Reciprocal vectors of a two dimensional lattice without borrowing from the third dimension

linear algebravectors

For a three dimensional lattice defined by $\bf a, \bf b, \bf c$ the reciprocal vectors (from this answer) are:

$$
{\bf A} = \frac{{\bf b}\times {\bf c}}{{\bf a}\cdot({\bf b}\times {\bf c})}
\quad
{\bf B} = \frac{{\bf c}\times {\bf a}}{{\bf b}\cdot({\bf c}\times {\bf a})}
\quad
{\bf C} = \frac{{\bf a}\times {\bf b}}{{\bf c}\cdot({\bf a}\times {\bf b})}.
$$

Calculating the diffraction of waves (e.g. electrons or X-rays) by a periodic two-dimensional surface I calculate the in-plane reciprocal vectors $\bf a, \bf b$ using these by "temporarily inventing" a third periodic direction, setting $\bf c$ to $\bf \hat z$ i.e. (0, 0, 1), then throwing away the third reciprocal vector afterward.

For some specific special cases of 2D lattices I cheat and just use for the magnitudes:

$$A, \ B = \frac{1}{a}, \ \frac{1}{b}; \ \ \gamma = \frac{\pi}{2} \text{ (rectangular)}$$

and

$$A, \ B = \frac{2}{\sqrt{3}a}; \ \ b=a,\ \ \gamma = \frac{\pi}{3} \text{ (hexagonal)}$$

and in both cases:

$$\bf a \perp \bf B, \ \bf b \perp \bf A$$

without having to resort to "borrowing from" the third dimension. But in this case I still have to stop and think about the direction of those two i.e. on which side of $\bf A$ to draw $\bf B$ because perpendicularity offers two choices, twice.

This answer to How much like reciprocals are reciprocal vectors? Is there a matrix division that allows $\mathbf{A} = 1 / \mathbf{a}$ in three or two dimensions? answers "No" for three dimensions. For two dimensions I don't know if it says "no" or "it depends…"

Question: Is there a way to calculate reciprocal vectors of a two dimensional lattice without borrowing from the third dimension?

Best Answer

Yes, the reciprocal vectors (also called "dual basis") to any list of vectors $v_1,\ldots,v_n$ in $\mathbb R^n$ can be computed by inverting and taking transpose of the $n$-by-$n$ matrix $M$ whose rows are those vectors. That is, the columns $w_j$ of $M^{-1}$ are the reciprocal/dual vectors. After all, the $i,j$ entry of $M\cdot M^{-1}$ is $v_i\cdot w_j$, and is $0$ for $i\not=j$ and $1$ for $i=j$.

In general, there are various formulas expressing $M^{-1}$ in terms of the entries of $M$, "minors" of $M$, and so on.

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