Stringy Things

# Notes on String Theory: Ward Identities, Noether’s Theorem, and OPEs

1.1. Example 1

In the last entry we derived both the quantum version of Noether’s theorem and the Ward identity given in Polchinski’s textbook. This means we obtained the idea of the existence of conserved currents and how Ward identities in general constrain the operator products of these same currents. Let us now elaborate on some examples. The solutions to these examples are given in Polchinski (p.43); however, a more detailed review of the computation and of some of the key concepts will be provided below.

We start with the simplest example, where we once again invoke the theory of free massless scalars. Following Polchinski, the idea is that we want to perform a simple spacetime translation ${\delta X^{\mu} = \epsilon a^{\mu}}$. The action will be left invariant under worldsheet symmetry. But as what we want to derive is the current, given what we have been discussing, this means we should pay special attention to the fact that we are required to add ${\rho (\sigma)}$ to the above translation. Recall that we defined ${\rho(\sigma)}$ in our derivation of the Ward identity. The important point to note is that, again, the action is still invariant and from past discussions we already understand ${\rho(\sigma)}$ has a compact or finite support. From this set-up, let us now rewrite the action for massless scalars,

$\displaystyle S_{P} = \frac{1}{4\pi \alpha^{\prime}} \int d^2\sigma \partial X^{\mu}\partial X_{\mu} \ \ (1)$

When we vary (1) we obtain the following,

$\displaystyle \delta S = \frac{\epsilon a_{\mu}}{2\pi \alpha^{\prime}} \int d^2 \sigma \partial^{a}X^{\mu} \partial_{a}\rho \ \ (2)$

There is a factor of 2 from varying ${\partial X}$ that gives us a reduced denominator. We have also used the identity stated in Polchinski’s textbook, namely ${\delta X^{\mu}(\sigma) = \epsilon \rho(\sigma) a^{\mu}}$, where we can treat ${\epsilon}$ and ${a^{\mu}}$ as constants and therefore pull them in front of the integral.

Before we can move forward, there is something we have to remember. Recall the path integral formulation from our last discussion, where we found the variation to be proportional to the gradient. The result is written again below for convenience,

$\displaystyle [d\phi^{\prime}]e^{-S[\phi^{\prime}]} = [d\phi]e^{-S[\phi]}[1 + \frac{i\epsilon}{2\pi} \int d^2\sigma J^{a}\partial_{a}\rho + \mathcal{O}(\epsilon)^2] \ \ (3)$

If there is no contribution from the metric, then the measure in brackets becomes ${- \delta S}$. What this tells us is that the variation must be something like,

$\displaystyle \delta S = -\frac{i\epsilon}{2\pi} \int d^2\sigma J^{a}\partial_{a}\rho \ \ (4)$

Now notice, in computing both (2) and (4) we may establish the following interesting relation,

$\displaystyle \partial^{a}X_{a} \partial_{a} \rho \frac{\epsilon a_{\mu}}{2\pi \alpha^{\prime}} = -\frac{i\epsilon}{2\pi}J^{a} \partial_{a}\rho \ \ (5)$

The first step is to simplify. Immediately, we can see that we can cancel the ${\partial_{a}\rho}$ terms on both sides,

$\displaystyle \partial_{a}X^{\mu} \frac{\epsilon a_{\mu}}{2\pi \alpha^{\prime}} = -\frac{i\epsilon}{2\pi}J_{a} \ \ (6)$

This still leaves us with a bit of a mess. What we need to do is recall another useful fact. In the last section we studied the invariance of the path integral under change of variables, which, at the time, enabled us to obtain Noether’s theorem as an operator equation. Explicitly put, we had something of the general form ${ \frac{\epsilon}{2\pi i} \int d^{d} \sigma \sqrt{g} \rho(\sigma) \langle \nabla_{a}J^{a}(\sigma) ... \rangle}$. Notice that we have all of the ingredients. Given the Noether current is,

$\displaystyle J_{a} = a_{\mu}J_{a}^{\mu}$

We may substitute for ${J_{a}}$ in (6) and then work through the obvious cancellations that appear, including a cancellation of signs. Once this is done, we go on to obtain the following expression for the current,

$\displaystyle J_{a}^{\mu} = \frac{i}{\alpha^{\prime}} \partial_{a}X^{\mu} \ \ (7)$

