Rank-one connection ($A-B=a\otimes N$)

Two constant deformation gradients are rank-one connected when their difference is an outer product $a\otimes N$. For a nonzero jump its matrix rank is one; equality $A=B$ is the trivial compatible case. The vector $N$ specifies a reference interface normal, and $a$ is the jump amplitude. The Hadamard jump condition identifies this algebraic condition with continuity of an affine deformation across a planar interface. The normal can be normalized by rescaling $a$ without changing the product.

The unrotated variants differ by $2\delta e_3\otimes e_1$, giving normal $e_1$. The rotation about $e_2$ with $\cos \theta =(1-{\delta }^2)/(1+{\delta }^2)$ and $\sin \theta =2\delta /(1+{\delta }^2)$ gives $U_{+}-\overline{R}{U}_{-}=[2\delta /(1+{\delta }^2)](e_1+\delta e_3)\otimes e_3$, giving the second normal $e_3$. If these are the only two compatible rotations, the only gradients in $SO(3)U_{+}$ compatible with $\overline{R}{U}_{-}$ are $U_{+}$ and $\overline{R}{U}_{+}$. Left multiplication by a rotation preserves rank, so the second classification follows directly from the first.

Suppose an affine deformation on each side of a plane with unit normal $N$ is continuous across it. Tangential derivatives agree, so $(A-B)t=0$ for every $t\perp N$. Decomposition of an arbitrary vector into its tangent component and $(x\cdot N)N$ then gives $(A-B)x=((A-B)N)(N\cdot x)$, proving the rank-one form. Conversely that form makes the two affine maps agree on the plane when their constants agree there.

Orientation preservation on both sides also requires $1-N\cdot A^{-1}a>0$: the matrix determinant lemma gives $\det B=\det A(1-N\cdot A^{-1}a)$. This inequality is additional geometric information supplied by the determinant signs, not by continuity alone.