I haven't seen any reference which explains these things and I am not sure of all the steps of the argument or the equations. I am trying to reproduce here a sequence of arguments that I have mostly picked up from discussions and I would like to know of references for the background details and explanations.

  • It seems that in general the Witten Index (${\cal I}$) can be written as $ Tr (-1)^F \prod_i x_{i}^{C_i}$ where $x_is$ are like the fugacities of the conserved quantity $C_is$ and $C_is$ form a complete set of simultaneously measurable operators.

    I would like to know the motivation for the above and especially about calling the above an ``index" even if it depends on the fugacities which doesn't seem to be specified by the Lagrangian of the theory or by the underlying manifold? Even with this dependence does it reproduce some intrinsic property of either the underlying manifold or the theory?

  • Say $\phi$ and $\psi$ are the bosonic and fermionic component fields of a superfield and ${\cal D}$ be the superderivative. Now when the above tracing is done over operators of the kind $Tr({\cal D}^n\phi)$ and $Tr({\cal D}^n\psi)$ it is probably called a "single trace letter partition function (STLP)" and when it also includes things like products of the above kind of stuff it is called a ``multi-trace letter partition function (MTLP)"

I would like to know about the exact/general definitions/references for the above terminologies and the motivations behind them.

  • Probably for adjoint fields, if $f(x)$ is a STLP then apparently in some limit ("large N"?) one has, MTLP = $\prod _{n=1} ^{\infty} \frac{1}{1-f(x^n)}$

  • One also defines something called the ``full STLP (FSTLP)" where one probably includes terms also of the kind $Tr({\cal D}^n \phi ^m {\cal D}^p \psi ^q)$ and then the MTLP can be gotten from that as MTLP = $exp [ \sum _{m=1} ^ {\infty} \frac{1}{m} FSTLP]$

  • The upshot all this is probably to show that MTLP is the same as Witten Index (in some limit?).

I don't understand most of the above argument and I would be happy to know of explanations and expository references which explain the above concepts (hopefully beginner friendly!).

(I see these most often come up in the context of superconformal theories and hence references along that might be helpful especially about the representations of the superconformal group.)


Dear Anirbit, a good question. To figure out the answer, it's a good idea to realize why people talk about indices in the first place. What is a particular formula called the index good for?

Well, it's good because it doesn't change under continuous deformations. Consequently, there may be many methods to calculate the same number - and, which is related, many ways to interpret this number.

Now, in your previous question about short multiplets,

Superconformal theories

all of us discussed what the index is and why it is invariant. All answers to your new question are already written there. More or less. The purpose of the index is that it is invariant under continuous changes of parameters. That includes the changes of your variables $x$.

The terms $\ln(x_i) C_i$ - which exponentiate to $x_i^{C_i}$ - may be absent in your Lagrangian or Hamiltonian but you may always add them to the Lagrangian or Hamiltonian. So they are examples of the continuous parameters that don't change the value of the index. As long as it is a full-fledged index, the result will be independent of all quantities $x_i$. The more quantities $x_i$ you add, the more ways to calculate the index may be found. So you may try to calculate them for $x=1$, $x=0$, $x=\infty$, or another value (or combination of values, to be more general).

It's useful to mention that the index is sometimes generalized to similar concepts such as "elliptic genus" which are only invariant under the deformation of some parameters but they explicitly depend on others, usually holomorphically or in a similar way. They're not just "numbers" describing the theory but whole "functions". Such expressions are useful in theories where the short multiplets may still possess some quantities that may change continuously. So the elliptic genus doesn't just "count" the short multiplets which is what the full-fledged index essentially does (with some weights); instead, it counts some function of the quantities that the short multiplets possess.

Of course, in each case, one has to prove that the quantity is invariant under the transformations of some kinds and it depends on others.

  • $\begingroup$ Thanks for your reply. I didn't get the point you are making about the independent of changes in $x_i$. The index is quite explicitly a function of $x_is$ and they look like rational polynomials in $x_is$ or with fractional powers etc. Also can you shed light about the letter partition function issue. I am quite in the dark about that. $\endgroup$ – user6818 Jan 30 '11 at 10:50
  • $\begingroup$ For example one can see between page 15 and 30 of this paper arxiv.org/pdf/hep-th/0310285v6 to see some usages of this letter partition function which I am puzzled about. $\endgroup$ – user6818 Jan 30 '11 at 11:04

