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$T^i_{~~jkl}$ is a tensor such that $T^i_{~~jkl}=3T^i_{~~ljk}$ is some coordinate system. Prove that $T^i_{~~jkl}=3T^i_{~~ljk}$ in all coordinate systems.

The given answer says:

\begin{align} \bar T^i_{~~jkl}-\underbrace{3\bar T^i_{~~ljk}}_{(*)}&= T^p_{~~rst}\frac{\partial \bar x^i}{\partial x^p}\frac{\partial x^r}{\partial \bar x^j}\frac{\partial x^s}{\partial \bar x^k}\frac{\partial x^t}{\partial \bar x^l}-\underbrace{3T^p_{~~rst}\frac{\partial \bar x^i}{\partial x^p}\frac{\partial x^r}{\partial \bar x^l}\frac{\partial x^s}{\partial \bar x^j}\frac{\partial x^t}{\partial \bar x^k}}_{(**)}\\ &=T^p_{~~rst}\frac{\partial \bar x^i}{\partial x^p}\frac{\partial x^r}{\partial \bar x^j}\frac{\partial x^s}{\partial \bar x^k}\frac{\partial x^t}{\partial \bar x^l}-\underbrace{3T^p_{~~trs}\frac{\partial \bar x^i}{\partial x^p}\frac{\partial x^t}{\partial \bar x^l}\frac{\partial x^r}{\partial \bar x^j}\frac{\partial x^s}{\partial \bar x^k}}_{(***)}\\ &=(T^p_{~~rst}-3T^p_{~~trs})\frac{\partial \bar x^i}{\partial x^p}\frac{\partial x^r}{\partial \bar x^j}\frac{\partial x^s}{\partial \bar x^k}\frac{\partial x^t}{\partial \bar x^l}=0 \end{align}

I would like to know how I can see that $(*)=(**)=(***)$. I've been staring at this for quite a while now.

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  • $\begingroup$ This question is a bit daft because in fact $T$ has to be $0$. This is because $T^i_{jkl}=3T^i_{klj}=9T^i_{ljk}=27T^i_{jkl}$ and so $26T^i_{jkl}=0$ $\endgroup$ Jun 21, 2016 at 1:40

1 Answer 1

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  • ($*$) $\to$ ($**$) is using the transformation law for tensors
  • ($**$) $\to$ ($***$) is just changing the names of the summation labels ($rst$) $\to$ ($trs$). This is just like saying that $\sum_{i=1}^n i = \sum_{j=1}^n j$.
  • ($***$) $\to$ the final result follows by using $T^i_{jkl} = 3T^i_{ljk}$.

The answer to this problem is really just a combination of two small facts: 1) if $T$ is a tensor then $A^{i}_{jkl} \equiv T^{i}_{jkl} - 3T^i_{ljk}$ is also a tensor and 2) If a tensor $A^{i}_{jkl} = 0$ in one coordinate system then it vanishes in all coordinate systems. The first fact is a direct consequence of the definition of a tensor and the second fact is a direct consequence of the transformation law.

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