Non-orientable Worldsheets

Transcription

1 TU Wien Institute for Theoretical Physics Student Project Non-orientable Worldsheets Alexander Durdik, BSc MatNr.: 9598 Supervisor: Univ.Ass. Dipl.-Ing. Dr.rer.nat Johanna Knapp December 5, 6

2 Contents The Orientifold operation Closed Strings 7. A quick look at the torus partition function Klein Bottle Amplitude The bosonic sector contribution The Neveu-Schwarz sector contribution The Ramond sector contribution Combining all sectors Open Strings 6 3. Chan-Paton Indices Open String Amplitudes The Annulus The bosonic sector contribution The Neveu-Schwarz sector The Ramond sector The Chan-Paton contribution Combining all sectors The Möbius Strip The Effects of Ω on Open Strings The Bosonic sector The Neveu-Schwarz sector The Ramond sector The Chan-Paton contribution Combining all sectors The transverse pictures and their implications 5 4. The Klein Bottle at tree-level The Annulus at tree-level The Möbius Strip at tree-level Tadpole cancellation Combining the Interactions A Appendix: The GSO-projection 3 B Appendix: The fermionic string 33 B. The fermionic action and the equations of motion B. Boundary conditions for fermionic strings

3 C Appendix: Light cone gauge 36 C. The bosonic string in light cone gauge C. The superstring in light cone gauge D Appendix: Proof of the Jacobi identity 38 E Appendix: The P -transformation 4 F Appendix: The problems with the tree-level Möbius Strip 4 G Appendix: List of used symbols 43

4 Abstract In this work I will start with giving a quick overview of how the orientation reversal is implemented in string theory by using the properties of the closed string. This I will use to calculate the non-orientable -loop amplitude for the closed string, the Klein-Bottle amplitude. Then we will take a look at the open string and work out how the orientation reversal acts on them. With this the full -loop amplitude of the open string, consisting of an orientable part,the annulus and a non-orientable part, the Möbius strip, will be computed. The next discussion will be about the simplest tree-level amplitudes for the unoriented closed string and for both oriented and unoriented open string. This will lead to notion of diverging tadpole diagrams and how they can be cancelled. The first half of the appendix consists of diverse smaller discussions necessary for understanding the main calculations as the GSO-projection, how fermionic strings are integrated in superstring theory and the light cone gauge and its fixing. The second part of the appendix consists of proofs, which are too lengthy to keep in the main sections, but are relevant.

5 THE ORIENTIFOLD OPERATION The Orientifold operation When working through the basic concepts of string theory, one usually looks at worldsheets of strings, which are called orientable. This means that when we define an arrow pointing in the normal direction and move arbitrarily around the surface until we reach our starting point, the orientation of our arrow will be unchanged. If this arrow can flip direction when wandering around the surface, the surface is called unorientable. It is easier to understand the concept by looking at an open string but calculating the open string is actually more difficult. Imagine a rigid open string moving through space-time. Its worldsheet will be a flat plane like a piece of paper. Now let us glue one of the ends of the paper strip together. This will be an annulus a cylinder without its top and bottom surface. It is an orientable surface with two boundaries. But what if we had twisted the piece of paper before gluing it together? Then it would represent an unorientable surface called the Möbius-Strip. If we mark a starting point with an arrow and then move one time around the loop, the resulting arrow s orientation will be flipped compared to the original. This surface has only one boundary but additionally it has also what is called a crosscap. A crosscap is a circle, where the points on opposite sides are identified. If we now take an oriented closed string Wick-rotate it to the Euclidean plane and impose periodicity conditions on the spatial and imaginary time variable, we get a -loop amplitude for the closed strings. Its worldsheet is then described by a torus a surface with one handle. Its unorientable cousin is called the Klein-Bottle one handle and two crosscaps. If we want to include such unorientable surfaces into our theory, we need an operator which generates orientation reversal of the string. Orientation reversal means that we need to flip the direction of the string, meaning that we reverse the direction in which the coordinate parameter increases. This can be achieved by transforming the space-like worldsheet coordinate σ into its own negative this transformation will be denoted by Ω, while leaving the time-like coordinate τ unchanged: Ω : σ σ = π σ, τ τ = τ A very important property of this transformation is that Ω. This is See also [BCoST p 5. In these equations the length of the string is normalized to π. Changing the length of the string coincides with a reparametrization of σ. It is therefore possible to fix it to any value. This normalization makes the equations simpler and shorter.

6 THE ORIENTIFOLD OPERATION easily verified by applying Ω again on σ : σ Ω σ = π σ = π π σ = σ This means that changing the orientation twice is the same as not changing the orientation at all and it implies that Ω has only two eigenvalues: ±. It is easier to use complex coordinates z, z. These coordinates are obtained by firstly performing a Wick rotation, which means going to an imaginary time. τ iτ 3 This gives us complex cylinder ccordinates, which can be transferred to the complex plane via: z = e τ+iσ, z = e τ iσ 4 This transforms the world sheet into the complex plane, where points with equal time parameter τ are mapped into a ring around the origin. The diameter of the ring depends on the specific value of τ: z = e τ e iσ ρτe iσ Figure : This depicts how the transformation of σ, τ to z, z The orientifold operator acts on these new variables like this: Ω : z = e τ+iσ z = e τ iσ = z ; z z = z 5

7 THE ORIENTIFOLD OPERATION 3 That means the Ω transformation exchanges z and z. It basically acts as a complex conjugation. Let us now take a look at the solution of the equations of motion of the closed bosonic string: X µ z, z = X µ L z + Xµ R z 6 Where X µ L z and Xµ R z are as follows : X µ xµ α L z = iα pµ L log z + i n X µ xµ α R z = iα pµ R log z + i n α µ n n z n α µ n n z n All symbols used are defined in Appendix G. Under Ω the string coordinate functions transform into: X µ L z Ω X µ xµ α L z = iα pµ L log z + i X µ R z Ω X µ xµ α R z = iα pµ R logz + i n n 7 α µ n n z n 8 α µ n n z n 9 By comparing the string coordinate functions with each other and imposing invariance under Ω, we can read off how Ω acts on the bosonic modes: p L Ω p R, α n Ω α n This relation makes it sensible to define a zero mode for the oscillators: α µ = α µ = α pµ L α pµ R Therefore Ω exchanges left and right movers. Things get a little more complicated when working with worldsheet fermions. The solutions for the field equations look like this 3 ψ µ z = ψ r µ z r+ ψ µ z r Z+ν = r Z+ν See [BCoST p 3f. 3 See [BCoST p 89. ψ µ r z r+

