# Simple indice question -- metric, traces

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##

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##

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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:

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