Periodicity of the dual scalar in $T$-duality

Question

This question concerns the dual scalar (Lagrange multiplier) field $\lambda$, which appears in discussions of $T$-duality (e.g. in section 7.5.1 of David Tong's notes where it is called $\tilde\phi$).

In particular, while it is clear that $\lambda$ cannot be non-compact, I am unable to understand why it must be $2\pi$-periodic.

Consider Lorentzian signature, with convention $\eta_{00}=1=-\eta_{11}$ and $\epsilon^{01}=1=-\epsilon^{10}$.

Consider a theory of maps $\phi : \mathbb{R} \times S^1 \to S^1$ with action functional \begin{equation} S[\phi] = \int d^2x \frac{R^2}{2}(\partial\phi)^2.\tag{7.49} \end{equation}

The aim is to identify conditions on the field $\lambda$ which yield equivalence of the functional integrals, $$ \int[d\phi]e^{iS[\phi]}=\int[db][d\lambda]e^{iS[b,\lambda]} $$ where \begin{equation} S[b,\lambda] = \int d^2x \left[ \frac{R^2}{2}b^2 + \frac{1}{2\pi} \epsilon^{\mu\nu}b_\mu\partial_\nu\lambda \right]. \end{equation} Contradiction argument: If $\lambda$ was a non-compact scalar, then functionally integrating over $\lambda$ produces a delta functional constraint $\delta[\epsilon^{\mu\nu} \partial_\mu b_\nu]$. This means field configurations can be restricted to the form $b_\mu = \partial_\mu \phi$ for a non-compact scalar $\phi$. Thus, $\lambda$ cannot be non-compact.

I'll now list two unsuccessful approaches I've explored, which the assume that $\lambda$ is $2\pi$-periodic.

Approach 1: Integrating by parts and using a lattice discretization on the relevant part of functional integral gives \begin{align} \int [d\lambda]\exp\left[i\int d^2 x \frac{1}{2\pi} \lambda(\epsilon^{\mu\nu}\partial_\mu b_\nu)\right] & \sim \prod_x \int_{0}^{2\pi} d\lambda(x) \exp\left[i\frac{1}{2\pi} \lambda(x) \epsilon^{\mu\nu}\partial_\mu b_\nu(x)\right] \\ & = \prod_x \frac{2\pi i}{\lambda(x) \epsilon^{\mu\nu}\partial_\mu b_\nu(x)}\left[1-e^{i\lambda(x) \epsilon^{\mu\nu}\partial_\mu b_\nu(x)}\right] \end{align} This approach seems to be on the wrong track!

Approach 2: Again working on the lattice, attempt to incorporate periodicity of $\lambda$ via the derivative operator: \begin{align} \int [d\lambda]\exp\left[i\int d^2 x \frac{1}{2\pi} \epsilon^{\mu\nu}b_\mu \partial_\nu \lambda\right] = \sum_{n_{x,\nu}}\prod_x \int_{-\pi}^\pi d \lambda_x \exp\left[i \sum_{x,\nu} \frac{1}{2\pi} \epsilon^{\mu\nu}b_\mu(\lambda_x - \lambda_{x+e_\nu}+2\pi n_{x,\nu})\right] \end{align} This appears more promising, although it is unclear how to proceed.

Approach 3 (following @ɪdɪətstrəʊlə):

Expand \begin{align} \lambda & = \sum_i \lambda_i \psi_i \\ b_\mu & = \sum_j \alpha_j Y_{\mu,j} + \sum_i \beta_i (\partial_\mu \psi_i) \end{align} where $\psi_i$ and $Y_{\mu,j}$ are orthonormalised (transversal) eigenfunctions.