8 THE ORIENTIFOLD OPERATION 4 Where ν = { for the Ramond sector / for the Neveu-Schwarz sector See Appendix B for details. We would like to have invariance under Ω of the product ψ ψ, because such terms arise in the Lagrangian. This means: Ω ψ µ r ψ r,µ Ω! = ψ µ r ψ r,µ 3 As we saw in the bosonic case, the orientation reversal operator exchanges the left and right moving modes. In the fermionic case, this does not mean Ω? to simply exchange ψ r ψ r, because such an action would not leave this product invariant. It would produce a negative sign: Ω ψ µ r ψ r,µ Ω = Ω ψ µ r Ω Ωψ r,µ Ω = ψ µ r ψ r,µ = ψ r,µ ψ µ r ψ µ r ψ r,µ Consequentially this means that Ω acting on fermionic modes has to behave differently whether they are left or right movers. One of them has to produce a negative sign. Therefore the action of the orientation reversal can be written like this: ψ r Ω ψ r, ψr Ω ψ r The main transformation that Ω generates is therefore an exchange between the left and right moving modes as well as sign changes in the fermionic case. In a theory with orientation reversal, it is necessary to only allow strings, which are invariant under this operation. If this mode exchange would also change the state of the string, meaning that it is asymmetric in its left and right moving modes, it would still be possible to generate a -loop amplitude. We would just have to perform another orientation reversal. But this is indistinguishable from not changing the orientation at all. As we integrate over all different interaction times we would make double counts. We therefore need to project onto states that are eigenstates of Ω and additionally have the eigenvalue +. Therefore we need an operator which gives a projection to this subspace. This projection is performed by: P Ω = + Ω 4 This is a projection operator since Ω has only two possible eigenvalues namely + and and therefore splits the entire space into two subspaces. It is

9 THE ORIENTIFOLD OPERATION 5 Ω Ω Ω τ Figure : This figure shows how the double peridicity of an unoriented string corresponds to an oriented string with the doubled interaction time. τ also unipotent: + Ω PΩ = = + Ω + Ω 4 = + Ω 4 = + Ω = P Ω 5 The aim of this work is to calculate the partition function for unorientable worldsheets. The full partition function for the closed string with both orientable and unorientable worldsheets will take the following form also taking the GSO-projection into account 4 : + Ω F Z = Tr q L F q L 6 Here I used q = e πiτ, where τ is the Teichmüller parameter. See [BCoST p 34 and section. for details. L and L are given by: L = : α nα µ n,µ : +c n Z L = : α µ n α n,µ : + c n Z 7 Here c and c are normal ordering constants. The partition function describes the -loop amplitude of a closed string. As we saw earlier, means we make the time variable periodic. This trace can be split into two parts: Z = T + K 8 4 See Appendix A for details.

10 THE ORIENTIFOLD OPERATION 6 Where: T = Tr F K = Tr Ω F q L q L F F q L q L 9 T describes the orientable part of the partition function and is therefore called the torus amplitude. In analogy, is K the unorientable part of the partition function and will be called the Klein-Bottle amplitude.

11 CLOSED STRINGS 7 Closed Strings This section will focus on calculating the partition functions for unorientable worldsheets of the closed string.. A quick look at the torus partition function All of the partition functions that are discussed in this work can be computed by looking at the torus amplitude and adding additional symmetries. It is therefore a good idea to make a short detour to the description of a torus to motivate a few features we are going to use later. For a more detailed discussion of the torus see [BCoST p. 34ff. The torus is described by a complex parameter τ = τ +iτ, which is called the Teichmüller parameter. It can be used to characterize the different tori and enables us to take the trace in the partition function. The partition functions for all four surfaces torus, Klein-Bottle, annulus and Möbius strip contain terms of the form: Tr q L q L This can be rewritten: Tr q L q L = Tr e πiτl e πi τ L = Tr e πiτ +iτ L e πiτ iτ L = Tr e πiτ L L e πτ L + L = Tr e πiτ P l.c. e πτ H l.c Here P l.c. = L L gives translations along the string in light cone coordinates and H l.c. = L + L gives time translations, which means its the Hamiltonian. Therefore the τ coordinate describes the direction along the length of the string and τ describes the time coordinate of the propagating string.

12 CLOSED STRINGS 8 With the level matching condition we get L = L and therefore: Tr q L q L = Tr e πτ H l.c 6 dτ dτ d d p = p N e πτ H l.c. N p 7 τ π d osc dτ d d p = A τ π d e πτ α p 4 p p N e πτ Hl.c. osc N = Aπ d δ d dτ = AV d dτ τ τ Tr q L q L Tr q L q L osc Here I used the shorthand notation for the Hamiltonian where Hl.c. osc is the Hamiltonian without the zero modes, which can be identified with the momentum. Also the q s in the last line are to be taken with the imaginary part of τ only. The constant factor we get from the τ integration, A, can be removed by normalization and will not be explicitly written. Finally the contraction p p gives us the volume of the d-dimensional space V d, which can be shown like this for one direction: p p = lim p p p p 3 = lim dx p x x p 3 p p = lim dxe ixp e ixp 33 p p = lim dxe ixp p 34 p p = lim p p πδp p 35 = πδ =! dxe = V 36 For a more detailed discussion of the torus amplitude and other underlying principles see [BCoST p 46ff. We are going to use equation 3 as a starting point for the computation of the partition functions for the Klein- Bottle and in section 3 for the annulus and the Möbius strip, which are -loop amplitudes of open strings 5. 5 The inclusion of the projection operators is irrelevant to the derivation shown here.

13 CLOSED STRINGS 9. Klein Bottle Amplitude Now we can turn our attention to the partition function associated with the Klein-Bottle 6. K = F Ω Tr q LB,R,NS F q B,R,NS L 37 We have to integrate over all possible Klein-Bottles, which are parametrized by τ. The Klein-Bottle can be described by using its covering torus. with F Tr Ω = = dτ τ dτ τ L,R L,R F q LB,R,NS F L, R Ω ˆR F R, L ˆR = { for the NS sector for the R sector q B,R,NS L = F q LB,R,NS q B,R,NS L L, R F q LB,R,NS B,R,NS L q L, R and L, R means a physical state, which is described by acting with any number of creation operators on the vacuum. This can also be split into fermionic and bosonic states: L, R = LR,NS, R R,NS LB, R B 39 The sum is to be understood to include all possible combinations of bosonic and fermionic states. The minus sign in the last line of equation 38 comes from the fact that the degenerate vacuum state of the Ramond sector is fermionic and induces a minus sign when the orientifold operator Ω acts on its right moving vacuum. As an easy description for the degenerate Ramond vacuum I introduce the i and the i vacuum for each independent 6 When I write something like B, R, NS I mean all neccessary contribution of each sector; bosonic, Ramond and Neveu-Schwarz respectively. As all contributions factorize it is unnecessary to write all of them explicitly but simply substitute the bracket with the sectors. 38

