Chern-Simons form of Euler class

Consider the Euler class for curvature $$F_{AB} = d\omega_{AB}+\omega_A^{~~~C}\wedge\omega_{CB}$$ where $$\omega$$ is the spin-connection given by $$\int_{\mathcal{M}} \epsilon^{ABCD}F_{AB} \wedge F_{CD} = \int_{\mathcal{M}}\star F^{CD}\wedge F_{CD}$$ where $$\epsilon^{ABCD}F_{AB} = \star F^{CD}$$. Now, I wish to find its Chern-Simon form. For that I expand one of the $$F$$'s in the above. $$\int_{\mathcal{M}}\star F^{CD}\wedge F_{CD} = \int_{\mathcal{M}}(d\omega_{AB}+\omega_A^{~~~C}\wedge\omega_{CB})\wedge \star F^{AB} \\= \int_{\partial\mathcal{M}}\omega_{AB}\wedge\star F^{AB}+ \int_{\mathcal{M}}\omega_{AB}\wedge d\star F^{AB}+ \int_{\mathcal{M}}\omega_A^{~~~C}\wedge\omega_{CB}\wedge \star F^{AB}$$ The first term in the last equality is the Chern-Simons form I am looking for but somehow the last two terms I do not seem to be able to find a way to cancel. Can someone provide any suggestions as to how to go about this? My expectation was that somehow the last two terms would combine to become $$\int_{\mathcal{M}}\omega_{AB}\wedge D\star F^{AB}$$ which vanishes by Bianchi identity. However, $$\int_{\mathcal{M}}\omega_{AB}\wedge D\star F^{AB} \neq \int_{\mathcal{M}}\omega_{AB}\wedge d\star F^{AB} + \int_{\mathcal{M}}\omega_A^{~~~C}\wedge\omega_{CB}\wedge \star F^{AB}$$ Can someone help with this?

Edit: I have noticed that

$$\int_{\mathcal{M}}\omega_{AB}\wedge d\star F^{AB} + \int_{\mathcal{M}}\omega_A^{~~~C}\wedge\omega_{CB}\wedge \star F^{AB} \\= \int_{\mathcal{M}} \omega_{AB}\wedge\star d(\omega^{AC}\wedge\omega_C^{~~~~B}) + \int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star d\omega^{AB} +\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star(\omega^{AD}\wedge\omega_D^{~~~~B})\\= \int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})+\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star(\omega^{AD}\wedge\omega_D^{~~~~B})\\ = \int_{\partial\mathcal{M}}\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB}+\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star(\omega^{AD}\wedge\omega_D^{~~~~B})$$ where I have simply substituted for $$F$$ in the above. It seems that the last term will not vanish by any means. Does that mean the Euler-Class above doesn't have a corresponding Chern-Simons form? Can anyone comment on this?

The Euler class in this Lorentz gauge gravity context is called the topological Gauss-Bonnet form.

However, rather than $$\int_{\mathcal{M}} \epsilon^{ABCD}F_{AB} \wedge F_{CD} = \int_{\partial\mathcal{M}}\omega_{AB}\wedge\star F^{AB} + \int_{\partial\mathcal{M}}\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB}$$ one is supposed to look for: $$\int_{\mathcal{M}} \epsilon^{ABCD}F_{AB} \wedge F_{CD} = \int_{\partial\mathcal{M}}\omega_{AB}\wedge\star F^{AB} -\frac{1}{3} \int_{\partial\mathcal{M}}\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB} \\=\int_{\partial\mathcal{M}}\omega_{AB}\wedge\star (d\omega^{AB}+\frac{2}{3}\omega^A_{~~~C}\wedge\omega^{CB})$$ The derivation in OP of $$\int_{\mathcal{M}} \omega_{AB}\wedge\star d(\omega^{AC}\wedge\omega_C^{~~~~B}) + \int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star d\omega^{AB}= \int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})$$ seems to be wrong. It should instead be $$\int_{\mathcal{M}} \omega_{AB}\wedge\star d(\omega^{AC}\wedge\omega_C^{~~~~B}) + \int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star d\omega^{AB}= -\frac{1}{3}\int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})$$ since you have to be mindful of the extra minus sign when you switch the order of Exterior product between one-forms such as $$d$$ and $$\omega$$.

