we can restore G-covariance by adding the following counter terms:

G cov . 1 α t ,ϕ = t ω ,ϕ + ∑ a (w)∂αϕ(0). (28) (ω−;w) ( ;w) α α ω ! D E | |≤ The task is to determine a from (27) and (22): 0 α α t,x (I g)(Vω α w) = (I D(g)g)a (29) − −| | − 3. TENSORIAL LORENTZ COVARIANCE

The ﬁrst cohomology group of L+↑ vanishes [Sch95][chapter 4.5 and references there]. We determine a from the last equation. The most simple solution appears in the case of Lorentz invarinance in one coordinate. This situation was completely 0 analyzed in [BPP99] for ∂αw(0) = δα. The following two subsections generalize the results to arbitrary w, w(0) = 0. 6 LORENTZCOVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 5

4 3.1. Lorentz invariance in R . If we expand the index α into Lorentz indices µ1,...,µn, (29) is symmetric in µ1,...,µn and therefore a will be, too. We just state our result from [BPP99] which is modiﬁed through the generalization for the choice of w:

n 1 [ − ] n 1 !! 2 n 2s !! (µ1...µn) ( ) ( ) (µ1µ2 µ2s 1µ2s a = − ∑ − η ...η − (n + 2)!! (n 2s 1)!! × s=0 − − β 0 2 s µ2s+1 µn 1 2∂µn) µn) ∂ t,(x ) x ...x − x x x β Vω nw , (30) × − − D E if we choose the fully contracted part of a to be zero in case of n being even. We used the notation

1 1 b(µ1...µn) = ∑ bµπ(1)...µπ(n) , b[µ1...µn] = ∑ sgn(π)bµπ(1)...µπ(n) , n! n! π Sn π Sn ∈ ∈ for the totally symmetric resp. antisymmetric part of a tensor.

3.2. Dependence on w. Performing a functional derivation of the Lorentz invariant extension with respect to w, only Lorentz invariant counter terms appear.

Deﬁnition. The functional derivation is given by:

δ . d F(g),ψ = F(g + λψ) . δg dλ λ=0

This deﬁnition implies the following functional derivativ es: δ ϕ ψ 1 αψ ∂ ϕ 1 δ t(ω;w)( ), = ∑ α t(ω;w),x α w− (0), (31) w − α ω ! | |≤ δ 1 α 1 S,Vωw ,ψ = ∑ S,W ω (x ψ) ∂αw− (0), (32) δw h i α! ( ;w) α ω | |≤ for any distribution S.

Proof. We show how to derive the first relation. Inserting the definition we find:

α α d 0 x 1 x 2 t ω λψ (ϕ) = t, ψ ∑ ∂α ϕw− (0) + w ∑ ∂α ϕψw− (0) , dλ ( ;w+ ) − α! α! λ=0 * α ω α ω + | |≤ | |≤

6 DIRK PRANGE using Leibnitz rule and rearranging the summation in the second term,

1 0 ψ α∂ ϕ 1 = ∑ α t, x α w− (0) α ω !* − | |≤ ν α∂ ϕ 1 x ∂ ψ 1 +wx α w− (0) ∑ ν ν w− (0) ν ω α ! + | |≤ −| | 1 0 α ψ ∂ ϕ 1 = ∑ α t,x W(ω α ;w) α w− (0) − α ω ! −| | | |≤ 1 αψ ∂ ϕ 1 = ∑ α t(ω;w),x α w− (0), − α ω ! | |≤ α ϕ αϕ where we used the relation x W(ω α ;w) = W(ω;w)(x ) on the last line. The second equation follows from a similar calculation.−| |

We calculate the dependce of a on w. With (32) we get:

n 1 [ − ] δ n 1 !! 2 n 2s !! (µ1...µn) ( ) ( ) (µ1µ2 µ2s 1µ2s a (w) = − ∑ − η ...η − δw (n + 2)!! (n 1 2s)!! × s=0 − − 1 β β 0 2 s µ2s+1 µn 1 2∂µn) µn) ∂ ψ ∂ 1 ∑ β t,(x ) x ...x − x x x β W(ω n;w)(x ) βw− (0). × β ω n ! − − | |≤ − D E β βψ To condense the notation we again use as a multiindex. Since W(ω n;w)(x ) is sufﬁcient regular, we can put the x’s and derivatives on the left and the same− calculation like in [BPP99] applies. The result is

