I am trying to show that $curl\,curl\,(\mathcal E)=0$, where $\mathcal E$ represents the symmetric gradient, i.e., for vector-valued function $u(x_1,x_2)=(u_1(x_1,x_2),u_2(x_1,x_2))$ $$ \mathcal E(u)= \begin{bmatrix} \partial_1 u_1 & \frac12(\partial_2 u_1+\partial_1 u_2)\\ \frac12(\partial_2 u_1+\partial_1 u_2) & \partial_2 u_2 \end{bmatrix} $$ I know there is a formula sayinng that $$ curl\,curl\,= \nabla div - \Delta $$ and I am tryinng to apply this formula. However, I got confused that, say $\mathcal E(u)$ is a 2 by 2 matrix, so $\Delta \mathcal E(u)$ should be a 2 by 1 vector. But $div(\mathcal E(u))$ is a 2 by 1 vector and $\nabla div \mathcal E(u)$ is then a 2 by 2 matrix... so their dimension does not match... where I am wrong? Also, I tried to explicitly write down the computation but they did not cancel to 0...

update: for a reference I found, they write $\nabla div - \Delta$ as a 2 by 1 vector, but no explicit formula can be found there...

Any help is really welcome!

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    $\begingroup$ How are you defining the curl of a matrix? $\endgroup$ – Connor Harris Apr 25 '18 at 20:29
  • $\begingroup$ @ConnorHarris this is one point I confused on. I can not find a definition of curl of a matrix online... $\endgroup$ – Covepe Apr 25 '18 at 20:43
  • $\begingroup$ If $\mathcal E(u)$ is a matrix then $\Delta \mathcal E(u)$ should be too. I guess the curl is probably taken columnwise? $\endgroup$ – Anthony Carapetis Apr 26 '18 at 8:01
  • $\begingroup$ @AnthonyCarapetis how you define $\Delta \mathcal E(u)$? $\endgroup$ – Covepe Apr 26 '18 at 8:02
  • $\begingroup$ The component-wise Laplacian is the obvious choice; if this is not what you mean then you should tell us your definition. Taking it to be component-wise is the same thing as taking it to be column-wise (if you use the component-wise defintion of the Laplacian of a vector), which means the identity $\operatorname{curl}^2 =\nabla \operatorname{div} - \Delta$ will apply columnwise. $\endgroup$ – Anthony Carapetis Apr 26 '18 at 8:03

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