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I'm currently studying the Chern Simons theory from the Hall effect lecture notes of David Tong. [1]

In page 5 the current is calculated where it is asserted that

$$J^{i} = \frac{\delta S_{CS}}{\delta A_i} = -\frac{k}{2 \pi}\epsilon_{ij}E_j $$

I'm confused with how is the computation is carred out. For example I tried

$$J^{1} = \frac{\delta}{\delta A_1}(\frac{k}{4 \pi} \int d^3x \epsilon^{\mu \nu \rho}A_{\mu} \partial_{\nu} A_{\rho} ) = \frac{k}{4 \pi}(\epsilon^{1\nu \rho}\partial_\nu A_\rho - \epsilon^{\mu \nu 1} \partial_\nu A_\mu) $$

$$= \frac{k}{4 \pi} (\epsilon^{1\nu \rho}\partial_\nu A_\rho + \epsilon^{1\nu \rho}\partial_\nu A_\rho )= \frac{k}{2 \pi} (\partial_2 A_0 - \partial_0 A_2) $$

Somehow the electric field appears from the last equality but I'm having a hard time making it appear. How does it appear?

Reference:

[1] http://www.damtp.cam.ac.uk/user/tong/qhe/five.pdf

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Recall that $F_{i0} = \partial_i A_0 - \partial_0 A_i = - E_i$. Hence, $\partial_2 A_0 - \partial_0 A_2 = -E_2$, as desired.

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