1 δ ∂βw− (0) β µ1 µn ψ µ1 µn ψ δ a ··· (w), = ∑ β t(ω;w),x x x + w β ω n ! " ··· | |≤ − D E 0, n odd, 2(n 1)!! 2 n β (µ µ µ µ ) − − t ω ,(x ) 2 x ψ η 1 2 η n 1 n , n even. ( (n+2)!! ( ;w) ··· − # Using this result and (31) we find: D E δ ω linv dn n t φ ψ ∑ 2 φ 0 δ (ω;w), , = ( ), w − n=0 n! D E n even . 2(n 1)!! 1 n β 2 2 ψ ∂ 1 dn = − ∑ β t(ω;w),(x ) x βw− (0), (n + 2)!! β ω n ! | |≤ − D E where we set d0 = 1. LORENTZCOVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 7 3.3. General Lorentz covariance. If the distribution 0t depends on more than one variable, 0txα will not be symmetric in all Lorentz indices in general. Since xα transforms like a tensor, it is natural to generalize the discussion to the case, where 0t transforms like a tensor, too. Assume rank(0t) = r, then D(g)g is the tensor representation of rank p = r + n,n = α , in (29). From now on we will omit the indices. So if 4m | | α t D(R 0 ), we denote by x – formerly x – a tensor of rank n built of x1,...,xm. ∈To solve\{ (29)} we proceed like in [BPP99]. Since the equation holds for all g we will I solve for a by using Lorentz transformationse in the infinitesimal neighbourhood of . θ θ If we take αβ = [αβ] as six coordinates these transformations read: 1 αβ g I + θαβl , (33) ≈ 2 with the generators αβ µ αµ β βµ α (l ) ν = η δν η δν . (34) − Then, for an infinitesimal transformation one finds from (29): m αβ . 0 [α∂β] αβ I I αβ B = 2 t,x ∑ x j j (Vω nw) =(l + + l )a, (35) − ⊗···⊗ ··· ⊗···⊗ j=1 α,β being Lorentze four-indices. In [BPP99] our ability to solve that equation heavily relied on the given symmetry, which is in general absent here. Nevertheless we can find an inductive construction for a, corresponding to equation (29) in [BPP99]. We build one Casimir operator on the r.h.s. (the other one is always zero, since we p are in a (1/2,1/2)⊗ representation). The case p = 1. Just to remind that p is the rank of xt, this occurs if either t is a vector and x = 1,(n = 0), or t is a scalar and x = x1,...,xm. (35) gives: 1 αβ 1 αβ e lαβB = lαβl a = 3Ia, (36) e 2 e2 − since the Casimir operator is diagonal in the irreducible (1/2,1/2) representation. The case p = 2. We get 1 αβ αβ (lαβ I + I lαβ)B =( 6I + lαβ l )a. (37) 2 ⊗ ⊗ − ⊗ Since a is a tensor of rank 2, let us introduce the projector onto the symmetric resp. antisymmetric part and the trace:

µν 1 µ ν ν µ µν 1 µ ν ν µ µν 1 µν P = (δρδσ + δρδσ), P = (δρδσ δρδσ), Pη = η ηρσ, (38) S ρσ 2 A ρσ 2 − ρσ 4 2 P = P, PS + PA = I, PS PA = τ, (39) − where τ denotes the permutation of the two indices. Using (34), we ﬁnd 1 αβ lαβ l = 4Pη τ. (40) 2 ⊗ − 8 DIRK PRANGE Now we insert (40) into (37). The trace part will be set to zero again. Acting with PA and PS on the resulting equation gives us two equations for the antisymmetric and symmetric part respectively. This yields:

1 αβ a = (PS + 2PA)(lαβ I + I lαβ)B . (41) −16 ⊗ ⊗

Inductive assumption. Now we turn back to equation (29). We note that any contrac- tion commutes with the (group) action on the rhs. Hence, if we contract (35), we ﬁnd on the rhs:

αβ αβ ηij(l I + + I l )a = ⊗···⊗ ··· ⊗···⊗αβ αβ (l I + + i + + j + + I l )(ηija), ⊗···⊗ ··· ··· ··· ⊗···⊗ where i, j denote the positions of the correspondingb indices,b and the hat means omis- sion. Therefore the rank of (35) is reduced by two and we can proceed inductively. With the cases p = 1, p = 2 solved, we assume that all possible contractionsc of a are known.

