# Simplest way to see that the Euler density is topological

I’m looking for a simple way to see that the integrated Euler density $$\sqrt{g} \, E_{2n}$$ is topological (i.e. metric-independent in general even dimension $$2n$$). I can see it it must be true in two dimensions since $$E_2 = R$$ and because of the two-dimensional relation between the Ricci and scalar curvature $$R_{\mu\nu} = (R/2)g_{\mu\nu}$$ which implies that the variation vanishes $$\begin{equation} \delta \int \mathrm{d}^{2}x \sqrt{g} \, E_2 = \int \mathrm{d}^{2}x \sqrt{g} \left(R_{\mu\nu} - \frac{1}{2}R g_{\mu\nu}\right) \delta g^{\mu\nu} = 0 \end{equation}$$ I suspect that the key to general dimension lies in the formula $$\begin{equation} E_{2n} = \frac{1}{2^n} {R^{i_1j_i}}_{k_1l_1} \cdots {R^{i_nj_n}}_{k_nl_n} \epsilon_{i_1 j_1 \cdots i_n j_n} \epsilon^{k_1l_1\cdots k_n l_n} \end{equation}$$ but it’s unclear how to proceed other than brute force.

• Look at [Anderson: The variational bicomplex], page 129, Example 4.14. : ncatlab.org/nlab/files/AndersonVariationalBicomplex.pdf Feb 21 at 14:49
• Thanks for the comment @BenceRacskó. That’s interesting. I now think you can also see it by direct computation using dimensional-dependent identities from cambridge.org/core/journals/… If I have time I’ll try to write an answer based on this later Feb 26 at 7:00