= Reciprocal crystallographic metric matrix
{title2=$G^*=G^{-1}$}
The \b[reciprocal crystallographic metric matrix] is the <Gram matrix> of the <reciprocal lattice> basis. With direct-basis matrix $A=(\mathbf a\ \mathbf b\ \mathbf c)$ and the convention $\mathbf a_i^*\cdot\mathbf a_j=\delta_{ij}$, the reciprocal-basis matrix is $A^*=A^{-T}$. Consequently,
$$
G^*=(A^*)^TA^*=A^{-1}A^{-T}=(A^TA)^{-1}=G^{-1},
$$
where $G$ is the direct <crystallographic metric matrix>. If direct lengths are in angstroms, the reciprocal metric entries have units of inverse squared angstroms.<Metric tensor in the IUCr Online Dictionary of Crystallography>[Definition, inverse metric and reciprocal-basis components]{ref}
For an original calculation, take
$$
G=\begin{pmatrix}4&3&0\\3&9&0\\0&0&16\end{pmatrix}
\,\mathrm{\mathring A^2},\qquad
G^*=\begin{pmatrix}1/3&-1/9&0\\-1/9&4/27&0\\0&0&1/16\end{pmatrix}
\,\mathrm{\mathring A^{-2}}.
$$
For a <Miller index> column $\mathbf h$, the corresponding reciprocal vector is $\mathbf g_{\mathbf h}=A^*\mathbf h$. Its squared magnitude determines the interplanar spacing:
$$
d_{\mathbf h}^{-2}=\mathbf h^TG^*\mathbf h.
$$
For $(110)$ this gives $d^{-2}=7/27\,\mathrm{\mathring A^{-2}}$, so $d\simeq1.964\,\mathrm{\mathring A}$.
The same matrix also gives angles between plane normals. In this example,
$$
\cos\angle(\mathbf a^*,\mathbf b^*)
=\frac{-1/9}{\sqrt{(1/3)(4/27)}}=-\frac12.
$$
The reciprocal angle is120degrees while the corresponding direct-cell angle is60degrees. In an oblique cell, reciprocal directions cannot be recovered by simply inverting the three edge lengths separately.
A convention using reciprocal vectors with dot products $2\pi\delta_{ij}$ multiplies this reciprocal metric by $(2\pi)^2$. In that convention the spacing formula includes the matching factor $(2\pi)^2$ in its numerator. Stating the convention prevents an otherwise systematic length error.
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