domain branching. The resulting scaling, proportional to $\varepsilon^{4/5}$, differs from the one derived for the scalar problem. This difference can be understood as a consequence of the degeneracy of the rank-one connection (in the same sense as for double roots of polynomials). To see this, let us consider a typical branching pattern, as illustrated in the top right part of Figure \[fig:PD2\], which consists of a fine-scale mixture of regions with $Du$ close to $\operatorname{SO}(2)A_j$ and regions with $Du$ close to $\operatorname{SO}(2)B_j$, with interfaces approximately horizontal. Since $u$ does not jump across the interface, if $u$ is sufficiently regular, the tangent component of the gradient does not jump either. In other words, given a vector $v\in S^1$ tangent to the interface, then the directional derivative $Duv$ must be the same on the two sides of the interface. If $Du\in \operatorname{SO}(2)A_j$ on one side, and $Du\in \operatorname{SO}(2)B_j$ on the other side, then necessarily $|A_jv|=|B_jv|$. For $j=1$, this equation is equivalent to $$(v_1-\alpha v_2)^2+v_2^2 = 
  (v_1+\alpha v_2)^2+v_2^2 \,,$$ which simplifies to $\alpha v_1v_2=0$. For $j=2$ the condition $|A_jv|=|B_jv|$ is equivalent to $$v_1^2 + (1-\alpha)^2 v_2^2
=  v_1^2 + (1+\alpha)^2 v_2^2\,,$$ which simplifies to $\alpha v_2^2=0$. It is easy to see that $v_2=0$ 