Chern-Simons Energy-Momentum Tensor

Question

I'm assuming the following statement is true. I'm not finding any reference which shows that explicitly.

Statement: Chern-Simons term is a topological one and does not contribute to the Energy-Momentum tensor.

My problem is that I'm not finding this! I'm making some mistake and I don't know where!

I'm taking the Chern-Simons term

$${\cal L}_{CS} = \frac{\kappa}{2} A_{\sigma} \varepsilon^{\sigma \mu \nu} \partial_{\mu} A_{\nu}$$

Via Noether theorem this term contributes to the Stress-Energy-Momentum tensor through

$$T^{\mu \nu}_{CS} = \frac{\partial {\cal L}_{CS}}{\partial \partial_\mu A_\alpha} \partial^\nu A_\alpha - \eta^{\mu \nu} {\cal L}_{CS}$$

Clearly

$$\frac{\partial {\cal L}_{CS}}{\partial \partial_\mu A_\alpha} = \frac{1}{2} \kappa A_{\sigma} \varepsilon^{\sigma \mu \alpha}$$

So that,

$$T^{\mu \nu}_{CS} = \frac{\kappa}{2}A_{\sigma} \varepsilon^{\sigma \mu \alpha} \partial^\nu A_\alpha - \eta^{\mu \nu} \frac{\kappa}{2} A_{\sigma} \varepsilon^{\sigma \lambda \gamma} \partial_{\lambda} A_{\gamma}$$

And it's not zero.

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I tried other thing... may be what is zero is the contribution to the energy-momentum conservation... So I compute the following

$$\partial_\nu T^{0 \nu}_{CS} = \frac{\kappa}{2}\partial_\nu \left[A_\sigma \varepsilon^{\sigma 0 \alpha} \partial^\nu A_\alpha - \eta^{0 \nu} A_\sigma \varepsilon^{\sigma \beta \alpha}\partial_\beta A_\alpha\right]$$

opening it

$$\frac{2}{\kappa}\partial_\nu T^{0 \nu}_{CS} = \partial_0 \left[A_\sigma \varepsilon^{\sigma 0 \alpha} \partial^0 A_\alpha\right] + \partial_i \left[A_\sigma \varepsilon^{\sigma 0 \alpha} \partial^i A_\alpha\right] - \partial^0 \left[A_\sigma \varepsilon^{\sigma 0 \alpha}\partial_0 A_\alpha\right] - \partial^0 \left[ A_\sigma \varepsilon^{\sigma i \alpha}\partial_i A_\alpha\right] \\= + \partial_i \left[A_\sigma \varepsilon^{\sigma 0 \alpha} \partial^i A_\alpha\right] - \partial^0 \left[ A_\sigma \varepsilon^{\sigma i \alpha}\partial_i A_\alpha\right]$$ Closing it back

$$\frac{2}{\kappa}\partial_\nu T^{0 \nu}_{CS} = \left[\varepsilon^{\sigma 0 \alpha}\partial^i - \varepsilon^{\sigma i \alpha} \partial^0 \right] (A_\sigma \partial_i A_\alpha)\\ \frac{2}{\kappa}\partial_\nu T^{0 \nu}_{CS} = \left[\varepsilon^{i \sigma \alpha} \partial^0 -\varepsilon^{0 \sigma \alpha}\partial^i \right] (A_\sigma \partial_i A_\alpha)$$

which is not something trivially zero. Where is my mistake?

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Let me follow now (based on suggestions of Jamals in the answers) the idea that we have to fix the gauge to notice there is no contribution to the Stress-Energy Tensor.

