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LORENTZ COVARIANCEIN EPSTEIN-GLASER RENORMALIZATION

DIRK PRANGE

II. INSTITUT FÜR THEORETISCHE PHYSIK UNIVERSITÄT HAMBURG LURUPER CHAUSSEE 149 D-22761 HAMBURG GERMANY

EMAIL: DÁÊúÈÊAÆGE@DEËYºDE

ABSTRACT. We give explicit and inductive formulas for the construction of a Lorentz covariant renor- malization in the EG approach. This automatically provides for a covariant BPHZ subtraction at totally spacelike momentum useful for massless theories.

INTRODUCTION Out introduction is rather brief since this paper mainly is a continuation of [?]. There we a gave a formula for a Lorentz invariant renormalization in one coordinate. We use the same methods to construct a covariant solution for more variables. We review the EG subtraction and give some useful formulas in section 1. The cohomological analysis that leads to a group covariant solution is reviewed in section 2. In sections 3, 4 we give an inductive construction for these solutions in the case of tensorial and spinorial Lorentz covariance. Since the symmmetry of the one variable problem is absent in general some permutation group calcu- lus will enter. Necessary material is in the appendix. In the last section 5 we give a covariant BPHZ subtraction for arbitrary (totally spacelike) momentum by fourier transformation.

1. THE EG SUBTRACTION We review the extension procedure in the EG approach [?]. A more general introduction can be ω found in [?], a generalization to manifolds in [?, ?]. Let D (Rd) be the subspace of test functions vanishing to order ω at 0. As usual, any operation on distributions is defined by the corresponding action on testfunctions. Define d ω d ϕ ϕ W(ω;w) : D(R ) D (R ), 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 ω then 0t can be ∈ ω6 d \{ } defined uniquely on D (R ),and ϕ . 0 ϕ t(ω;w), = t,W(ω;w) (2)

Work supported by the DFG Graduiertenkolleg “Theoretische Elementarteilchenphysik” Hamburg.

1 2 DIRK PRANGE

0 d defines an extension – called renormalization – of t to the whole testfunction space, t(ω;w) D0(R ). With the Leibnitz rule ∈

∂ γ 1 ∂ ∂ γ( fg)= ! ∑ ν µf νg, (3) µ+ν=γµ! ! 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.TheW-operation (1) is simplified if we require | |≤ w(0)=1and∂αw(0)=0, for 0 < α ω(this was our assumption in [?]). It can be achieved by the 1 | |≤ 1 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)=0for ,cresp. a are given by −| | | |≤ α α α α a = t(ω;w),x Vω α w , c = t(ω;w),x w . (9) −| | 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 arbitraty. They have to be chosen in such a way that Lorentz covariance of 0t is preserved 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. LORENTZ COVARIANCE IN EPSTEIN-GLASER RENORMALIZATION 3

2. THE G-COVARIANT EXTENSION We will first define the notion of a G-covariant distribution. So let G be a linear transformation d group on R 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 D0(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) 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 α (0) = ∑ t,x (I 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 [?], [?][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(g 1)+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) − 4 DIRK PRANGE 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 [?][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 0 one coordinate. This situation was completely analyzed in [?]for∂αw(0)=δα. The following two subsections generalize the results to arbitrary w, w(0) = 0. 6 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 [?] which is modified through the generalization for the choice of w:

n 1 [ − ] (n 1)!! 2 (n 2s)!! a(µ1...µn) = − ∑ − η(µ1µ2 ...ηµ2s 1µ2s (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) × − − if we choose the fully contracted part of aDto be zero in case of n being even. We used the notationE 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 derivatives:

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 + | |≤
| |≤

LORENTZ COVARIANCE IN EPSTEIN-GLASER RENORMALIZATION 5 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 [?] 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.

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)gis the tensor representation of rank p = r + n,n = α , in (29). From now on we will omit the 4m | | α indices. So if t D(R 0 ),wedenotebyx– formerly x – a tensor of rank n built of x1,...,xm. To solve (29)∈ we proceed\{ } like in [?]. Since the equation holds for all g we will solve for a by θ θ using Lorentz transformations in the infinitesimale neighbourhood of I.Ifwetake αβ = [αβ] as six 6 DIRK PRANGE 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 αβ B = 2 t,x ∑ x j j (Vω nw) =(l I+ +I l )a, (35) − ⊗···⊗ ··· ⊗···⊗  j=1  α,β being Lorentz four-indices.e In [?] 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 [?]. We build one Casimir operator on the r.h.s. (the other one is always zero, since we are in a p (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),ortis a scalar and x = x1,...,xm. (35) gives: 1 αβ 1 eαβ e lαβB = lαβl a = 3Ia, (36) e 2 2 − 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 ⊗ − 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 contraction 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 corresponding indices, and the hat means omission. Therefore b b 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 contractions of a are known. c LORENTZ COVARIANCE IN EPSTEIN-GLASER RENORMALIZATION 7

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 general π Sp the µ ...µ µπ 1 ...µπ 1 ∈ 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 summary of all necessary ingredients in . τ ∑ τ τ appendix A. So let k = τ Sp be the sum of all transpositions of Sp.Thenk 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 1 The sum runs over all partitions (m)=(m1,...,mr), i=1mi = p,m1 m2 mr and hτ = 2 p(p χ τ ≥ ≥···≥ − 1) is the number of transpositions in Sp. (m) is the character of in the representation generated by e which is of dimension f . We use (43), the ortogonality relation e e = δ and the (m) (m) (m) (m0) (m)(m0) completeness ∑(m) e(m) = I in (42). The expression in brackets on the l.h.s may be orthogonal to η ε ε some e(m). The 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 be set to zero. We arrive at

e(m) 1 αβ a = ∑ (lαβ + + lαβ)B + 8 ∑ Pη a , c m 2 I I 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 = 4asanexample: 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) − 8 DIRK PRANGE

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,soe4a∝ε.

