Can anyone tell me where can I read about the notion of "short" and "long" representations? Like what they are etc.
From where can I learn the arguments which show that the bosonic subalgebra of $\cal{N}=2$ and $\cal{N}=3$ Lie superalgebra are given by $SO(3,2)\times SO(2)$ and $SO(3,2)\times SO(3)$ respectively?
I have often seen the following statement being made about the above which I do not understand, that primary states of these algebras are labeled by $(\Delta,j,h)$ where $\Delta$ is the scaling dimension of the primaries, $j$ is its spin and $h$ is its $R-$charge (or R is its charge highest weight)
Can someone explain the above labeling scheme. I don't understand much of the above.
Somehow the above doesn't seem to fit with my elementary understanding of what are primary operators in 1+1 CFT which are defined using particular forms of the OPE with stress-energy tensor.
- Then one argues that for $j\neq 0$ unitarity forces $\Delta \geq \vert h\vert + j+1$ and for $j=0$ unitary representations occur when $\Delta = h$ or $\Delta \geq \vert h \vert + 1$
I haven't been able to trace the above argument in any reference book or exposition.
It is said that the isolated representations which saturate the above unitarity bound for $j=0$ are all short.
Apparently the Witten Index in this context is defined as $Tr(-1)^Fx^{\Delta +j}$ and it vanishes on all long representations but is nonzero on all short representations.
Further to understand the state content of all unitary representations in superconformal algebras people defined two different kinds of Witten indexes, $\cal{I}^+$ and $\cal{I}^-$ as,
$${\cal I}^+ = Tr(-1)^Fx^{\Delta+j}e^{-\beta(\Delta-j-h)}$$ and
$${\cal I}^+ = Tr(-1)^Fx^{\Delta+j}e^{-\beta(\Delta-j+h)}$$
The above are apparently independent of $\beta$. I would like to know why.
Apparently the first index above receives contributions only from states with $\Delta = j+h$ and all such states probably are annihilated in some sense and have something to do with the cohomology of the supercharge with charges $(\frac{1}{2},-\frac{1}{2},1)$ Similarly somehow the second index above receives contributions only from states with $\Delta = j-h$ and all such states probably are also annihilated in some sense and have something to do with thecohomology of the supercharge with charges $(\frac{1}{2},\frac{1}{2},1)$
- Eventually I see these indices to be writable as products (sometimes infinite) of rational polynomials in $x$. I don't know how it happens. This way of writing has something to do with "single/multitrace operators/primaries" (another concept which I don't understand!)
I would be grateful to hear of explanations of the above arguments and constructions and also if detailed references/expositions exist for the above.
Answer
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.
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