1 Polyakov path integral and BRST cohomology

Transcription

1 Week 7 Reading material from the books Polchinski, Chapter 3,4 Becker, Becker, Schwartz, Chapter 3 Green, Schwartz, Witten, chapter 3 1 Polyakov path integral and BRST cohomology We need to discuss now the details of the Polyakov path integral. The main idea is to do a gauge fixed version of the path integral over Riemann surfaces that we have discussed so far. The main problem is to try to understand the precise inner workings of this path integral from a global point of view, because we have an infinite symmetry to contend with. Locally, as we have discussed so far, we can always bring the string metric to lightsheet coordinates (for the Lorentzia metric), or equivalently, to complex analytic coordinates where only g z z 0. However, this is not global enough for our purposes. To understand a little better how things behave globally, we need to step back a little bit and discuss the symmetries of the action that we want to consider. We have Weyl invariance, and diffeomorphism invariance. Weyl invariannce is easy, because all we need to do is change the value of the metric keeping the coordinate points fixed: g(σ) exp(2w(σ))g(σ), or in a coordinate independent formulation g exp(w)g (1 + w)g for w a global (bounded function) on the manifold. The one that is more tricky is diffeomorphism invariance. How do we express the statement that we are allowed to make general coordinate transformations in a coordinate independent way? Well, we want to think about it for infinitesimal coordinate transformations. This in general moves points just a little bit. We can describe this by an infinitesimal global vector field on the Riemann surface V. In this language, 1

2 any coordinate function σ a varies locally as follows σ a σ a + V (σ a ) = σ a + V a (1) where V = V a a. Remember that vector fields are derivations: they take functions into functions, and at least locally, one can describe them in terms of a local basis for derivations a. This is the content expressed above. Similarly, we can see that g will be affected by these infinitesimal changes of coordinates in the following form g ab g ab + a V b + b V a (2) This involves lowering the indices of the vector field, so it depends explicitly on V. Also, derivatives are covariantized. This can be proved by explicitly noticing that what stays invariant is g ab dσ a dσ b = g a b dσa dσ b (3) This final result is not immediately obvious. We use the fact that there is a local coordinate system where g δ ab and the local Christoffel symbols vanish Γ = 0 (this just states that g = 0 to first order). For this coordinate system we get that the metric varies as b V a + a V b, just from the general infinitesimal coordinate transformation. Going to a general coordinate frame just covariantizes all the derivatives, and that way we get a general result. Also notice that we have lowered the indices of V with respect to the metric g, so the variation of the metric depends very explicitly on the metric. We find it convenient to absorb part of the action of V as a Weyl transformation, so that the variation of the metric is traceless: g ab g ab + a V b + b V a g ab c V c Diff (4) g ab g ab (1 + w) Weyl (5) This is just a linear change of variables in the V, w plane, and it s functional determinant is one (w w + V ). This is most easily visualized in a local complex coordinate system, where g zz = 0, g z z = 0 and g z z 1. The total variations are δg zz V z (6) δg z z V z (7) δg z z wg z z + V z + V z (8) 2

3 but we can make a linear change of variables in w to absorb V + V The process of gauge fixing the path integral over Riemann surfaces, is that we choose a complex coordinate system on the Riemann surface, so that g zz = g z z = 0; while we have g z z 0. Locally, we can choose g z z = Const, but this is not allowed globally. However, in two dimensions, we have Riemann s uniformization theorem that states that we can always choose g z z to have constant curvature. For different genus, we get different answers. For g = 0, the metric is a round sphere. For g = 1, the metric is flat. For g > 1 the metric has constant negative curvature. This means that for g > 1 all of the metrics result from isometries to the upper half plane with the Poincare metric and with identifications by the group of isometries of the upper half plane SL(2, R). This is the covering space of the Riemann surface. Selecting a metric amounts to selecting a finite subgroup of SL(2, R). This is described by some finite number of generators and hence, a finite number of functions mapping to SL(2, R). This means that the problem of classifying all Riemann surfaces up to Weyl invariance results into a finite dimensional space of equivalence classes (the so-called moduli space of metrics). This is fortunate for us. This means that up to a finite number of integrals, we are left over with a gauge fixing that eliminates all degrees of freedom from the metric. The metric with constant curvature is important for us. We will also normalize it so that the volume of the metric is one. In complex coordinates, making V traceless makes our life easier. That means that g z z does not vary under V transformations and the variations factorize neatly. g zz g zz + z V z (9) g z z g z z + z V z (10) g z z g z z (1 + w) (11) Our choice of gauge fixing is g zz = 0 = g z z, and g z z = ĝ z z where in the last case we are using a constant curvature metric with unit volume. So now we want to do a path integral with insertions of delta functions of this gauge condition, but we need to ensure that the result is gauge invariant. This is where the Faddeev-Popov determinant comes into place. 3

