9

When both "space" ($x$) and "time" ($y$) directions are periodic, the Laplacian on torus with coodinate $z=x+iy$ has a normalized zero mode $$ \varphi_0(z) = \frac 1 {\sqrt{{\rm Im}(\tau)}} $$ (Here $\tau$ is the modular parameter defining the torus.) As $$ -\nabla^2 \varphi_0=0. $$ the zero mode means that the Laplace operator is not 1-1 and so prevents ...


6

Anomalies is very popular and fast growing theme, so I will try collect main terminology and give major references. Anomalies with chiral fermions: David Tong: Lectures on Gauge Theory Jeffrey A. Harvey: TASI 2003 Lectures on Anomalies Adel Bilal: Lectures on Anomalies • An ABJ anomaly implies an explicit violation of the global symmetry. It is as bad ...


6

The constant term is needed because on a compact manifold with periodic boundary conditions there is a zero mode in the spectrum of the laplacian. This is easier to see on the circle, where $\frac{d^2}{dt^2}f = \lambda f$ has the periodic constant solution $f(t) =c$. This makes the operator not-inveritble. One should instead work in the space orthogonal to ...


6

You're almost there! Remembering how dilatations act on $\mathcal{O}_i(x_i)$, to demand covariance of the 3-pt funstion you need$^{(*)}$ $$\big\langle \mathcal{O}_1(x_1)\mathcal{O}_2(x_2)\mathcal{O}_3(x_3)\big\rangle = \frac{C_{123}}{x_{12}^a\,x_{23}^b x_{31}^c}, $$ where $x_{ij}:=|x_i-x_j|$ (I change your $X_{ij}$ notation to not carry an extra square ...


6

Simplest example that comes to mind: a linear sigma model (the kind you encounter in bosonic string theory). Its central charge $c$ is a positive integer and there are $c$ different primaries $\partial X^{\mu}$ ($\mu$ ranging from $0$ to $c-1$). These primaries are different fields, but they all have conformal weights of $(1, 0)$ hence the same scaling ...


5

In my opinion you have a very basic misunderstanding of what theories in physics are, and what relationship physics theories have with science. Physics is a science that studies nature using measurements and observations and mathematical models to fit these measurements and observations and predict what new data would be expected. The theories impose new ...


5

A descendant is not something that has derivatives in it. It is something that is a total derivative of something else $\mathcal{O}' = \partial_\mu \mathcal{O}$. More precisely, the definition of a primary operator is $$ K_\mu \mathcal{O}(x) = 0\,, $$ where $K_\mu$ is the generator of special conformal transformations. You can prove that you operator is a ...


5

The free scalar field can have any complex central charge, due to the linear modification of the energy-momentum tensor that you mention. The case $c=1$ is a bit special, in particular it allows compactification with an arbitrary radius, while for general $c$ the radius is quantized. See Section 4.1 of my review article for more details: https://arxiv.org/...


5

The energy momenum tensor is defined by $$ \delta S[\phi, g_{\mu\nu}]= \frac 12 \int d^dx \sqrt{g} T^{\mu\nu}\delta g_{\mu\nu}. $$ In this variation we vary the geometry but keep the fields $\phi(x)$ fixed. If we start in flat space where $g_{\mu\nu}=\delta_{\mu\nu}$ and make a diffeomorphism then $\delta g_{\mu\nu}= \partial_\mu \epsilon_\nu+ \partial_\nu\...


5

Before we even start talking about ladder operators, there is a much simpler test. Consider that $P_\mu$ is Hermitian. (You might have seen the relation $P^\dagger_\mu=K_\mu$. This is a source of a lot of confusion, and I will comment on it later. For now, let's simply agree that $P_\mu$ is the usual momentum and thus certainly $P^\dagger_\mu=P_\mu$.) Let's ...


5

This is a bit long, but I'm going to be super precise. Let's work in Poincare coordinates in AdS, $$ ds^2 = \frac{L^2}{z^2} ( dz^2 + dx^\mu dx_\mu ) . $$ $\Delta$ is the eigenvalue of scale transformations which acts as $z \to \lambda z$, $x^\mu \to \lambda x^\mu$. A scalar field of mass $m$ in the bulk satisfies an equation of the form $$ (\Box - m^2) \Phi(...


4

The introduction to this answers a part of my question in a concise manner. The following points are necessary in order for the CFT to have a holographic dual. The first AdS/CFT papers (Maldacena, Witten, GKP) state that the three point functions like $\langle TTT \rangle$ must have specific structures for a holographic dual to exist. (This means that any ...


4

I know topic this topic is 6 years old, but having just came across it, I feel the need to clarify some confusion around it, starting from the question, passing through the accepted answer and the ensuing comments. Most of this confusion sparks from two different but similar notions, that of a conformal transformatrion and that of a conformal isometry. Let ...


4

This is a very good question and one that is quite difficult to answer. In physics, the use of a group is usually to describe a symmetry of the physical system. What we know of the monster is that it has a non-trivial action in 196883 dimensions. In other words, the Monster group is the symmetry of a 196883 dimensional object. The hope of Monster moonshine ...


4

The starting point is to consider a CFT on a sphere, $S^{d-1}$ so the manifold under consideration is the "cylinder" $S^{d-1} \times {\mathbb R}$ with metric $$ ds^2 =- dt^2 + R^2 d\Omega_{d-1}^2 $$ The next step is to perform a Wick rotation. Here, we rotate $t \to - i \tau$. This step is useful for many reasons, not least of which is that it brings all ...


