1 Introduction
1.1 Basic Idea
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the local structure is discrete (i.e., there is a non-zero minimum separation between pairs of points in the structure);
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the structure has a symmetry determined by \(\boldsymbol{l}_{0}, \boldsymbol{l}_{1},\boldsymbol{l}_{2}\)—in the traditional view, \(\{ \boldsymbol{l}_{0},\boldsymbol{l}_{1},\boldsymbol{l}_{2} \}\) is a basis of \(L\), which is a discrete subgroup of \(\mathbb{R}^{3}\) with addition as group operation, denoted \((\mathbb{R}^{3},+)\).
1.2 Restriction to Two-Dimensional Crystal States
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The fact that \([ \boldsymbol{l}_{0}, \boldsymbol{l}_{1} ] (\cdot )\) is not a linear combination of \(\boldsymbol{l}_{0} (\cdot )\) and \(\boldsymbol{l}_{1} (\cdot )\) implies that the dislocation density tensor is non-constant in \(\varOmega \).
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Three-dimensional Lie groups and algebras have been extensively studied and classified, for example, by Bianchi [11] and Jacobsen [12]—this amounts to a classification of structure constants modulo change of basis. According to [12] there are four different classes: the abelian, nilpotent, solvable and simple classes of Lie algebras. The abelian case leads to traditional crystallography, and we considered the nilpotent case in [1]. The solvable case divides into unimodular and non-unimodular classes—we consider the unimodular case here, and hope to present the non-unimodular solvable and (so-called) simple case in forthcoming works.
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Let \(\boldsymbol{\mathfrak{g}}_{2}\) denote the algebra generated by Lie algebra elements of the form \([\boldsymbol{g},\boldsymbol{h}]\), \(\boldsymbol{g},\boldsymbol{h} \in \boldsymbol{\mathfrak{g}}\), and call it the derived algebra (or commutator algebra, or commutator ideal as \([\boldsymbol{\mathfrak{g}}, \boldsymbol{\mathfrak{g}}_{2} ] = \boldsymbol{g}_{2}\)). Define \(\boldsymbol{\mathfrak{g}}^{1} \equiv \boldsymbol{\mathfrak{g}}\), \(\boldsymbol{\mathfrak{g}}^{2} = [ \boldsymbol{\mathfrak{g}}, \boldsymbol{\mathfrak{g}}^{1}]\), \(\boldsymbol{\mathfrak{g}}^{3} = [ \boldsymbol{\mathfrak{g}}, \boldsymbol{\mathfrak{g}}^{2}] \ldots \boldsymbol{\mathfrak{g}}^{n} = [\boldsymbol{\mathfrak{g}},\boldsymbol{\mathfrak{g}}^{n-1} ]\). Then \(\boldsymbol{\mathfrak{g}}\) is nilpotent if \(\boldsymbol{\mathfrak{g}}^{n} = \{ \boldsymbol{0} \}\) for some \(n \in \mathbb{Z}\). Also define \(\boldsymbol{\mathfrak{g}}_{1} = \boldsymbol{\mathfrak{g}}\), \(\boldsymbol{\mathfrak{g}}_{2} = [ \boldsymbol{\mathfrak{g}}_{1}, \boldsymbol{\mathfrak{g}}_{1} ] \ldots \boldsymbol{\mathfrak{g}}_{n} = [ \boldsymbol{\mathfrak{g}}_{n-1}, \boldsymbol{\mathfrak{g}}_{n-1} ]\). Then \(\boldsymbol{\mathfrak{g}}\) is solvable if \(\boldsymbol{\mathfrak{g}}_{n} = \{ \boldsymbol{0} \}\) for some \(n \in \mathbb{Z}\). It will turn out, in the solvable case that we consider here, where \(\dim \boldsymbol{\mathfrak{g}} = 3\), that \([\boldsymbol{\mathfrak{g}}_{2} , \boldsymbol{\mathfrak{g}}_{2} ] = \{ \boldsymbol{0} \}\), so that the commutators of algebra elements commute with each other, and this simplifying factor facilitates many of the computations below.