Which is precisely what Polchinski gives in eqn. (2.3.13) on p.43 of his textbook. Automatically, we can see our currents are conserved. And, of course, we are free to switch to holomorphic and antiholomorphic indices and we can do so with relative ease,

$\displaystyle J_{a}^{\mu} = \frac{1}{\alpha^{\prime}}\partial X^{\mu}$

$\displaystyle \bar{J}_{a}^{\mu} = \frac{1}{\alpha^{\prime}}\bar{\partial} X^{\mu} \ \ (8)$

In the manner indicated above, we have successfully constructed the current following a spacetime translation. For this example the goal is to now use an operator to check the Ward identity and see if the overall logic is sound. What we require in the process are the appropriate residues, and to find these we will need to compute the OPEs. So to test some of the ideas from earlier discussions in the context of the given example.

Recall the formula for OPEs given the reverse of the sum of subtractions, namely the sum of contractions, as described in (11) of this post. In this formula recall that we have two operators which are normal ordered, $: \mathcal{F}:$ and $: \mathcal{G}:$. These are arbitrary functionals of X and typically the range of X is non-compact.

Now, in Polchinski’s first example we consider the case where ${\mathcal{F} = J_{a}^{\mu}}$ and ${\mathcal{G} = e^{ikX(z, \bar{z})}}$. In other words, we want to compute the product of the current and the exponential operator. As the product is normal ordered, there are no singularities and the classical equations of motion are satisfied. Instead, the singularities are produced from the contractions, or, in this case, the cross-contractions as ${z \rightarrow z_{0}}$. Furthermore, as a sort of empirical rule, it can be said that the most singular term in ${\frac{1}{z - z_{0}}}$ comes from the most cross-contractions. And we should recall that we compute the cross-contractions by hitting our operators with ${\delta / \delta X^{\mu}_{\mathcal{F}}}$ and ${\delta / \delta X^{\mu}_{\mathcal{G}}}$, respectively. Hence, from the master formula for cross-contractions,

$\displaystyle : \frac{i}{\alpha^{\prime}} :\partial X^\mu(z): :e^{ik X(z_0,{\bar z_{0}})}: = \exp [- \frac{\alpha^{\prime}}{2} \int d^2 z_1 d^2 z_2 \ln \mid z_{12}\mid^2 \frac{\delta}{\delta X_{\mathcal{F}}^{\mu}(z_1, \bar{z}_1)} \frac{\delta}{\delta X_{\mathcal{G} \mu}(z_2, \bar {z}_2)}] \ \ :\frac{i}{\alpha^{\prime}} \partial X^\mu(z) e^{i k X(z_{0},\bar{z}_{0})}:$

$\displaystyle = : \frac{i}{\alpha^{\prime}} \partial X^{\mu}(z) e^{i k X(z_{0},\bar{z}_{0})} : - \frac{i}{2} : \int d^2 z_1 d^2 z_2 \ln \mid z_{12} \mid^2 \frac{\delta(\partial X^{\mu}(z))}{\delta X_{\mathcal{F}}^{\mu}(z_1, \bar{z}_1)} \frac{\delta ( e^{i k X(z_0, \bar{z}_0)})}{\delta X_{\mathcal{G} \mu}(z_2, \bar{z}_2)} \ \ (9)$

Note that for ${\mathcal{G}}$, which, in this case is ${e^{ikX}}$, it is an eigenfunctional of ${\delta / \delta X_{\mathcal{G}} (z_{2}, \bar{z}_{2})}$. Likewise, for for ${\delta / \delta X_{\mathcal{F}}}$ we will end up with a delta function,

$\displaystyle = : \frac{i}{\alpha^{\prime}} \partial X^{\mu}(z) e^{i k X(z_{0},\bar{z}_{0})} : - \frac{i}{2} : \int d^2 z_1 d^2 z_2 \ln \mid z_{12} \mid^2 \partial (\delta^{\mu}_{\alpha} \delta^2(z_1, z)) i k^{\alpha} \delta^2(z_2, z_0) e^{i k X(z_{0}, \bar{z}_0)} : \ \ (10)$

Now, we can pull out the $ik^{\mu}$ which flips the sign,

$\displaystyle = : \frac{i}{\alpha^{\prime}} \partial X^{\mu}(z) e^{i k X(z_{0},\bar{z}_{0})}: + \frac{k^{\mu}}{2} : \partial (\int d^2 z_1 d^2 z_2 \ln \mid z_{12} \mid^2 \delta^2(z_1, z) \delta^2(z_2, z_{0}) e^{i k X(z_{0},\bar{z}_0)}) : \ \ (11)$

