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Is the result of the product of metric tensors $\eta^{\mu\nu} \eta_{\mu\nu} = 1$?

If so how would I prove this?

I know that tensors are represented as matrices but I don't know how I'd prove this (if it is true).

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    $\begingroup$ Evaluate the double sum and see. Sometimes brute force gives more insight than formalism does. $\endgroup$ – G. Smith Mar 29 at 17:53
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    $\begingroup$ Contractions are just sums. There are 16 terms. As knzhou’s answer shows, only four are nonzero. $\endgroup$ – G. Smith Mar 29 at 18:03
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    $\begingroup$ The formal way uses $\eta^{\alpha\gamma}\eta_{\gamma\beta}=\delta^\alpha_\beta$: inverse metric matrix times metric matrix equals identity matrix. $\endgroup$ – G. Smith Mar 29 at 18:08
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Just use the definition. All off-diagonal elements of the metric vanish, so $$\eta^{\mu\nu} \eta_{\mu\nu} = \eta^{00} \eta_{00} + \eta^{11} \eta_{11} + \eta^{22} \eta_{22} + \eta^{33} \eta_{33} = 1 + 1 + 1 + 1 = 4.$$ Also, the question you asked before this one is about the "permutation invariance of the maximal helicity violating gravitational amplitude in the spinor-helicity formalism". If you're still confused about how ordinary tensors work, you really need to back up and review the basics before continuing!

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