Euler density of two-dimensional manifolds

Question

I am asking this question after reading this post: What is Euler Density?.

For a two dimensional manifold, the Euler density is given by:

\begin{equation} E_2=2R_{1212} \end{equation}

(note that $R_{1212}$ is the only independent component of the Riemann tensor in 2d).

Now, integrating over the 2d manifold, we should get the Euler characteristic

\begin{equation} \chi=\int d^2x \sqrt {(\textrm{det }g)} E_2, \end{equation}

where $(\textrm{det }g)$ is determinant of the metric. But $E_2=2R_{1212}=R(g_{11}g_{22}-g_{12}g_{21})=R \textrm{ det }g$, where $R$ is the Ricci scalar of the 2d manifold. This gives

\begin{equation} \chi=\int d^2x (\textrm{det }g)^{\frac{3}{2}} R, \end{equation}

which contradicts the first term of equation 3.2.3b of Polchinski's 'String Theory', volume 1. What's the reason for this contradiction?

Answer 1

I) It seems the resolution to OP's question lies in the difference between

1. the Levi-Civita symbol, which is not a tensor and whose values are only $0$ and $\pm 1$; and

2. the Levi-Civita tensor, whose definition differs from the Levi-Civita symbol by a factor of $\sqrt{|\det(g_{\mu\nu})|}$.

II) The 2D Euler-density is

$$ E_2~=~ \frac{1}{8\pi} \epsilon_{\mu\nu}\epsilon_{\lambda\kappa} R^{\mu\nu\lambda\kappa}~=~ \frac{1}{8\pi} \epsilon^{\mu\nu}\epsilon^{\lambda\kappa} R_{\mu\nu\lambda\kappa}~=~\frac{S}{4\pi}~=~\frac{K}{2\pi},$$

where $S=2K$ is the scalar curvature and $K$ is the Gaussian curvature. The Euler characteristics is

$$\chi~=~\int d^2x \sqrt {| \det g_{\mu\nu} |} E_2.$$