We can compute that the form I've found to $T_{CS}^{\mu \nu}$ is not gauge invariant, in fact, doing $A^\mu \rightarrow A^\mu + \partial^\mu \chi$ we have

$$T_{CS}^{\mu \nu} \rightarrow T_{CS}^{\mu \nu} - \kappa (\varepsilon^{\mu \sigma \alpha} \partial_\sigma A_\alpha)\partial^{\nu}\chi $$

If we look specifically to $T^{00}$ we have

$$T_{CS}^{00} = - \kappa A_0 \varepsilon^{ij} \partial_i A_j$$

So that with a gauge transformation,

$$T_{CS}^{00} \rightarrow T_{CS}^{'00} = - \kappa A_0 \varepsilon^{ij} \partial_i A_j - \kappa (\varepsilon^{0 i j} \partial_i A_j)\partial^{0}\chi $$

We can choose a gauge where $\partial^{0}\chi = - A^0$ and then $T_{CS}^{'00} = 0$, as we where looking for.

Anyway, for $T_{CS}^{'\mu \nu}$

$$T^{'\mu \nu}_{CS} = \frac{\kappa}{2}A_{\sigma} \varepsilon^{\sigma \mu \alpha} \partial^\nu A_\alpha - \eta^{\mu \nu} \frac{\kappa}{2} A_{\sigma} \varepsilon^{\sigma \lambda \gamma} \partial_{\lambda} A_{\gamma} - \kappa (\varepsilon^{\mu \sigma \alpha} \partial_\sigma A_\alpha)\partial^{\nu}\chi $$

Things don't look fine.

Let's take a look on the momentum vector,

$$T^{'i 0}_{CS} = \frac{\kappa}{2}A_{\sigma} \varepsilon^{\sigma i\alpha} \partial^0 A_\alpha + \kappa (\varepsilon^{i\sigma \alpha} \partial_\sigma A_\alpha)A^{0} $$

It's pure gauge! Every term involve $A^0$ and we fixed that $A^0 = \partial^0 \chi$

Answer 1

Recall that the Chern-Simons action in terms of differential forms is given by,

$$S= \frac{k}{4\pi}\int_M \mathrm{Tr} \left[ A\wedge \mathrm{d}A + \frac{2}{3}A \wedge A \wedge A\right]$$

where $A$ is our gauge connection. We now employ a definition of the stress-energy tensor which we would normally also apply when varying the Einstein-Hilbert action back in general relativity, namely,

$$T^{\mu\nu}=\frac{2}{\sqrt{|g|}}\frac{\delta S}{\delta g^{\mu\nu}}$$

But the action is completely independent of the metric,$^{\dagger}$ hence the variation is zero and $T^{\mu\nu}=0$.

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Consider instead Maxwell-Chern-Simons theory which in tensor notation is given by,

$$\mathcal{L}=\frac{1}{2e^2}E^2_i - \frac{1}{2e^2}B^2_i + \frac{\kappa}{2}\epsilon^{ij} \dot{A}_i A_j + \kappa A_0 B$$

Remember $A_0$ is not dynamical, and imposes an analogous Gauss' law constraint. In the $A_0=0$ gauge, the conjugate momentum is given by,

$$\Pi^i=\frac{1}{e^2}\dot{A}_i + \frac{\kappa}{2}\epsilon^{ij}A_j$$

The Hamiltonian is given by the standard Legendre transform of the Lagrangian, i.e.

$$\mathcal{H}=\Pi^i \dot{A}_i - \mathcal{L}= \frac{e^2}{2} \left( \Pi^i-\frac{\kappa}{2}\epsilon^{ij}A_j\right)^2+\frac{1}{2e^2}B^2+ + A_0(\partial_i \Pi^i + \kappa B)$$

With some manipulation, the expression can be shown to equal,

$$\mathcal{H}=\frac{1}{2e^2}(E^2+B^2)$$

when expressed in terms of the electric and magnetic fields. If we consider pure Chern-Simons with the Maxwell term removed, $\mathcal{H}=0$. The aforementioned are suggesting to us $T^{\mu\nu}=0$.

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$\dagger$ If the Chern-Simons term depended on the metric tensor, it would imply it was also dependent on the geometry of the manifold, and hence not topological.