Differentiation of tensor product

differential-geometryindex-notationkronecker producttensor-productstensors

I have a tensor equation

$$\frac{\partial A_{ij}}{\partial B_{kl}}=\frac{\partial A_{ij}}{\partial C_{pq}}\frac{\partial C_{pq}}{\partial B_{kl}}
$$
$C_{pq}$ can be written as $C_{pq}=B_{pq}+aB_{mm}\delta _{pq}$, where $a$ is scalar.
So I get,

$$\frac{\partial A_{ij}}{\partial B_{kl}}=\frac{\partial A_{ij}}{\partial C_{pq}} \bigg( \frac{\partial B_{pq}}{\partial B_{kl}}+a\frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}} \bigg)$$

Question 1: What is result of $$\frac{\partial B_{pq}}{\partial B_{kl}} \quad \textrm{and} \quad \frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}$$

Looks like it is
$$\frac{\partial B_{pq}}{\partial B_{kl}} = \frac{1}{2}(\delta _{pk} \delta _{ql} + \delta _{pl} \delta _{qk}) $$

and

$$
\frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}= \delta _{pq} \delta _{kl}
$$
Can someone comment whether above 2 results are correct and how to prove them?

After some simplification,
$$\frac{\partial A_{ij}}{\partial C_{pq}}= \delta _{ip} \delta _{jq}$$
(This result is got after putting values of tensors and not by itself.)

Question 2: So, based on results above, is the following product correct ?

$$(\delta _{ip} \delta _{jq})(\delta _{pq} \delta _{kl})=\delta _{iq} \delta _{jq} \delta _{kl}$$

The left hand side has $i,j,k,l$ as free indices and above equation also has them with $q$ summed over index.

Thanks in advance.

Best Answer

Looks like it is $$\frac{\partial B_{pq}}{\partial B_{kl}} = \frac{1}{2}(\delta _{pk} \delta _{ql} + \delta _{pl} \delta _{qk}) $$ and $$ \frac{\partial (B_{mm} \delta_{pq})}{\partial B_{kl}}= \delta _{pq} \delta _{kl} $$

Partial derivatives are only well-defined when you specify what is held constant. We usually adopt a convention obtaining $$\frac{\partial B_{pq}}{\partial B_{kl}}=\delta_{pk}\delta_{ql}$$for general rank-$2$ tensors and $$\frac{\partial B_{pq}}{\partial B_{kl}}=\frac12(\delta_{pk}\delta_{ql}+\delta_{pl}\delta_{qk})$$for symmetric rank-$2$ tensors. In either case, your second result follows by contracting the repeated index $m$.

based on results above, is the following product correct ? $$(\delta _{ip} \delta _{jq})(\delta _{pq} \delta _{kl})=\delta _{iq}\delta _{jq} \delta _{kl}$$ The left hand side has $i,j,k,l$ as free indices and above equation also has them with $q$ summed over index.

That follows if you contract $p$. I assume you're using it in $$\frac{\partial A_{ij}}{\partial B_{kl}}=\delta_{ip}\delta_{jq}\left(\frac{\delta_{pk}\delta_{ql}+\delta_{pl}\delta_{qk}}{2}+a\delta_{pq}\delta_{kl}\right)=\frac{\delta_{ik}\delta_{jl}+\delta_{il}\delta_{jk}}{2}+a\delta_{ij}\delta_{kl}.$$

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