Area of a surface in terms of the densitized triad

Question

Hi, I need to know if this relation is correct for a metric:

$g_{a[b}g_{c]d}=\frac{1}{2}\epsilon_{ace}\epsilon_{bdf}gg^{ef}$

I know that :

$\frac{1}{2}\epsilon_{ace}\epsilon_{bdf}g^{ef}=g_{b[a}g_{c]d}$

but I don't see how the determinant $g$ of the metric could appear.

Edit:

Ok so the previous relation emerged when computing the area of a surface $S$ in terms of the "densitized" triad $E_{i}^{a}=ee_{i}^{a}$ where $a,b,c,...$ are the spatial coordinates and $i,j,k,...$ are $SU(2)$ coordinates, e the determinant of the triad matrix defined by $g_{ab}=e_{a}^{i}e_{b}^{j}\delta_{ij}$ where $g_{ab}$ is the spatiale metric. So, since the computation of the area uses the determinant of the the metric $h_{\alpha\beta}$ induced by $g_{ab}$ on $S$: ($\alpha,\beta,... =1,2\;and\; a,b,..=1,2,3$)

$h_{\alpha\beta}=g_{ab}\frac{\partial x^{a}}{\partial\sigma^{\alpha}}\frac{\partial x^{b}}{\partial\sigma^{\beta}}$

So in computing the determinant $h$ explecitely on finds the term

$g_{a[b}g_{c]d}$ which needs to equal to $\frac{1}{2}\epsilon_{ace}\epsilon_{bdf}gg^{ef}$ in order to obtain the final result:

$h=E_{i}^{a}E_{j}^{b}\delta^{ij}n_{a}n_{b}$ where $n$ are normal vectors $n_{a}=\epsilon_{abc}\frac{\partial x^{b}}{\partial\sigma^{1}}\frac{\partial x^{c}}{\partial\sigma^{2}}$

EDIT2:

After the notification of Willie Wong, I decided to put my original problem as a question, i.e: deriving the expression of the determinant of the induced metric on $S$ in terms of the densitized triad.

Answer 1

Edit 2:

Contrary to your description in your question, I am going to assume that your E's are tensor densities of weight 1, not 2 (otherwise the units don't work out right). Then a sketch of the argument goes something like this:

Let $\sigma^\alpha$ be a coordinate system on $S$, extend this with the normal vector $n$ so that we get a local basis in the tangent space. In this basis $f_0 = n, f_1 = \partial/\partial \sigma^1, f_2 = \partial/\partial\sigma^2$ the metric tensor $g$ looks like $$ (g) = \begin{pmatrix} g_{00} & 0 \\\\ 0 & (h) \end{pmatrix} $$ So in this coordinate system $|h| g_{00} = |g|$, or $|h| = (g^{-1})_{00} |g|$ since $(g)$ is block diagonal.

Now, using that $E_i^a = e e_i^a$ where $e_i^a$ are an orthonormal basis, we have that the inverse metric $g^{-1}$ is equal to $\sum_i e_i^a\otimes e_i^b$. So the right hand side of your formula reads $$ E_i^aE_j^b \delta^{ij}n_a n_b = e^2 g^{ab}n_an_b$$ So if $e$ is chosen to be the square root of the metric determinant, then the expression agrees with what is shown above.

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Original answer:

Take a look at the Wikipedia entry for Levi-Civita symbols http://en.wikipedia.org/wiki/Levi-Civita_symbol

Now, I assume that when you say metric you mean a (pseudo-)Riemannian metric tensor. Note that the Levi-Civita symbol by itself is not a tensor. Nor is the metric determinant a scalar (they are both frame/coordinate dependent). But with a choice of an orientation, the object which sometimes is written as $\sqrt{|g|}\epsilon_{abc}$ and sometimes just $\epsilon_{abc}$ is a covariant object: it is a way of writing the volume form.

So I suspect you are just confused about the "frame" expression of the volume form (in an orthonormal frame the metric determinant is just 1, and the volume form agrees with the Levi-Civita symbol) and the coordinate expression of the volume form...

EDIT (after OP's edit):

Something is wrong with the relation you are trying to prove: Take trace relative to $g^{ac}$, The left hand side gives $-2 g_{bd}$ and the right hand side vanishes. So I would start by checking your computation for the determinant of $h$.