With the definition of the tensor \begin{equation} J_{ij} = I_{ij} - \tfrac{1}{3}\delta_{ij}I^{k}_{k}. \end{equation} I have seen the quantity \begin{equation} J_{ij}J_{ij} \end{equation} written as \begin{equation} I_{ij}I_{ij} - \tfrac{1}{3}I_{ii}I_{jj}. \end{equation} How is this possible?
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Let's define $I=I^k_{~k}$ to make things look nicer. We have: $$J_{ij}J^{ij}=(I_{ij}-\frac{1}{3}\delta_{ij}I)(I^{ij}-\frac{1}{3}\delta^{ij}I)$$ $$=I_{ij}I^{ij}-\frac{1}{3}I^{ij}\delta_{ij}I-\frac{1}{3}I_{ij}\delta^{ij}I+\frac{1}{9}\delta_{ij}\delta ^{ij}I^2$$ The second two terms are equal (they're just scalars), so: $$J_{ij}J^{ij}=I_{ij}I^{ij}-\frac{2}{3}I^{ij}\delta_{ij}I+\frac{1}{9}\delta_{ij}\delta ^{ij}I^2$$ Now, $\delta_{ij}\delta ^{ij}=D$ where $D$ is the number of dimensions of the manifold. I assume we're using a 3D manifold, so $\delta_{ij}\delta ^{ij}=3$. This simplifies the above to: $$J_{ij}J^{ij}=I_{ij}I^{ij}-\frac{2}{3}I^{ij}\delta_{ij}I+\frac{1}{3}I^2$$ Since apparently the metric is Euclidean (as was discussed in the comments above), $g_{ij}=\delta_{ij}$ and therefore $I^{ij}\delta_{ij}=I^i_{~i}=I$. So everything reduces to: $$J^{ij}J_{ij}=I^{ij}I_{ij}-\frac{1}{3}I^2$$ |
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Just square and work out the different contributions: $$\left(\frac{1}{3}\delta_{ij}I_{kk}\right)^2 = \frac{1}{9} \delta_{ij}^2 I_{kk}^2 = \frac{1}{3} I_{ii}I_{jj}$$ since $\delta_{ij}^2 = 3.$ The cross terms yield $$-\frac{2}{3}\delta_{ij}I_{ij}I_{kk} = -\frac{2}{3} I_{ii}I_{jj}.$$ Voilà! |
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