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I originally asked a similar question in the physics section, but now I realize that it is more appropriate to ask it here. I actually another post which directly gives that specific answer here, but I want to understand how I can manage this algebra.

I need to symmetrize a not fully symmetrized tensor (e.g. $T^{i\{jk\}}$). There are some related answers that I found such as this, but again it does not say anything if two of the indices of the rank 3 tensor is already symmetrized.

Kaan Güven
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  • Do you know how to symmetrise $T^{ijk}$? – Michael Albanese Jul 29 '22 at 13:38
  • I don't know the whole theory, but when I see for example this answer I can understand what a symmetric tensor look like. On the other hand, since I don't know about the symmetric group yet, I can't understand this: $$ (T_S){i_1,\dots,i_p} = \frac{1}{p!} \sum{\sigma \in \mathfrak{S}p} T{i_{\sigma(1)} , \dots i_{\sigma(p)}} , , $$ – Kaan Güven Jul 30 '22 at 12:42
  • If $\sigma$ is an element of the symmetric group on $p$ elements, then $\sigma(1), \dots, \sigma(p)$ is just a permutation of $1, \dots, p$. So for example, $(T_S){i_1,i_2} = \frac{1}{2}(T{i_1,i_2} + T_{i_2,i_1})$. Do you think you can write out the formula for the $(T_S)^{ijk}$? – Michael Albanese Jul 31 '22 at 14:15
  • Yes, in that case it should be: $(T_S)^{ijk} = \frac{1}{3!}(T^{ijk}+T^{ikj}+T^{jik}+T^{jki}+T^{kij}+T^{kji})$ – Kaan Güven Jul 31 '22 at 14:28
  • Correct, and because $T$ is already symmetric in the last two indices, some of those terms are equal, e.g. $T^{ijk} = T^{ikj}$. – Michael Albanese Jul 31 '22 at 14:33
  • So, in his case it becomes $(T_S)^{ijk} = \frac{1}{3}(T^{ijk} + T^{jik} + T^{kij})$. Thank you :)) – Kaan Güven Jul 31 '22 at 15:21
  • Yes. Something similar happens for antisymmetrisation. – Michael Albanese Jul 31 '22 at 15:25

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