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If I have to prove that $A_{kh}=g_{ij}\frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^h}$ is symmetric in $k$ and $h$, knowing that $g_{ij}=g_{ji}$, for the term $\frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^h}$ it is allowed to rename $i$ with $j$ and viceversa and so can I write the following? $$g_{ij}\frac{\partial x^i}{\partial y^k}\frac{\partial x^j}{\partial y^h}=g_{ij}\frac{\partial x^j}{\partial y^k}\frac{\partial x^i}{\partial y^h}$$ I am not so sure that this is only a matter of renaming the indeces, since for instance if I rename the indeces for a very general (no symmetric) matrix $s_{kj}$ (I rename $k$ with $j$ and viceversa) this becomes $s_{jk}$ but this would mean that this $s$ is symmetric.

So how can I prove really the symmetry of $A_{kh}$?

Arctic Char
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cely
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1 Answers1

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You can rename indices that are summed over ($\sum_i f(i) = \sum_j f(j)$), so your derivation of $A_{ij}=A_{ji}$ is correct. One is not allowed to rename so-called free indices, whose value is fixed -- doing that would lead to the wrong conclusion that every matrix is symmetric, as you pointed out.

Blazej
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  • Would it be possible for you to verify the following question, please? This is the link: https://math.stackexchange.com/questions/4073674/schouten-theorem-and-doubts-in-the-calculation-of-the-tensor-derivative?noredirect=1#comment8418089_4073674 – Ozni Mar 25 '21 at 16:27
  • @Blazej So can I rename the index as I have done but obviously I can not do this for $g_{ij}$, right? – cely Mar 25 '21 at 17:17
  • You should rename both instances of the index $i$ at the same time. For example $a_i b^i$ and $a_j b^j$ both mean $a_1 b^1 + a^2 b_2 + \ldots + a^n b_n$. As for your question: let $J^i_j=\frac{\partial x^i}{\partial y^j}$. Expressions $J^i_k J^j_h g_{ij}$ and $J^j_k J^i_h g_{ji}$ both mean the same sum (over two indices this time). Then in the second you use $g_{ij} = g_{ji}$ (this is not relabeling of indices -- this time we use the symmetry of $g$). – Blazej Mar 25 '21 at 17:32
  • Ok so the fact that now I can do this for $g_{ij}$ (but not for $s_{jk}$ of my question) is due by the fact that now we are dealing with a sum over $i$ and $j$? – cely Mar 25 '21 at 17:41
  • Yes. This is completely analogous to the reason why $a_i b^i = a_j b^j$ -- just think about the sum that is really meant by these abbreviations. Note also that the expression $a_i b^i$ does not depend on $i$ in any sense, but rather it is the sum of terms in which $i$ goes through all of its allowed values. On the other hand expression $a_i$ depends on $i$ -- if we plug in $i=1$ we obtained $a_1$, while plugging in $i=2$ gives $a_2$ - two different components of $a$. – Blazej Mar 25 '21 at 17:55
  • Thanks a lot now it is all clear! – cely Mar 25 '21 at 18:08