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Volume 183, number 1 PHYSICS LETTERS B 1 January 1987

D I L A T O N T A D P O L E F O R T H E O P E N B O S O N I C S T R I N G "

Michael R. D O U G L A S and Benjamin G R I N S T E I N 1

California Institute of Technology, Pasadena, CA 91125, USA

Received 15 September 1986

The amplitude for emission of a single dilaton for world-sheets with the topology of a disk is computed, using Polyakov's formulation of string theory. Combined with previous results this gives the complete amplitude to first non-trivial order in the string coupling constant. The total amplitude vanishes for gauge group SO(8192), while the vacuum energy is negative and independent of the gauge group.

One important unsolved issue in string theory is that of finiteness. It is currently believed that all divergences are infrared in nature, arising when massless particles go on-shell in tadpole contributions. In the open superstring these divergences can lead to anomalies [ 1,2]. In this letter we calculate the dilaton tadpole amplitude for the open bosonic string to lowest non-trivial order. Polyakov's path integral formulation [3-7] offers a convenient approach to string perturbation theory. Moreover it provides the means for calculating directly zero- and one-particle S-matrix elements. Using this method, a one-loop calculation of the vacuum energy in the closed bosonic string was recently carried out by Polchinski [8]. In the open stirng there are two topologies which contribute to the first order beyond tree level [2], the projective plane P2 and the disk D 2. The vacuum energy and one-dilaton amplitude were computed for world-sheets with the topology of the sphere and projective plane by Grinstein and Wise [9]. The ampfitudes for P2 are the same for the open string as for the unoriented closed string, so we need only calculate the amplitudes for D 2.

Our starting point is the gauge-fixed Polyakov path integral with a sum over all world-sheet topologies. The partition function is, to the order of interest,

Z ~ - ~ [de t 'P?e ] 1/2 = f [dx] exp [ - (S + Set)]. (1) S2,P2,D2 V(CKV)d

In eq. (1)P is a differential operator that maps vectors v d into second-rank tensors

+ ~a ~b -- gab gccl) Vc Od,

and det' denotes the determinant excluding the zero modes. The factor V(CKV) in the denominator ofeq. (1) is the volume of the group generated by the conformal killing vectors. ~ is the order of the group D of diffeomor- phism classes [8]. The elements of D represent the connected components of the group of conformal diffeomor- phi sms, including those which do not preserve orientation.

The action is

s=½ f d2}x/~gab + x u ( } ) ~ x u ( } ) , (2a)

and the local counterterms are

Work supported in part by US Department of Energy under contract DE-AC 03-81-ER40050. a Tolman Fellow.

52 0370-2693/87/$ 03.50 © Elsevier Science Publishers B.V. (North-Holland Physics Publishing Division)


The renormalized value of X is the string coupling constant, which enters as X -x , where X is the Euler characteris- tic, so to first order only manifolds with X > 0 will contribute. The/a 2, p and p' counterterms are not conformally invariant (ds is the length element along the boundary and k is its extrinstic curvature) and are chosen to cancel non-conformally invariant terms arising from the regularization of the quantum theory.

To d&me the general open string theory we must include Chan-Paton factors [10]. We define the sum over topologies for the open string as a sum over "topologies with indices". Each connected component of the boundary of a surface is given a group index, which ranges over the values 1 toN. In the sum over topologies we include each surface with specified indices once. This procedure reproduces the Chan-Paton rules for the gauge group SO(N) :as defined in the operator formalism and in string field theory. Boundaries are considered distinguishable, even for a surface with automorphisms which permute the boundaries. This gives the right answer because in the case of a ~urface with symmetry, we have already included the proper symmetry factor in order (D).

The coefficients which weigh the contributions to the partition function from different world-sheet topologies are uniquely determined by requiring a unitary S-matrix. In eq. (1) we have assumed that these coefficients are unity for each topology. One can argue heuristically as follows [11]. Any dependence of these coefficients on the world-sheet topology not already contained in (2) would correspond t0non-local terms in the action. Such a modi- fication would spoil unitarity in the two-dimensional field theory on the world-sheet.

To compute the scattering amplitudes for on.shell strings, we insert vertex operators on the world-sheet. These are to be regarded not as fundamental elements of the theory but as convenient shortcuts to doing the path integral with specified boundary conditions on incoming and outgoing strings. The coupling constants for the vertex operators are determined uniquely in terms of X by requiring factorization of tree amplitudes [12,13].