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In the case that \(\dim \boldsymbol{\mathfrak{g}}=3\), \(\boldsymbol{\mathfrak{g}}\) is simple if and only if \(\dim \boldsymbol{\mathfrak{g}}_{2} = 3\).
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We shall construct canonical forms of the bracket relations, (1.20) or (1.21), explicitly, as many have done before, in cases where \(\dim \boldsymbol{\mathfrak{g}}_{2} \neq 3\), arriving at Jacobsen’s results, but follow a route which is a little different to Jacobsen’s, just to illustrate the method.
1.3 Outline of Content
1.4 Mechanical Context and Motivation
2 Lie Groups and Homogeneous Spaces
2.1 Lie Groups and Algebras
2.2 Right Invariant Fields
2.3 Lie Algebras of Right Invariant Fields
2.4 Group and Algebra Homomorphisms
2.5 Left Action of a Group on a Manifold
2.6 Homogeneous Space
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If \(H\) is a closed subgroup of \(G\), then \(G/H\) can be given the structure of a manifold, with \(\boldsymbol{\lambda } \colon G \times G/H \to G/H\) defined by \(\boldsymbol{\lambda } ( \boldsymbol{g}, \boldsymbol{k}H)=\boldsymbol{\psi }(\boldsymbol{g},\boldsymbol{k})H\) smooth and transitive. Then \((G,G/H)\) is a homogeneous space.
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If \((G,M)\) is a homogeneous space, and \(m \in M\), then \(I_{m}\) is a closed subgroup of \(G\).
2.7 Flow Along Infinitesimal Generators
3 Solvable Groups and Algebras
3.1 \(\dim \boldsymbol{\mathfrak{g}}_{2}=1\)
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If \(\boldsymbol{h} = s_{i} \boldsymbol{f}_{i}\) is a non-zero element of the Lie algebra (3.7), with \(\boldsymbol{u}=\boldsymbol{f}_{0}\), \(\boldsymbol{v}=\boldsymbol{f}_{1}\), \(\boldsymbol{w} \equiv \boldsymbol{f}_{2}\), and if \(L\) is an automorphism of the form (3.9), thenwhere$$ L \boldsymbol{h} = (L_{ij} s_{j}) \boldsymbol{f}_{i} \equiv t_{i} \boldsymbol{f}_{i}, $$(3.10)Then if \(s_{0} \neq 0\) we can choose \(L_{10},L_{20}\) so that \(t_{1} = t_{2} = 0\), so that \(\langle \boldsymbol{h} \rangle \) is mapped by an algebra automorphism to \(\langle \boldsymbol{f}_{0} \rangle \). If \(s_{0} = 0\), we can choose \(L_{11},L_{22}\) to rescale coordinates, and bring \(\langle \boldsymbol{h} \rangle \) to one of the forms \(\langle \boldsymbol{f}_{1} \rangle \), \(\langle \boldsymbol{f}_{2} \rangle \), \(\langle \boldsymbol{f}_{1} + \boldsymbol{f}_{2} \rangle \).$$ t_{0} = s_{0}, \qquad t_{1} = L_{10} s_{0} + L_{11} s_{1}, \qquad t_{2} = L_{20} s_{0} + L_{22} s_{2}. $$(3.11)
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If one replaces \(\boldsymbol{w}\) by \(\tilde{\boldsymbol{w}}= \boldsymbol{w}+\boldsymbol{v}\) in (3.7), the commutation relations becomeand, with this choice of basis, these relations have the form (1.21).$$ [\boldsymbol{u},\boldsymbol{v}] = \boldsymbol{v}, \qquad [ \boldsymbol{u},\bar{ \boldsymbol{w}}] = \boldsymbol{v}, \qquad [ \boldsymbol{v},\bar{\boldsymbol{w}}]= \boldsymbol{0}, $$(3.12)