4

You are indeed missing some pieces. You can immediately see that there is no way your formula works in general since the full expansion reads $$ \begin{aligned} e^{ix\cdot P} K_\mu e^{-ix\cdot P} &= \sum_{n,m=0}^\infty \frac{i^{n-m}}{n!m!} (x\cdot P)^n\,K_\mu\,(x\cdot P)^m\,. \end{aligned}\tag{1}\label{ini} $$ The problem is that this is a mess because ...


4

In the free theory the answer is really simple, $$ X(\sigma)X(\sigma') = - \frac{\alpha'}{2} \log(\sigma - \sigma') ~ + : X(\sigma)X(\sigma'): $$ where $:~:$ denotes normal ordering.


4

CFTs at non integer values of the dimension $d$ can be defined by the axioms of crossing as all other CFTs. This is possible because the blocks are known analytically in $d$ $[1]$. However when $d$ is not an integer, these theories are necessarily non unitary, in that they must have states with negative norm. See $[2]$ and also $[3]$ for further details. In $...


4

If you take the limit of $\frac{\Delta_n}{R}$ for $R \to \infty$ with fixed $n$ you do get zero, but keep in mind that you also have infinitely many $\Delta_n$. Thus it might be possible (and is in fact true) that by taking the limit $n \to \infty$ at the same time in appropriate way you end up with a nonzero result for infinite $R$. Then you can immediately ...


4

Short answer is: the mass of the string and the mass of the excitation of the worldsheet CFT are two different masses. To elaborate a bit. The Hilbert space of the string is not the same as the Hilbert space of the CFT due to the presence of constraints. In fact, string states are parametrized by CFT states with conformal dimensions $h = \bar{h} = 1$. In ...


4

Scale invariance is common in physical systems at phase transitions. If the characteristic length of a system is small in one phase (disordered) and infinite in another phase (ordered), then typically the system is scale invariant at the transition between the two phases. A non-trivial physical observation is that in many cases scale invariance (together ...


4

This follows because the $d$-dimensional (global) conformal group ${\rm Conf}(d)$ is locally isomorphic to the proper Lorentz group $G:=SO(d\!+\!1,1)$ in Minkowski space $$\mathbb{R}^{d+1,1}~\cong~\mathbb{R}^d\times \mathbb{R}^{1,1}, \qquad \mathbb{R}^{1,1} ~\cong \underbrace{\mathbb{R}}_{\ni x^+}\times \underbrace{\mathbb{R}}_{\ni x^-},\tag{1} $$ cf. e.g. ...


4

I would say that the claim that there are no non-trivial CFTs in $d>6$ is just a speculation for which there isn't much evidence. The belief is that above six spacetime dimensions, the only unitary CFTs are simply free theories and that all non-trivial fixed points can be described by mean field theory. In addition to what you said, that there are fewer ...


4

Actually... There is some evidence of CFTs in $d>6$. In [1] they construct a solution in AdS$_8$ implies the existence of a CFT in $d=7$. This is not a definitive answer because there are still some issues about the solution. One has to prove full nonperturbative stability and also there is a region in spacetime where the coupling becomes big and one has ...


3

For the Ising model, we have the power of exact solutions. In particular, the one-dimensional transverse-field quantum Ising model, defined by the Hamiltonian $$ H = - J \sum_{i} \sigma^z_i \sigma^z_{i+1} - h \sum_i \sigma^x_i, $$ turns out to be exactly solvable. One can use the Jordan-Wigner transformation to map it to free fermions, after which you can ...


3

The character of a representation of the Virasoro algebra depends on the whole structure of the representation, not just of the conformal dimension $h$ of the highest-weight state. For a Verma module, the character is simply $$ \chi_{c,h}^\text{Verma}(\tau) = \frac{q^{h+\frac{1-c}{24}}}{\eta(\tau)}$$ as you wrote. However, for generic $c$ there are two ...


3

The conjugation formula you have in general dimensions is true only for scalar operators. In $d=2$, this corresponds to an operator with $h={\bar h} = \frac{\Delta}{2}$. We therefore need to prove $$ {\cal O}^\dagger(z,{\bar z}) = \frac{1}{(z {\bar z})^{\Delta} } {\cal O} \left( \frac{1}{ {\bar z} } , \frac{1}{z} \right) \qquad \qquad (1) $$ To derive this, ...


3

They have put all the anti-holomorphic dependence into $\bar{v}(\bar{z})$. So for holomorphic modes of stress energy tensor $[L_n, \bar{v}(\bar{z})]=0$. For anti-holomorphic generators $[\bar{L}_n, \bar{v}(\bar{z})]\neq0$. Field $X(z, \bar{z})$ have holomorphic and anti-holomorphic parts in Laurent expansion (2.89). Then from equation (2.40) with $h=\bar{h}=...


3

The standard reference for torus parttion functions is E. and H. Verlinde Chiral Bosonization, Determinants and the String Partition Function, Nuc Phys B 288 (1987) 357-396. They show that partition function for the $++$ spin structure is zero. The correlators are not necessarily zero as the zero eigenvalue in the denominator of the fermion Green ...


3

If you wish to determine the stress tensor from the action, you will need to know how the $bc$ ghost system couples to gravity. To figure this out, we recall that the action arose from the BRST gauge fixing of the string path integral. If you follow through that procedure, we find the action $$ S_{gh} = \frac{1}{2\pi} \int d^2 \sigma \sqrt{g} b_{ab} \nabla^a ...


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