14 CLOSED STRINGS direction 7. They can be transformed into each other by the operators: ψ+ i = ψ i + ψ i ψ i = ψ i ψ i 4 4 These then satisfy: ψ i + i = i 4 ψ i i = i 43 ψ i + i = ψ i i = 44 The negative sign induced by Ω can be shown easily. From the left moving fermions we get: Ω i,l = Ωψ+ i i,l = Ω ψ i + ψ i+ i,l = ψi + = ψ i + i,r = i,r i+ ψ Ω i,l and from the right moving fermions: Ω i,r = Ω ψ + i i,l = Ω ψi + i+ ψ i,l = ψ i + ψ i+ Ω i,l = ψ i + i,r = i,r As the operators L B,R,NS B,R,NS and L only count the number of creation operator insertions and F and F count the number of fermions, this can be 7 The index i runs from to d. For example, for superstring theory, the index i has a range of to 4. In contrast to the fermions, whose indices run from to 8 in superstring theory. See Appendix C for details, especially C.

15 CLOSED STRINGS reduced to sum over only the left movers. Therefore with K dτ τ K Σ : K Σ = L;NS,α L, L F q LB,NS L;R,α F B,NS L q L, L L, L q LB,R B,R L q L, L 45 As the left and right moving sectors need to be identical by the level matching condition, it can be written more compactly like this: K Σ = L;NS,α L F q q LB,NS L L q q LB,R L L;R,α The easiest way to calculate this is to compute the fermionic and bosonic contributions separately... The bosonic sector contribution Let us start with the calculation for the bosonic sector: d d p L q q LB L = π V dq q α p d 4 q q d 4 N q q n,i αi n αi n N L;α = V d q q d 4 = V d q q d 4 = V d q q d 4 N= d d p α e πiτ τ π d d d p α p e πτ π d N= N n,i p 4 N= N n,i N N= n,i q q αi n αi n N q q αi n αi n N q q αi n αi n N d d p α p e πτ π d The integral over the momenta is a d-dimensional Gaussian integral in momentum space. The factor q q d /4 comes from the normal ordering constant in L B of the bosonic sector, which is given by c B = 4.

16 CLOSED STRINGS The integral over momentum space can be carried out: d d p α p e πτ = dp π d π d i e πτ α p i = π d i i = πτ π d α i dp i e πτ α p i dx e x = π d π d πτ α d = 4π τ α d Also the sum over the bosonic spectrum can be computed: N= N n,i q q αi n αi n N = = n,i n,i N= N q q αi n αi n N N q q N n N N= = n,i = q q n n q q d 4 q q n d ηiτ d Here I used the definition of the η-function as found in [BCoST p 47: ητ = e πiτ e πiτn n= Putting those pieces together, we get for the bosonic sector: L;α L q q LB L = V d 4π τ α d ηiτ d 46

17 CLOSED STRINGS 3.. The Neveu-Schwarz sector contribution Now it is time to commit ourselves to the fermionic sectors. Let us begin with the NS-sector: L F q q LNS L = L q q LNS L L F q q LNS L L;NS L;NS L;NS To avoid making the same computation twice both sums are identical apart from the F operator we will instead compute this sum like this: N b = L;NS L bf q q LNS L where b is either or /. With c NS = we get: 48 N b = q q d 48 q q rψi r ψi r N N=, N bf r,i As these are fermionic operators it is only possible to insert one creation operator of the same mode, hence the summation over only and. Let us also not forget at this point that the indices of fermionic modes in the NS-sector are half-integer 8. N b = q q d 48 = q q d 48 = q q d 48 = q q d 48 = q q d 48 = q q d 48 q q d 48 N bf N=, r,i + b q q r r,i n,i + b q q n n= n= n= q q 4 q q rψi r ψi r N d + b q q n d + b q q n + b q q n + b q q n + b q q n q q n d q q n ϑ [ d b iτ ηiτ 8 This means their indices can be written as r = n where n Z >.

18 CLOSED STRINGS 4 N b = ϑ [ b iτ ηiτ d The definition of the above used ϑ-function is as follows taken from [BCoST p 8: ϑ [ a b τ = e πiab+ a τ n= e πiτn + e πiaτ+b+τn τ + e πiaτ+b τn+ τ 47 Setting the correct parameter b and inserting this result back, yields: N + N / = ϑ [ iτ ηiτ d ϑ [ / iτ ηiτ d..3 The Ramond sector contribution The last sector we have to compute is the Ramond sector. Here we have to be careful to take the degenerate vacua into account. For a d-dimensional space we get d different vacua. Also, we have to consider the GSO-projection. It removes half of the vacua. Therefore we get 9 R = L;R L q q LR L = d = d = d = d = d d q q 4 d q q 4 d q q 4 d q q 4 d q q 4 N q q rψi r ψi r N r,i N r,i N=, + q q r r,i N q q rψi r ψi r N d + q q r r n= d + q q n + q q n 9 For the Ramond sector r Z and the normal ordering constant is given by c R = 4.

19 CLOSED STRINGS 5 R = d = d = d d q q 4 d q q 4 d q q 4 = d q q 4 = + q q n + q q n q q n q q n n= + q q n= d + q q n + q q n q q n q q n + q q n + q q n q q n q q n n= q q [ / ϑ ηiτ iτ 4 ϑ [ / ηiτ d..4 Combining all sectors iτ d If all of those results are put together, we get: K Σ = V d 4π τ α d ηiτ 3 d V d = 4π τ α d ηiτ d d ϑ [ iτ d ϑ [ / iτ d ϑ [ iτ d ηiτ d ϑ[ / iτ d ηiτ d ϑ [ / d iτ d ϑ[ / iτ d ηiτ d This implies: K = V d 4π α d dτ 4τ d+ ϑ [ d ϑ [ d / ϑ [ / ηiτ 3d The ϑ- and η-functions satisfy a Jacobi identity: [ ϑ 4 [ ϑ 4 [ / / ϑ 4 = + η η η d iτ 48 Therefore for d = the Klein-Bottle amplitude vanishes. This is shown in Appendix D.

