Some questions about the compact boson in David Tong's notes on Gauge Theory

Question

The notes can be found at http://www.damtp.cam.ac.uk/user/tong/gaugetheory.html.

In Sec. 7.5.1, T-Duality, around Eq. 7.51, it says that the Bianchi identity $\partial_\mu(\epsilon^{\mu\nu}\partial_\nu\phi)=0$ may not hold. Does this indicate $\partial_\mu\partial_\nu\phi\neq \partial_\nu\partial_\mu\phi$. If so, how can this happen?

In addition, since $\int d^2x$ is an integration over an infinite-long tube, the boundary integration $\oint d x^\mu$ should be over two circles at opposite ends of the tube. Each circle gives a winding number, so the final results will be the difference of the two winding numbers, which is indeed an integer. However, one can continuously move one circle along the tube to the other while the winding number cannot vary in a continual way, so there should be no difference between the two windings, and the integer should always be zero. Where is my argument wrong?

Another question is about how to integrate out the compact boson. I can only try to understand it with a naive analogy $\int_0^{2\pi} d\theta \, e^{i\theta a}=\frac{e^{i 2\pi a}\;\;-1}{ia}$, and I fail to make a further step.

Answer 1

In Sec. 7.5.1, T-Duality, around Eq. 7.51, it says that the Bianchi identity $\partial_\mu(\epsilon^{\mu\nu}\partial_\nu\phi)=0$ may not hold. Does this indicate $\partial_\mu\partial_\nu\phi\neq \partial_\nu\partial_\mu\phi$. If so, how can this happen?

Since $\phi$ always lies in the range $[0,2\pi)$ (see Eq 7.50), at places where $\phi$ goes from $2\pi-\epsilon$ to $\epsilon$, the derivative will be infinite. In other words, the non-trivial winding prevents the naive Bianchi identify from working.

However, one can continuously move one circle along the tude to the other while the winding number cannot vary in a continual way, so there should be no difference between the two windings, and the integer should always be zero. Where is my argument wrong?

Let's just take a finite cylinder of length $1$; let $z$ be the coordinate along the height of the cylinder ($0\leq z\leq 1$). At $z=0$ let's say $\phi=0$ (no winding), and at $z=1$ suppose $\phi=2 \theta \mod 2\pi$, where $0\leq \theta \leq 2\pi$ is the angular coordinate along the cylinder (this corresponds to a winding number of $2$). Here is a function on the cylinder that transitions from winding number $0$ at $z=0$ to winding number $2$ at $z=1$ \begin{equation} \phi(z,\theta) = 2 z \theta \mod 2\pi \end{equation} The winding number in fact does change discontinuously as you vary $z$; it changes from $0$ to $1$ at $z=1/2$ and from $1$ to $2$ at the boundary $z=1$.

Another question is about how to integrate out the compact boson

I don't understand what you are asking here, unfortunately. The propagator is given after Eq 7.58, so you can use that expression as a cross-check when doing a Gaussian path integral over $\phi$.

Answer 2

I had the exact same questions when reading this section of these notes. While it is always true that $\partial_\mu \partial_\nu \phi =0$ if there is a vortex then $\phi$ won't be well defined at that point and so this quantity won't necessarily vanish. Then in this case the loop integral $\oint dx_\mu \partial_\mu \phi$ will say by how much it doesn't vanish by Stokes theorem.

As for integrating out the compact boson, the answer is essentially what you would expect for if $\tilde \phi$ were non-compact: it just sets the coefficeint on $\tilde \phi$ to be zero. Here is a sketch of what I worked out modulo factors of $2\pi$. First we Fourier transform over spatial coordinates (it shouldn't matter here if space is compact or non-compact) \begin{align} Z[\phi] &= \int \mathcal D \tilde \phi \exp\left[{i \int d^2 x \ \tilde \phi \epsilon_{\mu \nu} \partial_\mu \partial_\nu \phi}\right] \\ &= \int \mathcal D \tilde \phi \exp \left[i \epsilon_{\mu \nu} \int _k k_\mu k_\nu \tilde \phi(k) \phi(k) \right] \end{align} Then consider the integral over just one spatial mode: \begin{align} Z[\phi(k_0)] = \int_0^{2\pi} d \tilde \phi(k_0) \exp \left[i \epsilon_{\mu \nu} k_0^\mu k_0^\nu \tilde \phi(k_0) \phi(k_0)\right] \end{align} Since $\tilde \phi$ is periodic, the integral from 0 to $2\pi$ will be the same as the integral from $2\pi$ to $4\pi$ etc., so taking the integral to be over all values of $\tilde \phi$ rather than just $0$ to $2\pi$ just multiplies the partition function by a constant. But the integral over a periodic funciton is the sum over discrete modes which allows us to apply the Poisson summation formula \begin{align} Z[\phi(k_0)] &= \sum_{\tilde \phi_n(k_0)}\exp \left[i \epsilon_{\mu \nu} k_0^\mu k_0^\nu \tilde \phi_n(k_0) \phi(k_0)\right] = \sum_n \delta (\epsilon_{\mu \nu} k_0^\mu k_0^\nu \phi(k_0)-2\pi n) \end{align} This is true for every mode $k_0$ so \begin{align} Z[\phi] = \sum_{n(x) } \delta(\epsilon_{\mu \nu} \partial_{\mu} \partial_\nu \phi- 2\pi n(x)) \end{align} In other words, the integral imposes the constraint that $\frac 1 {2\pi }\text d ^2 \phi \in \mathbb Z$ locally which means that its integral will also be in $ \mathbb Z$.

While I think this results is what Tong is pointing to in his notes, there are still a few points I'm unsure about. First, the above analysis only holds if $\phi$ is well defined but it is exactly at the vorticies where $n \ne 0$ that it isn't well defined. Also, if the theory is on a torus and $\phi$ winds nontrivially then $\partial_\mu \partial_\nu \phi = 0$ everywhere. And, since $\phi$ is well-defined and differentiable everywhere, by Stokes theorem the integral over the whole space also vanishes: $\int_T \partial_\mu \partial_\nu \phi = 0 $. However, if we choose a path along the torus, it is still the case that $\oint dx_\mu \partial_\mu \phi \in \mathbb Z $. So in this picture the fact that (1) $\text d^2 \phi$ vanishes everywhere and (2) the winding number is quanitzed are seperate statements.