Notice that we have delta functions inside the integrand, so we are left with,

$\displaystyle : \frac{i}{\alpha^{\prime}} \partial X^{\mu}(z) e^{i k X(z_{0},\bar{z}_{0})} : + \frac{k^{\mu}}{2} : \partial ( \ln \mid z - z_{0} \mid^2 e^{i k X(z_0,\bar{z}_0)}) :$

$\displaystyle = : \frac{i}{\alpha^{\prime}} \partial X^{\mu}(z) e^{i k X(z_{0},\bar{z}_{0})} : + \frac{k^{\mu}}{2 (z - z_{0})} e^{i k X(z_0,\bar{z}_0)} \ \ (12)$

And so we obtain the following result,

$\displaystyle \frac{i}{\alpha^{\prime}} : \partial_{a}X^{\mu} : :e^{ikX}: \sim \frac{k^{\mu}}{2 (z - z_{0})} : e^{ikX}: \ \ (13)$

Where ${\sim}$ means the most singular pieces. We can also perform the same calculations for the antiholomorphic term,

$\displaystyle \frac{i}{\alpha^{\prime}} : \bar{\partial}X^{\mu}: : e^{ikX}: \sim \frac{k^{\mu}}{2(\bar{z} - \bar{z}_{0})} :e^{ikX}: \ \ (14)$

As Polchinski notes and as we see, the OPE is in agreement with the Ward identity. But we can still carry on a bit further. To conclude this example, recall explicitly to mind the Ward identity and our residues. Switching back to the holomorphic case and evaluating the LHS of (14) notice we find, picking out only the residues,

$\displaystyle \frac{i}{\alpha^{\prime}} : \partial X^{\mu} e^{ikX}: = (\frac{k^{\mu}}{2} + \frac{k^{\mu}}{2}) G = k^{\mu} G \ \ (15)$

So, we have that it must be equal to ${k^{\mu}}$ times the operator ${\mathcal{G}}$ from above. Now, given that ${\mathcal{G} = \mathcal{A}}$, the right-hand side of the Ward identity tells us that,

$\displaystyle k^{\mu} \mathcal{A} = \frac{1}{i \epsilon} \delta \mathcal{A} \ \ (16)$

And, again, from the Ward identity we can see in (15) that with a bit of algebra the variation of the operator must be,

$\displaystyle \delta \mathcal{A} = ik^{\mu}\epsilon \mathcal{A} \ \ (17)$

Where we are assuming the variation is only in one direction. Interestingly, as an aside, what is actually happening are the following transformation properties,

$\displaystyle \mathcal{A} = e^{ikX} \rightarrow e^{ikX + ik^{\mu}\epsilon}$

$\displaystyle X^{\mu} \rightarrow X^{\mu} + \epsilon \ \ (18)$

1.2. Example 2

In the first example we considered a spacetime translation. We can now look to the second example in Polchinski’s textbook, where we want to consider a worldsheet translation, particularly how the ${a}$ of the ${\sigma}$ coordinates transforms as ${\delta \sigma^{a} = \epsilon v^{a}}$. Here ${v^{a}}$ is a constant vector. It follows that from the action for free massless scalars is invariant under this transformation, with the above symmetry clearly understood given ${X}$ is a scalar and how ${\delta \sigma^{a}}$ does not change the measure of integration. And so, just as in the first example, what we want to do is investigate the construction of the conserved current as a result of this worldsheet symmetry transformation and then test the Ward identity.

The first step is to note that because we are dealing with a scalar theory we may write explicitly,

$\displaystyle \sigma^{a} \rightarrow \sigma^{\prime a} = \sigma + \epsilon v^{a} \ \ (19)$

Where, for any worldsheet symmetry transformation, the scalar fields simply transform as follows,

$\displaystyle X^{\prime \mu}(\sigma^{\prime}) = X^{\mu}(\sigma) \ \ (20)$

From which it also follows that,

$\displaystyle X^{\prime \mu}(\sigma + \delta \sigma) = X^{\mu}(\sigma) \implies X^{\prime \mu}(\sigma) = X^{\mu}(\sigma - \delta \sigma) \ \ (21)$

Where we should recognise that in brackets on the left-hand side of the first equality, ${\sigma + \delta \sigma = \sigma^{\prime}}$.

