Reciprocal basis

For a basis $e_1,e_2,e_3$ with scalar triple product $\Delta$, the reciprocal vectors are cyclic cross products divided by $\Delta$ and satisfy $f_i\cdot e_j={\delta }_{ij}$. This construction is used for the basis of a reciprocal lattice.

In a real finite-dimensional inner product space, let two bases satisfy $⟨u_i,w_j⟩={\delta }_{ij}$. Taking inner products with $u_i$ in $v={\sum }_j{b}_j{w}_j$ gives $b_i=⟨u_i,v⟩$. If also $v={\sum }_j{a}_j{u}_j$, then $b_i={\sum }_j⟨u_i,u_j⟩a_j$. Thus the Gram matrix $G_U$ converts the original coordinates into reciprocal basis coordinates, and $G_W=G_U^{-1}$ converts back. With Euclidean Cartesian column basis matrices $U,W$, reciprocity is $U^{T}W=I$, hence $W=U^{-T}$ and $W^{-1}U=U^{T}U=G_U$. This is why dot products with nonorthogonal basis vectors are coordinates in the reciprocal basis, rather than coordinates in the original basis.

For a three-dimensional basis with nonzero scalar triple product $\Delta$, its reciprocal basis satisfies ${\hat{e}}_2\times {\hat{e}}_3=e_1/\Delta$ and the cyclic analogues. Consequently its scalar triple product is $1/\Delta$, and taking the reciprocal basis again recovers the original basis. The coordinates of $V$ in the original basis are $V\cdot {\hat{e}}_i$; in the reciprocal basis they are $V\cdot e_i$.

For a real basis matrix $E$ with basis vectors as rows, the reciprocal basis has row matrix $E^{-T}$. If either basis is orthonormal, both are, and $E^{-T}=E$. Thus $E$ is an orthogonal matrix, with determinant either $1$ or $-1$. It is a rotation matrix precisely when the original ordered basis is positively oriented relative to the reference coordinates. For example, $\mathrm{diag}(1,1,-1)$ is its own orthonormal reciprocal basis matrix but is an improper orthogonal transformation.