# Show that point group of the reciprocal lattice is the same as the point group of the direct lattice

I have searched days for the answer of this question, but without success. Some background information:

A point group must fulfil the mathematical requirements of a group:

• The product of two operations is another operation in the group
• The identity element (I) is present
• The existence of an inverse operator $$RR^{-1} = 1$$
• Associative multiplication of operations*

The definition of the reciprocal lattice is: All vectors $$\vec{G}$$ for which $$e^{i\vec{G}*\vec{T}}=1$$ for all translation vectors $$\vec{T}$$ of the translation lattice.

Let P be a point-group operation of the direct lattice (To stress, the original lattice $$\vec{R}=a_1\vec{n_1}+a_2\vec{n_2}+a_3\vec{n_3},$$ from which we start), i.e., $$\vec{T'}=P\vec{T}$$ are elements of the direct lattice for all elements $$\vec{T}$$ of the direct lattice. The inner product has the following property: $$\vec{G}*({P\vec{T}})=(P^{-1}\vec{G})*{\vec{T}}$$.

The question: Argue now that the point group of the reciprocal lattice is the same as the point group of the direct lattice. In other words, show that if P is a point-group operation of the direct lattice, then P is also a point-group operation of the reciprocal lattice.

Hint: For all $$g$$ in the group there is an element $$g^{-1}$$, the inverse of $$g$$, such that $$g•g^{-1}=g^{-1}•g=e$$.

I hope I have given sufficient information to answer the question. Any help would be much appreciated.

The job is to show that if $$g$$ is an element of the group, then $$g(\vec{G})$$ is an element of the reciprocal lattice. To do this, we will compute $$e^{i(g(\vec{G}))\cdot\vec{T}}$$ and show that it is equal to 1 for all elements $$\vec{T}$$ of the direct lattice. We use the property of the inner product to say that $$e^{i(g(\vec{G}))\cdot\vec{T}} = e^{i\vec{G}\cdot(g^{-1}(\vec{T}))} =e^{i\vec{G}\cdot\vec{T}'}\,,$$ where $$\vec{T}'=g^{-1}(\vec{T})$$ is a direct lattice vector because $$g^{-1}$$ is an element of the group of symmetries of the direct lattice. Finally, $$e^{i(g(\vec{G}))\cdot\vec{T}} =e^{i\vec{G}\cdot\vec{T}'} = 1\,,$$ and since we started with an arbitrary $$\vec{T}$$, this is true for all $$\vec{T}$$, and hence $$g(\vec{G})$$ is an element of the reciprocal lattice.
• Hi march, I have two confusions. The first one is what "property of the inner product" is used in the proof? The second one is can the symmetry operations be restricted to be on $(hkl)$ planes, and $G$ restricted to the reciprocal vector normal to the planes? And so the conclusion is the symmetry of $G$ is equivalent to the symmetry on $(hkl)$ planes? Nov 15, 2022 at 5:59
• @meTchaikovsky. The "property of the inner product" is mentioned in the OP, it being the one that allows us to move the $g$ acting on $\vec{G}$ to $g^{-1}$ acting on $\vec{T}$. As for your second question, I don't really get your what you're asking, but if you can expand on it and be more precise maybe, it might be worth it's own question. Nov 15, 2022 at 16:23
• Thank you, I just thougt this property of inner product is not so obvious and should be proved instead of using as a prerequsite. The second question is can I narrow the symmetry operations to those that leave $(hkl)$ planes invariant? And $g$ becomes the symmetry operations that leaves $G$ in $[hkl]$ direction in reciprocal space invariant. Nov 16, 2022 at 0:51