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Take $\mathbf{e}={\mathbf{e}}_x$ and the surface normal ${\mathbf{e}}_z$. Under reflection in the free surface, polar vectors transform by $M=\mathrm{diag}(1,1,-1)$ and an axial torque transforms by $\det (M)M$. A parallel torque therefore reverses sign, while the separation direction ${\mathbf{e}}_x$ does not. The free-surface image of a rotlet dipole is a dipole of strength $-D$ at $z=+h$.

The observation point at the cell is directly below the image, so $\mathbf{e}\cdot \mathbf{r}=0$ and the image velocity vanishes. A symmetry proof of the vanishing normal drift is also useful: reflection in the vertical plane containing $\mathbf{e}$ reverses the axial dipole moment, preserves the normal component of a polar velocity, and leaves the geometry unchanged. Linearity would make that same component change sign with $D$, forcing it to be zero. Thus this singularity alone produces no attraction or repulsion.

Write $X,Y,Z$ relative to the image, so the cell is at $(0,0,-2h)$. The image field isAlthough its value at the cell is zero, its velocity gradient is not. In particular, the surface-induced yaw of a rotlet dipole hasA spherical torque-free body rotates with the local fluid angular velocity ${\Omega }_z={\omega }_z/2$. Its swimming direction therefore turns in the surface plane and, at fixed height and speed, it traces a circle with radius $V_{\mathrm{swim}}/|{\Omega }_z|$. The sign of $D$ sets the sense of turning in this convention. An elongated swimmer also responds to strain; its rotation coefficient depends on its shape and need not equal one half of the vorticity. When the stresslet is included, its changing height produces a changing turning rate rather than an exactly fixed-radius circle.