4 This relies on the observation that for finite integrals (or for finite group invariant integrals) one has the following identity: dθδ(f(θ)) = 1/f (θ) f=0 (12) This is proved by changing variables from θ to f, and using dyδ(y) = 1. Or more generally d k θδ k (f i (θ)) = 1/ det( j f i (θ)) f=0 (13) where d k θ is a Haar-metric for the group, that near the identity is just dθ i. The determinant is just the Jacobian from changing of variables from θ to the f i. The determinant found there is the Faddeev-Popov determinant. Clearly we have the identity det( j f i (θ)) f=0 d k θδ k (f i (θ)) = 1 (14) Now, in the above, we can take path integral limit, and consider this as an identity for a path integral over the gauge orbit with a gauge fixing condition f = 0. What we then do, is to write the above identity in the full Polyakov path integral: we insert one. The functional integral over the dθ is just the volume of the gauge orbit (group action). So we have D[g]DX Diff Weyl Dθ I δ(f θ ) det ( ) δf exp( S) (15) δθ f=0 where f are the complete set of gauge fixing functions. Now, we integrate over D[g], which is where we set our gauge conditions. Modulo a finite number of parameters associated to global choices, which we will discuss in more details later, the local integral gives us det( z ) det( z ) (16) from the variations with respect to V. For w, we get the functional determinant of g over the Riemann surface. This is ultralocal (does not depend 4

5 on derivatives), and can be renormalized away or absorbed in the volume of the surface (which we have chosen fixed and proportional to the curvature). The cancellation of the variation of the action with respect to w, once the matter is included, also leads to the condition c total = 0. The other two pieces involve functional determinants of nice differential operators on the worldsheet. This determinant can be written as ( ) DbDc exp d 2 zb z z z c z (17) and if we lower and raise indices b z z b zz, c z c z, while the z becomes z. A similar result holds for the other variation. Note here that b zz and c z transform as 2 tensors and ( 1) tensors respectively, and that the action is that of a CFT of b, c type. What we find, by matching conformal weights to the b-c theory, that c ghost = 26, with a stress tensor given by T bc A : b c : +B : c b : (18) We also need the contractions that arise from the Greens function of and it s give by b(z)c(w) = 1 (19) z w Matching the weight of c gives us B = 1, and getting the full OPE for the stress tensor with c right gives us A + B = 1. Again, this is a free field theory on the string worldsheet, and we can quantize it in a straightforward way. If we take mode oscillators on the cylinder worldsheet, we have zero modes and non-zero modes. Quantizing the non-zero modes is easy. We define a ground state by b m 0 = c m 0 = 0 for m > 0. However, we get an ambiguity at m = 0, where {b 0, c 0 } = 1 (20) and we have two states 0 and 0. These differ in their ghost charge (if c has positive ghost charge, then b has negative ghost charge). The current generating ghost charge in the cylinder is j g =: bc : (21) 5

