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Let $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$ and $\hat{\mathbf{r}}=\mathbf{r}/r$. Its Jacobian isThe chain rule therefore yieldsTaking its cross product with $\mathbf{r}$ kills the radial term and proves that spherical inversion preserves angular derivatives:This is also the invariance of the differential expression underlying the angular momentum operator.

A scalar spherical harmonic $Y$ depends only on direction, so $Y(\hat{\mathbf{r}})=Y(\widehat{\overline{\mathbf{r}}})$ and its gradient is tangential. Thus $\hat{\mathbf{r}}\cdot {\nabla }_{\overline{\mathbf{r}}}Y=0$, andEach of the three normalized vector spherical harmonic families is consequently unchanged as an angular basis function:The normalization requires $n\ge 1$ for $\mathbf{B}$ and $\mathbf{C}$; at $n=0$ the gradient vanishes and only $\mathbf{P}$ remains.

This substitution into the basis functions is distinct from the pushforward of a vector field of a physical vector field by inversion. Under that Jacobian map the radial vector picks up $-a^2/r^2$ and the two tangential vectors pick up $+a^2/r^2$. Likewise, the weighted Kelvin transform of a scalar includes the extra factor $a/r$. Neither operation changes the angular differential identity just proved.

For $\overline{\mathbf{r}}=a^2\mathbf{r}/r^2$, the chain rule gives ${\nabla }_{\mathbf{r}}=(a^2/r^2)(I-2\hat{\mathbf{r}}{\hat{\mathbf{r}}}^T){\nabla }_{\overline{\mathbf{r}}}$. The radial term is killed by crossing with $\mathbf{r}$, giving the identity. Purely angular vector spherical harmonics are therefore unchanged as angular basis functions. This is distinct from pushing forward a physical vector field by the inversion Jacobian.