# How to prove $g^{\mu\nu}\Lambda^{\rho}{}_{\mu}\Lambda^{\sigma}{}_{\nu}=g^{\rho\sigma}$ for the inverse metric?

In Srednicki's book, we have \begin{align*} g_{\mu\nu}\Lambda^\mu{}_\rho\Lambda^\nu{}_\sigma=g_{\rho\sigma} \end{align*} and let $$\Lambda \to \Lambda^{-1}$$, use the relationship $$(\Lambda^{-1})^\rho{}_{\nu}=\Lambda_{\nu}{}^\rho$$, we have \begin{align*} g_{\mu\nu}(\Lambda^{-1})^\mu{}_\rho(\Lambda^{-1})^\nu{}_\sigma =g_{\mu\nu}\Lambda_{\rho}{}^{\mu} \Lambda_{\sigma}{}^{\nu} =g_{\rho\sigma} \end{align*} But how to do after? Do I have $$g_{\mu\nu}\Lambda_{\rho}{}^{\mu} = \Lambda_{\rho\nu}$$? In another book, it gives $$(\Lambda^{-1})^{\mu}{}_{\rho}\equiv g^{\mu\beta}g_{\rho\alpha}\Lambda^{\alpha}{}_{\beta}$$, and then \begin{align*} \begin{aligned} g^{\alpha\beta}& =g^{\alpha\rho}g^{\beta\sigma}g_{\rho\sigma} =g^{\alpha\rho}g^{\beta\sigma}g_{\mu\nu}(\Lambda^{-1})^{\mu}{}_{\rho}(\Lambda^{-1})^{\nu}{}_{\sigma} =g^{\alpha\rho}g^{\beta\sigma}g_{\mu\nu}g^{\mu\gamma}g_{\rho\delta}\Lambda^{\delta}{}_{\gamma}g^{\nu\phi}g_{\sigma\tau}\Lambda^{\tau}{}_{\phi}\\& =\delta^{\alpha}{}_{\delta}\delta^{\beta}{}_{\tau}\delta^{\gamma}{}_{\nu}g^{\nu\phi}\Lambda^{\delta}{}_{\gamma}\Lambda^{\tau}{}_{\phi} =g^{\nu\phi}\Lambda^{\alpha}{}_{\nu}\Lambda^{\beta}{}_{\phi} \end{aligned} \end{align*} However, why the $$g^{\alpha\rho}g^{\beta\sigma}g_{\mu\nu}g^{\mu\gamma}g_{\rho\delta} = g^{\alpha\rho} g_{\rho\delta}g^{\beta\sigma}g_{\mu\nu}g^{\mu\gamma}$$, I mean they can change with each other?

I can write \begin{align*} g^{\alpha\rho}g^{\beta\sigma}g_{\mu\nu}g^{\mu\gamma}g_{\rho\delta} = g^{\alpha\rho}g^{\beta\sigma}g_{\nu\mu}g^{\mu\gamma}g_{\rho\delta} = g^{\alpha\rho}g^{\beta\sigma}\delta_{\nu}{}^{\gamma} g_{\rho\delta} = g^{\alpha\rho}g^{\beta\sigma}g_{\rho\delta}\delta_{\nu}{}^{\gamma} \end{align*} So do I have $$g^{\beta\sigma}g_{\rho\delta} = g_{\rho\delta}g^{\beta\sigma}$$

Similarly, in $$g^{\beta\sigma}g_{\mu\nu}g^{\mu\gamma}g_{\rho\delta}\Lambda^{\delta} {}_{\gamma}g^{\nu\phi}g_{\sigma\tau} = g^{\beta\sigma}g_{\sigma\tau}g_{\mu\nu}g^{\mu\gamma}g_{\rho\delta}\Lambda^{\delta}{}_{\gamma}g^{\nu\phi}$$, do I have $$g_{\rho\delta}\Lambda^{\delta} {}_{\gamma}g^{\nu\phi}g_{\sigma\tau} = g_{\sigma\tau}g_{\rho\delta}\Lambda^{\delta} {}_{\gamma}g^{\nu\phi}$$

Could you tell me the rule of abstract index notation, such as what is the difference of right and left of index?

The $$g_{ij}$$ are the components of the metric tensor with respect to some local coordinate basis, which means they are scalars and can be imterchanged, i.e.

$$g_{ij} g_{kl} = g_{kl} g_{ij}$$

This is of course true for all tensor components.

Moreover, the metric tensor $$g$$ is symmetric which means

$$g_{ij} = g_{ji}$$

This is not true for tensors generally, however, the metric tensor is defined such that this holds.

The $$g_{ij}$$ are the components of the metric tensor. They are called covariant components. The so-called contravariant components are $$g^{ij}$$ which are defined (!) as the inverses, so that

$$g_{ij} g^{jk} = \delta_i^k$$

Given another tensor, say $$T_{ij}$$, then we define the quantity

$$T^i{}_{j} := g^{ik} T_{kj}$$

and similarly for tensors of higher or lower order and for the other indizes.

Let $$\{ b_i \}_{i=1}^n$$ be a basis of a vector space, let $$X,Y$$ be arbitrary vectors within that space and let $$f$$ be a bilinear map. The components $$f_{ij}$$ of $$f$$ are then the quantities

$$f(X,Y) = f\Big( X^i b_i, Y^j b_j \Big) = X^i Y^j f(b_i, b_j) =: X^i Y^j f_{ij}$$

The metric tensor is a bilinear map, its components are obtained analogously.

This was omly a very shallow and application-oriented summary. The metric tensor and its properties are essentiel to differential geometry and general relativity. It is really an interesting topic, I suggest diving into it. I found it insightful to start with classical differential geometry and only then move on to modern differential geometry and general relativity.

• Thank you @Octavius.
– liZ
Commented Jun 20 at 13:10

Let me use condition $${\Lambda^{-1}}^{\rho}_{}{\nu}= \Lambda_{\nu}{}^{\rho}$$ We can raise $$\nu$$ index left side as well as right side, $${\Lambda^{-1}}^{\rho\nu}= \Lambda^{\nu\rho}$$ Now contract above result with $$\Lambda_{\nu\rho}$$, $${\Lambda^{-1}}^{\rho\nu}{\Lambda_{\nu\rho}}= \Lambda^{\nu\rho}\Lambda_{\nu\rho}$$ And actually $${\Lambda^{-1}}^{\rho\nu}{\Lambda_{\nu\rho}}=\delta{}^{\mu}_{}{\mu}=4.$$ Therefore, $$\Lambda^{\nu\rho}\Lambda_{\nu\rho}=4$$ Redefine the indices, $$\Lambda^{\sigma\mu}\Lambda_{\sigma\mu}=4$$ Since $$g_{\rho \sigma}\Lambda^{\rho}_{}{\mu}= \Lambda_{\sigma\mu}$$, using this with previous one, $$g^{\mu\nu}g_{\rho\sigma}\Lambda^{\rho}{}{\mu}\Lambda^{\sigma}{}{\nu}=4$$ Contract (roughly multiply) both side by $$g^{\rho\sigma}$$ and use $$g^{\rho\sigma}g_{\rho\sigma}=4$$. We arrived at desire result, $$g^{\mu\nu}\Lambda^{\rho}{}{\mu}\Lambda^{\sigma}{}{\nu}=g^{\rho\sigma}$$