In David Tong's CFT notes, he argues that the OPE for the holomorphic stress tensor with itself must take the form $$T(z)T(w)=\frac{c/2}{(z-w)^4}+2\frac{T(w)}{(z-w)^2}+\frac{\partial T(w)}{z-w}+finite$$ I'm a bit confused about this point. He argues that in a unitary CFT, no operators can have negative scaling dimensions, suppressing the higher order poles allowed by symmetry from appearing in the expansion. From this OPE, he derives the Virasoro algebra. Then later, when examining consequences of unitarity, he uses the Virasoro algebra to prove that no operators can have negative scaling dimension. Isn't this argument a little circular then? Is there a way to argue for the general form of the $TT$ OPE that doesn't invoke such circular reasoning?
[Physics] Stress Tensor OPE in CFT
conformal-field-theoryoperatorsquantum-field-theorystress-energy-momentum-tensorstring-theory
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Dear Anirbit, great questions.
Long and short representations are not difficult to be defined and most introductory texts to supersymmetry explain them. However, some of them don't use this particular terminology, so let me tell you: long multiplets are representations that transform nontrivially under all supercharges. So you can't find a supercharge that annihilates the whole representation. Consequently, if there are $N$ real supercharges, the dimension of the representation has to be $2^{N/2}$. On the other hand, short multiplets - also known as BPS multiplets - are annihilated by a subset of the supercharges, some holomorphic or chiral subset of the spinor(s), so the irreducible short multiplets have a lower dimension than $2^{N/2}$. Because some supercharges annihilate the whole multiplet, their anticommutator also annihilates it, and these anticommutators are given by energy minus some central charges (bosonic generators that appear on the right hand side of the SUSY algebra, together with the energy-momentum). That's why the short multiplets also minimize the energy among the charged states - they saturate the BPS bound and the short multiplets are also called BPS states. Massless particles' supermultiplets are often BPS but there can also be massive short multiplets. "Most" of the multiplets in the spectrum of a theory are long. The subsets of supercharges that annihilate a representation may be "really large" in which case the representation is even smaller, "hypershort". This terminology is often used but depends on the context (the group).
The groups you write are relevant only for superconformal theories in certain dimensions, namely $d=2+1$ in your case. In 3 dimensions, the Lorentz group is $SO(2,1)$ and the conformal group - the first factor in the bosonic subgroup of the superconformal group - is obtained by adding 1 spatial and 1 temporal dimension (this is a universal rule in any dimension, a rule that becomes self-evident in AdS/CFT where you add $1+1$ dimensions because the larger group is the isometry of an anti de Sitter space which is a hyperboloid inserted to a space with an extra time coordinate and the extra holographic spatial coordinate), so it is $SO(3,2)$ in both cases. The spinors in $d=2+1$ dimensions are real (Majorana), so if you have $N=2$ spinors or $N=3$ spinors, the rotation group that rotates the internal index labeling the different spinors of the supercharges is inevitably $SO(2)$ or $SO(3)$ which is the other factor you mentioned. The bosonic subgroup of the superconformal groups is simply the product of the "spacetime part" and the "internal part" (R-symmetry): they have to be separated in this way by the Coleman-Mandula theorem.
The primary fields or operators or highest-vector states etc. are operators or states such that you may obtain all other elements of the representation by the raising and lowering operators. For each algebra, there is a different set of raising and lowering operators - and different number of them. The more "factors" your algebra has, the more labels you need. That's why your actual labels you learned from the $2$-dimensional conformal algebra are obviously not applicable for different algebras in different dimensions. The dimension of an operator is a label of a conformal primary field in any spacetime dimension (or it is a sum of other labels). However, in $d=2+1$ dimensions, you also need to know how the field behaves under the rotation of spatial dimensions in $SO(2,1)$, that's given by the spin, and how it behaves under the $SO(2)$ or $SO(3)$ R-symmetry group, which is given by the R-charge.
