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Given a 4-dimensional compact manifold (torsion free), the Euler characteristic is defined as:

$$E_4 ~=~ \int \epsilon_{abcd}R^{ab} \wedge R^{cd}$$

with $R^{ab}$ is the curvature 2-form. Perturb the connection 1-form (represent by $\delta \omega^{ab}$), $E_4$ should be unchanged. How can I proof that, and what will be change if the manifold is not torsion free anymore?

Given that:

Connection 1-form from torsion-free condition:

$$T^a ~=~ De^a ~=~ de^a + \omega^a_b \wedge e^b ~=~ 0$$

and

$$\omega^{ab} ~=~ \delta^{ac} \omega^b_c.$$

Curvature 2-form:

$$R^{ab} ~=~ D\omega^{ab} ~=~ d\omega^{ab} + \omega^a_c \wedge \omega^{cb}.$$

Gauss-Bonnet term appears as a topological term in some theoretical gravitational actions. I don't see why it's unchanged, so I post this question of mine here.

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    $\begingroup$ Well the point of a topological invariant is that it depends only on the topology of the space, and not the geometry (metric, connection, etc.). $\endgroup$ Nov 18 '13 at 21:51
  • $\begingroup$ Also, $E_4 ~=~ \int \epsilon_{abcd}R^{ab} \wedge R^{cd}$, is the $\wedge$ not redundant here? I would suggest you start from $R = D \omega$ $\endgroup$ Nov 18 '13 at 21:57
  • $\begingroup$ Nope, the wedge should be there since I'm using Cartan's notation. $\endgroup$
    – user109798
    Nov 19 '13 at 1:41
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    $\begingroup$ @lionelbrits To elaborate on user109798's last comment, the symbol $R^{ab}$ is a two-form for each pair of its indices. Having said that, user109798, would you mind giving a reference for the expression you have written for $E_4$? I tried finding one and failed. $\endgroup$ Nov 19 '13 at 2:38
  • $\begingroup$ You can find it in some papers about Lovelock's gravity. For example, this paper: iopscience.iop.org/0264-9381/8/8/018/pdf/0264-9381_8_8_018.pdf You might be more familiar with the following form $S \sim \int d^4x \sqrt{-g} (R^{\mu \nu \rho \sigma}R_{\mu \nu \rho \sigma} + 4 R^{\mu \nu} R_{\mu \nu} + R^2)$. You can derive it from the given Cartan form of $E_4$ by using $R^{ab} = \frac12 R^{ab}_{cd} \wedge e^c \wedge e^d$ and then $\int e^a \wedge e^b \wedge e^c \wedge e^d = \epsilon^{abcd} dV$ ($dV = \sqrt{-g} d^4 x$). $\endgroup$
    – user109798
    Nov 19 '13 at 3:53
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I think you can try a variation with respect to $\omega_{ab}$. First of all $R^{ab} \neq D\omega^{ab}$ because $\omega^{ab}$ is not a tensor, the correct statement actually is $$\delta R^{ab} = D\delta\omega^{ab}$$ Using this, we may do a variation of $E_4$ with respect to $\omega^{ab}$, to obtain

$$\delta E_4 = -2\int \epsilon_{abcd} D R^{ab}\wedge \delta\omega^{cd}$$ I have used integration by parts to shift the covariant exterior derivative on $R^{ab}$. However, already $DR^{ab} = 0$ by Bianchi identity (https://en.wikipedia.org/wiki/Curvature_form). So, this shows that of $E_4$ is independent of any variation in $\omega^{ab}$. Notice I have not used any assumption about torsion here. My proof is independent of that.

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