Induction step. Multiplying (35) with the generator and contracting the indices yields:

1 αβ 3pI + 2 ∑ τ a = (lαβ I + + I lαβ)B + 8 ∑ Pη a. (42) −2 ⊗···⊗ ··· ⊗···⊗ ij τ Sp i< j p ∈ ≤ The transposition τ acts on a by permutation of the corresponding indices. For a gen- µ ...µ µπ 1 ...µπ 1 eral π Sp the action on a is given by: πa 1 p = a − (1) − (p). In order to solve this equation∈ we consider the representation of the symmetric groups. We give a brief . τ ∑ τ summary of all necessary ingredients in appendix A. So let k = τ Sp be the sum ∈ of all transpositions of Sp. Then kτ is in the center of the group algebra ASp . It can be decomposed into the idempotents e(m) that generate the irreducible representations of

Sp in ASp . 1 kτ = hτ ∑ χ (τ)e . (43) f (m) (m) (m) (m) ∑r The sum runs over all partitions (m)=(m1,...,mr), i=1 mi = p,m1 m2 mr 1 χ ≥ ≥···≥ and hτ = 2 p(p 1) is the number of transpositions in Sp. (m) is the character of τ − in the representation generated by e(m) which is of dimension f(m). We use (43), the ortogonality relation e e = δ and the completeness ∑ e = I in (m) (m′) (m)(m′) (m) (m) (42). The expression in brackets on the l.h.s may be orthogonal to some e(m). The η ε ε corresponding e(m)a contribution will be any combinations of ’s and ’s – being the totally antisymmetric tensor in four dimensions – transforming correctly and thus can LORENTZCOVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 9 be set to zero. We arrive at

e(m) 1 αβ a = ∑ (lαβ I + + I lαβ)B + 8 ∑ Pη a , c m 2 ij (m) ( ) − ⊗···⊗ ··· ⊗···⊗ i< j p ! ≤ c(m)=0 6 χ τ r . (m)( ) (m) (m) (m) (m) c(m) = 3p + p(p 1) = 3p + ∑ bi (bi + 1) ai (ai + 1) , (44) − f(m) i 1 − = with a =(a1,...,ar),b =(b1,...,br) denoting the characteristics of the frame (m), see appendix A. Let us take p = 4 as an example: idempotent Young frame dimension character

χ τ e(4) f(4) = 1 (4)( ) = 1

χ τ e(3,1) f(3,1) = 3 (3,1)( ) = 1

χ τ e(2,2) f(2,2) = 2 (2,2)( ) = 0

e f = 3 χ (τ) = 1 (2,1,1) (2,1,1) (2,1,1) −

e f = 1 χ (τ) = 1 (1,1,1,1) (1,1,1,1) (1,1,1,1) −

We ﬁnd for (42) 1 a = (2e + 3e + 4e + 6e ) r.h.s(42). (45) 48 (4) (3,1) (2,2) (2,1,1) ×

We see that no e(1,1,1,1) appears in that equation. It corresponds to the one dimensional sgn-representation of S4, so e4a ∝ε.