You're asking a lot. This answered the first three points. OK, but let me continue with your other questions and complaints:
- The inequality (and all similar inequalities) may be derived by writing $\Delta\pm h-j-1$ as a square a real supercharge, or a sum of such squares, so it cannot be negative. This is a general trick in all of supersymmetric theories. For $j=0$, $|h|+j+1$ reduces to $|h|+1$ which explains that there are unitary representations for such values of $\Delta$ if the values are allowed by the algebra (integer or half-integer spin etc.). I don't understand your representations that are supposed to exist for $\Delta=|h|$ which violates the bound. However, there may exist some exotic or irregular representations of similar groups with "smaller dimensions", too. The representation theory of $SL(2,R)$ itself - which is isomorphic to $SO(2,1)$ - is very subtle and has many kinds of representations and even many types of unitary reps.
There are three more points then:
Representations that saturate the BPS bound are usually short because the bound's total quantity (which is non-negative, the quantity of the $\Delta-h$ type) may be written as the squared real supercharges, or their sum (such as $QQ^\dagger$ for a non-hermitean $Q$), and if the quantity vanishes, it follows that the supercharges have to vanish across the representation as well - of course, assuming a positive-definite Hilbert space and (which is related) unitary representation. The invariance of the whole rep under a subset of supercharges is how we defined "shortness".
Witten's index vanishes for all long representations because one may always pair the basis vectors of the long multiplet to objects of the form $|\psi\rangle$, $Q|\psi\rangle$, for a well-chosen $Q$ and a basis of $\psi$'s, and the contribution of the two vectors in the pair to the Witten's index is opposite up to the sign (which is flipped by the addition of the $Q$), so the pair's contribution cancels. On the other hand, when you construct the index properly, short representations will contribute nonzero because the sign of the Witten index contribution is constructed so that it is positive for the whole short representation or at least there's no cancellation. The "would-be" partners in the pairs are zero because $Q|\psi\rangle=0$ across the short multiplet for the same $Q$ on the short representation, because it's short.
You need two indices, with the opposite signs in front of $h$, because, as explained above, you want to show that $\Delta$ is bigger than $|h|+j+1$ which is equivalent to showing it's bigger than $h+j+1$ as well as $-h+j+1$. In general, indices are independent of $\beta$ or any other continuous parameter because they're indices. If you change $\beta$ or any other continuous parameter of the theory, the long multiplets may shift their mass but their contribution to the index will still vanish. The short multiplets either keep their mass, or they may shift it - away from the BPS bound - but that's possible only if they combine with the "mirror" short multiplet(s) into a long multiplet and those two (or many) short multiplets' contribution to the Witten's index was zero before the deformation, anyway. So because the (short) multiplets that contribute nonzero may only "go long" and disappear in pairs (or, on the contrary, they may appear by decomposing a long multiplet that previously existed and went to zero mass where it can split), the Witten's index is invariant under all continuous deformations of the theory, including the changes of $\beta$. That's its main property making it a powerful calculational tool.
You wrote exactly what bosonic generators the two types of short multiplets are annihilated by. For those generators of the $j\pm h$ type, you also find different sets of supercharges that square to $j+h$ or $j-h$, respectively.
- I didn't understand what you meant by $x$ but it's plausible that the reason is that this was your proposal and it was based on a misunderstanding.
At any rate, I think that I have answered all your questions.
The central charge itself appears in the commutation relations because the quantum theory allows not only linear, but projective representations of the conformal Witt algebra to be physically admissible. Such projective representations are in bijection to linear representations of the central extensions of the Witt algebra $$[L_m,L_n] = (m-n)L_{m+n}$$ which are the Virasoro algebras $$ [L_m,L_n] = (m-n)L_{m+n} + \frac{c}{12}(m^3-m)\delta_{m,-n} $$ for $c\in\mathbb{R}$. For more on projective representations and central extensions, see this Q&A of mine.