Plugging into the action, integrating by parts and using antisymmetry of the permutation symbol and using the fact that $\partial_\nu$ kills the zero mode of $\lambda$, I get

\begin{align} \int d^2 x \frac{1}{2\pi} \epsilon^{\mu\nu}b_\mu \partial_\nu \lambda & = \frac{1}{2\pi}\sum_{i\neq 0} \sum_{j} \alpha_j \lambda_i \int d^2 x \, \epsilon^{\mu\nu} Y_{\mu,j}\partial_\nu \psi_i \\ & = \frac{1}{2\pi}\sum_{i\neq 0} \lambda_i \int d^2 x \, \epsilon^{\mu\nu} b^\perp_\mu(x)\partial_\nu \psi_i \end{align}

Then

\begin{align} &\int[d\lambda]\exp\left[i\int d^2 x \frac{1}{2\pi} \epsilon^{\mu\nu}b_\mu \partial_\nu \lambda\right] = \\ & \int d\lambda_0 \prod_{i\neq 0}\left[\int d\lambda_i \exp\left(i\frac{1}{2\pi}\lambda_i \int d^2 x \, \epsilon^{\mu\nu} b^\perp_\mu(x)\partial_\nu \psi_i \right)\right] \end{align}

Finiteness of the functional integral seems to reaffirm the fact that $\lambda_0$ is compact. It's not clear to me how to proceed to show $2\pi$-periodicity, however.

Approach 4: An interesting observation is that if we integrate by parts on $\lambda$ and then demand that the functional $e^{i S[b,\lambda]}$ is invariant under the transformation $\lambda(x) \mapsto \lambda(x) + 2\pi$ then we obtain $$\exp\left[-\frac{i}{2\pi}\int d^2 x \partial_\mu (\epsilon^{\mu\nu}b_\nu)\right]=1$$ which is indeed equivalent to Tong's (7.51). I'm not sure what to make of this.

Answer 1

If λ was a non-compact scalar, then functionally integrating over λ produces a delta functional constraint $\delta[\epsilon_{\mu\nu}\partial_{\mu}b_\nu]$ . This means field configurations can be restricted to the form $b_\mu=\partial_\mu\phi$ for a non-compact scalar $\phi$. Thus, $\lambda$ cannot be non-compact.

First of all, the last sentence, the conclusion, is literally wrong. Though depending on what OP really means, it could be a careless slip of tongue or genuine misunderstanding.

As long as $\phi$ is twice differentiable, Bianchi identity always holds and $\epsilon_{\mu\nu}\partial^{\mu}\partial^{\nu} \phi = 0$ identically. Compactification or not has no consequence on derivatives, where only an infinitesimal neighborhood of $\phi$ is concerned.

Now, I know that Tong specifically says he wants to relax Bianchi identity in order to allow for winding. That is his slip of tongue. What he really means is not a compact $\phi$ winding around the periodic spatial direction. Rather, he wants to introduce instantons, which are singular configurations where the winding number around the spatial direction changes along the time axis. An instanton in (1+1)-D Minkowski spacetime is topologically the same as a vortex in 2D Euclidean space, and that is where the meaning of "winding" gets mixed up.

The supposed modification to Bianchi identity reads $$ \epsilon_{\mu\nu}\partial^{\mu}\partial^{\nu} \phi = 2\pi \sum_{i} n_i \delta(r - r_i), $$ where $i$ labels the individual instanton, $n_i$ is its strength and $r_i$ its location in spacetime.

Go to imaginary time, and discretize the 2D Euclidean space on a square lattice. The LHS $\epsilon_{ij}\partial_i\partial_j \phi$ is now the circulation of $\partial\phi$ around a plaquette, and should be an integer multiple of $2\pi$. This fixes the periodicity of the corresponding Lagrange multiplier.

Edit: just want to mention that $\int_{-\pi}^{\pi} d\lambda \,e^{i2\pi\lambda X} \sim \sum_{n=-\infty}^{\infty} \delta(X - n)$, applied to every plaquette.

If anyone knows of a way to argue for this result without going through the lattice, I'd really love to know.

