# Free boson twisted boundary condition and $T^2$ partition function

Many CFT textbooks discuss free boson theory and free fermion theories on the torus.

The partition function for the boson theory (without compactification and orbifold) is obtained by summing over the Verma modules from all possible highest weight states $$|\alpha \rangle$$ coming from the vertex operators $$e^{i \alpha X}$$. The result reads

$$\sqrt{\frac{1}{\operatorname{Im}\tau}} \frac{1}{\eta(\tau) \overline{\eta(\tau})} \ ,$$ where the $$\operatorname{Im}\tau$$ factor comes from integrating over all possible "momenta" $$\alpha$$, while the $$\eta(\tau)$$ comes from summing over Virasoro descendants.

For free fermion theory, the torus partition function takes in all four sectors, (R, R), (R, NS), (NS, R), (NS, NS), corresponding to different boundary conditions of the field $$\psi$$.

The uncompactified free boson theory also has a twisted sector as well, with an anti-periodic boundary condition $$X(e^{2\pi i}z) = - X(z)$$. In the big yellow book, this situation is discussed, where the two-point function of $$X$$ and the stress tensor is computed. But what is left untouched is the representation theory in this sector.

I wonder what is the $$T^2$$ partition function of the uncompactified is we also consider anti-periodic boundary condition?

• Perhaps I'm misunderstanding the question, but we do take into account the twisted sector while compactifying on an orbifold, whereupon the "real" Hilbert space only consists of $\mathbb Z_2$-invariant states (you can do this with a projection operator in the trace) Jan 18 at 16:07
• @NiharKarve You are right, when $\mathbb{Z}_2$ orbifolding, people do consider the twisted sector. But what about before orbifolding and compactification, just the vanilla free boson theory? Jan 18 at 17:10
• @NiharKarve Ok I see. In the textbooks at hand, people always impose the twisted condition together with the compactification $X \sim X + 2\pi R$. I guess what I want to ask is if the compactification is optional, just do orbifold, and if so, what is the Hilbert space. I'm guessing one simply removes the $\mathbb{Z}_2$ non-invariant states from the original boson theory? Jan 19 at 1:22
• @NiharKarve Asked differently, what is the $T^2$ partition function if we also consider the anti-periodic boundary condition? For periodic condition, the partition function is as written in the post. Jan 19 at 1:50
• 1) Only the condition $X\sim X + 2\pi R$ is compactification on the circle, orbifold compactification needs the additional $\mathbb Z_2$ symmetry. 2. Naïvely all you have to do is start with the circle-compactified theory and then remove the non-$\mathbb Z_2$-invariant states, but modular invariance of the partition function forces inclusion of states generated by half-integer moded oscillators: that's the twisted sector. So the actual partition function is $\frac12 Z_\text{circ}+\left|\frac\eta{\vartheta_2}\right|+\left|\frac\eta{\vartheta_3}\right|+\left|\frac\eta{\vartheta_4}\right|$. Jan 19 at 9:31

$$\renewcommand{\Im}{\operatorname{Im}}$$Let's dissect the question a bit. First, the periodic BCs on the free boson on a torus correspond to (R,R). But of course, no one stops you from imposing (R,NS), (NS,R) and (NS,NS) BCs and compute the partition function. The physical meaning of this is having turned on a background $$\mathbb{Z}_2$$ gauge field along either or along both cycles of the torus. So to rephrase your question, you essentially ask the following:

We know that $$Z_{\text{(R,R)}}[\tau] = \frac{1}{\sqrt{\Im(\tau)}}\frac{1}{\left|\eta(\tau)\right|^2}.\tag{0.1}\label{1}$$ What is $$Z_{(\bullet,\circ)}[\tau]$$ with $$(\bullet,\circ)\in\{\text{(R,NS), (NS,R), (NS,NS)}\}?$$

There are two ways to answer, and both of them use the following fact of life:

The partition function of the non-compact scalar on a torus $$\mathbb{T}^2_\tau:=\mathbb{C}/\left(\mathbb{Z}\oplus\tau\mathbb{Z}\right)$$, with $$\tau\in\mathbb{H}$$ (the upper half plane), can be read from the partition function of the compact scalar by sending the compactification radius to infinity, $$R\to\infty$$.

### Way 1

The simplest way to obtain the answer is as follows.