It seems that the manipulations of indices such $$A/B/C/D$$ and $$\epsilon^{ABCD}$$ ($$\star$$) operation could be at times confusing/daunting. There is actually a way to get rid of (or hide) these nuisances and do all the calculations in a much simplified/elegant way: all you have to do is write the spin connection one-form in terms of gamma operators (see reference here) $$\omega = \frac{1}{4}\omega^{AB}\gamma_A\gamma_B$$ Then a lot of formulations can be simplified. For instance, the 4-$$\omega$$ term can be rewritten as $$\omega_A^{~~~C}\wedge\omega_{CB}\wedge\star(\omega^{AD}\wedge\omega_D^{~~~~B})\sim Tr[\gamma_5 \omega \wedge\omega \wedge\omega \wedge\omega ]$$ where $$Tr[...]$$ is the trace of gamma matrices, and $$\gamma_5 = i\gamma_0\gamma_1\gamma_2\gamma_3$$. And the proof of it being identical to zero is straight forward, since: $$Tr[\gamma_5 \omega \wedge\omega \wedge\omega \wedge\omega ] \\= Tr[\gamma_5 \omega \wedge (\omega \wedge\omega \wedge\omega) ] \\= Tr[\omega \wedge \gamma_5(\omega \wedge\omega \wedge\omega) ] \\= -Tr[\gamma_5 (\omega \wedge\omega \wedge\omega)\wedge \omega ] \\= -Tr[\gamma_5 \omega \wedge\omega \wedge\omega \wedge\omega ]$$ where we have used the fact that $$\gamma_5 \omega = \omega \gamma_5$$ and $$Tr[F\wedge G] = -Tr[G\wedge F]$$ if both $$F$$ and $$G$$ are odd-forms.

Other identities could also be easily proved. For example: $$d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB}) \\ \sim d(Tr[\gamma_5\omega\wedge\omega\wedge\omega])$$ and $$d(Tr[\gamma_5\omega\wedge\omega\wedge\omega]) \\=Tr[d(\gamma_5\omega\wedge\omega\wedge\omega])] \\=Tr[\gamma_5d\omega\wedge\omega\wedge\omega - \gamma_5\omega\wedge d(\omega\wedge\omega)] \\=Tr[\gamma_5d\omega\wedge\omega\wedge\omega - \gamma_5\omega\wedge d\omega\wedge\omega + \gamma_5\omega\wedge\omega\wedge d\omega] \\=3Tr[\gamma_5d\omega\wedge\omega\wedge\omega]$$

I will leave you an exercise to prove that the cosmological constant term as shown below is NOT identical to zero (contrary to the 4-$$\omega$$ term being identically zero as proved above)

$$CC \sim Tr[\gamma_5 e \wedge e \wedge e \wedge e ]$$ where $$e= e^A \gamma_A$$ is the veirbein/tetrad. Hint: $$\gamma_5 e = -e \gamma_5$$

• I would be very grateful if you can explain how you get the $-1/3$. Dec 16, 2023 at 4:41
• $\int_{\mathcal{M}} \omega_{AB}\wedge\star d(\omega^{AC}\wedge\omega_C^{~~~~B}) = -\frac{2}{3}\int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})$ and $\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star d\omega^{AB}= \frac{1}{3}\int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})$ Dec 18, 2023 at 15:07
$$\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star(\omega^{AD}\wedge\omega_D^{~~~~B}) = \int_{\mathcal{M}} \epsilon^{ABCD}\omega_A^{~~~E}\wedge\omega_{EB}\wedge\omega_{CF}\wedge\omega_{~~~~D}^{F}$$ If we look at $$\omega_1^{~~~E}\wedge\omega_{E2}\wedge\omega_{3F}\wedge\omega_{~~~~4}^{F} =(\omega_1^{~~~3}\wedge\omega_{32}+\omega_1^{~~~4}\wedge\omega_{42})\wedge(\omega_{31}\wedge\omega_{~~~~4}^{1}+\omega_{32}\wedge\omega_{~~~~4}^{2}) = 0$$ Therefore, the last term is zero by default. Hence, we then finally have
$$\int_{\mathcal{M}} \epsilon^{ABCD}F_{AB} \wedge F_{CD} = \int_{\partial\mathcal{M}}\omega_{AB}\wedge\star F^{AB} + \int_{\partial\mathcal{M}}\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB}$$ which is what I was looking for.