20 3 OPEN STRINGS 6 3 Open Strings Now we will focus on open strings. The open string has a few features that a closed string does not have. These new features come from the fact that an open string has a boundary. As a consequence it only has one set of oscillator modes because the left moving excitations are reflected on the boundary and become right movers. Therefore they are not independent of each other and must be equivalent. Also the end of the string has no periodicity constraint and can obey two different boundary conditions : σ X µ σ=,π = δx µ σ=,π = Neumann boundary condition Dirichlet boundary condition 49 The Neumann boundary condition describes moving ends of the string. The Dirichlet boundary condition fixes the endpoint of the string to a certain value. The string itself can still vibrate but the endpoint will stick to where it was at the beginning. It is also possible that the different endpoints of the string obey different boundary conditions and, what s more, also the different space-time directions can obey different conditions, which means that the endpoint is fixed in certain directions while it is free to move in others. Now a problem arises with the Dirichlet boundary conditions: If the string cannot move in this direction but the oscillations are reflected by the boundary, this means that momentum flows off the end of the string and momentum conservation is lost. One can save momentum conservation by actually thinking of the hypersurface the string is attached to as a dynamical object the so-called Dp-brane, which is a p + -dimensional object. The momentum that flows off the end of the string is then absorbed by the Dp-brane. A string attached to a Dp-brane can move freely in p+ directions it obeys Neumann boundary conditions. For a target space with d-dimensions this means that the endpoint of the string is fixed in d p + directions. 3. Chan-Paton Indices Open string theory shows us a picture of strings which are attached to different branes. A priori there is no reason to believe there has to be any specific number of branes. The endpoints of two different strings can only interact The length of the open string is usually normalized to π in contrast to the normalization of π of the closed string. So I will keep this convention here.

21 3 OPEN STRINGS 7 when they are attached to the same brane. It can be thought of as a charge. One can give an endpoint a label that carries the information on which brane the string ends. 3. Open String Amplitudes In this section I will calculate the one-loop amplitude for the open strings. This is only relevant for NN open strings as the DD strings cannot perform a closed loop because they are fixed in space. Z open = Tr +Ωq LB,R,NS There are two contributions: The Annulus: A = Tr q LB,R,NS and the Möbius Strip: M = Tr 3.3 The Annulus = Tr q LB,R,NS Ωq LB,R,NS + Ωq Tr LB,R,NS 5 As before with the closed string, this calculation will be split into the different sectors: Bosonic, Neveu-Schwarz, Ramond and, as a new feature to the open strings, the Chan-Paton contribution The bosonic sector contribution We first are going to compute Tr q LB. As with the closed strings, we can start with equation 3 but for open strings the right movers are not independent of the left movers and therefore do not separately enter the trace as it would just lead to double count every state. Also bear in mind that in This is, of course, only true for the directions in which they are fixed to a point in spacetime and therefore obey the Dirichlet boundary condition. As I only look at strings that can move freely, I am not going to mix up Dirichlet and Neumann conditions for different directions and will therefore not explicitly write the dimensionality of a Dpbrane but call it a D-brane.

22 3 OPEN STRINGS 8 case of the open string α = α p. Tr q LB = q d 4 Vd B = q d 4 Vd = = q d 4 d V d 4π α τ d q d 4 d V d 4π α τ d = d V d 4π α d d d p π d qα p N d d p α p e πτ π d q n n= q n n= N n= q α n α n N q d N n n= d d 3.3. The Neveu-Schwarz sector The next contribution is Tr bf q LNS : 3 Tr bf q LNS NS N τ d ηiτ d 5 = q d 48 = q d 48 = q d 48 = q d 48 N=, r N+ r N+ n= [ ϑ b iτ = ηiτ N r N+ d bf q rψ r ψ r N d bn q r N N=, + b q r d + b q n d d 3 Again with the insertion of bf we avoid making the same calculation twice. 5

23 3 OPEN STRINGS The Ramond sector The next sector is the Ramond sector Tr q LR : Tr q LR = d R q d 4 = d q d 4 = d q d 4 = r= N=, N N=, r= d q r N d + q r r= iτ [ / ϑ ηiτ d d q rψ r ψ r N The Chan-Paton contribution As stated before, the Chan-Paton indices describe on which D-brane the string ends. As we have no information yet as to how many of them exist, we will simply compute this generally. Let us assume there are N different D-branes. Therefore there are also N different values for Chan-Paton index. The trace over those indices takes the form: N Tr = Tr ij ij = ij ij = N N = N 54 CP Combining all sectors A = d N V d 4π α d τ d+ i,j= Taking the results of the previous sections and taking the GSO-projection into account gives us: dτ ϑ [ d ϑ [ d / ϑ [ / d iτ η 3 55 d As it is the case with the Klein-Bottle, this vanishes for d = but only because there are two divergent expressions, namely ϑ [ [ / [ and ϑ +ϑ / for τ, which cancel each other, as shown in appendix D. 3.4 The Möbius Strip The Möbius Strip includes again the orientation reversal operator Ω. It is therefore necessary to take a look at the effects it has on open strings.

24 3 OPEN STRINGS 3.4. The Effects of Ω on Open Strings To find the action of Ω, we need to take a look at the classical solution of the open string 4. For the bosonic string these are 5 : α X µ NN z, z = xµ α p µ α µ k logz z + i z k + z k k k X µ DD z, z = cµ + α πi cµ c µ α log + i z z µ k z k z k 56 k c µ and c µ are the positions of the D-branes where the two ends of the strings are attached to. The here used conditions are Xσ = = c and Xσ = π = c. As the open string s length is fixed here to π the orientation reversal operator acts on open string coordinates like this: σ Ω π σ, τ Ω τ 57 Therefore on z, z: k z = e τ+iσ z = e τ iσ Ω e τ+iπ σ = e τ iσ = z Ω e τ iπ σ = e τ+iσ = z 58 Using these transformations on the coordinate functions will give us the transformations of the oscillator modes: X Ω NN x µ α p µ log z z α α µ k + i z k + z k k k α = x µ α p µ k α µ k logz z + i z k + z k 59 k That means that for the NN string the oscillator modes behave like this: k α k Ω e iπk α k 6 4 Here N stands for Neumann boundary conditions and D for Dirichlet. A NN string satisfies Neumann boundary conditions on both ends. It is not necessary to look at DN or ND strings as the Ω operation transforms the DN sector into the ND and vice versa. Therefore they cannot form eigenstates and so they don t play a role. 5 See [BCoST p 7f.