We will use the conventions of ref. [9]. As shown there, for a tachyon vertex

VT@) = f V xptip'x( )],

the three- and four-point tachyon scattering amplitudes are

An(P1 ..... Pn) = (27026 ~ (171 + "" +Pn)an, (3a)

a3(Pl ,P,2 ,P3) = X-2 @Te2)3 Qs, (3b)

a4(Pl,P2,P3,P4 ) = 7rX_2(KT e2~4 ,q i"(3 -- (1/8r 0 (t + s))P(t/8rr -- 1)I ' (s/87r" 1) ' ~esp((1/S~r) (s + t) 2 23r(2 - t/8~r)r(2 - slS~r) ' (3c)

where s = (Pl +P2) 2 and t = (Pl +P3) 2,

Qs = 212 (3/16zr) 3 [det 'P?P] 1/2 [det '(_V2)]-13

is a normalization factor including the volume of conformal Killing vectors, zero mode contributions, and a factor of 1/2 for d', and e is a short-distance cutoff. Qs depends on our conventions for the functional measures, but physical quantities only depend on the combination Xp = XQ~ 1/2 . The normalization of/~T iS determined by com- paring (3b) with the tachyon pole in (3@

~T = (87r2) 112 XQs 112 e - 2 .

The dilaton vertex operator is

( ~__~_.24 R x / ~ exp(ip . x ) , (4) xu ~ b xv 16rr 7re-2 ]

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where [14]

e~V (p) = (1 /X/r~ ( ~ v _ p ~ V _ pV~U ) ,

and p •/) = 1,p2 = if2 = 0. The counterterms in V D are needed to preserve conformal invariance in scattering amplitudes involving the dilaton. A calculation similar to the tachyon case gives

/~D = 8~'2 ~ Q ~ l / 2

To define thepath integral on the disk, we must specify the boundary conditions. The physical condition that momentum does not flow out of the boundary leads to Neumann boundary conditions on the coordinates x~,

na l)aX~t [ aM = O,

where n a is the normal vector to the boundary. In terms of the tangent vector to the boundary t a,

n a = e a b g b c t c .

Note that n a depends on the world-sheet metric. Unless we choose our boundary conditions on the metric variations correctly, the boundary conditions will couple x~ and gbc. The correct choice is

8gab [aMnat b = O .

This insures that metric variations do not change the definition of the normal vector n a . This condition can also be expressed as a boundary condition on v a, the vector fields which generate infinitesimal diffeomorphisms. Since ~gab = V (a°b ), we have

n a t b V(aOb) = O.

To completely define the operator P ~ P we need a boundary condition for each component of v a. The second boundary condition comes from the requirement that the infinitesimal diffeomorphisms preserve the boundary

naoa = O .

These two boundary conditions ensure that the operator P ~ P is self-adjoint and positive [5 ]. If we work in the critical dimension, the conformal factor of the metric drops out, and we do not need a boundary condition for it [151.

The calculation of scattering amplitudes for the disk now proceeds in parallel to that for the projective plane. In fact the two calculations are so similar that we will be able to produce a single explicit expression for the amplitude which contains both possibilities simultaneously. We represent the disk as the Riemarm sphere with points identified if they are images under reflection in a plane passing through the equator. Thus the equator corresponds to the boundary of the disk. In complex coordinates this map is z ~ 1/~. We can represent the projective plane in a similar way by identifying z with - l [ z . In both cases we use the metric induced by stereographic projection from the sphere in euclidean three-space,

ds 2 = 4dzdz/(1 + Iz 12) 2 •

We will use the symbol D 2 to refer to the disk, P2 to the projective plane, and M tO refer to either. The parameter k will be +1 for D 2 and - 1 for P2, so our identification is now z with k/'~.

We first discuss the group of globally defined conformal transformations. For P2 this is SO(3) [9]. For the D 2 case this is well known to be SL(2;R). This is a non-compact group, and has infinite volume. We therefore have the result that the zero-point amplitude, or vacuum energy, due to the disk is zero. Therefore, the entire contribution to the vacuum energy in order X is from the P2 topology, and we f'md that the vacuum energy is non-zero and independent of the choice of gauge group.

We now consider the one-point dilaton amplitude. The relevant group here is the group of globally defined conformal transformations which preserve the location of a single point, the location of the vertex operator. If we

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PHYSICS LETTERS B 1 January1987 Volume 183, number 1

place the vertex at the center of the disk, we see that the remaining transformations are simple rotations, a group with finite volume. We thus reach the surprising conclusion that the one-point amplitude is non-zero * 1. This is in contradiction with the idea that the amplitude for a diagram with a zero-momentum dilaton insertion can be derived by differentiating the zero-point amplitude with respect to the slope parameter.