Answer 2

The dual of compact boson is a "discrete" boson. For a full lattice treatment, I will Wick rotate to Euclidean space. The compact boson becomes the familiar XY model. On the lattice, the action is typically discretised with the corresponding measure (up to a multiplicative constant): $$ S[\phi] = -J\sum_{\langle x,y\rangle}\cos(\phi_x-\phi_y) \\ d\mu \propto e^{-S[\phi]} \prod d\phi_x $$ with $\phi_x\in(-\pi,\pi)$. However, the mathematical analogy is clearer with the Villain model: $$ S[\phi] = -\frac J2\sum_{\langle x,y\rangle}(\phi_x-\phi_y+2\pi n_{xy})^2 \\ d\mu \propto e^{-S[\phi]} \prod d\phi_x $$ where there is an additional field on the edges $n_{xy}\in\mathbb Z$. You can now set the gradient: $$ b_{xy} = \phi_x-\phi_y+2\pi n_{xy} $$ Essentially, $n$ is there to allow integer winding numbers while $\phi$ encodes the curl free part. To switch entirely to $b$, the periodic Bianchi identities need to be imposed on all loops: $$ \sum_{(x,y)\in\square}b_{xy} \in 2\pi\mathbb Z $$ This is done by inserting a Dirac comb using the Poisson summation formula: $$ \sum_{n\in\mathbb Z} \delta(t-2\pi n) = \frac1{2\pi}\sum_{n\in\mathbb Z} e^{in t} $$ so the measure becomes: $$ d\mu \propto e^{-\hat S[b,\lambda]}\prod_{\langle x,y\rangle} db_{xy} $$ with: $$ \begin{align} \hat S[b,\lambda] &= \frac J2\sum_{\langle i,j\rangle}b_{xy}^2+i \sum_\square \tilde\phi_\square \sum_{(x,y)\in\square}b_{xy} \\ &= \frac J2\sum_{\langle i,j\rangle}b_{xy}^2+i\sum_{\langle X,Y\rangle}\epsilon_{xy,XY}(\tilde\phi_X-\tilde\phi_Y)b_{xy} \end{align} $$ with $X,Y$ represent faces (i.e. dual vertices), $\tilde\phi_X\in\mathbb Z$ and $\epsilon$ is there to impose that $(X,Y)$ and $(x,y)$ are dual edges (orthogonal). I integrated by parts to get the second line.

To finish the analogy, you can even integrate out the $b$ field. You therefore get: $$ \mu \propto e^{-\tilde S[\tilde\phi]} \\ \tilde S[\tilde\phi] = \frac1{2J}\sum_{\langle X,Y\rangle}(\tilde\phi_X-\tilde\phi_Y)^2 $$ Had you continued with the usual cosine XY model, you would have ended up with the modified dual action: $$ \tilde S[\tilde\phi] = \sum_{\langle X,Y\rangle}\ln\left(I_{\tilde\phi_X-\tilde\phi_Y}(J)\right) $$ and $I$ the modified Bessel function of the first kind. Had you kept the Minkowski signature, the result would have been the same, you will just need to keep track of the signs.

Thus, viewing $\tilde\phi$ as the dual boson, $\tilde\phi$ is not $2\pi$-periodic but rather discrete with step $1$. In statistical physics, this is a well known fact. Integral loop curves of $\partial\phi$ are level lines of $\tilde\phi$. The discretisation of the circulation translates into the discrete steps $\tilde\phi$ can take. This is an example of a solid-on-solid model typically used for the surface roughening of crystals. More generally, a $T$ periodic boson is dual to a $2\pi/T$ discrete boson. This is consistent with the limit $T\to\infty$: a boson is dual to itself.

You can convert this to usual quantum mechanics by using continuous time but discrete space. If you start with the Hamiltonian ($V$ is a periodic potential): $$ H = \sum\frac12p_k^2+V(\phi_k-\phi_{k-1}) $$ with $\phi$ $2\pi$-periodic and $p$ the conjugate discrete canonical momentum satisfying the Weyl uncertainty: $$ e^{-i\phi_k}p_ke^{i\phi_k} = p_k+1 $$ Duality amounts to the change: $$ \tilde\phi_k = p_{k+1}-p_k \quad p_k = \tilde\phi_k-\tilde\phi_{k-1} $$ Now, it is $\tilde\phi$ which is discrete and $\tilde p_k$ which is periodic with the same uncertainty principle: $$ e^{i\tilde p_k}\tilde \phi_ke^{-i\tilde p_k} = \tilde \phi_k+1 $$ And the Hamiltonian is now: $$ H = \sum V(\tilde p_k)+\frac12(\tilde\phi_{k+1}-\tilde\phi_k)^2 $$ For example, to recover the cosine XY model, you just need to take $V=-J\cos$.