# Identity for the inverse metric tensor using its determinant

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.

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.