I have a confusion regarding a calculation given below : $$ \begin{split} \int_M C^{ij}\nabla_i f \nabla_j f d\mu & = \frac{1}{3}\int_M g^{ij}g_{ij} C^{ij}\nabla_i f \nabla_j f d\mu \\ &= \frac{1}{3}\int_M g_{ij}C^{ij}g^{ij}\nabla_i f \nabla_j f d\mu \\ &= \frac{1}{3}\int_M g_{ij}C^{ij}\nabla f^2 d\mu =0 \end{split} $$ since the Cotton tensor is traceless. This is done in dimensions three. I think this calculation is erroneous. Can we insert $g^{ij}g_{ij} $ inside?
$\begingroup$
$\endgroup$
2

$\begingroup$ In its current form the formulas are at best ambiguous. Perhaps you should write everything out explicitly without using any summation conventions. $\endgroup$– Deane YangCommented Jul 17, 2020 at 17:23

$\begingroup$ In fact, you are misusing the summation convention. You are using the $ij$pair twice and then crossbinding them incorrectly. The first term on the righthand side under the integral would be more correctly written as $g^{kl}g_{kl}C^{ij}\nabla_if\nabla_jf$, but then in the next line you using it as though it were $g^{kl}g_{ij}C^{ij}\nabla_kf\nabla_lf$, but it is clear that these two quantities are not at all the same. Reusing indices that are already bound in a summation is a misuse of the summation convention, and it can easily lead (as it does here) to incorrect formulae. $\endgroup$– Robert BryantCommented Jul 23, 2020 at 16:12
Add a comment
