Gauge invariance of $\theta$-term in QCD

I have a problem with the $\theta$-term in the QCD-Lagrangian

$$\mathcal{L} = \overline{q}(x)i\gamma_\mu{D^\mu} q(x)-\overline{q}(x)\mathcal{M}q-\frac{1}{4}F^a_{\mu\nu}F_a^{\mu\nu}+ \theta \frac{g^2}{32 \pi^2} F_{\mu \nu}^a \tilde{F}^{\mu \nu}_a$$

I guess the last term is gauge invariante by itselfe since it is proportional to the trace:

$$tr(F^{\mu\nu}\tilde{F}_{\mu\nu})=2\epsilon_{\mu\nu\rho\sigma}tr[(\partial^\mu A^\nu+igA^\mu A^\nu)(\partial^\rho A^\sigma +igA^\rho A^\sigma)]$$

But on the other hand the $\theta$-term is topological non-trivial and the integral

$$\int d^4x \ \ tr[F^{\mu\nu}\tilde{F}_{\mu\nu}] =: \int d^4x \ \ \partial_\mu K^\mu$$ is proportional to $n \in Z$ (the winding number) and not invariant under large gauge transformations.

Which step is wrong? And if the $\theta$-term is not gauge invariant how can we fix this problem in the Lagranian?

The problem is that $K^{\mu}$ is not a true vector on the manifold. It depends on the gauge potential $A$ as well as the gauge field $F$. It is alright to use the Stokes theorem for these types of objects locally on coordinate pathches but not globally on the whole manifold.
Suppose that the integration is over $S^4$. We divide the integration into two parts, on the upper hemisphere and on the lower hemisphere, touching at the equator $S^3$. We know that on each hemisphere (I am using a coordinate free notation) $$\mathrm{tr}(F\wedge F) = d\mathrm{CS}(A)$$ Where $CS(A)$ is the Chern Simons class. Using the Stokes theorem $$\int_{S^4} \mathrm{tr}(F\wedge F) = \int_{D_{up}^4} d\mathrm{CS}(A_{up}) + \int_{D_{down}^4} d\mathrm{CS}( A_{down}) = \int_{S^3} \mathrm{CS}( A_{up}) - \mathrm{CS}( A_{ down })$$ The local gauge potentials $A_{up}$ and $A_{down}$ are connected by a transitigauge transformation $$A_{up} = gA_{down}g^{-1}+ddg^{-1}$$ It is known that the Chern-Simons classes are not gauge invariant, they change by a Wess-Zumino-Witten form under gauge transformation:
$$\mathrm{CS}( gA_{down}g^{-1}+ddg^{-1})- \mathrm{CS}( A_{ down }) = \mathrm{WZW}(g)$$ Thus we obtain: $$\int_{S^4} \mathrm{tr}(F\wedge F) = \int_{S^3}\mathrm{WZW}(g)$$ For nontrivial configurations like instantons, the transition gauge transformations are large gauge transformations, for which the Wess-Zumino-Witten is nonvanishing (It is the winding number of the transition group configurations).
• Thanks for your answer. For $tr(F\wedge F)=dCS(A)\neq 0$ we obtain a non-invariant Lagrangian under $SU(3)_c$ but how can one fix this issue? – Alpha001 Nov 8 '17 at 13:47
• cont. Moreover, by adding the $\theta$ term we superficially break the large gauge symmetry because the Lagrangian is not invariant. However, since the equations of motion are invariant, this is a quasi-symmetry and after quantization, we can find a Hilbert space with an invariant action of the large gauge transformation. This is the Hilbert space built over the theta vacuum. – David Bar Moshe Nov 8 '17 at 14:20