I want to calculate the energy-momentum tensor from the Chern-Simons Lagrangian in index notation. I know the Chern-Simons Lagrangian does not depend on the metric because it can be written in the coordinate-free form $\mathcal{L} = A\wedge dA$ as is e.g. also explained in this post. However, I want to see the same result when I work explicitly in index notation. $$ S = \sqrt{-g}\int dx^3 \left( \frac{\epsilon^{\mu\nu\rho}}{\sqrt{-g}} A_\mu\partial_\nu A_\rho - J^\mu A_\mu \right) $$

Edit When I posted this question I made a stupid mistake, I forgot to include a factor of $\frac{1}{\sqrt{g}}$ in the definition of the Levi-Cevita tensor $\frac{\epsilon^{\mu\nu\rho}}{\sqrt{g}}$. This factor will cancel the $\sqrt{g}$ in the measure of the integral and will obviously make the Chern-Simons term independent of the metric. Written this way, the Lagrangian does obviously does not depend on the metric as the factors of $\sqrt{-g}$ will cancel everywhere.

Using that the energy-momentum tensor is given by the variation of the Lagrangian with respect to the metric, it follows that the energy-momentum tensor will vanish if the Lagrangian doesn't depend on the metric. $$ T^{\mu\nu} = \frac{2}{\sqrt{-g}} \frac{\delta \mathcal{L}}{\delta g^{\mu\nu}} $$


1 Answer 1


Lol, why is the determinant of the metric not under the integral?

The correct action is $S=\int \left( \mathrm{d}^3 x \sqrt{-g}\right) \left( \frac{1}{\sqrt{-g}} \epsilon^{ijk}\right) A_i \partial_j A_k$, where $\epsilon^{ijk}$ is a tensor density, not a tensor. So we see that the metric dependence cancels.

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    $\begingroup$ @JgL There are no object as $\partial^a \phi$, $D^a \phi$ or $A^a$. They are meaningful only if the indices are lowered. So one has to do it before taking derivatives with respect to the metric. $\endgroup$ Commented Mar 18, 2017 at 2:10
  • $\begingroup$ Yes, so doesn't that mean I have to do the same for $\epsilon^{ijk}$? This is the point that is confusing me. $\endgroup$
    – Jeroen
    Commented Mar 18, 2017 at 23:20
  • $\begingroup$ @JgL No. The easiest way to see it, is to rewrite all the stuff in the form notation, in which it is obvious. Here $\epsilon^{ijk}$ is not a tensor, but a tensor density with components $\pm 1$, it has the same value as $\epsilon_{ijk}$. $\endgroup$ Commented Mar 19, 2017 at 1:30

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