Metric tensor and Kronecker delta

[Burnstryk]

Homework Statement

I am trying to derive the following relation using inner products of vectors:

Homework Equations

$$g_{\mu\nu} g^{\mu\sigma} = \delta_{\nu}^{\hspace{2mm}\sigma}$$

The Attempt at a Solution

What I have done is take two vectors and find the inner products in different ways with contravariant and covariant components:

$$\textbf{v}.\textbf{w}$$

I have obtained the following relations:

$$g_{\mu\nu} v^{\mu} w^{\nu} = g^{\mu\nu}v_\mu w_\nu = v_\nu w^\nu = v^\nu w_\nu$$

Using these relations I decided to take a vector with an arbitrary component (sigma) and multiply it by the metric and inverse considering the lowering and operating nature:

$$g_{\mu\nu} g^{\mu\sigma} v_{\sigma} = g_{\mu\nu} v^{\mu} = v_\nu = \delta_{\nu}^{\hspace{2mm}\sigma}v_\sigma$$

and hence obtain the original result.

I wanted to see if these arguments and method make sense or if I'm confusing everyone.

 

[andrewkirk]

Your calculations are correct, but they do not constitute a proof that $g_{\mu\nu}g^{\mu\nu}=\delta_\nu{}^\mu$ because it does not follow from the fact that two sums are equal that their individual summands are pairwise equal.

I'm a bit rusty on this but my recollection is that the result you are seeking to prove is an assumption, rather than something to be proved.

 

[strangerep]

[...] but my recollection is that the result you are seeking to prove is an assumption, rather than something to be proved.

That's correct.

$g^{\rho\sigma}$ is defined to be the (components of) the matrix inverse to $g_{\mu\nu}$, and it always exists because $g_{\mu\nu}$ is assumed to be non-singular.