3.2 \(\dim \boldsymbol{\mathfrak{g}}_{2}=2\)
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For later use, note that if \(\boldsymbol{h}=s_{i} \boldsymbol{f}_{i}\) is a non-zero element of the Lie algebra with \(\sigma = \left ( {\scriptsize\begin{matrix}{} 1&0 \cr 0&-1 \end{matrix}} \right ) \), and \(L\) has the form (3.30), then \(L \boldsymbol{h} = t _{i} \boldsymbol{f}_{i}\) whereor$$t_{0} = s_{0}, \qquad t_{1} = \alpha s_{1} + \ast s_{0}, \qquad t_{2} = \beta s_{2} + \ast s_{0}, $$where \(\alpha \beta \neq 0\). Hence \(\langle \boldsymbol{h} \rangle \) coincides with the span of one of the following choices of \((s_{0},s _{1},s_{2})^{\top }\):$$t_{0} = -s_{0}, \qquad t_{1} = \alpha s_{2} + \ast s_{0}, \qquad t_{2} = -\beta s_{1} + \ast s_{0}, $$$$ \left ( \textstyle\begin{array}{c} 1 \\ 0 \\ 0 \end{array}\displaystyle \right ) , \left ( \textstyle\begin{array}{c} 0 \\ 1 \\ 0 \end{array}\displaystyle \right ) , \left ( \textstyle\begin{array}{c} 0 \\ 1 \\ 1 \end{array}\displaystyle \right ) . $$(3.32)
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If \(\sigma = \left ( {\scriptsize\begin{matrix}{} 0&1 \cr -1&0 \end{matrix}} \right ) \), then similarlyor$$t_{0} = s_{0}, \qquad t_{1} = \alpha s_{2} + \beta s_{2} + \ast s_{0}, \qquad t_{2} = - \beta s_{1} + \alpha s_{2} + \ast s_{0}, $$So \(\langle \boldsymbol{h} \rangle \) coincides with the span of one of the following choices of \((s_{0},s_{1},s_{2})^{\top }\):$$t_{0} = -s_{0}, \qquad t_{1} = \beta s_{1} + \alpha s_{2} + \ast s_{0}, \qquad t_{2} = \alpha s_{1} - \beta s_{2} + \ast s_{0}. $$(because \(\left ( {\scriptsize\begin{matrix}{} \alpha &\beta \cr -\beta &\alpha \end{matrix}} \right ) \left ( {\scriptsize\begin{matrix}{} s_{1} \cr s_{2} \end{matrix}} \right ) =\left ( {\scriptsize\begin{matrix}{} 1 \cr 0 \end{matrix}} \right ) \) has a solution for \(\left ( {\scriptsize\begin{matrix}{} \alpha &\beta \cr -\beta &\alpha \end{matrix}} \right ) \) with \(\alpha \beta \neq 0\), given \(\left ( {\scriptsize\begin{matrix}{} s_{1} \cr s_{2} \end{matrix}} \right ) \neq \boldsymbol{0}\)).$$ \left ( \textstyle\begin{array}{l} 1 \\ 0 \\ 0 \end{array}\displaystyle \right ) , \left ( \textstyle\begin{array}{l} 0 \\ 1 \\ 0 \end{array}\displaystyle \right ) $$(3.33)
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If \(\sigma = \left ( {\scriptsize\begin{matrix}{} 1&0 \cr 0&-1 \end{matrix}} \right ) \), the corresponding commutation relations have the form \([\boldsymbol{u},\boldsymbol{v}]=\boldsymbol{v}\), \([\boldsymbol{u}, \boldsymbol{w}]=-\boldsymbol{w}\), \([\boldsymbol{v},\boldsymbol{w}]= \boldsymbol{0}\). They may be put in the form of (1.21) by the change of basis \(\boldsymbol{v} \to \tilde{\boldsymbol{v}} \equiv \boldsymbol{v} + \boldsymbol{w}\), \(\boldsymbol{w} \to \tilde{\boldsymbol{w}} = \boldsymbol{v} - \boldsymbol{w}\). The commutation relations in the case \(\sigma = \left ( {\scriptsize\begin{matrix}{} 0&1 \cr -1&0 \end{matrix}} \right ) \) are in the form (1.21).