25 3 OPEN STRINGS The fermionic classical open string solutions of the equations of motion are 6 NN ψ α σ, τ = r ψ α σ, τ = r ψ α r e irτ+σ 6 ψ α r e irτ σ 6 DD ψ i σ, τ = r ψ i re irτ+σ 63 ψ i σ, τ = r ψ i re irτ σ 64 Acting on them with Ω yields: It follows: ψσ, τ Ω ε r ψ re irτ+π σ = ε r ψ re irπ e irτ σ ψσ, τ Ω ε r ψ r e irτ π+σ = ε r ψ r e irπ e irτ+σ 65 Ω εe irπ ψ r ψ Ω 66 r εe irπ ψr ψ r Where ε = + for NN strings and ε = for DD strings. interested only in NN strings we will from here on use ε = +. As we are 3.4. The Bosonic sector Now we can begin our computation of the Möbius Strip again with the bosonic oscillator calculation: d d p Tr Ωq LB = V d N Ωe πiτ α p + k;i α i k αi k 4 N 67 B π d N We have to be careful about the signs introduced by the orientation reversal. As we get a negative sign for each of the inserted oscillator modes, we get for Ω: N Ω = α n α n α nk Ω = e iπn n α n α n α nk = e iπn n N 68 With n = k j= n j. Also there is only one relevant direction for the computation of the closed loop. We can again set τ = iτ. Inserting this into 67 6 See [BCoST p 9.

26 3 OPEN STRINGS yields: Tr Ωq LB = V d B = q d 4 Vd = q d 4 = q d 4 d πi = e 4 d d p π d eπiτα p N V d d 4π α τ d V d d 4π α τ d d d p π d e πτ α p k V d d 4π α τ d The Neveu-Schwarz sector N Ωe πiτ k;i N α i k αi k 4 N d e πin k+iτ 69 7 d 7 e πik+iτ k e πiiτ + d 4 ηiτ + d 7 ηiτ + d 73 Here we have a bit of a problem. We do not know how the orientation reversal acts on the ground state of the open fermionic string. For now we will assume that it does not change the state but gives an overall factor which we will call NS for now: Tr NS Ω bf q LNS = NS q d 48 = NS q d 48 = NS q d 48 = NS e d πi 48 = q d 48 r=n+ N=, N=, N Ω bf e πτ r rψ r ψ r N e πir N bn e πτ r N d + b e πir iτ + r= e πiiτ + d 48 ϑ [ b iτ + ηiτ + ϑ [ b iτ + d ηiτ + d d 74

27 3 OPEN STRINGS The Ramond sector As with the Neveu-Schwarz sector we have no possible way of knowing how Ω acts on the ground states. Again we will assume it is just a numerical factor which is identical for all vacua but can be different from the Neveu-Schwarz sector s constant. We will call it R : Tr Ωq LR = q d 4 R d = R q d 4 = R q d 4 = Re N=, d d d πi 4 r= N Ωe πτ r rψ r ψ r N d e πir Niτ + N=, d + e πiriτ + r= ϑ [ / iτ + ηiτ The Chan-Paton contribution d 75 Here the orientation reversal shows a different effect. As Ω exchanges the two endpoints it also exchanges the Chan-Paton indices and potentially gives a negative sign: Ω ij = ɛ ji 76 Here ɛ = ±. The trace over the Chan-Paton factors is therefore given by: Tr Ω = ij Ω ij = ɛ ij ji = ɛ ij ij = ɛn 77 CP i,j i,j i=j Combining all sectors Combining the results of the previous sections gives us the following sum of ϑ-functions under the integral: NS e d πi 48 [ ϑ d ϑ [ d d πi / + R e 4 ϑ [ / d 78 We require that this gives us again zero for d =. Also we would like to d πi cancel the overall factor of e 4 given by the bosonic sector contribution.

28 3 OPEN STRINGS 4 d πi We therefore choose NS = e 6 and R = to obtain: M = ɛn V d d 4π α d dτ ϑ [ d ϑ [ / τ d+ d ϑ [ / ηiτ + 3d d iτ + 79

29 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS 5 4 The transverse pictures and their implications 4. The Klein Bottle at tree-level Up until now, we have looked at the closed string only in such a way that it undergoes a twist and moves in a loop. But one can also look at the Klein-Bottle differently. The first diagram of figure 3 shows the way we have looked at the Klein- Bottle the previous calculation. The closed string moves along the vertical axis and then inverts its orientation. This describes the loop amplitude or direct channel. πτ 4πτ arranging scaling π π πl Figure 3: The two different ways of looking at the Klein-Bottle. Picture based on [LoOD p 7. First we cut the surface along the dashed line in its center. Then we attach the separated piece at the top of the remaining part. In order to do that, we have to flip the surface around. This is shown by the big arrow. Next, we can rescale to lengths of the surface due to conformal invariance. As reparametrization freedom is given on the string worldsheet, it is possible to now see the horizontal axis as the time axis. Then it looks like a closed string which starts out with a crosscap, then propagates before terminating on a second crosscap. It defines a tree-level amplitude or transverse channel. The vertical time, τ, enters into the definition of the trace. If one switches to the transverse picture it is necessary to use the horizontal time,which is given by τ = 4l.7 Also then integration measure needs to be transformed: dτ = dl 4l This new time parameter, l, gives us the interaction time between two objects, known as O-planes. This O-plane can be seen as a hyper-surface 7 This equation is given by the necessity of the closed string to have a fixed length of π π. It then follows: τ = πl as seen in figure 3. π

30 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS 6 where the closed string couples to and which identifies opposite points of the string. Using the transformations of the η- and ϑ-functions 8 η = iτητ 8 τ ϑ [ a b = iτϑ [ a τ b τ 8 this can be written in a more compact form by using the transformations of the η and ϑ functions, given by 8 and 8 9 : ϑ [ a b w [ τ η iτ 3w w ϑ ab τ w iτ 3w ητ 3w τ [ ϑ ab τ w = 8 iτ w ητ 3w The Klein-Bottle amplitude can therefore be rewritten like this: ϑ [ d ϑ [ d / K = = V d 4π α d V d 4π α d = d V d 4π α d dl d+ 4l 4 l dl d l d l d dl ϑ [ d ϑ [ d / ϑ [ / η il ϑ [ d ϑ [ / ϑ [ / ηil 3 d d 3d d d il ϑ [ / ηil 3d il d il 83 For d = this is again zero, due to the Jacobi identity. But this zero comes from two cancelling infinite expressions for large l. The ϑ-function shows exponential decline for large l as well as the η-function. The latter leads to divergences as it is in the denominator. Expanding these functions for large l and setting the dimension to d = leads to: K = 4 V 4π α 5 dl e πl O e πl These divergences are an artefact of perturbation theory. They describe a particle being generated from the vacuum and they appear in higher loops. They are called tadpoles. 8 All relations of the η and ϑ-functions used here can be found in [BCoST p 9. For further reading see [TLoT and [TLoT. 9 Using the shorthand notation w = d See appendix D, p.38 for details

