Identity for the inverse metric tensor using its determinant

Question

I would like to prove this relation:

$$g^{\mu\nu} = \frac{1}{3!} \frac{1}{g} \epsilon^{\mu\rho\sigma\kappa}\epsilon^{\nu\alpha\beta\gamma} g_{\rho\alpha} g_{\sigma\beta} g_{\kappa\gamma}, \tag{1}$$

with $g_{\mu\nu}$ the metric tensor, $g^{\mu\nu}:=(g^{-1})^{\mu\nu}$ its inverse and $g:= \det g_{\mu\nu}$ its determinant given by:

$$g = \frac{1}{4!} \epsilon^{\mu\rho\sigma\kappa}\epsilon^{\nu\alpha\beta\gamma} g_{\mu\nu} g_{\rho\alpha} g_{\sigma\beta} g_{\kappa\gamma}. \tag{2}$$

I can derive $(1)$ using the trace $g_{\mu\nu} g^{\mu\nu} = 4$, but I am pretty sure that this is not a valid derivation since several matrices can have the same trace. Instead, I would like to use the definition of $g_{\mu\nu}$, which is:

$$g_{\mu\nu} g^{\nu\rho} = \delta_\mu^\rho. \tag{3}$$

But I still have trouble to obtain the identity, since I cannot see where to use it really. I noticed that I can write:

$$1 = \frac{1}{g} g = \frac{1}{g} \frac{1}{4!} \epsilon^{\mu\rho\sigma\kappa}\epsilon^{\nu\alpha\beta\gamma} g_{\rho\alpha} g_{\sigma\beta} g_{\kappa\gamma} g_{\mu\nu} =: \frac{1}{g} \frac{1}{4!} A^{\mu\nu} g_{\mu\nu}, \tag{4}$$

which in matrix notation would be equivalent to:

$$1 = \frac{1}{g} \frac{1}{4!} \text{Tr} (A G). \tag{5}$$

Multiplying both sides by $G^{-1}$, I get:

$$G^{-1} = \frac{1}{g} \frac{1}{4!} \text{Tr} (A G) G^{-1}. \tag{6}$$

Looking at the desired result $(1)$, it seems that the trace should evaluate as $\text{Tr} (AG) G^{-1} = 4 AG G^{-1} = 4 A$, but I don't see how to show that. Doing the same manipulation in index notation, I get the same annoying "lack of connection" between the indices:

$$g^{\lambda\delta} = \frac{1}{g} \frac{1}{4!} \epsilon^{\mu\rho\sigma\kappa}\epsilon^{\nu\alpha\beta\gamma} g_{\rho\alpha} g_{\sigma\beta} g_{\kappa\gamma} g_{\mu\nu} g^{\lambda\delta}. \tag{7}$$

Any suggestion would be appreciated.

Answer 1

Here is one elementary approach to prove eq. (1):

1. Argue that the RHS transforms as components of a symmetric (2,0) tensor.

2. Argue that it is enough to prove eq. (1) for $g_{\mu\nu}=\delta_{\mu\nu}$.

3. Now consider fixed indices $\mu$ and $\nu$.

4. Case $\mu\neq\nu$: Argue that the RHS must be zero.

5. Case $\mu=\nu$: Argue that only $3!=6$ combinations of the 6 summation variables $(\alpha,\beta,\gamma,\rho,\sigma,\kappa)$ on the RHS can give non-zero contributions. This explains the $3!=6$ normalization factor on the RHS.