6 and Q = : bc : dz (22) Notice also that the differential operators z involved in general can have zero modes: modes that are annihilated by it. j g however is not a primary operator. This has consequences (the current is only anomalously conserved), so we will need insertions of zero modes in the path integral to make it nonvanishing. The ones associated to c are related to global holomorphic infinitesimal transformations of coordinates (conformal isometries, or conformal Killing vectors). These show that the gauge fixing procedure can have redundancies left over. For example, on the sphere, we get that c(z) 1, z, z 2 are all global zero modes. The zero modes for b are harder to interpret. What we have to remember is that the gauge condition was written as g = fixed values. Thus having a zero mode for b is related directly to something about g, as b is in the dual space to g. The interpretation is that we can consider some kind of Dirac-type operator on variations of g, paired with the things that are gauge variations of g as we found above. δg αβ ( α V β ) sym. traceless (23) The pairing is via a first order differential operator. Variations δg orthogonal to infinitesimal gauge variations are related to the moduli (choices of inequivalent metrics) of the Riemann surface. This is what the zero modes of b end up capturing. The total number of zero modes (b c) = 3(g 1), where g is the genus of the Riemann surface, is controlled by index theory (it is a topological number), moreover, it is well known that the number of zero modes does not jump on the Moduli-space of Riemann surfaces (this space is therefore smooth, except for global identifications). We find that #(zero modes of c on sphere) is exactly three, so that there are no b zero modes: there are the global holomorphic transformations of the sphere to itself. This tells us that the sphere has no moduli! But we have a redundancy of the choice of holomorphic coordinates. The path integral is formally zero, unless we have extra insertions of c. 6

7 Indeed, three insertions suffice, and we define for the b, c theory on the sphere c(0) c(0) 2 c(0) b,c = 1 (24) How do we insert these values of c? Well, in our gauge fixing procedure we might also have the coordinates of the vertex operators associated to string states. We can choose three of these coordinates to be at any three points we want z 1, z 2, z 3. So in our gauge fixing procedure we will also get d 2 wδ(w z i ), and then δw V a contributes to the F.P. determinant. This would involve three extra b ghosts associated to these insertions. The F.P. action would then have an extra piece that would look like S FP = bi δ c z z=zi (25) but δ c z = c. If we integrate the Grassmann numbers b i, we would get c(z 1 )c(z 2 )c(z 3 ). We end up only picking the zero mode contributions in the path integral. For each z i we get c(z i ). A quick evaluation shows that this ends up multiplying the target space X dependent vertex operator amplitude by i k (z i z k ). One can get this by noticing that we are computing a threepoint function of primaries of conformal weight one, so this is governed by the conformal identities. Another way to prove this, is to use the Taylor expansion of c up to order z 2 i in the normalization of the path integral equation 24. Similarly, on the cylinder, we have one zero mode for c and one for b as above, but only one of them annihilates the vacuum (either b 0 or c 0. We choose the vacuum to be annihilated by b 0, as we have a zero mode killing vector associated to rotations of the cylinder. We will explain this better in what follows. 2 BRST quantization Well, the path integral we have written so far is of the form ( D[g]DbDcδ(F θ )DX exp S b δf ) δθ c (26) where F is a gauge fixing function, and we have written the determinant with Faddeev-Popov ghosts. 7

8 Since the action is no longer gauge invariant, there are technicalities that make renormalization etc more problematic. It is also convenient to write the δ functional in terms of a functional Fourier trasnform representation: δ(f θ ) = DB exp(i BF θ ) (27) The final form of the action is therefore: S total = S 0 + i BF θ bδ c F = S 0 + S 1 + S 2 (28) c appears as the infinitesimal gauge variation of F. This motivates looking for a symmetry that encodes the fact that the original system is gauge invariant. This is the BRST symmetry. It consists of gauge variations with c as parameter. c clearly transforms in the adjoint, and in the Fadeev Popov procedure it is associated to derivatives with respect to the gauge parameters, namely F i θ α cα δ c F i (29) Thus, for the original fields in the path integral (X and g) δ BRST X δ c X (30) Similarly, since c is in the adjoint, it should transform in an obvious way δ BRST c c c ic a c b fc ab /2 (31) where fc ab are the structure constants of the Lie-alegbra symmetry (in this case, this is the Virasoro algebra). The factor of 2 is because we have two c s on the right hand side, so it would seem like a double contribution to the variation. S 0 is gauge invariant, so it is invariant under the BRST symmetry too. Also, variation of BF θ gives us a term proportional to Bδ c F, and δbf. The first term can be cancelled from the variation of the ghost lagrangian if δ BRST b ib. Making everything vanish is done by imposing δ BRST B = 0. Indeed, it can be easily shown that the symmetry associated to BRST variations is nilpotent: 8