Of course, like the first example, we’re interested in how operators transform. And so we want to consider,

$\displaystyle \delta X^{\mu}(\sigma) = X^{\prime \mu}(\sigma) - X^{\mu}(\sigma) = X^{\prime \mu} (\sigma^{a} - \epsilon v^{a}) - X^{\mu}(\sigma) \ \ (22)$

Expanding and only keep the 1st terms, what we end up with is precisely an expression for how our operators transform,

$\displaystyle \delta X^{\mu}(\sigma) = -\epsilon(\sigma) v^{a}\partial_{a}X^{\mu} \ \ (23)$

Now, what we want to do is check with the Ward identity. So, like before, let’s start by varying the action and then build from there,

$\displaystyle \delta S = \delta [\frac{1}{4\pi \alpha^{\prime}}\int d^2\sigma \partial^{a}X^{\mu}\partial_{a}X_{\mu}]$

$\displaystyle = \frac{1}{2\pi \alpha^{\prime}} \int d^2\sigma \partial^{a}X^{\mu}\partial_{a}\delta X_{\mu} \ \ (24)$

Where ${\delta X_{\mu} = -\epsilon(\sigma)v^{a}\partial_{a}X_{\mu}}$. The implication is as follows. From (24) we can substitute for ${\delta X_{\mu}}$,

$\displaystyle \delta S = \frac{1}{2\pi \alpha^{\prime}} \int d^{2}\sigma \partial^{a}X^{\mu}\partial_{a}(-\epsilon(\sigma)v^{a}\partial_{a}X_{\mu})$

$\displaystyle = -\frac{\epsilon}{2\pi \alpha^{\prime}} \int d^2\sigma \partial^{a}X^{\mu}\partial_{a} v^{b}\partial_{b}X_{\mu} + \partial^{c}X^{\mu}\partial_{c}v^{d}\partial_{d}X_{\mu}$

$\displaystyle = -\frac{\epsilon}{2\pi \alpha^{\prime}} \int d^2\sigma \partial^{a}X^{\mu}\partial_{a} v^{b}\partial_{b}X_{\mu} \ [1] + \partial_{d}(\frac{1}{2}v^{d} \partial^{c}X^{\mu}\partial_{c}X^{\mu}) \ [2] \ \ (25)$

Where, for pedagogical purposes, the first and second integrands have been labelled [1] and [2] respectively. The reason is because it will be useful to recall these pieces separately in order to highlight some necessary computational logical and procedure. Before that, however, we should think of the conserved current. It follows, as we have already learned,

$\displaystyle -\delta S = \frac{i}{2\pi} \int d^2\sigma \sqrt{-g}J^{a}\partial_{a}\epsilon \ \ (26)$

Remember, looking at (2) in a previous entry, we can see clearly that ${J^{a}(\sigma)}$ is the coefficient of ${\partial_{a}\rho (\sigma)}$. In the first example we become more familiar with this fact. And what Polchinski is referencing in the single passing sentence that he provides prior to eqns. (2.3.15a) and (2.3.15b) is that we need to make contact with this formalism. It is convenient to now reassert the ${\rho(\sigma)}$ term,

$\displaystyle = -\frac{\epsilon}{2\pi \alpha^{\prime}} \int d^2\sigma [ (\partial^{a}X^{\mu}\partial_{b} v^{b}X_{\mu}) \partial_{b}(\rho(\sigma)) \ [1] + (\rho(\sigma)) \partial_{c}(\frac{1}{2}v^{c} \partial^{d}X^{\mu}\partial_{d}X^{\mu})] \ [2] \ \ (27)$

Now, let’s look at both pieces of (27). Piece [1] above looks fine and, on inspection, seems quite manageable. Piece [2], on the other hand, is not very nice. In taking one step forward, what we can do is integrate the second piece by parts. This has the benefit that we can eliminate the total derivative that arises and eliminate the surface terms. To save space, the result is given below,

$\displaystyle \delta S = \frac{\epsilon}{2 \pi \alpha'}\int d^2\sigma v^b\partial^a X^{\mu} \partial_b X_{\mu} \partial_a - \partial_b (\frac{1}{2}v^b \partial^a X^{\mu} \partial_a X_\mu)$

