Planarity test across a cubic-to-tetragonal twin ($p=(p_1,p_2,0),\Pi =p^{\perp }\text{ remains plane}⟺p\parallel N$)

Let $B=\mathrm{diag}(b,a,a)$, with $a,b>0$, $a\ne b$, and $A=R\mathrm{diag}(a,b,a)$ be a twin-related gradient with $\cos \theta =2ab/(a^2+b^2)$, $\sin \theta =\kappa (b^2-a^2)/(a^2+b^2)$, $\kappa =\pm 1$. Their jump is $(b^2-a^2)(-b,\kappa a,0)\otimes (1,\kappa ,0)/(a^2+b^2)$.

A plane through the origin with normal $p=(p_1,p_2,0)$ is spanned by $e_3$ and $t=(-p_2,p_1,0)$. Both gradients map $e_3$ to the same direction, so the two image half-planes are coplanar exactly when $At$ and $Bt$ are parallel in their transverse section. Direct calculation givesIts nonzero prefactor makes planarity equivalent to $p\parallel (1,\kappa ,0)$, the twin normal. On that plane the two affine maps agree, proving sufficiency as well. The restriction on the third normal component matters; this statement is not a classification of every possible material plane.