The globally defined conformal transformations on the sphere are the linear fractional transformations

z'=(Az +B)/(Cz +D), AD-BC= 1. (5)

The transformations on M are those which preserve the relationship between a point and it image, so

k/~' = ( k / e ) ' ,

This is true if

A = D , C = k B , [AI2 - k I B I 2 = I ,

or, for k = +1, if

A = - / ) , C = - B , IB[ 2 - 1 A I 2 =1.

The two disconnected components for the disk imply d (D 2) = 2 while d(P2) = 1, We can let A = (1 + k l B I 2 ) 1/2

× exp(ia)and parameterize the group by B and a. The invariant measure is then proportional to d 2 B da. Since in the case of the disk the integration region for B is the entire complex plane, we see that the group volume is infinite.

We normalize the group measure by finding conformal Killing vectors normalized with respect to our standard metric, and checking that the measure reduces to the unit measure near the identity. The normalized Killing vectors are

k z 2 ] , [16~r~__l_kZ2/ , i [~-~-\_ff]

and so the properly normalized measure is

(~'g~Tr) 3/2 d2B do~,

where o~ is integrated from 0 to ~r, and the region of integration for B is the entire complex plane for D 2, and the unit disk for P2'

The one-dilaton amplitude is

[det' p t p ] 1/2 AD(p) = V(CKV) f [dx] exp [ - (S +Sct)] VD(P),

where VD(P) is the dilaton vertex defined in (4). The integral over the zero modes gives a factor of (2rr)266(p). In order to perform the functional integral we need the propagator on M, defined by

1) We have subtracted the zero mode since we integrate over it separately. The propagator can be computed using the method of images. It is

~1 Here we disagree with ref. [16].

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-- 1.J-6 [in(-[Zl --Z2 i2 +l e2 exp [--dPl (zl +z2)]) ( ~ +In

<x~(zl)xv(z2)) = 47r uu (1 + Iz 112) (1 + Iz2[ 2) (1 + Iz 112) (1 + 1z212)! + 2 .

The e term was inserted as the effect o f a covafiant short-distance cutoff, e is a length on the world-sheet, which in coordinate distance is e/~/gzF = e e x p ( - ¢ ) .

Performing the functional integral we obtain

[det' P'~P] 1/2 [de t ' (_V2) ] -13 [ _ % / ~ ] AD(P) = (21r)26 8 (P)~-I/~D V(CKV)~ ~ ] aD' (6a)

where

[) ,2=kf . ( 6 b ) f d2z lim ~ z 0 z l n l l - k S z ' 1 a D ='ffnit disk z'--,z unit disk (1 -- klz 12) 2

Note that the factor o f k in a D means that the contributions to the dilaton tadpole from P2 and D 2 have opposite signs. In the case of D 2 the integral in (6b) is divergent, as is V(CKV), but the ratio is finite. To calculate it we trade the z integration for an integration over d 2 B, The jacobian is

l~(z,~)/a(g,B)l = IOz/~BI 2 - [ Oz/0BI 2 = (1 - k l z l 2 ) 2 ,

so that

= k i , jd2B" a D R

The region of integrat ionR is such that the transformations (5) map the point z = 0 onto the entire unit disk:

R = IB[2<~o f o r D 2,

= IBI 2 < 1 / 2 f o r P 2 .

Therefore

a D = (1/Tr) (3/16rr)3/2 V(CKV) × 1 for D 2 ,

X - 1 / 2 f o r P 2 .

We now turn to the computation of the determinants on D 2 . The eigenvectors of P tP are essentially vector spherical harmonics. We have

(Pt eO)b = _2va Vao b -- 2o b ,

so the eigenvectors on S 2 are

o a =gab(3[3~b)Ylm , o a = (1/%/g)eab(~/~b)Ykn , (7a,b)

with eigenvalues 2(l - 1) (l + 2). In (7b) eab is the antisymmetric symbol. Our boundary conditions are nao a = 0 and nat b V(aOb) = 0. In angular coordinates these are o °= 0 and Y0o~ + V¢o 0 = 0 at 0 = ~r[2. For the solutions (7a) these are

~o Ylm = ~o~3e~Y1m = O,

which imply that 1 - m is even. For solutions (7b) these are

-- (ao 2 - a )rt = o ,

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which imply that I -- m is odd. Thus for any l, rn we get only one eigenvector. Therefore

(det' PtP )D 2 = [(det' Pt t")s2 ] 1/2.

On P2 the eigenvectors are restricted by v ° (p) = - v ° ( - p ) and v~(p) = ve(-p). This also has one solution for each l and m, and the result is the same.

