# Contraction of metrices

1. Aug 15, 2013

### Dreak

I'm starting to doubt about something:

In 4 dimension, what value has δ$^{λ}_{μ}$ in following equation:
ημ$\nu$η$\nu$λ = δ$^{λ}_{μ}$

is it 4 or 1?

and IF it's 1, what is the difference that this equation:

ημσημσ

= 4?

2. Aug 15, 2013

### CompuChip

$\delta_\mu{}^\lambda = 1$ if $\mu = \lambda$ and 0 otherwise.

Only repeated indices are contracted, perhaps writing out the summations explicitly makes it clearer:

$$\eta_{\mu\nu} \eta^{\nu\lambda} \equiv \sum_{\nu = 1}^4 \eta_{\mu\nu} \eta^{\nu\lambda}$$
which is equal to 1 if $\mu = \lambda$ and 0 otherwise. In matrix notation, it basically says $\eta^{-1} \eta = \mathbb{I}$.

If you also contract the two free indices, you get
$$\eta_{\mu\nu} \eta^{\nu\mu} \equiv \sum_{\mu = 1}^4 \sum_{\nu = 1}^4 \eta_{\mu\nu} \eta^{\nu\mu}$$
Replacing the inner sum by the delta, this is
$$\cdots = \sum_{\mu = 1}^4 \delta_\mu{}^\mu$$
and since the upper- and lower index are the same, the delta evaluates to 1 for every value of mu:
$$\cdots = \sum_{\mu = 1}^4 1 = 4$$

As $$\eta_{\mu\nu} \eta^{\nu\lambda}$$ is the $(\mu, \lambda)$ component of the matrix $\eta^{-1} \eta$, in matrix notation the double sum this basically says
$$\sum_{i = 1}^4 (\eta^{-1} \eta)_{ii} = 4$$
which is indeed the trace of the 4x4 identity matrix.

3. Aug 15, 2013

### Dreak

Wow thanks, very clear explenation.
Thank you very much! :)