4. SPINORIAL LORENTZ COVARIANCE

In this section we follow [?]. 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) where g is a 2 2-matrix in the SL(2,C) representation of L+↑ . For the complex conjugated represen- tation we use the× dotted indices, i.e. ΨX˙ X˙ ΨY˙ = g Y˙ , (47) X˙ X ε with g Y˙ = g Y in the SL(2,C) representation. The indices are lowered and raised with the -tensor. ε ε . ε AB = A˙B˙ = A˙B˙, (48) εABε εBAε δB AC = CA = C. (49) . µ We define the Van-der-Waerden symbols with the help of the Pauli matrices σµ and σµ = σ :

AX˙ . 1 AX . 1 T σµ = (σµ) , σµ ˙ = (σ )AX . (50) √2 AX √2 µ e 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) ⊗ ⊗ LORENTZ COVARIANCE IN EPSTEIN-GLASER RENORMALIZATION 9

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˙ e x = x σµ . (59) 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 consists of u sum- αβ mands with one Sαβ and v summands with one S with u,v > n. Multiplying 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 ⊗···⊗ ··· ⊗···⊗

= 3(u + v)I + 2 ∑ (4Pεij I)+2 ∑ (4Pεij I) a. (61) − 1 i< j u − 1 i

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

8 ∑ Pεij + ∑ Pεij a . (62) 1 i< j u 1 i

5. GENERAL COVARIANT BPHZ SUBTRACTION

In this section we shrink the distribution space to S 0 since we are dealing with Fourier transforma- 4m iq tion. Let x,q, p R . BPHZ subtraction at momentum q corresponds to using w = e · in the EG subtraction [?]. ∈

. 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 always possible for \( ;e ·) 2 | |≤ q totally spacelike, (∑ j I q j) < 0, I 1,...,m [?, ?]. In massive theories one can put q = 0and 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 α α ∑ i| | BPHZ subtraction for momentum q by adding α ω α! a pα to (64), according to equation (28). For | |≤ 0 β β ω+1, tx is a well defined distribution on S and so is ∂β0t. | |≥ 5.1. Lorentz invariance on R4. We have b k iq iq 1 m iq iq 1 k Vk e · = e · ∑ ( iqx) , ∂σVke · = iqσe · ( iqx) . (65) m=0 m! − k! − 10 DIRK PRANGE

Inserting this into (30) we find:

n 1 n ω+1 [ −2 ] µ µ i ( ) (n 1)!! (n 2s)!! ρ µ µ a( 1... n) = qσ ...qσ ∑ qρ∂ ∂( 1 q( 1 ω− − 1 ω n −  ( n)! (n + 2)!! − s 0 (n 2s 1)!! − × − = − −   µ2µ3 µ2sµ2s+1 µ2s+2 µn) s σ1 σω n0 η ...η ∂ ...∂ ∂ ...∂ − t(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 pµpνa restores Lorentz invariance of the setting sun graph subtracted at q. − 2 b 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). γ m q α β αβ n ω+1 [ ∂ ]∂ ∂ 0 B = 2i ( ) ∑ ∑ γ q j j γ t(q). (67) − j=1 γ =ω n ! | | − Here, q j are the m components of q hence γ is a 4m index and α,β aree b 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 translates 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 [?]) 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-GLASER RENORMALIZATION 11

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 [?, ?, ?]. 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 sum- mation 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 ∑ ej = I. (73) j=1 ∑k α α The center of ASp consists of all elements je j, j C. 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

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

m1 boxes ... m2 boxes ... . . 12 DIRK PRANGE called a Young frame (m).Letustakep=5asanexample:

(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.Givena 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 different tableaux in (78) lead to the same idempotent (77) which is just the sum of all permutations. 1 e = +(12)+(13)+(23)+(123)+(132) . (3) 6 I
For the frame we only need the column permutations in (77). We find 1 e(1,1,1) = I (12) (13) (23)+(123)+(132) . 6 − − − 1 2 1 3 2 1 2
3 3 1 3 2 The frame has two standard tabelaux. For the tableaux 3 , 2 , 3 , 1 , 2 , 1 we find: 22 e(2,1) = (I +(12))(I (13)) + (I +(13))(I (12)) + (I +(12))(I (23)) + (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. LORENTZ COVARIANCE IN EPSTEIN-GLASER RENORMALIZATION 13

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. Interchanging 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 characters of the transpositions in [?]: (m) (m) Define the rank r of a frame to be the length of the diagonal.g Let ai and bi be number of boxes g a ... a below and to the right of the i’th box, reading from lower right to upper left. Call 1 r the b1 ... br characteristics of (m),e.g.   X X 034 r=3,characteristics = . X 025  

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