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Use the paper's velocity gradient convention $A_{ij}={\partial }_i{v}_j$, so $\Omega =(A-A^T)/2$ and $D=(A+A^T)/2$. In a rigid motion there is rotation but no rate-of-strain tensor: $D=0$. A polar order parameter attached to the material must rotate exactly with it. In this convention its material derivative is $-\Omega \mathbf{p}$, and its corotational derivative of a polar vector must vanish.

Thus the compensating term is $+\Omega \mathbf{p}$ with coefficient exactly one. This follows from an objective time derivative and is independent of material properties. By contrast, the response to $D\mathbf{p}$ describes deformation and flow alignment of a polar order parameter; its coefficient $\xi$ can depend on the material. The transposed velocity gradient convention is essential to the sign of $\Omega$.

Choose unequal positive numbers $a,b$. The planar ellipseis a simple closed curve and is not a circle. The nonidentity proper rigid motionis rotation through $\pi$ about the $z$-axis, and $R\gamma (t)=\gamma (t+\pi )$. It therefore preserves the curve as a set.

The tangent vectors ofareHence the first fundamental form has coefficientsPutThe upward unit normal is $N=(-f_u,-f_v,1)/W$, so the second fundamental form has coefficientsThus the two forms are

The Gaussian curvature isSince $EG-F^2=W^2$, the graph formula isas in Gaussian curvature of a graph surface.

For the final claim, fix a point of $\gamma$ and make a rigid motion taking $P$ to the plane $z=0$. The common tangent plane is horizontal, so locally $\Sigma$ is the graph $z=f(u,v)$. Along the projected curve $c(s)$ of tangency,Differentiating the second identity givesBecause $c$ is a smooth curve, $c^{\prime }(s)\ne 0$, so the Hessian is singular. Its determinant is zero, and the graph formula gives $K=0$ at every point of $\gamma$. This is tangency to a plane along a curve forces zero Gaussian curvature.

The relation isso $K$ is the Klein bottle group. Let it act on ${\mathbb{R}}^2$ byThese are Euclidean isometries and satisfy $ab{a}^{-1}=b^{-1}$. Every element has a normal form $b^m{a}^n$. The orbit of $(0,0)$ is discrete, and a rectangle of finite size meets every orbit, so the action is proper and cocompact.

The squareis a vertical translation, while $b$ is a horizontal translation. They commute, andas an isometry only when $m=n=0$. Hence $⟨a^2,b⟩\cong {\mathbb{Z}}^2$. The normal form shows that every element lies in either $⟨a^2,b⟩$ or $a⟨a^2,b⟩$, so this subgroup has index two in $K$.