3.3 Group Multiplication
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If \(G\) is a simply connected, connected, non-compact, non-nilpotent three-dimensional Lie group with maximal nilpotent subgroup \(N\) and discrete subgroup \(D\) such that \(G/D\) is compact, then \(\dim N = 2\).
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The matrix group \(\{ e^{\sigma ^{\top } g} \colon (g,\boldsymbol{g}) \in D \}\) is generated by a single element \(e^{\sigma ^{\top } \theta }\), \((\theta ,\boldsymbol{\theta }) \in D\), \(\theta \neq 0\). \(D \cap \{(0,\mathbb{R}^{2})\}\) is a lattice generated by two independent elements \((0,\boldsymbol{e}_{1}),(0,\boldsymbol{e}_{2})\), \(\boldsymbol{e}_{1},\boldsymbol{e}_{2} \in \mathbb{R}^{2}\) say. Moreover the lattice is preserved by \(\phi _{\theta } \colon \mathbb{R}^{2} \to \mathbb{R}^{2}\), so thatAlso \(\{ e^{\sigma ^{\top } t} \colon t \in \mathbb{R} \}\) must be a one parameter subgroup of the unimodular group, so$$ \phi _{\theta } (\boldsymbol{e}_{\alpha } ) = \gamma _{\beta \alpha } \boldsymbol{e}_{\beta }, \quad \alpha ,\beta = 1,2,\ \gamma \equiv (\gamma _{\beta \alpha }) \in \textrm{GL}_{2} ( \mathbb{Z}). $$(3.51)which implies that$$ \det \bigl(e^{\sigma ^{\top } t}\bigr)=e^{(\operatorname{tr} \sigma ^{\top })t}=1, $$(3.52)So \(D\) is generated by three distinct elements \((0,\boldsymbol{e}_{1}),(0, \boldsymbol{e}_{2}),(\theta ,\boldsymbol{\theta })\). Note that the inner automorphism \(\xi _{\boldsymbol{h}} \colon G \to G\) defined by$$ \operatorname{tr} \sigma = 0, \qquad \gamma \in \textrm{SL}_{2} ( \mathbb{Z}). $$(3.53)(3.54)and can be chosen such that$$ \xi _{\boldsymbol{h}} \bigl(0,\boldsymbol{\theta }'\bigr)=\bigl(0, \boldsymbol{\theta }'\bigr), $$(3.55)provided$$ \xi _{\boldsymbol{h}} (\theta ,\boldsymbol{\theta })=(\theta , \boldsymbol{0}), $$(3.56)(3.57)
3.4 Catalogue of Discrete Subgroups
4 Canonical Forms of Vector Fields in \(\pmb{\mathbb{R}}^{2}\) with Three-Dimensional Solvable Lattice Algebras
4.1 Homogeneous Space Construction
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The elements of the one parameter subgroup \(H=\{ \boldsymbol{x}(t); t \in \mathbb{R} \}\) corresponding to \(\langle \boldsymbol{e}_{0} \rangle \) coincide with the points of the integral curve of the right invariant field whose components at the group identity are \(\boldsymbol{e}_{0} \equiv (1,0,0)^{\top }\). Sowhere \(\boldsymbol{\psi }\) is given by (3.45). This gives \(H=\{ t \boldsymbol{e}_{0} ; t \in \mathbb{R} \}\). If \(\boldsymbol{g}=(z,x,y) \in G\), the left cosets have the form$$ \frac{d}{dt} \boldsymbol{x} (t) = \nabla _{1} \boldsymbol{\psi } \bigl(0, \boldsymbol{x}(t)\bigr) \boldsymbol{e}_{0}, $$(4.1)So elements of the