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Edit: This is a secondary question on how Ralph's solution can be simplified by choosing an appropriate origin in Euclidean space.

Ralph's solution to my original question, in the context of space groups, states that an isometry $(x,q)$ in a space group $G$ with linear part $q\in Q< GL(Z^n)$, must have a translational part x for which

$X_q=\lbrace x\in\mathbb R^n: (q^{1}+\cdots+q^{m})\cdot x\in \mathbb Z^n\rbrace =Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

Note first that from the composition of isometries we find that

$(t_1,q_1)(t_2,q_2)=(t_1+q_1\cdot t_2,q_1\cdot q_2)$

$\Leftrightarrow X_{q_{1}\cdot q_{2}}=X_{q_{1}}+(q_{1}-1)\cdot X_{q_{2}}$

This means that we must only consider the $X_q$ for the generators of the finite group $Q< GL(Z^n)$ (i.e. the point group).

After a shift of origin in Euclidean space, i.e. an affine transformation $(v,1)$ with $v\in \mathbb R^n$, we can write that

$X_q'=Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)+(q-1)\cdot v$

Since $(q-1)\cdot v\in Col(q-1)$, we can find for every $u\in Col(q-1)$ a vector $v\in \mathbb R^n$ for which $(q-1)\cdot v=-u$. Thus for a proper choice of origin we can write for a generator q

$X_q=\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

so that $t_q = X_q\ mod\ \mathbb Z^n$ is a rational number with maximal denominator $|Q|$ (which is the maximal possible m). The question is now, can we find one $v\in \mathbb R^n$ so that this simplification can be done for all $X_q$? For this, the column spaces $Col(q-1)$ for generators q of Q should be linear independent. If we call $S$ the generating set of Q, then this can be expressed as

$\forall q_i,q_j\in q,p\in S: Col(q_i-1)\cap Col(q_j-1)=\lbrace Col(q-1)\cap Col(p-1)=\lbrace 0 \rbrace$

Is this true?

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Edit: This is a secondary question on how Ralph's solution can be simplified by choosing an appropriate origin in Euclidean space.

Ralph's solution to my original question, in the context of space groups, states that an isometry $(x,q)$ in a space group $G$ with linear part $q\in Q< GL(Z^n)$, must have a translational part x for which

$X_q=\lbrace x\in\mathbb R^n: (q^{1}+\cdots+q^{m})\cdot x\in \mathbb Z^n\rbrace =Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

Note first that from the composition of isometries we find that

$(t_1,q_1)(t_2,q_2)=(t_1+q_1\cdot t_2,q_1\cdot q_2)$

$\Leftrightarrow X_{q_{1}\cdot q_{2}}=X_{q_{1}}+(q_{1}-1)\cdot X_{q_{2}}$

This means that we must only consider the $X_q$ for the generators of the finite group $Q< GL(Z^n)$ (i.e. the point group).

After a shift of origin in Euclidean space, i.e. an affine transformation $(v,1)$ with $v\in \mathbb R^n$, we can write that

$X_q'=Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)+(q-1)\cdot v$

Since $(q-1)\cdot v\in Col(q-1)$, we can find for every $u\in Col(q-1)$ a vector $v\in \mathbb R^n$ for which $(q-1)\cdot v=-u$. Thus for a proper choice of origin we can write for a generator q

$X_q=\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

so that $t_q = X_q\quad X_q\ mod\ \mathbb Z^n$ is a rational number with maximal denominator $|Q|$ (which is the maximal possible m). The question is now, can we find one $v\in \mathbb R^n$ so that this simplification can be done for all $X_q$? For this, the column spaces $Col(q-1)$ for generators q of Q should be linear independent. If we call $S$ the generating set of Q, then this can be expressed as

$\forall q_i,q_j\in S: Col(q_i-1)\cap Col(q_j-1)=\lbrace 0 \rbrace$

Is this true?

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So to summarize

Edit: This is a secondary question on how Ralph's solution can be simplified by choosing an appropriate origin in Euclidean space.

Ralph's solution to my original question: for each , in the context of space groups, states that an isometry with linear part $q$ (x,q)$in a space group (cfr. edited question)$G$with linear part$q\in Q< GL(Z^n)$, all possible translation parts are given bymust have a translational part x for which$X_q=\lbrace x\in\mathbb R^n: Ax\in (q^{1}+\cdots+q^{m})\cdot x\in \mathbb Z^n\rbrace =Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$where Note first that from the composition of isometries we find that$Col$and (t_1,q_1)(t_2,q_2)=(t_1+q_1\cdot t_2,q_1\cdot q_2)$

$Null$ are column and null space \Leftrightarrow X_{q_{1}\cdot q_{2}}=X_{q_{1}}+(q_{1}-1)\cdot X_{q_{2}}$This means that we must only consider the$X_q$for the generators of the finite group$(q-1)$. If we now shift Q< GL(Z^n)$ (i.e. the point group).

After a shift of origin in Euclidean space, i.e. an affine transformation $(v,1)$ with trivial linear part and $v\in \mathbb R^n$, then we get can write that

$X_q'=Col(q-1)+\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)+(q-1)\cdot v$

Since $(q-1)\cdot v\in Col(q-1)$, we can find for a particular choice of every $x$ in u\in Col(q-1)$a vector$X_{qi}$where v\in \mathbb R^n$ for which $i$ runs over the number (q-1)\cdot v=-u$. Thus for a proper choice of$q$in the finite group origin we can write for a generator q$Q$X_q=\frac{1}{m}(Null(q-1)\cap \mathbb Z^n)$

so that $x_i t_q = u_i + X_q\quad mod\ frac{z_i}{m_i} + (q_i-1)v\quad\quad\quad 1\leq i\leq |Q|$

where $u_i \in Col(q_i -1)$ and mathbb Z^n$is a rational number with maximal denominator$z_i \in |Q|$(Null(q_i -1)\cap \mathbb Z^n)$which is the maximal possible m). I expect there must be a The question is now, can we find one $v\in \mathbb R^n$ so that this simplification can be done for which all $x_i \in \mathbb Q^n$ X_q$? For this, the column spaces$Col(q-1)$for all igenerators q of Q should be linear independent. Any ideasIf we call$S$the generating set of Q, then this can be expressed as$\forall q_i,q_j\in S: Col(q_i-1)\cap Col(q_j-1)=\lbrace 0 \rbrace\$

Is this true?

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