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This suggests, as a shortcut notation, the concept of lowering indices; from any vector we can construct a (0, 1) tensor defined by contraction with the metric: $$A_\nu ≡ g_{\mu\nu}A^\mu$$

so that the dot product becomes $$g_{\mu\nu}A^\mu B^\nu = A_\nu B^\nu$$ We also define the inverse metric $g^{\mu\nu}$ as the matrix inverse of the metric tensor: $$g^{\mu\nu}g_{\nu\rho} = δ^\mu_\rho$$

where $δ^\mu_\rho$ is the (spacetime) Kronecker delta. (Convince yourself that this expression really does correspond to matrix multiplication.) Then we have the ability to raise indices: $$A^\mu = g^{\mu\nu}A_\nu$$ Note that raising an index on the metric yields the Kronecker delta, so we have $$g^{\mu\nu}g_{\mu\nu} = δ^\mu_\mu = 4$$

I don't understand why $δ^\mu_\mu = 4$. Shouldn't it equal 1?

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    $\begingroup$ The quantity $\delta_\nu^\nu$ is called the trace, and the trace of the Kronecker delta is indeed 4. can you clarify what is confusing you about this? Is that you think $g^{\mu\nu}g_{\mu\nu}$ should be equal to 1? If so, I suspect you're mixing up $g^{\mu\nu}g_{\mu\nu}$, which gives the trace of $g$, with $g^{\mu\nu}g_{\nu\rho}$, which multiplies the two matrices and gives the matrix $I$. $\endgroup$ Commented May 23, 2015 at 7:44
  • $\begingroup$ @JohnRennie Thanks! That was exactly what I was thinking. $\endgroup$
    – Jimmy360
    Commented May 23, 2015 at 7:45
  • $\begingroup$ In that case the question (v3) is essentially a duplicate of physics.stackexchange.com/q/66394/2451 $\endgroup$
    – Qmechanic
    Commented May 23, 2015 at 8:12
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    $\begingroup$ A repeated index on $\delta^\mu{}_\mu$ implies a summation (summation convention). Then what we truly mean is $\delta^\mu_\mu = \sum_{\mu=0}^3 \delta^\mu_\mu = 1+1+1+1 = 4$. More generally, it is equal to $d$, the dimension of space-time. $\endgroup$
    – Prahar
    Commented May 23, 2015 at 12:25

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