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$g^{ab}$ is the metric tensor (Minkowski)

Im trying to understand if this is true:

$g^{ab}g_{ab}=?=g^{aa}g_{aa}$

Have a nice day.

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As a third, but equivalent, perspective to the existing answers: When in doubt, you can always put the summation back explicitly. (Just take it back out once you understand the answer and before getting evil looks from your peers for not having used Einstein summation!)

In your case, you would have $$ \sum_a \sum_b g^{ab} g_{ab} = ? = \sum_a \sum_a g^{aa} g_{aa} \ \ \mathrm{(Nonsense!)} $$ which you'll probably agree is nonsense. Using the same dummy index on both summations just doesn't make sense. If you instead, intended it to be $$ \sum_a \sum_b g^{ab} g_{ab} = ? = \sum_a g^{aa} g_{aa} \ \ \mathrm{(Nonsense!)}$$ with the understanding that one $a$ pair is fixed from having done the sum on $b$ and one is the dummy index of the remaining sum, that's not an accepted interpretation of the notation with the explict summation or in the Einstein convention. It's inherently ambiguous what's to be summed and what's not if you try to do it like that.

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No. The right side of your equation is meaningless. You have to contract indices two at a time, not four at a time. Otherwise you don’t get a Lorentz-invariant result.

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  • $\begingroup$ I dont understand you sorry. In the left side, first I contract a pair of a's and then another pair getting a delta of a's who is obviusly 1. For the left side, I contract b's and then a's getting the same (1). Am I right? $\endgroup$ – Isaac Domínguez Larrañaga Jan 20 at 17:17
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    $\begingroup$ You cannot have a tensor expression with four of the same index, such as four $a$’s. It violates the rules of tensor algebra. $\endgroup$ – G. Smith Jan 20 at 17:21
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As @G.Smith pointed out in his answer, $g^{aa}g_{aa}$ doesn't make sense.

You can calculate $g^{ab}g_{ab}$ as a special case of raising and lowering indices. Contracting the $b$ indices first you get $$g^{ab}g_{ab}=\delta^a_a=...$$ I leave out the final step as an exercise for you.

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