Contracting one index of a metric with the inverse metric

[shinobi20]

Homework Statement

Show $g_{\mu\nu}g^{\nu\lambda} = \delta^\lambda_\mu$ using the orthogonality and completeness condition.

Relevant Equations

$g_{\mu\nu} = \vec{e_\mu} \cdot \vec{e_\nu}$
$g^{\mu\nu} = \vec{e^\mu} \cdot \vec{e^\nu}$
$\vec{e_\mu} \cdot \vec{e^\nu} = \delta^\nu_\mu \quad \rightarrow \quad \sum_\alpha (e_\mu)_\alpha (e^\nu)_\alpha = \delta^\nu_\mu ~$ (orthogonality)
$\vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \quad (e_\mu)_\alpha (e^\mu)_\beta = \delta_{\alpha\beta}~$ (completeness)

Since $\nu$ is contracted, we form the scalar product of the metric and inverse metric,

$g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda}) = \vec{e_\mu} \cdot (\vec{e_\nu} \cdot \vec{e^\nu}) \cdot \vec{e^\lambda} = \delta^\lambda_\mu$

I am not sure how to use the completeness condition here, can anyone point me in the proper direction?

 

[nrqed]

Homework Statement:: Show $g_{\mu\nu}g^{\nu\lambda} = \delta^\lambda_\mu$ using the orthogonality and completeness condition.
Relevant Equations:: $g_{\mu\nu} = \vec{e_\mu} \cdot \vec{e_\nu}$
$g^{\mu\nu} = \vec{e^\mu} \cdot \vec{e^\nu}$
$\vec{e_\mu} \cdot \vec{e^\nu} = \delta^\nu_\mu \quad \rightarrow \quad \sum_\alpha (e_\mu)_\alpha (e^\nu)_\alpha = \delta^\nu_\mu ~$ (orthogonality)
$\vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \quad (e_\mu)_\alpha (e^\mu)_\beta = \delta_{\alpha\beta}~$ (completeness)

Since $\nu$ is contracted, we form the scalar product of the metric and inverse metric,

$g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda}) = \vec{e_\mu} \cdot (\vec{e_\nu} \cdot \vec{e^\nu}) \cdot \vec{e^\lambda} = \delta^\lambda_\mu$

I am not sure how to use the completeness condition here, can anyone point me in the proper direction?

You have to be very explicit with all the steps. First, note that when you wrote
$g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda})$
it is confusing since you are using a dot symbol to have more than one meaning here. You should start by writing
$g_{\mu\nu}g^{\nu\lambda} = \sum_\nu (\vec{e_\mu} \cdot \vec{e_\nu}) (\vec{e^\nu} \cdot \vec{e^\lambda})$
Now write these dot products in terms of summation over the other types of indices (that you wrote as \alpha,\beta). Also, note that to be precise, the completeness relation is

$\vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \sum_\mu \quad (e_\mu)_\alpha (e^\mu)_\beta = \delta_{\alpha\beta}~$

 

[shinobi20]

You have to be very explicit with all the steps. First, note that when you wrote
$g_{\mu\nu}g^{\nu\lambda} = (\vec{e_\mu} \cdot \vec{e_\nu}) \cdot (\vec{e^\nu} \cdot \vec{e^\lambda})$
it is confusing since you are using a dot symbol to have more than one meaning here. You should start by writing
$g_{\mu\nu}g^{\nu\lambda} = \sum_\nu (\vec{e_\mu} \cdot \vec{e_\nu}) (\vec{e^\nu} \cdot \vec{e^\lambda})$
Now write these dot products in terms of summation over the other types of indices (that you wrote as \alpha,\beta). Also, note that to be precise, the completeness relation is

$\vec{e_\mu} \otimes \vec{e^\mu} = \vec{I} \quad \rightarrow \sum_\mu \quad (e_\mu)_\alpha (e^\mu)_\beta = \delta_{\alpha\beta}~$

$g_{\mu\nu}g^{\nu\lambda} = \sum_\nu (\vec{e_\mu} \cdot \vec{e_\nu}) (\vec{e^\nu} \cdot \vec{e^\lambda})$

$= \sum_\nu (\sum_\alpha (\vec{e_\mu})_\alpha (\vec{e_\nu})_\alpha) (\sum_\beta (\vec{e^\nu})_\beta (\vec{e^\lambda})_\beta)$

$= \sum_\alpha \sum_\beta (\vec{e_\mu})_\alpha (\sum_\nu (\vec{e_\nu})_\alpha (\vec{e^\nu})_\beta) (\vec{e^\lambda})_\beta$

$= \sum_\alpha \sum_\beta (\vec{e_\mu})_\alpha \delta_{\alpha \beta} (\vec{e^\lambda})_\beta$

$= \sum_\alpha (\vec{e_\mu})_\alpha (\vec{e^\lambda})_\alpha$

$= \delta^\lambda_\mu$