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I would like to show the following (we are in 2D, we apply Einstein's notation, and the norm is defined by $|n|=n_in_i$)

$$\mathbf{n} = n_x \mathbf{e}_x+n_y\mathbf{e}_y$$

$$(\nabla\cdot\mathbf{n})^{2}+(\mathbf{n}\times\nabla\times\mathbf{n})^{2}+(\mathbf{n}\cdot\nabla\times \mathbf{n})^{2}=|\nabla \mathbf{n}|^2:=\sum_{i=1}^2\sum_{j=1}^2(\partial_i n_j)^2$$

where $|\mathbf{n}|=1$, $\nabla\times\bf n=\partial_x n_y-\partial_y n_x$ and I didn't use Einstein's notation for the last term, in order to emphasise how it is to be intended.

what would be the quickest way possible, using Einstein's index notations or ...?

usumdelphini
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  • Is $|\mathbf{n}| = 1$? – user66081 Jul 27 '16 at 12:08
  • Yes, sorry. I've fixed it. – usumdelphini Jul 27 '16 at 12:10
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    What's the difference between the left-most term and the right-hand side? – joriki Jul 27 '16 at 12:13
  • in the right hand side there are the squares of the mixed derivatives, e.g., $(\partial_x n_y)^2+...$, whereas in the left-most term there are only $\partial_i n_i\partial_j n_j$ terms. – usumdelphini Jul 27 '16 at 12:20
  • I don't understand how $(\partial_x n_y)^2$ arises. Could you spell out the right-hand side explicitly? What's $\nabla\mathbf n$? (Apparently not $\nabla\cdot\mathbf n$?) What norm is being applied? – joriki Jul 27 '16 at 12:52
  • @joriki I removed my original response to you because it was inaccurate. I think now your confusion arises, as mine did, from the fact that $\bf n$ represents a vector. This should help. –  Jul 27 '16 at 17:08
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    So the answer to my question what norm is being applied seems to be: the Frobenius norm. That's even less self-evident than using $\nabla\mathbf n$ to denote a matrix rather than the divergence of $\mathbf n$. – joriki Jul 27 '16 at 17:55
  • I apologise for this, I didn't specify it because it is common practice in the physics literature to use these notations without explanation. But I do agree with you, out of context they should be explained. – usumdelphini Jul 27 '16 at 18:09
  • in 2d, what do you mean by $\nabla \times \mathbf{n}$? – user66081 Jul 27 '16 at 21:34
  • $\partial_xn_y-\partial_yn_x$, a scalar. – usumdelphini Jul 28 '16 at 07:53

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