4. SPINORIAL LORENTZ COVARIANCE

In this section we follow [SU92]. The most general representation of L+↑ can be built of tensor products of SL(2,C) and SL(2,C) and direct sums of these. A two component spinor Ψ transforms according to ΨA A ΨB = g B , (46) 10 DIRK PRANGE C where g isa2 2-matrix in the SL(2, ) representation of L+↑ . For the complex conjugated representation× we use the dotted indices, i.e. ΨX˙ X˙ ΨY˙ = g Y˙ , (47) X˙ X C with g Y˙ = g Y in the SL(2, ) representation. The indices are lowered and raised with the ε-tensor. ε ε . ε AB = A˙B˙ = A˙B˙, (48) εABε εBAε δB AC = CA = C. (49)

We deﬁne the Van-der-Waerden symbols with the help of the Pauli matrices σµ and . µ σµ = σ :

AX˙ . 1 AX . 1 T σµ = (σµ) , σµ ˙ = (σ )AX . (50) e √2 AX √2 µ They satisfy the following relations e σ AX˙ σ η σ σµ ε ε µ νAX˙ = µν µAX˙ BY˙ = AB X˙Y˙ (51) With the help of these we can build the inﬁnitesimal spinor transformations

1 αβ g I + θαβS , (52) ≈ 2 with the generators ˙ αβ A σ[αAX σβ] (S ) B = BX˙ . (53)

AX˙ XA˙ Note that the σ’s are hermitian: σµ = σµ . Again we deﬁne the projectors for the tensor product. But we have only two irreducible parts:

AB 1 A B A B AB 1 AB P = (δ δ + δ δ ), Pε = ε ε , (54) S CD 2 C D D C CD 2 CD 2 P = P,PS + Pε = I. (55) We get the following identities: αβ αβ S Sαβ = S Sαβ = 3I, (56) − αβ αβ S Sαβ = S Sαβ = 4Pε I, (57) ⊗ ⊗ − αβ αβ S Sαβ = S Sαβ = 0. (58) ⊗ ⊗ In order to have (29) in a pure spinor representation we have to decompose the tensor x into spinor indices according to

AX˙ . µ AX˙ x = x σµ . (59) e LORENTZ COVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 11 Assume tx transforms under the u-fold tensor product of SL(2,C) times the v-fold tensor product of SL(2,C) then, for inﬁnitesimal transformations, (29) yields: e αβ αβ αβ B =(S I + + I S )a, (60) ⊗···⊗ ··· ⊗···⊗ with Bαβ from equation (35) in the corresponding spinor representation. The sum con- αβ sists of u summands with one Sαβ and v summands with one S with u,v > n. Multi- plying again with the generator and contracting the indices gives twice the Casimir on the r.h.s. Inserting (56-58) yields:

αβ (Sαβ I + + I Sαβ)B ⊗···⊗ ··· ⊗···⊗ I I I = 3(u + v) + 2 ∑ (4Pεi j ) + 2 ∑ (4Pεi j ) a. (61) − 1 i< j u − 1 i< j v − ≤ ≤ ≤ ≤ u The sum over u runs over 2 (u 1) possibilities and similar for v, so we ﬁnd the induction: −

1 αβ a = (Sαβ I + + I Sαβ)B + u(u + 2) + v(v + 2)"− ⊗···⊗ ··· ⊗···⊗

8 ∑ Pεi j + ∑ Pεi j a . (62) 1 i< j u 1 i< j v # ≤ ≤ ≤ ≤ It already contains the induction start for a(AB),a(XY) and aAX˙ .

5. GENERAL COVARIANT BPHZ SUBTRACTION

In this section we shrink the distribution space to S ′ since we are dealing with Fourier transformation. Let x,q, p R4m. BPHZ subtraction at momentum q cor- iq ∈ responds to using w = e · in the EG subtraction [Pra99]. . ip t\ω iq (p) = t ω iq ,e (63) ( ;e ·) ( ;e ·) · D E α 0 ip (p q) ∂q iq = t,e · ∑ −α αe · . (64) * − α ω ! + | |≤

