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?


2 Answers 2


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 $\omega$.

Added note:

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 $

  • $\begingroup$ I would be very grateful if you can explain how you get the $-1/3$. $\endgroup$ Dec 16, 2023 at 4:41
  • 1
    $\begingroup$ $\int_{\mathcal{M}} \omega_{AB}\wedge\star d(\omega^{AC}\wedge\omega_C^{~~~~B}) = -\frac{2}{3}\int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})$ and $\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star d\omega^{AB}= \frac{1}{3}\int_{\mathcal{M}}d(\star\omega_{AB}\wedge\omega^A_{~~C}\wedge\omega^{CB})$ $\endgroup$
    – MadMax
    Dec 18, 2023 at 15:07
  • 1
    $\begingroup$ To help you understand the mathematical manipulations, I have added a note to the answer. $\endgroup$
    – MadMax
    Dec 18, 2023 at 18:53

I think I have have figured out how it works. It was just a simple matter of expanding the components

$$\int_{\mathcal{M}} \omega_A^{~~~C}\wedge\omega_{CB}\wedge\star(\omega^{AD}\wedge\omega_D^{~~~~B}) = \int_{\mathcal{M}} \epsilon^{ABCD}\omega_A^{~~~E}\wedge\omega_{EB}\wedge\omega_{CF}\wedge\omega_{~~~~D}^{F}$$ 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.


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