Is the Curl $\nabla \times$ symmetric under inner product？

Question

If $\nabla\cdot u=0,$ and $\nabla \times(\nabla \times u)=\nabla(\nabla \cdot u)-\nabla^2 u,$ we know that $-\Delta u=\nabla \times(\nabla \times u).$ But I have a question, why $$(-\Delta u,v)_{L^2}=(\nabla \times u,\nabla \times v)_{L^2}?$$ Is the Curl $\nabla \times$ symmetric under inner product？

Answer 1

It comes from the divergence of the cross product identity in the following way: \begin{eqnarray} \nabla\cdot[v\times (\nabla\times u)] &=& (\nabla\times v)\cdot(\nabla\times u) -v\cdot[\nabla\times(\nabla\times u)] \\&=& (\nabla\times v)\cdot(\nabla\times u) + v\cdot\Delta u-v\cdot\nabla(\nabla\cdot u) \end{eqnarray} Integrating over all space makes the LHS vanish by the divergence theorem and $\nabla\cdot u$ vanishes by assumption, leaving $$ (\Delta u, v) + (\nabla\times u, \nabla\times v) = 0 $$ and the result follows.