$\displaystyle =\frac{\epsilon}{2 \pi \alpha'}\int d^2\sigma [v^b(\partial^a X^\mu \partial_b X_{\mu} -\frac{1}{2}\delta^{a}_{b} \partial_b X^\mu \partial^b X_\mu) \partial_a] \ \ (28)$

How to interpret (28)? Notice something very interesting. We have the stress-energy tensor plus some additional terms outside the small brackets. If we make the appropriate substitution for the stress-energy tensor we therefore obtain,

$\displaystyle \delta S = -\dfrac{\epsilon}{2 \pi}\int d^2\sigma \, (v^c\,T_c^a)\partial_a \ \ (29)$

If we bring the constant epsilon back into the integrand, we have an integral over the worldsheet times a derivative of the parameter of an infinitesimal transformation. Whatever is left can be interpreted as a conserved current. Hence, then, if we go back and inspect (2) in this post we come to establish what Polchinski states in eqn. (2.3.15a). Our indices are slightly different up to this point, but this is merely superficial and when we rearrange things we find,

$\displaystyle J^{a} = iv^{b}T_{b}^{a}$

And then lowering the index on ${J}$,

$\displaystyle J_{a} = iv^{b}T_{ab} \ \ (30)$

This is our conserved current. In certain words, it is natural to anticipate a conserved current on the string worldsheet and also for this current to be related to the stress-energy tensor. Just thinking of the physical picture gives some idea as to why this is a natural expectation. But we are not quite done.

What we want to do, ultimately, is define the stress-energy tensor as an operator with full quantum corrections. But, as we are working in conformal field theory, there is an ambiguity about how we might define it related to normal ordering. Let’s explore this for a moment.

We should think of stringy CFTs by way of how we will define a set of basic operators, and then from this show what is the stress-energy tensor. Moreover, it is a property of the stress-energy tensor and the basic operators we utilise that will give definition to the CFT. In CFT language, it is given that the stress-energy tensor can be written as,

$\displaystyle T_{ab} = \frac{1}{\alpha^{\prime}} : \partial_{a}X^{\mu}\partial_{b}X_{\mu} - \frac{1}{2}\delta_{ab}(\partial X)^2 : \ \ (31)$

This is what Polchinski cites in eqn. (2.3.15b). We can still go a step further and discuss the topic of conformal invariance in relation to this definition. For instance, from the principles of conformal invariance, it remains the case that as discussed much earlier in these notes,

$\displaystyle T_{a}^{a} = 0$

Which is to say, as we should remember, the stress-energy tensor is traceless. This condition of tracelessness tells us how, if we were to go to holomorphic and antiholomorphic coordinates,

$\displaystyle T_{a}^{a} = 0 \rightarrow T_{z\bar{z}} = 0 = T_{\bar{z}z} \ \ (32)$

Where one may recall, also, the non-vanishing parts ${T_{zz}}$ and ${T_{\bar{z}\bar{z}}}$ from an earlier discussion in this collection of notes. It follows that if the stress-energy tensor is, in fact, traceless, we may invoke the conservation of the current such that,

$\displaystyle \nabla^{a}J_{a} = 0 = \nabla^{a} T_{ab} = 0 \ \ (33)$

Which is to say that we have full conservation for the full stress-energy tensor. We can write this in terms of holomorphic and antiholomorphic coordinates as expected,

$\displaystyle \bar{\partial}T_{zb} + \partial T_{\bar{z}b} = 0 \ \ (34)$

This gives us two choices:

$\displaystyle b = z \implies \partial T_{zz} = 0$

$\displaystyle b = \bar{z} \implies \partial T_{\bar{z}\bar{z}} = 0 \ \ (35)$

Where, as it was discussed some time ago, ${T_{zz} = T(z)}$ is a holomorphic function and ${T_{\bar{z}\bar{z}} = \bar{T}(\bar{z})}$ is an antiholomorphic function.