9 δbrst 2 = 0 and moreover, in this formulation, it is independent of F the gauge choice. The gauge fixing term together with the ghost action can be written nicely as δ BRST (bf) (32) so obviously δ 2 BRST (bf) = δ BRST(S 1 + S 2 ) = 0 (33) At this level, the BRST symmetry a formal statement that can be made independent of the choice of gauge fixing functions. Also, one can make an infinitesimal change in the gauge function. This brings down a δ BRST bg down from the path integral, and it s correlation functions should vanish. In the Hamiltonian language, the BRST charge should be described by an operator, and the variation of a field by the variation of a quantum operator. This is given by a commutator. Thus, we should insert a commutator of the BRST charge with G, and the BRST charge should be hermitean. Therefore, for gauge invariant quantities/amplitudes it should be the case out {Q BRST, G} in = 0 (34) for all physical states out, in and for all G. Obviously this will vanish for all G operators only if Q BRST in = 0 = out Q BRST This shows that the answer is independent of the gauge choice. Also, since Q 2 BRST = 0, it is easy to show that states of the form phys + Q BRST χ have the same physical correlatorrs as phys, so they can not be distinguished by any physical measurement, and therefore they should be identified. What we find is that physical states are annihilated by Q, and states of the form Q χ don t matter. Also, gauge invariant operators should commute with Q BRST (this makes the answer independent of the gauge choice, and moreover keeps physical states physical. Also operators of the form [Q BRST, O] are trivial, as they act by zero on physical correlators.) Via the operator state-correspondence, these two cohomologies should be the same! So we can do everything in the operator language. The BRST charge should be generated by a contour integral of a current, and we make sure that it is covariant under conformal transformations by 9

10 making sure that it is a (1, 0) operator for b, c. If we have integrated the lagrange multiplier B, it would seem that δb is illl-defined. However, what we require then is that the BRST charge is consistent on everything else, and then the action on b is whatever it needs to be (one is evaluating the matrix elements of B on shell). This gives us j BRST (z) = ct m : ct gh : c (35) and Q BRST = dzj BRST (z) (36) A relatively simple (but very tedious homework calculation) shows that c matter = 26 is required to make Q 2 BRST = 0. The term c is there to make sure that j is a conformal primary (integrated over a contour it vanishes trivially). The interesting thing we find is that δ BRST b = [Q, b] T (37) so the BRST charge acting on the b ghosts produces directly the total stress tensor on the worldsheet. This means that all the matrix elements of the worldsheet stress tensor satisfy T = 0 for physical states (those that are annihilated by Q BRST ). This is exactly how we check that the BRST procedure gives us the Virasoro constraints on the worldsheet. The condition for BRST invariance of a vertex operator is then Q BRST V = dzj BRST (z)v (0) = 0 (38) for a contour around zero. As we have argued already, on the sphere we need to fix the position of various vertex operators and then we have to attach an extra c c to those insertions (one from left movers and one from right movers). Let us take V = c cv matter It is easy to do the contour integral. One can show that vanishing of the contour integral implies that V matter is a primary of conformal weight (1, 1). These are the vertex operators we need: physical states. 10

11 Notice also that if we take d 2 zv 1,1 (39) then we get an object that transforms covariantly under reparametrizations of the z variables. These are the so called integrated vertex operators. One can show in a straightforward way that if we use only integrated vertex operators, then the gauge fixing procedure on the sphere (the complex plane) will fix three of them to be wherever we choose them to be and it will attach combinations of c, c when we integrate over the positions and calculate the delta functions. The idea of having to insert vertex operators in integrated form is from diffeomorphism invariance. This is required in order to have something that is independent of the choices of coordinates when we consider that position is not a gauge invariant statement in a theory where the metric varies (combinations of fluctuations on the metric plus coordinate transformations can get you to a manifold that has the same metric, but where the insertion position has changed). The fact that one needs (1, 1) vertex operators is to have a Weyl invariant statement: usually we would integrate a density on the worldsheet etav (40) and the rescalings of η need to be balanced by the rescalings of V. 11