31 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS 7 4. The Annulus at tree-level As already discussed, the one-loop amplitude of the Annulus looks like this: N V d dτ ϑ [ d ϑ [ d / ϑ [ / d A = 4π α d τ d+ η 3 d iτ To get the tree-level amplitude, we need to rescale the surface as seen in figure 4. This gives us the new time parameter for the tree-level amplitude of the annulus. It is given by τ = l and dτ = dl l. πτ π scaling π πl Figure 4: The annulus as a propagating closed string. [LoOD p 6. Picture based on This gives us a different pre-factor as the integral measure will transform as well: dτ τ d+ dl d+ l = d d l l dl 84 Using the transformation law given by 8 lets us calculate the tree-level amplitude of the Annulus: Ā = N V d dl ϑ [ d ϑ [ d 4π α d l l d+ / ϑ [ / d η 3 d il ϑ [ d ϑ [ d / ϑ [ / d = N V d 4π α d = d+ N V d 4π α d dl d dl ϑ [ d ϑ [ / η 3 d d η 3 d ϑ [ / d il il 85

32 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS The Möbius Strip at tree-level The one-loop amplitude of the Möbius Strip was given by: M = ɛn V d 4π α d dτ ϑ [ d ϑ [ / τ d+ d ϑ [ / ηiτ + 3d d iτ + 86 To obtain the correct new time parameter, l, for the Möbius strip, we need to perform similar steps as we had to for the Klein-Bottle. πτ π 4πτ arranging π π scaling πl Figure 5: The tree-level interpretation of the Möbius strip. Picture based on [LoOD p 8. Figure 5 teaches us that the parameter l needs to satisfy τ = and also 8l dτ = dl. 8l As the η- and ϑ-functions contain a shift by in the arguments the connection between loop and tree-level amplitudes is not a simple S-transformation, defined by S : τ τ =, but a combination of S- and T -transformations τ with T : τ τ = τ +, which is given by: P = T ST ST 87 Using this transformation on iτ yields iτ P 4iτ 88 The effects of this transformation can be seen in detail in appendix E

33 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS 9 Inserting the equations gives: ϑ [ w [ / w [ ϑ ϑ w iτ / + iτ w iτ 3w iτ w ηiτ + 3w e πiw ϑ [ w [ / ϑ w πiw e 4 ϑ [ / e 3πiw 4 ηiτ + 3w e πiw ϑ [ w [ / ϑ w πiw e 4 ϑ [ / w iτ + w iτ + e 3πiw 4 ηiτ + 3w 89 Using this transformation and the substitution τ = 8l Möbius Strip tree-level amplitude: lets us calculate the V d M = ɛn 4π α d d πi e 4 ϑ [ / = ɛn V d 4π α d d πi e 4 ϑ [ dl Two things must now be shown: dl d+ 4l 4l 4l d d ϑ [ d d πi e / 8 ϑ [ / d πi3 e 4 ηil + 3d 8 d ϑ [ d d πi e d il + 8 ϑ [ / d πi3 e 8 ηil + 3d. The Jacobi identity is still satisfied with d = and d il + 9. The ϑ and η-functions still satisfy the same expansion for l as it is a priori not clear what happens due to addition of this factor. Both of these problems are addressed in the appendix F. 4.4 Tadpole cancellation The tadpoles we encountered cause divergences for large l. In a simple toy model example this becomes a little bit easier to comprehend:

34 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS 3 Let us consider a field theory of the form: S = d d x µ φ µ φ + Qφ This is a theory, where the field φ interacts with a field Q; which can also be viewed as a source term as Q has no dynamics. Therefore Q represents an immobile charged particle. If we expand around φ =, this theory leads to Feynman diagrams which look like this: Q k Q Figure 6: Two charges Q exchanging a φ-particle and exchanging momentum k. Picture based on [LoOD p 3 Mathematically written this means terms like: k = dle k l 9 Let us consider the limit k. On the left hand side this diverges obviously due to the vanishing momentum: k k 9 On the right hand side this divergence can be viewed as having a different origin, namely the long interaction time l between the charges: dle k l k dl 93 As we will shortly see, we can interpret the charged particle as O-planes and D-branes and the strings cause interactions between them. See [LoOD p 3.

35 4 THE TRANSVERSE PICTURES AND THEIR IMPLICATIONS Combining the Interactions The Klein-Bottle, the Annulus and the Möbius Strip tree-level amplitudes we encountered provide for different interactions between O-planes and D- branes. The Klein-Bottle can be interpreted as a closed string connecting two O-planes. The Annulus can also be looked at as a closed string that connects two D-branes. The endpoints of the open string performing the closed loop are then viewed as the origin of a closed string which is produced by a D-brane. The other endpoint of the open string is then the terminating point of the closed string on another D-brane. This amplitude therefore is identical to a tree-level diagram that connects two D-branes exchanging a closed string. The Möbius Strip on the other hand includes an orientation reversal which then can be perceived as a closed string connecting a D-brane and an O-plane. The Klein-Bottle, Annulus and Möbius strip amplitudes therefore give us the four different interactions between the D-branes and O-planes: Annulus Möbius Strip Klein-Bottle D-brane with D-brane D-brane with O-plane O-plane with D-brane O-plane with O-plane In the calculations, these tree-level diagrams can be obtained by interpreting the worldsheet differently and therefore exchanging the role of the time and space coordinates. For long interaction times between the O-planes and D-branes we again find our amplitudes becoming a divergent factor multiplying a vanishing factor. By adding up the tree-level amplitudes of the Annulus, the Klein-Bottle and the Möbius Strip, we find a way to eliminate the divergences altogether with d = : lim l K + Ā + M N + ɛn = N + 5 ɛn = 6 ɛn + 3 If we can find values for ɛ and N such that 94 ɛn + 3 = 95 All the potential divergences are cancelled. And indeed: for the choice ɛ = and N = 3 this equation is satisfied. The interactions between the O-planes and the D-branes are then in such a way that they cancel the tadpoles. That is why equation 95 is called the tadpole equation.

36 A APPENDIX: THE GSO-PROJECTION 3 A Appendix: The GSO-projection It can be shown that a fermionic string theory which uses the complete spectrum of the string is inconsistent 3. It is therefore necessary to remove some of the states from the spectrum. The GSO Gliozzi-Scherk-Olive projection removes states the following way:. Only one set the spinorial or the conjugated ones of the degenerate vacua in the R-sector shall be used. Every state with an even number of fermionic creation operators shall be omitted in the NS-sector These conditions can be incorproated in the theory by inserting a new projection operator: P GSO = F F 96 It is easy to show that this is a projection operator. First it is clear that the operator F has only two eigenvalues ±. Also P GSO = P GSO: F F PGSO = 4 4 = F + F F + F 4 4 = F F 4 4 = F F = P GSO 97 Obviously F and F commute. This projection operator eliminates every state with an even number of fermionic operators, because F n and n = =. 3 See also [BCoST p 3ff for details.

