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I would like to know why the following ensemble :

$$\{ \overrightarrow{G} / \forall \overrightarrow{R_n}, e^{i \overrightarrow{G}.\overrightarrow{R_n}}=1\}$$

where $\overrightarrow{R_n}$ is a vector of a lattice, is a lattice ?

The justification that I have found is that it is because if $\overrightarrow{G_1}$ and $\overrightarrow{G_2}$ are in the ensemble, then $\overrightarrow{G_1}+\overrightarrow{G_2}$ is also inside but I don't understand this explanation.

Indeed for me a lattice is a set of points such as any point of the system can be written as $\overrightarrow{R}=\sum_i \alpha_i \overrightarrow{a_i}$ where $\alpha_i$ are relative integers.

Any sum of points written like this has the same form, but how is the reciprocal of this affirmation true ?

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  • $\begingroup$ If the direct lattice is a Bravais lattice, the definition of the reciprocal lattice directly implies that it, too, is a lattice. I don't quite see where your confusion arises. $\endgroup$ – Jon Custer Nov 1 '16 at 14:52
  • $\begingroup$ According to the definition I have given with the ensemble I don't see why the reciprocal lattice should be a lattice. $\endgroup$ – StarBucK Nov 1 '16 at 14:54
  • $\begingroup$ I think you are being too broad on your allowable $R$. A Bravais lattice is a discrete set of vectors (for 3D not all in a plane) closed under vector addition and subtraction. Now, it becomes clear that $G$ must also be a discrete set of vectors closed under addition and subtraction, and thus becomes a Bravais lattice itself, the reciprocal lattice. $\endgroup$ – Jon Custer Nov 1 '16 at 15:05
  • $\begingroup$ I think it's enough to come up with the reciprocal primitive lattice vectors $\vec{b}_i$ by solving the system of equations $\vec{b}_i\cdot\vec{a}_j = 2\pi \delta_{ij}$, because then it's straight-forward to show that the general form for $\vec{G}$ is as a linear combination of the $\vec{b}_i$'s. $\endgroup$ – march Nov 1 '16 at 15:41
  • $\begingroup$ I will reformulate my question : Why is an ensemble of discrete translations closed under addition a Bravais Lattice ? The definition of a Bravais Lattice for me is the wikipedia definition : en.wikipedia.org/wiki/Bravais_lattice I agree that if I have this definition then I will have an ensemble of discrete points closed under addition, but why is the reciprocal of this sentence true ? $\endgroup$ – StarBucK Nov 1 '16 at 16:00
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As other people have pointed out, the fact that the reciprocal of a lattice is a lattice follows by definition.

Definition. A lattice in $\mathbb R ^n$ is a subgroup $L$ of $ (\mathbb R^n,+)$ which is discrete.

The word "discrete" means that every point $l\in L$ has a neighbourood which doesn't contain any other point of $L$. It is a theorem (see, e.g., [1]) that every lattice has generators, that is, there exist $k$ linearly independent vectors $\lbrace \mathbf e_i,i=1,2,\dots k\rbrace $ with $1\leq k \leq n$ such that the linear combinations $$\sum _{i=1} ^k m_i \boldsymbol e _i,$$with entire coefficients $m_i$ exhaust the lattice, so that the definition you are using is actually equivalent to the above one.

In any case for the three dimensional case, you can explicitly verify the "generators definition". Let $\mathbf a_1 ,\mathbf a_2,\mathbf a _3$ be generators of the direct lattice. Then: $$\mathbf b_3=2\pi\frac{\mathbf a _1 \times \mathbf a _2}{(\mathbf a _1 \times \mathbf a _2) \cdot \mathbf a _3} \ \, \text{and cyclic permutations}$$ are generators for the direct lattice. To see this, note that this three guys belong to the reciprocal lattice, are linearly independent, and any non-entire linear combination of them doesn't belong to the reciprocal lattice.

For a nice and simple introduction to solid state physics, I suggest you [2].


[1] Arnold V.I., "Mathematical Methods of Classical Mechanics", §49, D. [2] Ashcroft N.W., Mermin D.N, "Solid State Physics"

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You can think of a lattice as a discretized vector space. In a vector space, one can define it by, say, the set described by $\Sigma a_i \mathbf{e_i}$. However, the more fundamental definition, roughly speaking, is given by a set of vectors closed under linear combinations. In the same way, the more fundamental definition of a lattice is given by the set of discrete translations closed under discrete linear transformations.

Now, to answer your question more directly, why does the closure of the set imply that we can describe the set in the form $\Sigma n_i \mathbf{e_i}$? Let us simplify to the finite dimensional case. Consider the sum $\Sigma n_i \mathbf{v_i}$, where the sum is over the entire set of possible translations. Clearly, there are relations between the $\mathbf{v_i}$ because certain combinations are linearly dependent, so we can restrict to a linearly independent subset $\mathbf{e_i}$ (aka a basis). Hence, we get exactly what you have asked for.

Can this be done for an infinite dimensional case? I'm not entirely sure, you might run into issues of lattices and limits in the general case, cause I can imagine translating along an infinite series of vectors to get a net displacement that is infinitesimally small, meaning you could ostensibly "get back" to the regular vector space. In that case, can you really even think of it as a lattice anymore? But I digress.

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Another way of seeing it is that there is a complete symmetry in $G$, and $R$ in the equation $ e^{iG\cdot R}=1$. That is that if you can imagine $ R= \sum_i \alpha_i a_i $ for a fixed $G$ which you agree can be made into a lattice then for a fixed $R$ you can certainly imagine $G = \sum_i \beta_i b_i$. In other words all that matters is that the dot product eventually gives you a multiply of $ 2\pi $.

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