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Is the following true that if I have a symmetric 4-th order tensor $A_{ijkl}$? $$ A_{ijkl}=A_{klij}\\ A_{ijkl}(\delta_{mk}+\delta_{ml}-\delta_{mi}-\delta_{mj})= \\ A_{ijml}+A_{ijkm}-A_{mjkl}-A_{imkl}=0 $$ Thanks in advance!

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    What does $A_{ijkl}$ mean ? A matrix is usually indexed by two indices, not four. – 2017-01-10
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    Included what $A_{ijkl}$ is. – 2017-01-10
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    $A_{ijkl}$ is not a matrix but a 4-th order tensor. – 2017-01-10
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    Yes that is a more correct description of what $A_{ijkl}$ is. – 2017-01-10
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    What does $\delta$ mean? How the multplication $A_{ijkl}\delta_{mk}= A_{ijml}$ is defined? – 2017-01-10
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    $\delta$ is the Kronecker delta $\delta_{mk}=1$ if $m=k$, zero otherwise. – 2017-01-10
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    Yes, it is correct. The rule is simply swap out the common index that appears on the kronecker delta and the contracted term with the other index on the Kroncecker delta. – 2017-01-10

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Yes, it is correct. The rule is to simply swap out the common index that appears on the Kronecker delta and the contracted term with the other index on the Kronecker delta.

Eg. your first term $$ A_{ijkl}\delta_{mk} = A_{ijml} $$ The reason is that (assuming $k$ to run from 1 to 3,)

$$ A_{ijkl}\delta_{mk} = A_{ij1l}\delta_{m1} + A_{ij2l}\delta_{m2} + A_{ij3l}\delta_{m3} $$ Now only one of the three $\delta_{m}$ terms can be non 0. For example if $m=2$, $$ A_{ijkl}\delta_{2k} = A_{ij2l}\delta_{22} = A_{ij2l} $$ and so on.

So what sits in place of the contracted index k in $A_{ijkl}$ is precisely m.