This appears to be related to twisted or broken supersymmetry. For unbroken supersymmetry the total Hamiltonian ${\cal H}~=~N{\cal H}_b~+~N{\cal H}_f$ for $N$ strings with the bosonic operators $a,~a^\dagger$ and the fermionic operators $b,~b^\dagger$ for N strings is: $$ {\cal H}~=~{N\over 2}\sum_k (\omega(k)a_k^\dagger a_k~+~{N\over 2}\sum_k \omega(k)b_k^\dagger b_k. $$ There exists an operator $\lambda$ so that $\omega(k)/2~=~\omega log(\lambda_k)$. Thus by supersymmetry we may write the Hamiltonian as $$ {\cal H}~=~N\sum_k (log(\lambda_k)a_k^\dagger a_k~+~N\sum_k log(\lambda_k)b_k^\dagger b_k. $$

Now Wick rotate the integrand in the path integral $e^{-it H_b}$ under the substitution $it~\rightarrow~\beta$ and find that $$ Tre^{-NH_b\beta}~=~\sum_{i=1}^\infty\lambda_i^{-N\omega\beta}~=~\zeta(N\omega\beta). $$ This is a generalized zeta function, and for both bosons and fermions, $$ Tr\big((-1)^{N_b}e^{-N{\cal H}_b\beta}e^{-N{\cal H}_f\beta}\big)~=~e^{-N({N}_b~-~{N}_f)\omega\beta}~=~1 $$ since ${N}_b~=~{N}_f$ in unbroken supersymmetry. As such we then have that $$ Tr\big((-1)^{N_F}e^{-N{\cal H}_f\beta}\big)~=~{1\over{\zeta(N\omega\beta)}}. $$ and the fermion partition function may be written with the Mobius function $\mu$ as $$ Tr\big((-1)^{N_f}e^{-NH_f\beta}\big)~=~\sum_i{{\mu(\lambda_i)}\over{\lambda_i}^{N\omega\beta}}. $$

The supersymmetry is partially broken, only half the stringy bosons have superpairs then the bosonic Hamiltonian takes the form $$ {\cal H}_B~=~2N\sum_klog(\lambda_k)c^\dagger_k c_k~+~N\sum_k log(\lambda_k)f_k^\dagger f_k. $$ where $c_k~=~a_ka_k$ and $c_k^\dagger~=~a_k^\dagger a_k^\dagger$. The operators $f_k$ and $f^\dagger_k$ are fermionic operators \lq\lq recovered\rq\rq$~$from the bosonic Hamiltonian. Now if we compute $Z_F(\beta)$ we find that $$ Z_F(\beta)~=~ Tr\big((-1)^{N_F}e^{-N\sum_k log(\lambda_k)f^\dagger_k f_k\beta} \big)\times 1 $$ $$ ~=~ Tr\big((-1)^{N_F}e^{-N\sum_k log(\lambda_k)f^\dagger_k f_k\beta} e^{-2N\sum_k log(\lambda_k)c^\dagger_k c_k\beta}e^{-2N\sum_k log(\lambda_k)b^\dagger_k b_k\beta}\big) $$ $$ ~=~Tr\big((-1)^{N_f}e^{-\big(2N{\cal H}_F~+~N{\cal H}_B\big)\beta}\big). $$ where the unity involves the old fermion and boson operators for unbroken supersymmetry. Now $Z_F(\beta)$ is $$ Z_F(\beta)~=~ \sum_{n=1}^\infty{{\mu(n)}\over{n^{2N\omega\beta}}}~=~ {{\zeta(N\omega\beta)}\over{\zeta(2N\omega\beta)}}. $$ Here the $c,~c^\dagger$ boson fields have superpartners and the $f,~f^\dagger$ fields do not.

  • $\begingroup$ Thanks for your reply. Can you explain how you took the trace in the third displayed equation? I guess you wanted to evaluate, $Tr e^{\sum _k a_k^{\dagger}a_k log(\lambda _k ^{-N\beta \omega})}$. Will this not produce a k-dependent coefficient to the terms $\lambda _k ^{-N\beta \omega}$ terms? (coming from the expectation values of the operator $a_k^{\dagger}a_k$) ? Is the LHS of the 5th and the6th displayed equation supposed to be the same? (Is $N_F$ in the 5th the same as $N_f$ in the 6th?) I didn't understand how their RHS are equal. $\endgroup$ – user6818 Jan 31 '11 at 5:04
  • $\begingroup$ Also if you could explain how what you said is related to the letter partition function that I was enquiring about. If you could give some references. $\endgroup$ – user6818 Jan 31 '11 at 5:07

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