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With the definition of the tensor:

\begin{equation} J_{ij} = I_{ij} - \tfrac{1}{3}\delta_{ij}I^{k}_{k}, \qquad i,j,k\in\{1,2,3\}, \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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You're using rather confusing notation. Are you summing over i and j? The repeated lower indices are pretty ambiguous. – jld Mar 12 '13 at 21:59
Some sources use $J^{ij}J_{ij}$ and some use $J_{ij}J_{ij}$ to represent the same thing. I don't understand why. – user12345 Mar 12 '13 at 22:06
In Euclidean space, it's not ambiguous and of course summation is implied. – Vibert Mar 12 '13 at 22:08
But it's only unambiguous when the metric is Euclidean, so getting into the habit of writing everything with lower indices is a bad idea. It also looks sloppier. – jld Mar 12 '13 at 22:11
Yes. Flat 3D space is usually called 'Euclidean.' – jld Mar 13 '13 at 18:39
up vote 4 down vote accepted

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:


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:


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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

$$-\frac{2}{3}\delta_{ij}I_{ij}I_{kk} = -\frac{2}{3} I_{ii}I_{jj}.$$


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Why do you have $I_{kk}$? – user12345 Mar 12 '13 at 22:14
@user16307 The difference between upper and lower indices vanishes for using $(x,y,z)$ coordinates (and pretty much nothing else, as mentioned elsewhere), so $I_{kk}$ is another way of writing $I^k_k$ in this special case. So as long as your source has all $i$'s and $j$'s lowered (and is therefore ruining a beautiful aspect of this notation, imo), consistency dictates you should use $I_{kk}$ rather than $I^k_k$ (or $I^k{}_k$), as Vibert has done. – Chris White Mar 13 '13 at 0:34

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