Chern-Simons Energy-Momentum Tensor 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. 

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?

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 $A^0 = \partial^0 \chi$
 A: 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 $T^{\mu\nu}=0$.

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 $T^{\mu\nu}=0$.

$\dagger$ 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.