37 B APPENDIX: THE FERMIONIC STRING 33 B Appendix: The fermionic string B. The fermionic action and the equations of motion First let us take a look at the classical fermionic string. The fermionic equivalent of the classical bosonic string is 4 : S F = i d σe 4π Ψ µ γ α α Ψ µ 98 Here I introduced the two-component spinor Ψ and a special coordinate system which basis is given by the zwei-bein e a α. The α denotes an Einstein index, which are defined by the worldsheet and a is a tangent space index. The zwei-bein satisfies e α ae β b g αβ = η ab and therefore also satisfies e α ae β b ηab = g 5 αβ. Here g αβ is the worldsheet metric and η ab is the Minkowski metric. Therefore g = dete a α e. The next thing to do is to compute the equations of motion for the fermionic field Ψ: δ Ψ S F = i d σe 4π Ψ µ γ α α δψ µ = i d σe Ψµ α γ α δψ µ α 4π Ψµ γ α δψ µ 99 Therefore the equations of motion, δs =, look like this: γ α α Ψµ = This equation of motion also holds for the Ψ field, as can be read off of the action. As this is written in the string worldsheet variables τ, σ, we would like to know how it translates into z, z. The derivatives translate into: z = τ z τ + σ z σ z = τ z τ + σ z σ Inverting the definitions of the equations z = e τ+iσ and z = e τ iσ gives: log z + log z τ = log z log z σ = i 4 This is written in superconformal gauge. For details see [BCoST p 78f. 5 e α a e β b g αβ = η ab : Contracting the e a αs with their inverse e α a gives: g αβ = η ab e a αe b β and rearranging the indices then gives: g αβ = η ab e α a e β b

38 B APPENDIX: THE FERMIONIC STRING 34 Inserting these relations into equation gives us: z = z τ + iz z = z τ i z σ σ If we rewrite this again, we get the relations: 3 τ = z z + z z + z z z σ = i z z i z z z 4 These relations, together with the definitions ψ Ψ µ µ = ψ µ Ψ µ = Ψ µ γ γ = i γ = i 5 allow us to rewrite the equations of motion: z i γ α α Ψ µ = + z + i z z i ψ µ ψ µ z ψ µ = z ψ µ = 6 As both of the requirements must hold for any value z and z we get the two equations of motion in these coordinates: ψ µ = ψ µ = 7 We can therefore state that ψ µ = ψ µ z and ψ µ = ψ µ z. This leads to the mode expansions given in equation.

39 B APPENDIX: THE FERMIONIC STRING 35 B. Boundary conditions for fermionic strings We have to be careful about the boundary conditions induced by the surface term in 99. Let us take a closer look and assume that the equations of motion are satisfied: δ Ψ S F = i 4π Evaluating the two easy integrals gives: δ Ψ S F = i σ=σ +π dτe Ψµ γ δψ µ + 4π σ=σ dσdτe Ψµ γ δψ µ + Ψµ γ δψ µ 8 σ τ τ=τ dσe Ψµ γ δψ µ 9 If we assume that δψτ = τ = δψτ = τ =, which means that the starting and end state of the field is fixed, we can simplify this into: δ Ψ S F = i σ=σ +π dτe Ψµ γ δψ µ 4π σ=σ Using the definitions 5 leads to: δ Ψ S F = 4π dτe ψ µ, ψ µ δψµ δ ψ µ τ=τ σ=σ +π σ=σ As we still need δ Ψ S F =, we get the condition 6 : ψ δψ ψ δ ψ σ = ψ δψ ψ δ ψ σ + π This can be satisfied by ψ µ σ = ±ψ µ σ + π and δψ µ σ = ±δψ µ σ + π ψ µ σ = ± ψ µ σ + π and δ ψ µ σ = ±δ ψ µ σ + π 3 The anti-periodicity is only possible, because they are fermions on the worldsheet and therefore the action is quadratic in Ψ, which means the action is periodic independently of our choice. The periodic choice the positive sign in 3 is called the Ramond R sector and the anti-periodic ones the negative sign is referred to as the Neveu-Schwarz NS sector. This periodicity conditions carry over into the quantized fermionic string. As the signs of the two spinor components can be chosen independently, there are four different choices, namely R,R, NS,NS, NS,R and R,NS. The first two choices lead to spacetime bosons and the latter two to spacetime fermions. 6 Here I used the notation: α β α µ β µ

40 C APPENDIX: LIGHT CONE GAUGE 36 C Appendix: Light cone gauge C. The bosonic string in light cone gauge An easy way to quantize the bosonic string is to use the so-called light cone gauge. The Virasoro constraints can be explicitly solved in this gauge and the physical degrees of freedom become the only parameters of the theory. The light cone coordinates are defined as 7 : X ± = X ± X d 4 The remaining d coordinates are called the transverse coordinates and will be written as X i, where the index runs from to d. This gauge is possible as it is a conformal gauge and the residual gauge freedom is fixed by X + = α p + τ. All the other oscillator modes vanish in this gauge for X +. The inner product needs to be rewritten in this gauge and follows the rules: X + = X 5 X = X + 6 X i = X i 7 X Y = X µ Y µ = X i Y i + X + Y + + X Y = X i Y i X + Y + X Y 8 Let us take a look at the Virasoro constraints in light cone coordinates. These are T ±± = and can be rewritten into: τ X µ ± σ X µ = 9 Inserting our definition of the light cone coordinates, we get or the + and directions: τ X ± ± σ X ± = τ X ± ± σ X ± τ X ± ± σ X ± = τ X ± ± σ X ± τ X ± σ X Also we find that σ X + = from our choice of the light cone gauge. Using this and equation in equation 9, we get: τ X + τ X ± σ X = τ X i ± σ X i 7 See [ST p 7ff.