Well, it suffices to go look at the compactified case and stare at the orbifold computation. For example stare at equation (8.24) in Ginsparg's notes. You will see that it is only the (R,R) sector that contributes to the $$R$$ dependence. Therefore, the (R,NS), (NS,R) and (NS,NS) are identical in the uncompactified case (when you sent $$R\to\infty$$). So we have \begin{align} Z_\text{(R,NS)}[\tau] &= \left|\frac{2\eta(\tau)}{\vartheta_2(\tau)}\right| \tag{1.1}\label{1.1} \\ Z_\text{(NS,R)}[\tau] &= \left|\frac{\eta(\tau)}{\vartheta_4(\tau)}\right| \tag{1.2}\label{1.2}\\ Z_\text{(NS,NS)}[\tau] &= \left|\frac{\eta(\tau)}{\vartheta_3(\tau)}\right|. \tag{1.3}\label{1.3} \end{align}

### Way 2

Another way, is to do the computation from scratch. Namely, go back to the path integral and compute, say in the (R,R) case $$Z_\text{(R,R)}[\tau] = \frac{\operatorname{vol}(\text{zero-modes})}{\sqrt{\operatorname{det}_\text{(R,R)}'\!\big(\partial\bar\partial\big)}}.\tag{2.1}\label{2.1}$$ Up to factors of $$2$$ and $$\pi$$, you can then see the following: $$\operatorname{vol}(\text{zero-modes}) = \sqrt{\Im(\tau)}\tag{2.2}\label{2.2}$$ and the non-zero eigenvalues of $$\partial\bar\partial$$ on a torus with (R,R) BCs are simply $$\lambda^\text{(R,R)}_{n,m} = \frac{1}{\Im(\tau)^2}\left|n+\tau m\right|^2, \qquad (n,m)\in\mathbb{Z}^2\setminus(0,0),$$ giving $$\operatorname{det}'_\text{(R,R)}(\partial\bar\partial) = \Im(\tau)^2\left|\eta(\tau)\right|^4.\tag{2.3}\label{2.3}$$ Altogether plugging \eqref{2.2} and \eqref{2.3} in \eqref{2.1}, gives \eqref{1}.

Now, for the other boundary conditions, all you have to do is observe that an NS BC along either cycle shifts either (or both) $$n$$ or $$m$$ by $$\frac{1}{2}$$, therefore, e.g. for (R,NS) BCs you have $$\lambda^\text{(R,NS)}_{n,m} = \frac{1}{\Im(\tau)}\left| n+\tau\left(m+\frac{1}{2}\right) \right|^2, \qquad n,m\in\mathbb{Z}^2.$$ Note that now you don't have a zero-mode anymore. So all you have to do now is compute the determinant of $$\partial\bar\partial$$ with these boundary conditions and find $$Z_{(\bullet,\circ)}[\tau] = \frac{1}{\sqrt{\det_{(\bullet,\circ)}(\partial\bar\partial)}}.$$ Doing this should land you on \eqref{1.1}-\eqref{1.3}.

• Thanks for your answer! I didn't realize the $R$ independence of $R$ in the other sectors previously. Nice observation. So from your answer, somehow the Hilbert space of the "twisted sector" in the $R \to +\infty$ limit is still like some direct sum of $|m, n\rangle$ subsectors? How to understand this at the level of allowed primaries? Jan 19 at 13:44
• As a comparison, for the uncompactified boson theory, the untwisted sectors would consider $|\alpha\rangle$ sectors with continuous $\alpha$. Jan 19 at 13:46
• Actually, the RR sector partition function of the $\mathbb{Z}_2$ orbifold theory is computed in Ginsparg, eq (8.6). It's not obvious to me that the $R \to +\infty$ limit of the double infinite sum reproduces the expected factor $1/\sqrt{\operatorname{Im\tau}}$. Could you clarify a bit? Jan 20 at 5:58
• In (8.6) you can't directly take a limit $R\to\infty$ because the exponent contains both a $\propto R$ piece and a $\propto 1/R$ piece. To take the limit you must first Poisson resum one of the two sums so that you end up with something only $\propto R$ and then only the $(0,0)$ term contributes. The Poisson resummation also spits out a factor of $1/\sqrt{\mathrm{Im}(\tau)}$ from the Fourier transform of the exponential. Jan 20 at 12:45
• I see, I guess you are referring to Ginsparg's eq (8.3), where the sum is the Poisson resummation of eq (8.6). Indeed, each term (except when $n, n' = 0$) goes to zero individually when $r \to \infty$. Thanks! Jan 21 at 3:23