A primitive cell of a crystal lattice is a set $A$ such that two copies of $A$ which are translated by a lattice vector do not overlap and such that $A$ tiles the entire crystal.

Some primitive cells in an scc grid

I have read (for example in the german “Festkörperphysik” by Gross, Marx), that all primitive cells have the same size/volume.

Intuitively, this seems plausible, but is there a proof?

My precise, measure theoretic, interpretation of this statement is: If $a_1, \ldots, a_n$ is a basis of $\mathbf{R}^n$ and $A, B \subset \mathbf{R}^n$ are sets such that $(\cup (A+\alpha_1 a_1+\ldots+\alpha_n a_n))C$ and $(\mathbf{R}^n \cup (B+\alpha_1 a_1+\ldots+\alpha_n a_n))^C$ (where the union is over all $\alpha_1, \ldots,\alpha_n ∈ \mathbf{Z}$) are Lebesgue null sets and such that for all $\alpha_1,\ldots,\alpha_n∈\mathbf{Z}$: $(A+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap A$ and $(B+\alpha_1 a_1 + \ldots \alpha_n a_n) \cap B$ are Lebesgue null sets, then $A$ and $B$ have the same Lebesgue measure.

  • $\begingroup$ How formal should the answer be? If you are only interested in simply connected primitive cells (maybe with smooth boundaries), i.e. what physicists care about, the answer is very simple and intuitive. $\endgroup$ Jan 28, 2015 at 11:37
  • $\begingroup$ @NorbertSchuch I was looking for an answer like lemon's, but I would also be interested in a more intuitive answer if you can provide one. $\endgroup$ Jan 28, 2015 at 13:37

2 Answers 2


Let $\mathcal{L}\subseteq\mathbb{R}^n$ be a lattice with a basis $B\in\mathcal{R}^{n\times n}$ and $F\subseteq\text{span}(\mathcal{L})$ be measurable. $F$ tiles $\mathcal{L}$ iff

  1. $(x+F)\cap(y+F)=\emptyset\,\forall x\neq y\in\mathcal{L}$, and
  2. $\mathcal{L}+F=\text{span}(\mathcal{L})$

It is trivial to show (I'll leave it as an exercise) that 1. implies: $$ |(\mathcal{L}+x)\cap F|\leq 1 $$ while 2. implies $$ |(\mathcal{L}+x)\cap F|\geq 1 $$ and therefore $$ |(\mathcal{L}+x)\cap F|= 1 $$

for $x\in\text{span}(\mathcal{L})$. Then we have

\begin{align} \text{vol}(F) &= \int_{\mathbb{R}^n} 1_F(x)dx \\ &= \int_{B[0,1)^n}\sum_{y\in\mathcal{L}} 1_F(x+y)dx\\ &=\int_{B[0,1)^n}|(\mathcal{L}+x)\cap F|dx=\text{vol}(B[0,1)^n) \end{align}


The quotient forming map $\Bbb R^n\to\Bbb R^n/\Lambda$ is a local isometry (as translations by elements of the lattice $\Lambda$ are isometries without fixed points) onto a torus, whose volume is equal to the (absolute value of the) determinant of the basis of $\Lambda$.

The preimage of each point of the torus has exactly one point in the primitive cell, by definition of the primitive cell, so (as soon as the primitive cell has a well-defined volume) its volume must be equal to that of the torus.


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