The energy-momentum tensor classically has conformal weight $2$, so its conformal mode expansion is $T(z) = \sum_n T_n z^{-n+2}$. The question now becomes how one shows that $T_n = L_n$. Since the energy-momentum tensor is the conserved current for the translations generated by $L_{-1} = \partial_z$ (since $L_n = z^{-n+1}\partial_z$), its integral has to be the generator $L_{-1}$ itself (the integral of the conserved current is the Noether charge, which is the generator of the symmetry in the Hamiltonian formulations): $$ L_{-1} = \frac{1}{2\pi\mathrm{i}}\int T(z)\mathrm{d}z$$ and inserting the mode expansion we get $L_{-1} = T_{-1}$. More generally, for any conformal transformation $z\mapsto z+\epsilon(z)$, we get a conserved current $\epsilon(z)T(z)$. The $L_n$ generate the transformations $z\mapsto z + \epsilon_n z^{n+1}$. As above, it follows that $$ L_n = \frac{1}{2\pi\mathrm{i}}\int z^{n+1}T(z)\mathrm{d}z$$ and thus $L_n = T_n$. Therefore, the energy-momentum tensor of a conformal field theory is $T(z) = \sum_z L_n z^{-n+2}$.
Here we only considered the holomorphic parts, but the arguments for the anti-holomorphic pieces are exacctly the same
Best Answer
Let's start with the Ward Identities ((4.11) in Tong's lecture). In radial quantization for a 2D theory they read
$\delta_{\epsilon,\bar\epsilon}\Phi(w,\bar w)= \frac{1}{2\pi {\rm i}}\left( \oint_\gamma dz R\left[ T(z)\epsilon(z)\Phi(w,\bar w) \right] - \oint_\gamma d\bar z R\left[ \bar T(\bar z)\bar \epsilon(\bar z)\Phi(w,\bar w) \right] \right)$.
We know how a primary field transform under the a conformal transformation:
$\delta_{\epsilon}\Phi(w)= h (\partial\epsilon) \Phi(w) + \epsilon\partial\Phi(w) $.
(I have taken only the holomorphic part, a similar formula goes for the right moving sector. From now on, I omit the anti-holomorphic part.)
This, using Cauchy, we see that for a general conformal transformation, i.e. $\epsilon=z^n$, the OPE between $T$ and $\Phi$ has to be
$T(z)\Phi(w)= \frac{h}{(z-w)^2}\Phi(w) + \frac{1}{z-w}\partial \Phi(w)+\cdots$,
where it is understood that we are working in radial quantization, so we omit the $R$ in front of the OPE. Now, what happens if $\Phi$ is a quasi-primary? Quasi-primaries transform under the global part of the conformal group, $SO(3,1)$. This is the part that is preserved in $D>2$. The global part of the conformal group is given by $\epsilon=z^n$, $n=0,1,2$. In this case, what we obtain is
$T(z)\Phi(w)= \sum_{p\ge4} \frac{a_p(w)}{(z-w)^p}+\frac{h}{(z-w)^2}\Phi(w) + \frac{1}{z-w}\partial \Phi(w)+\cdots.$
What Tong is telling you now, is that $T$ is actually a quasi primary (actually, it is a descendant of the identity!) When he allows a central charge extension, he is stating that, at the quantum level, you find an anomaly determined by $c\neq0.$ classically, $T$ is a primary, but not quantically. This has nothing to do with unitarity. This is just the transformation of fields. Note that $h$ hasn't been set positive.
Now, just as you defined a symmetry transformation using Noether's currents, you can define symmetry generators for the conformal algebra (Virasoro generators) using radial quantization with $T$ being the conserved current. From there, you can compute the Virasoro algebra and you find its central-extension.
Using representation theory you can see that $h\ge0$ (or from the 2 point function and cluster decomposition) by looking at the first descendant of $\Phi$, $L_{-1}\Phi$. By looking at $L_{-2}\Phi$ you find $c\ge0$.
As you see, this is a linear argument. Not a cyclic one!