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I am learning General Relativity from Leonard Susskind's Lectures. In Lecture three, he introduces to covariant derivatives, and I understood it's meaning. But when he applies it to a Tensor, I am having some confusion. If we consider the Tensor $T_{mn}$ , it's covariant derivative $D_sT_{mn}$ , he says is given as $$D_sT_{mn}=\partial_sT_{mn}-\Gamma^t_{ms}T_{tn}-\Gamma^t_{ns}T_{tm}$$ Where $\partial_sT_{mn}$ represents the derivative of the Tensor with respect to $x^s$, while the Gamma's are the Christoffel Symbols. My question is that why are the contravariant indices of both the Christoffel symbols (The index t in this case) the same.

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    $\begingroup$ It's a dummy index. You can rename them if it confuses you. $\endgroup$
    – Noone
    Commented Apr 18, 2020 at 11:09
  • $\begingroup$ @ApolloRa I know that it is a dummy index, but the equation won't remain the same if the said indices were different. If they were different, say t and u, then the sum (the above represents sums of equations) would contain terms like t=0 with u=1, but if there were only t, it wouldn't have such terms $\endgroup$
    – SK Dash
    Commented Apr 18, 2020 at 11:27

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The left hand side has three indices: $m,n,s$. These are free indices. In order for this equation to be correct the right hand side must have the same $m,n,s$ free indices. Indeed it does. The $t's$ are dummy indices. It doesn't matter how you name them. You can change the $t's$ (of course both $t's$ of each term or else they will not mean summation) to whatever you like. Examples:

$$D_sT_{mn}=\partial_sT_{mn}-\Gamma^u_{ms}T_{un}-\Gamma^t_{ns}T_{mt}$$ $$D_sT_{mn}=\partial_sT_{mn}-\Gamma^u_{ms}T_{un}-\Gamma^u_{ns}T_{mu}$$ $$D_sT_{mn}=\partial_sT_{mn}-\Gamma^a_{ms}T_{an}-\Gamma^b_{ns}T_{mb}$$

All the above equations are the same.

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    $\begingroup$ Although, shouldn't the last term have $T_{mt}$ instead of $T_{tm}$, for the covariant derivative to be correct? $\endgroup$ Commented Apr 18, 2020 at 12:02
  • $\begingroup$ @Elias Riedel Gårding Indeed yes! I didnt check the equation in the question! Thanks for pointing it out! For an arbitary tensor $T_{mn} !=T_{nm}$ $\endgroup$
    – Noone
    Commented Apr 18, 2020 at 12:09
  • $\begingroup$ I would still take my stand, this is because, the dummy indices are summed over, so if we had separate variables, in this case a and b, we sum over a separately and b separately, so there are cases in which a is not the same as b, but if they had the same contravariant indices, there wouldn't be any cross terms, a and b must be the same. $\endgroup$
    – SK Dash
    Commented Apr 18, 2020 at 13:22
  • $\begingroup$ a and b are summed over the same "numbers". Consider a four dimensional spacetime, then a = 0,1,2,3 and b =0,1,2,3. Does this makes things clearer??? $\endgroup$
    – Noone
    Commented Apr 18, 2020 at 13:44
  • $\begingroup$ Hmm okay I got the point you are trying to make, thank you $\endgroup$
    – SK Dash
    Commented Apr 19, 2020 at 0:02

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