# Homework Help: Raising and Lowering Indices of the Metric

1. Apr 3, 2013

### Shmi

1. The problem statement, all variables and given/known data

"Evaluate: $g^{\mu \nu} g_{\nu \rho}$ where $ds^2 = g_{\mu \nu} dx^\mu dx^\nu , ds^2 = c^2 dt^2 - dx^2 - dy^2 - dz^2$"

2. Relevant equations

None necessary, just a notation issue

3. The attempt at a solution

Just using raising and lowering rules, I imagine each raising and lowering the others non-summed index or "contracting" the dummy index as I've heard it called.

$$g^{\mu \nu} g_{\nu \rho} = g_\mu^\rho$$

But that's a bit easy.

In matrix form, this new object $g_\mu ^\rho$ is a four by four identity matrix when you take the product of the two matrices.

This leads to me to think that $g^{\mu \nu} g_{\nu \rho} = g_\mu^\rho = \delta _\mu ^\rho$ is some intrinsic identity of this combination of metric tensors, and a quick google search does indeed reveal that to be the case.

What I'm really struggling with is that most sites present this as an obvious consequence of a matrix and its inverse. I can dig a matrix and it's inverse resulting in an identity, but I don't see how these matrices are inverses. Furthermore, this action produces a (1,1) tensor. Of course, given the manipulation rules, I see why it is a (1,1) but I don't see why me make the distinction for an identity matrix, or if it's just a matter of consistency.

If you could shed some insight on what's going as we raise and lower indices, I'd love to see why this result is so obvious to the authors of other literature on index notation.

2. Apr 3, 2013

### Dick

The g with the upper indices is DEFINED to be the inverse of the g with lower indices. That's really all there is to it. For other tensors the raising and lowering goes through the metric tensor.

Last edited: Apr 3, 2013
3. Apr 4, 2013

### Staff: Mentor

Another simple way of deriving your result is to make use of the coordinate basis vectors $\vec{a_i}$ and the basis one forms $\vec{a^j}$. These are related by $$\vec{a_i}\centerdot \vec{a^j}=\delta_i^j$$. If we use this to express the coordinate basis vectors in terms of the basis one forms, we get:
$$\vec{a_i}=g_{ij}\vec{a^j}$$ where

$$g_{ij}=\vec{a_i}\centerdot\vec{a_j}$$

If we now dot this equation with $\vec{a^k}$, we get:

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