set of left cosets may be identified by the two coordinates \((x,y) \in \mathbb{R}^{2}\). Define the projection \(\boldsymbol{\pi } \colon G \to G/H \equiv \mathbb{R}^{2}\) by$$ \boldsymbol{g} H = \bigl\{ \boldsymbol{\psi }\bigl((z,x,y),(t,0,0)\bigr); t \in \mathbb{R} \bigr\} = \bigl\{ (z+t,x,y) \colon t \in \mathbb{R} \bigr\} . $$(4.2)and choose a corresponding section \(\boldsymbol{\sigma } \colon G/H \equiv \mathbb{R}^{2} \to G\) by$$ \boldsymbol{\pi } (z,x,y) = (x,y), $$(4.3)(For consistency with (3.41), we should write these last two equations as \(\boldsymbol{\pi }(z,x,y)=\left ( {\scriptsize\begin{matrix}{} x \cr y \end{matrix}} \right ) \), \(\boldsymbol{\sigma } \left ( {\scriptsize\begin{matrix}{} x \cr y \end{matrix}} \right ) =(0,x,y)\), but is convenient to allow a little notational abuse in this subsection.)$$ \boldsymbol{\sigma } (x,y) = (0,x,y). $$(4.4)Then according to (2.45), with \(\boldsymbol{g}_{1} \equiv (z_{1},x _{1},y_{1})\), the corresponding group action \(\boldsymbol{\lambda } \colon G \times \mathbb{R}^{2} \to \mathbb{R}^{2}\) iswhere$$\begin{aligned} \boldsymbol{\lambda } (\boldsymbol{g}_{1},\boldsymbol{g}H)= \boldsymbol{\pi } \bigl(\boldsymbol{g}_{1} \boldsymbol{\sigma }( \boldsymbol{g}H)\bigr) &= \boldsymbol{\pi }\bigl(\boldsymbol{\psi } \bigl((z_{1},x _{1},y_{1}),(0,x,y)\bigr)\bigr) \\ &= \boldsymbol{\pi } \bigl((z_{1},x_{1}+\alpha , y_{1}+\beta )\bigr) \\ &=(x_{1}+\alpha ,y_{1}+\beta ), \end{aligned}$$(4.5)The three infinitesimal generators \(\boldsymbol{l}_{0}(\cdot ), \boldsymbol{l}_{1}(\cdot ),\boldsymbol{l}_{2}(\cdot )\) (corresponding to the choices \(\boldsymbol{v}= \boldsymbol{e}_{i}\), \(i=0,1,2\)) have components given by$$\left ( \textstyle\begin{array}{c} \alpha\\ \beta \end{array}\displaystyle \right ) = e^{\sigma ^{\top } z_{1}} \left ( \textstyle\begin{array}{c} x \\ y \end{array}\displaystyle \right ) , \quad \sigma = \left ( \textstyle\begin{array}{c@{\quad}c} 1&0 \\ 0&-1 \end{array}\displaystyle \right ) . $$where \((\lambda _{1},\lambda _{2})\) are the components of \(\boldsymbol{\lambda }\). It follows that$$ \frac{\partial \lambda _{r}}{\partial (\boldsymbol{g}_{1})_{i}} \bigl( \boldsymbol{0},(x,y)\bigr) \equiv \frac{\partial \lambda _{r}}{\partial ( \boldsymbol{g}_{1})_{i}} ( \boldsymbol{0},\boldsymbol{g} H), \quad r=1,2, \, i=0,1,2, $$(4.6)$$ \boldsymbol{l}_{0}(x,y)=(x,-y), \qquad \boldsymbol{l}_{1}(x,y)=(1,0), \qquad \boldsymbol{l}_{2}(x,y)=(0,1). $$(4.7)