It is perhaps quite obvious at this point that we may also write,

$\displaystyle T(z) = -\frac{1}{\alpha^{\prime}} : \partial X^{\mu}\partial X_{\mu}:$

$\displaystyle \bar{T}(\bar{z}) = -\frac{1}{\alpha^{\prime}} : \bar{\partial} X^{\mu}\bar{\partial} X_{\mu}: \ \ (36)$

Now, returning to our current (30), we can be completely general in our study of the current,

$\displaystyle J_{z} = iv(z)T(z)$

$\displaystyle \bar{J}_{\bar{z}} = i\bar{v(z)}\bar{T}(\bar{z}) \ \ (37)$

If we have conservation of the current, then the above is the same as,

$\displaystyle \nabla_{a}J^{a} = \bar{\partial}J_{z} + \partial J_{\bar{z}} = 0 \ \ (38)$

Which is to say that the new currents are conserved provided ${v(z)}$, previously considered a constant vector, is holomorophic. Additionally, the current is of course associated with symmetries; but what are these symmetries? They are the conformal transformations.

If, in the bigger picture, what we want to do is find ${\delta X}$ due to symmetries ${J_{z} = ivT(z)}$, to proceed recall the Ward identity ${\mathcal{A} = X^{\mu}}$. It follows we need to compute an OPE for the stress-energy tensor with our scalar field (complete computation is given in the Appendix of this chapter, along with other important and useful OPEs),

$\displaystyle :T(z) : :X^{\mu}(z_{0}, \bar{z}_{0}): = -\frac{1}{\alpha^{\prime}} : \partial X\partial X: : X^{\mu}(z_{0}, \bar{z}_{0}) :$

$\displaystyle \sim -(\frac{2}{\alpha^{\prime}}) \cdot (-\frac{\alpha^{\prime}}{2}\partial_{z}\ln \mid z - z_{0}\mid^2) : \partial X(z_{0}) :$

$\displaystyle T X^{\mu} \sim \frac{1}{z - z_{0}}\partial X(z_{0})$

$\displaystyle \bar{T}X^{\mu} \sim \frac{1}{\bar{z} - \bar{z}_{0}}\bar{\partial}X(\bar{z}_{0}) \ \ (39)$

Which is what Polchinski states in eqn. (2.4.6). And now we can use the Ward identity and take the residue of the current with the coefficients of the OPE for the holomorphic and antiholomorphic pieces,

$\displaystyle iv(z_{0}) \partial X(z_{0}) + i \bar{v}(z_{0} \bar{\partial} X(\bar{z}_{0}) = \frac{1}{i\epsilon} \delta X \ \ (40)$

And so what we find is that, for the current we have constructed, we have a symmetry transformation of the following form,

$\displaystyle \delta X = -iv(z_{0})\partial X(z_{0}) - i\bar{v}(z_{0} \bar{\partial}X(\bar{z}_{0}) \ \ (41)$

For ${z_{0} \rightarrow z_{0} + \epsilon v(z_{0})}$. If we drop ${z_{0}}$ and generalise,

$\displaystyle \delta X^{\mu} = -\epsilon v(z)\partial X - \epsilon\bar{v}(z)\bar{\partial}X \ \ (42)$

For ${z \rightarrow z + \epsilon v(z)}$ which is an infinitesimal transformation, where the only constraint is that ${z \rightarrow z^{\prime} = f(z)}$ is holomorphic.

The reason the transformation is so simple,

$\displaystyle \delta X^{\mu} = X^{\prime \mu}(z^{\prime}, \bar{z}^{\prime}) - X^{\mu}(z, \bar{z}) = X^{\mu}(z - \epsilon v, z - \epsilon\bar{v}) - X^{\mu} \ \ (43)$

Where, after Taylor expansion,

$\displaystyle \delta X^{\mu} = - \epsilon v\partial X - \epsilon\bar{v}\bar{\partial}X \ \ (44)$

It is important to point out that ${f(z) = \xi(z)}$ represents a global rescaling (but can also represent a local rescaling). If ${\mid \xi \mid = 1}$ then we have a simple rotation, and in general no scaling.

To conclude, from the very outset of this chapter, we may also recall to mind that in the context of the conformal group we are working in 2-dimensions. When we ask, ‘what is the analogue of this symmetry in higher dimensions?’, the answer is that in higher dimensions we can construct scale invariance as well. Indeed, in D-dimensions, if you have ${y^{\mu} \rightarrow \lambda y^{\mu}}$ you have additional special conformal transformations.

References

Joseph Polchinski. (2005). ‘String Theory: An Introduction to the Bosonic String’, Vol. 1.

Blumenhagen, R. and Plauschinn. (2009). ‘Introduction to Conformal Field Theory’.

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