- #1
shinobi20
- 267
- 19
- 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?
##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?