Eigenfunctions of the scalar laplacian on the disk correspond to eigenfunctions on the sphere which_are even under reflection in 0 ~ zr - 0. Such functions have zero normal derivative on the boundary. They are V~Ylm with eigenvalues X =l(l + 1) for l - m even. Thus

[det'(--V2)]D2 - HI[l(/+ 1)]l+1 = I ~ l .

~[det '(-V2)]s2}l/2 IIl[l(l+ 1)](2l*1)/2 l

This is divergent, but the divergence can be shown to be proportional to the length of the boundary [5]. We can absorb this divergence into our renormalization constant p. The finite part we are interested in is exactly the conformally invariant part, which we can evaluate using zeta-function regularization. Therefore,

[de t ' ( - ,V2)] D2 oo

t = = l_S , 1

!n~[det ' (-V2)]s }1/2= 1>0 ds = - - $'R ( 0 ) = ~ In 27r,

so that

[de t ' ( - V2)]D2/{ [det ' ( -V2)] s2}l/2 = x/~--~.

This was hardly a rigorous calculation, but a more careful treatment (regulating the determinants separately) gives the same answer.

On P2 scalar eigenfunctions have Elm (p) = Ylm (-P), which is satisfied for l even, m arbitrary. A similar calculation [9] gives

[det ' ( -V2)]p2/( [det ' ( -V 2)] S2 }112 = (7r/2)112,

Finally putting the pieces together we get

An(p ) = (2~r)266(p) [(3- 27)fir13] 1/2 X 2 -13 for D2,

X - 1 for P2

The total result for an open string with gauge group SO(N) is the sum of the results for P2 and D2, with D 2 weighed by a factor of N. Thus to this order the dilaton tadpole vanishes fo rN = 213.

In conclusion, we have computed the vacuum energy and dilaton tadpole to lowest non-trivial order for the open bosonic string. While the vacuum energy is non-zero and independent of the gauge group, the dilaton tadpole is zero for a unique choice of gauge group, SO(226/2). It is tempting to conjecture that the same mechanism is at work in the superstring, leading to the analogous choice of gauge group, SO(210/2). W o r k on this question is in progress. If this is so, then it seems fair to say that the cancellation of this contribution to the dilaton tadpole is not a consequence of supersymmetry, but is a more general consequence of some unknown principle of string theory.

We further expect this result to generalize to the divergences in the planar and non-orientable loops and to higher genus surfaces with boundary. This is because one can in principle factorize a diagram with boundarY by drawing a circle around the boundary whose size is shrinking to zero, and inserting a complete set of intermediate states. The divergence comes from the pole in the propagator for an on-shell intermediate dilaton coupling the rest of the diagram to the piece with the boundarY, whose contribution should be given by our calculation. One can represent a non-orientable surface in a similar way as an orientable surface with a disk cut out and a crosscap put

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in its place, and a simihir argument relates its divergence to that of P2' What is needed to make this argument rigorous is an explicit relation between the moduli spaces of the two surfaces.

Of course there is an independent contr ibut ion to the vacuum energy and the dilaton tadpole at one-loop order, coming from the toms diagram. This contr ibut ion will generally vanish only in a theory which has space- t ime supersymmetry.

We thank Ted Allen, Sumit Das, Ryan Rohm and John Schwarz for useful discussions.

References

[1] M.B. Green and J.H. Sehwazz, Phys. Lett. B 149 (1984) 117. [2] M.B. Green and J.H. Schwarz, Phys. Lett. B 151 (1985) 21. [3] A.M. Polyakov, Phys. Lett B 103 (1981) 207. [4] D. Friedan, Lectures 1982 Les Houches Summer School, eds. J. Zuber and R. Stora (North-Holland, Amsterdam, 1984). [5] O. Alvarez, Nucl. Phys. B 216 (1983) 125. [6] G. Moore and P. Nelson, Nucl. Phys. B 266 (1986) 58. [7] E.D'Hoker and D. Phong, Nucl. Phys. B 269 (1986) 205. [8] J. Polchinski, Commun. Math. Phys. 104 (1986) 37. [9] B. Grinstein and M.B, Wise, Caltech preprint CALT-68-1375 (1986).

[10] J. Paton and Chan Hong-Mo, Nucl. Phys. B 10 (1969) 519. [11] J. Polchinski, private communication. [12] S. Weinberg, Phys. Lett B 156 (1985) 309. [13] W. Weisberger, preprint ITP-SB-86-22 (1986), unpublished. [14] J. Scherk andJ. Schwarz, Nucl. Phys. B 81 (1974) 118. [15] A. Cohen et al., Nucl. Phys. B 267 (1986) 143. [16] C. Burgess and T. Morris, Princeton preprint (1986), unpublished.

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