41 C APPENDIX: LIGHT CONE GAUGE 37 Adding up the two equations with different signs gives: τ X = τx i + σ X i τ X + It is also necessary to mention that τ X + = α p + is just a constant. This shows us that X is indeed also determined by this choice and the remaining d coordinate functions carry all the information. Therefore the number of physical degrees of freedom reduces to d. C. The superstring in light cone gauge The light cone condition for the bosonic part of the string remains unchanged. The fermionic part is given by: Ψ + = 3 As before Ψ ± = Ψ ± Ψ d. The Virasoro constraints take the form of T ±± = and T F ± = and produce 8 : ± X = p + α ±X i ± X i + iψ i + ψ i + i ψ i ψi 4 ψ = α p + ψi + X i 5 ψ = α p ψ i + X i 6 Here I used the shorthand notation ± = τ ± σ. This shows that also in the case of the superstring, the ± directions can be gauged out so that the remaining d directions carry all the physically relevant information. 8 See [BCoST p 8ff for details. In principle the derivation is the same as given by the solely bosonic string, but now more complicated as more fields are involved.

42 D APPENDIX: PROOF OF THE JACOBI IDENTITY 38 D Appendix: Proof of the Jacobi identity The important bit of this calculation are the terms proportional to q and q as the η-function in the denominator generates a factor of q, because of ηiw w q 4 q = q 4 7 and the fact that is taken to the th power 9. This factor needs to be cancelled as q as w grows. We can therefore also ignore all factors with higher powers of q. The definition of the ϑ-function is: ϑ [ a b iw = e πiab+ iaw e πwn + e πiaiw+b+inw iw + e πiiaw+b inw+ iw n= Written with q e πw : 8 ϑ [ a b iw = e πiab q a n= q n + b q n+a + b q n a 9 With the approximation w we can ignore all terms where n > and get: ϑ [ a w b iw e πiab q a + b q a+ + b q a+ 3 This leads to the three approximated ϑ-functions: ϑ [ w iw + q ϑ [ / w iw q 8 + q = q 8 ϑ [ w / iw q 9 Here w is simply a real parameter and can either stand for τ as used in sections and 3 or for l as in section 4. Of course, q is here e πiw.

43 D APPENDIX: PROOF OF THE JACOBI IDENTITY 39 These taken to the 4 th power gives us: ϑ [ iw 4 ϑ [ iw 4 w ϑ [ 4 w iw w 6q q 8! = k!8 k! q k k= 8 q k 8! = k!8 k! q k The factor q of the ϑ [ -function will be cancelled by the factor given from ϑ [ / which has the opposite sign. For the factors q, we get two factors of 8 from ϑ [ [ / and ϑ result is therefore: ϑ [ 4 ηiw iw k=, which exactly cancel the 6 of ϑ [ /. The final 4 iw ϑ [ 4 / iw w + 6q + Oq ϑ [ / q 3 This result would be divergent if the numerical factors would not exactly cancel each other out. As they do the end result is. 3 3 Mostly the factor of order q is left out because it is already cancelled by the functions ϑ [ [ / and ϑ at any order of q and is therefore not really part of the Jacobi identity. I included it only for completeness and transparency of this calculation.

44 E APPENDIX: THE P -TRANSFORMATION 4 E Appendix: The P -transformation Using every transformation one at a time gives us 3 iτ T iτ + = iτ + S iτ + T iτ + + = S iτ + 4iτ T iτ + 4iτ = 4iτ = 4iτ iτ + iτ + 3 To get the final result for the Möbius Strip amplitude, it is easier albeit lengthier to use each of these transformations separately on the amplitude and the η- and ϑ-functions: ηiτ + T ηiτ + = e πi ηiτ S e πi η iτ = e πi iiτηiτ T e πi πi iiτ + ηiτ + = e 4 iiτ + ηiτ S e πi 4 i iτ + η iτ = e πi 4 iτηiτ T e πi 4 The same for the ϑ-functions: iτ + ηiτ + = e πi 4 iτηiτ + ϑ [ iτ + T ϑ [ [ iτ + = ϑ / iτ S ϑ [ / iτ = iiτϑ [ / iτ T iiτ + ϑ [ / πi [ / iτ + = e iiτ + ϑ iτ S e πi i iτ + ϑ[ / iτ = e [ πi iτϑ / iτ T [ e πi iτϑ / iτ + 3 The necessary relations can be found in [BCoST p 8f.

45 E APPENDIX: THE P -TRANSFORMATION 4 ϑ [ / iτ + T ϑ [ / πi iτ + = e S e πi 4 ϑ [ / 4 ϑ [ / iτ iτ = e πi 4 iiτϑ [ T [ e πi 4 iiτ + ϑ S e πi 4 T e πi / i iτ + ϑ[ / 4 iτϑ [ / / iτ iτ + = e πi 4 iiτ + ϑ [ / iτ iτ + iτ = e [ πi / 4 iτϑ iτ ϑ [ / iτ + To summarize: T ϑ [ [ / iτ + = ϑ iτ S ϑ [ iτ = iiτϑ [ iτ T iiτ + ϑ [ [ iτ + = iiτ + ϑ iτ S i iτ + ϑ[ iτ = iτϑ [ iτ T iτϑ [ iτ + ηiτ + P e πi 4 iτηiτ + 33 ϑ [ iτ + P [ e πi iτϑ / iτ + 34 ϑ [ / ϑ [ / iτ + P [ e πi / 4 iτϑ iτ + 35 iτ + P iτϑ [ iτ + 36

46 F APPENDIX: THE PROBLEMS WITH THE TREE-LEVEL MÖBIUS STRIP4 F Appendix: The problems with the treelevel Möbius Strip Let us start with the Jacobi identity. The equation we start from is: d ϑ [ d d πi e 8 ϑ [ / d il + M d πi e 4 ϑ [ / dl For d = it follows: e M πi ϑ [ dl = = ϑ [ / dl ϑ [ / dl d πi3 e 8 ηil + 3d 4 [ / ϑ 4 e πi ϑ [ / 4 il + e 3πi ηil + 4 [ ϑ 4 [ / 4 + ϑ il + ηil + 4 [ + ϑ 4 [ / ϑ ηil + 4 il + 37 This expression is again the Jacobi identity. The second problem is the expansion for large l. The expansions of the ϑ- and η-functions remain the same but the identification of how exactly q looks like changes to: q = e πiil+ = e 4πl e πi = e 4πl = q 38 As with the expansions for the Annulus and the Klein-Bottle, this q becomes small for large l, but with the opposite sign. Therefore we can use the result of Appendix D with the only exception of q becoming negative: ηil + ϑ [ 4 ϑ [ / 4 ϑ [ / l q = q 4 il q + O q + 6 q + O q 39 There seems to be a slight problem at q, but this factor is, as stated before, completely cancelled simply by the expansions of only two of the ϑ-functions and I included it only for making the calculation more transparent. Because of this, it is usually not explicitly written out. For q the minus sign is irrelevant. Therefore it is okay to use the same expansion as for the annulus and the Klein-Bottle.