# Simple indice question -- metric, traces

1. Dec 18, 2014

### binbagsss

Mod note: OP warned about not using the homework template.
Let $g_{ac}$ be a 3-d metric.
So the trace of a metric is equal to its dimension so I get $g_{ac}g^{ac}=3$

But I'm a tad confused with the expression : $g^{ac}g_{ad}$=$delta^{c}_{d}$
I thought it would be $3delta^{c}_{d}$#

Because $delta^{c}_{d}$ implies that $g^{ac}g_{ad}$ is only non-zero when c=d, in which case don't you just get: $g_{ac}g^{ac}=3$

Last edited by a moderator: Dec 18, 2014
2. Dec 18, 2014

### DeIdeal

The summation rule applies, in the last expression c is being summed over, but in delta-expression it is not. If you contract the indices in it, you get trace of the identity matrix, which is 3 in three dimensions:

$3=\delta^c_c=g^{ac}g_{ac}\neq\delta^c_d=g^{ac}g_{ad}$

The first is a scalar, the second is a (1,1) tensor. Btw, you need a blackslash \ in front of delta you want to write $\delta$

3. Dec 18, 2014

### Staff: Mentor

$$g^{ac}g_{ad}=g^{ca}g_{ad}=g^c_d=\delta^c_d$$