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For $r=|x|>0$, the gradient satisfies ${\partial }_{i}r=x_i/r$. The product rule and Einstein notation then give the displayed divergence formula in dimension $n$. In dimension three, its derivative matrix is symmetric, so the curl vanishes. Zero divergence forces $f(r)=C{r}^{-n}$ on each positive radial interval. This formula does not assert regularity at the origin.

Away from the origin, differentiate $r^2=x_j{x}_j$ using Einstein notation:HenceThis gradient is the radial unit vector. Division by $r$ is legitimate on the punctured domain; the distance function is not differentiable at the origin.

In Einstein notation, $(\mathbf{n}\cdot \mathbf{x})n_i=n_i{n}_j{x}_j$, so the matrix and, for $\alpha \ne 0$, its matrix inverse are

Using Einstein notation andwe obtainThus

The scalar triple product isPut $\Delta =[a,b,c]$. When $\Delta \ne 0$, the three vectorsform the reciprocal basis to $a,b,c$, so $e_i^{\prime }\cdot e_j={\delta }_{ij}$. Their scalar triple product is $1/\Delta$. Thereforesince the cyclic permutation preserves orientation. The identity also holds when $\Delta =0$, either by continuity or directly because the cross products are then linearly dependent.

For the given basis $e_1,e_2,e_3$, the same identities giveIf ${\sum }_i{\alpha }_i{e}_i^{\prime }=0$, dotting with $e_j$ gives ${\alpha }_j=0$, so the $e_i^{\prime }$ are linearly independent and hence form a basis. Moreover, $e_i\cdot e_j^{\prime }={\delta }_{ij}$, so the original basis is reciprocal to the primed basis. Uniqueness of a reciprocal basis gives

Every vector $K$ has a unique expansion $K={\sum }_i{\kappa }_i{e}_i^{\prime }$, and thenThis is an integer for every integer triple $(n_1,n_2,n_3)$ exactly when every ${\kappa }_i$ is an integer. Hence all such points areThey form the reciprocal lattice in the convention without a factor of $2\pi$.

In Einstein notation,This is

Similarly,after cyclically relabelling the first term and swapping two indices in the second. Hence