It is normalized at the subtraction point q, i.e.: ∂αt\ω iq (q) = 0, α ω. This is al- ( ;e ·) 2 | | ≤ ways possible for q totally spacelike, (∑ j I q j) < 0, I 1,...,m [EG73, Düt]. In massive theories one can put q = 0 and has∈ the usual∀ subtraction⊂{ at} zero momentum which preserves covariance. But this leads to infrared divergencies in the massless case. There we can use the results from above to construct a covariant BPHZ subtrac- α α ∑ i| | tion for momentum q by adding α ω α! a pα to (64), according to equation (28). | |≤ 0 β For β ω + 1, tx is a well deﬁned distribution on S and so is ∂β0t. | | ≥ b 12 DIRK PRANGE 5.1. Lorentz invariance on R4. We have k iq iq 1 m iq iq 1 k Vke · = e · ∑ ( iqx) , ∂σVke · = iqσe · ( iqx) . (65) m=0 m! − k! − Inserting this into (30) we ﬁnd:

n 1 n ω+1 [ −2 ] i ( ) (n 1)!! (n 2s)!! ρ (µ1...µn) ∂ ∂(µ1 (µ1 a = − − qσ1 ...qσω n ∑ − qρ q (ω n)! (n + 2)!! − (n 2s 1)!! − × − s=0 − − σ σ ηµ2µ3 ...ηµ2sµ2s+1 ∂µ2s+2 ...∂µn)s∂ 1 ...∂ ω n 0t(q). (66) × − 3 Example. Take the setting sun in massless scalar ﬁeld theory: 0t = i D3 ω = 2. 6 F ⇒b µ i ρ σ µ µ σ a = (qσqρ∂ ∂ ∂ q qσ∂ )0t(q), −3 − µν 1 ρ µ ν µ ν a = (qρ∂ ∂ ∂ q( ∂ ))0t(q),b 4 − µ 1 µν and adding ipµa 2 pµ pνa restores Lorentz invarianceb of the setting sun graph sub- tracted at q. − αβ iq 5.2. General induction. We only have to evaluate B with w = e · and plug the result into the induction formulas (62) and (44). m γ αβ n ω 1 q [α β] B = 2i ( ) + ∑ ∑ q ∂ ∂γ∂ 0t(q). (67) γ! j j − j=1 γ =ω n | | − Here, q j are the m components of q hence γ is a4m indexe andb α,β are four indices. The tensor (spinor) structure of ∂ is given by x in (35).

6. SUMMARY AND OUTLOOK e e The subtraction procedure in EG renormalization makes use of an auxiliary (test) function and hence breaks Lorentz covariance, since no Lorentz invariant test function exists. But this symmetry can be restored by an appropriate choice of counterterms. We give an explicit formula for their calculation in lowest order and an inductive one for higher orders. Using the close relationship to BPHZ subtraction this directly trans- lates into a covariant subtraction at totally spacelike momentum. We expect our solution to be useful for all calculations for which the central solution (w = 1, see [Sch95]) does not exist, namely all theories that contain loops of only massless particles.

7. ACKNOWLEDGMENT I would like to thank Klaus Bresser and Gudrun Pinter for our short but efﬁcient teamwork and K. Fredenhagen for permanent support. LORENTZ COVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 13

APPENDIX A. REPRESENTATION OF THE SYMMETRIC GROUPS Everything in this brief appendix should be found in any book about representation theory of ﬁnite groups. We refer to [Sim96, Boe70, FH91]. The group algebra ASp consists of elements a = ∑ α(g) g, b = ∑ β(g) g, (68) · · g Sp g Sp ∈ ∈ where α,β are arbitrary complex numbers. The sum of two elements is naturally given by the summation in C and the product is deﬁned through the following convolution: . ab = ∑ α(g1)β(g2) g1g2 = ∑γ(g) g with (69) g1,g2 · g · . 1 1 γ(g) = ∑ α(g1)β(g2) = ∑α(g1)β(g1− g) = ∑α(gg2− )β(g2). (70) g1g2=g g1 g2 The group algebra is the direct sum of simple twosided ideals:

AS = I1 Ik, (71) p ⊕···⊕ and k is the number of partitions of p. Every ideal Ij contains f j equivalent irreducible representations of Sp. Ij is generated by an idempotent e j AS : ∈ p 2 Ij = ASp e j e j = e j. (72) These idempotents satisfy the following orthogonality and completeness relations: k e jei = δ ji ∑ e j = I. (73) j=1 k The center of AS consists of all elements ∑ α je j,α j C. p j ∈ Every permutation of Sp can be uniquely (modulo order) written as a product of disjoint cycles. Since two cycles are conjugated if and only if their length is the same, the number of conjugacy classes is equal to the number of partitions of p. Denoting the j’th conjugacy class by c j we build the sum of all elements of one class . π k j = ∑ ASp (74) π c j ∈ ∈ which is obviously in the center of ASp , too. So we can expand ki in the basis e j: k 1 ki = hi ∑ χ j(ci)e j, (75) j=1 f j where χ j(ci) is the character of the class ci in the representation generated by e j and hi is the number of elements of ci. The dimension of that representation is equal to the multiplicity f j. The constuction of the idempotents can be carried out via the 14 DIRK PRANGE

Young taubleaux. A sequence of integers (m)=(m1,...,mr),m1 m2 mr ∑r ≥ ≥···≥ with j=1 = p gives a partition of p. To every such sequence we associate a diagram with

m1 boxes ... m2 boxes ... . . called a Young frame (m). Let us take p = 5 as an example:

(5) (4,1) (3,2) (3,1,1) (2,2,1) (2,1,1,1) (1,1,1,1,1) An assignment of numbers 1,..., p into the boxes of a frame is called a Young tableau. Given a tableau T , we denote (m) by (m)(T). If the numbers in every row and in every column increase the tableau is called standard. The number of standard tableaux for the frame (m) is denoted by f(m). It is equal to the dimension of the irreducible representation generated by the idempotent e(m). We will now answer the question

How to construct e(m). Set . R (T ) = π Sp π leaves each row of T setwise ﬁxed , . { ∈ | } C(T ) = π Sp π leaves each column of T setwise ﬁxed , { ∈ | } and build the following objects: . . P(T ) = ∑ p, Q(T ) = ∑ sgn(q)q, (76) p R (T) q R (T) ∈ ∈ then . f(m) e(T ) = P(T )Q(T) (77) p! is a minimal projection in ASp (generates a minimal left ideal). The central projection (generating the simple twosided ideal) is given by

. f(m) e = ∑ e(T ). (78) (m) p! T (m)(T)=(m) | Example p = 3. The frame has only one standard tableau 1 2 3 . All dif- ferent tableaux in (78) lead to the same idempotent (77) which is just the sum of all permutations. 1 e = I +(12)+(13)+(23)+(123)+(132) . (3) 6 LORENTZ COVARIANCE IN EPSTEIN-GLASERRENORMALIZATION 15

For the frame we only need the column permutations in (77). We ﬁnd 1 e = I (12) (13) (23)+(123)+(132) . (1,1,1) 6 − − − 1 2 1 3 2 1 2 3 3 1 The frame has two standard tabelaux. For the tableaux 3 , 2 , 3 , 1 , 2 , 3 2 1 we ﬁnd: 22 e = (I +(12))(I (13))+(I +(13))(I (12))+(I +(12))(I (23)) + (2,1) (3!)2 − − − +(I+(23))(I (12))+(I +(13))(I (23))+(I +(23))(I (13)) − − − 1 = 2I (123) (132) . 3 − − Up to order p = 4 the central idempotents are given by the sum of minimal projectors of the standard tableaux – they are orthogonal. The characters in the irreducible (m) representation can be computed through

f(m) χ (s) = ∑ ∑ sgn(q). (m) p! T (m)(T)=(m) p R (T) | q∈C(T) ∈pq=s Many other useful formulas can be derived from the Frobenius character formula. In- terchanging rows and columns in a frame (m) leads us to the dual frame (m). For the characters one ﬁnds: χ (s) = sgn(s)χ (s). There is a nice formula for the charac- (m) (m) ters of the transpositions in [FH91]: g g Deﬁne the rank r of a frame to be the length of the diagonal. Let ai and bi be number of boxes below and to the right of the i’th box, reading from lower right to upper left. a ... a Call 1 r the characteristics of (m), e.g. b1 ... br X X 0 3 4 r = 3, characteristics = . X 0 2 5

Then f r χ τ (m) (m)( ) = ∑(bi(bi + 1) ai(ai + 1)). p(p + 1) i=1 − 16 DIRK PRANGE

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