This question is about obtaining the boundary action from Chern-Simons theory.

While reading David Tong's chapter 6 on quantum Hall effect, I cannot derive an equation between (6.9) and (6.10) of the note. The note goes as follows:

Consider the Chern-Simons theory on a spacetime $M = \Sigma \times \mathbb R$, where $\Sigma = \{y<0\}$ is a half-infinite plane. The action is $$S = \frac{m}{4\pi} \int_M a \wedge da.\tag{p.159}$$

Tong imposed the condition $a_t-va_x=0$ at the boundary $y=0$, extended this to hold at the bulk. Introducing a new coordinates $t'=t, x'=x+vt, y'=y$, we have $a'_{t'} = a_t-va_x$ so that the condition simplfies to $a'_{t'}=0$. This gives a constraint $f'_{x'y'}=0$, so that the (classical) solution can be written as $a_i' = \partial_i \phi$ for $i=x',y'$, where $\phi = \phi(x,y,t)$ is a scalar function.

Inserting $a_{t'}'=0$ and $a_i' = \partial_i \phi$, the Chern-Simons action becomes $$S = \frac{m}{4\pi} \int d^3x' (\partial_{x'} \phi \partial_{t'} \partial_{y'}\phi - \partial_{y'}\phi \partial_{t'}\partial_{x'} \phi).\tag{1}$$ Up to this point, I am fine. However, Tong claims that the above simplifies to $$ S= \frac{m}{4\pi} \int_{y=0} d^2x' \: \partial_{t'} \phi \partial_{x'} \phi.\tag{2}$$ Considering the integration domain, I think integration by parts is performed, but I cannot derive this from the previous equation.


1 Answer 1


There are various integrations by parts taking place to get this correct. $\newcommand{\d}{\mathrm{d}}\newcommand{\R}{\mathbb{R}}\newcommand{\pd}{\partial}\newcommand{\ieq}{\overset{\scriptsize \int}{=}}$Note that since along the $t'$- and $x'$-directions there is no boundary, for any quantities $\bullet,\circ$ you have that $$\bullet\ (\pd_{t'}\circ) \ieq -(\pd_{t'}\bullet)\ \circ \qquad\text{and}\qquad \bullet\ (\pd_{x'}\circ) \ieq -(\pd_{x'}\bullet)\ \circ,\tag{$*$}$$ where I denote by $\ieq$ equality when integrated over $\R\times\Sigma$. Therefore your equation (1) can be rewritten as $$S_\mathrm{CS} = \frac{m}{2\pi} \int_\R \d t' \int_\Sigma \d x'\;\d y' \ \ \pd_{y'}\!\left( \pd_{t'}\phi\right)\ \pd_{x'}\phi, \tag{$1^\prime$}$$ but now, observe that $$ \pd_{y'}\!\left( \pd_{t'}\phi\right)\ \pd_{x'}\phi = \pd_{y'}\Big[ \pd_{t'}\phi\ \pd_{x'}\phi\Big] - \pd_{t'}\phi\ \pd_{y'}\pd_{x'}\phi \ieq \pd_{y'}\Big[ \pd_{t'}\phi\ \pd_{x'}\phi\Big] - \pd_{x'}\phi\ \pd_{t'}\left(\pd_{y'}\phi\right), $$ by using $(*)$ twice on the second term. So you see that $$ \pd_{y'}\!\left( \pd_{t'}\phi\right)\ \pd_{x'}\phi \ieq \frac12 \pd_{y'}\Big[ \pd_{t'}\phi\ \pd_{x'}\phi\Big].$$ And you're done, just plug this back into $(1')$, use Stokes' theorem to pull the total derivative to the boundary and you get \begin{align}S_\mathrm{CS} &= \frac{m}{4\pi} \int_\R \d t' \int_\Sigma \d x'\;\d y'\ \ \pd_{y'}\Big[ \pd_{t'}\phi\ \pd_{x'}\phi\Big] \\ &=\frac{m}{4\pi} \int_\R \d t' \int_{\pd\Sigma=\left\{y=0\right\}} \!\!\!\!\!\!\!\d x'\ \ \pd_{t'}\phi\ \pd_{x'}\phi, \end{align} which is precisely your equation (2).

  • $\begingroup$ Wow! Thanks for the trick involving integration by parts three times, with respect to each of $x',y',t'$, to get back to the original expression! $\endgroup$
    – Laplacian
    Commented Nov 14, 2022 at 11:29

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