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For the subalgebra \(\langle \boldsymbol{e}_{1} + \boldsymbol{e}_{2} \rangle \), one obtains \(H=\{(0,r,r)\colon r \in \mathbb{R}\}\), and so with \(\boldsymbol{g}=(t_{0},t_{1},t_{2})\),For fixed \(t_{0},t_{1},t_{2}\), (4.8) represents a line through \(\left ( {\scriptsize\begin{matrix}{} t_{1} \cr t_{2} \end{matrix}} \right ) \) also passing through the point \(\frac{t_{1} e^{-t_{0}} - t _{2} e^{t_{0}}}{e^{2t_{0}} + e^{-2t_{0}}} \left ( {\scriptsize\begin{matrix}{} e^{-t_{0}} \cr -e^{t_{0}} \end{matrix}} \right ) \). Note that this line has signed distance \((t_{1} e^{-t_{0}} - t_{2} e^{t_{0}}) / \sqrt{e^{2t_{0}} + e^{-2t _{0}}}\) from the origin in the \(t_{0}\) plane. So we can identify \(\boldsymbol{g}H\) as a point of \(\mathbb{R}^{2}\) with coordinates$$ \boldsymbol{g}H = \left \{ (t_{0},\tau _{1},\tau _{2}) \colon \left ( \textstyle\begin{array}{c} \tau _{1} \\ \tau _{2} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c} t_{1} \\ t_{2} \end{array}\displaystyle \right ) + r \left ( \textstyle\begin{array}{c} e^{t_{0}} \\ e^{-t_{0}} \end{array}\displaystyle \right ) , r \in \mathbb{R} \right \}. $$(4.8)and choose as projection mapping \(\boldsymbol{\pi } \colon G \to G/H \equiv \mathbb{R}^{2}\),$$ \boldsymbol{g}H \equiv \bigl(t_{0},t_{1} e^{-t_{0}} - t_{2} e^{t_{0}}\bigr), $$(4.9)As choice of section \(\boldsymbol{\sigma } \colon G/H \equiv \mathbb{R}^{2} \to G\) we take$$ \boldsymbol{\pi } (t_{0},t_{1},t_{2}) = \bigl(t_{0},t_{1} e^{-t_{0}} - t _{2} e^{t_{0}}\bigr). $$(4.10)as \(\sigma \) maps \(\boldsymbol{\pi }(t_{0},t_{1},t_{2})\) to the point identified after (4.8), which is the closest point, on the line represented by (4.8), to the origin in the plane \(t_{0}=\) constant. We calculate the group action, with \(\boldsymbol{g}_{1} = (z_{0},z_{1},z _{2})\),$$ \boldsymbol{\sigma }(x,y) = \biggl( x, \frac{ye^{-x}}{e^{2x} + e^{-2x}}, \frac{-ye^{x}}{e^{2x}+e^{-2x}} \biggr) , $$(4.11)where we used$$\begin{aligned} \boldsymbol{\lambda } (\boldsymbol{g}_{1} \boldsymbol{g}H) &= \boldsymbol{\pi }\bigl(\boldsymbol{g}_{1} \boldsymbol{\sigma }( \boldsymbol{g}H)\bigr)=\boldsymbol{\pi } \biggl( \boldsymbol{\psi } \biggl( (z _{0},z_{1},z_{2}), \biggl( x, \frac{ye^{-x}}{e^{2x}+e^{-2x}}, \frac{-ye ^{x}}{e^{2x}+e^{-2x}} \biggr) \biggr) \biggr) \\ &=\boldsymbol{\pi }(z_{0}+x,z_{1}+\tau _{1},z_{2}+\tau _{2}) \\ &=\bigl(z_{0}+x,(z_{1}+\tau _{1}) e^{-(z_{0}+x)}-(z_{2}+\tau _{2})e^{z_{0}+x}\bigr) \\ &=\bigl(z_{0}+x,y+z_{1} e^{-z_{0}-x} - z_{2} e^{z_{0}+x}\bigr), \end{aligned}$$(4.12)One finds that the infinitesimal generators are:$$\left ( \textstyle\begin{array}{c} \tau _{1} \\ \tau _{2} \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c@{\quad}c} e^{z_{0}}&0 \\ 0&e^{-z_{0}} \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{c} ye^{-x} \\ -ye^{x} \end{array}\displaystyle \right ) \big/ \bigl(e^{2x}+e^{-2x}\bigr) . $$$$ \boldsymbol{l}_{0} = (1,0), \qquad \boldsymbol{l}_{1}= \bigl(0,e^{-x}\bigr), \qquad \boldsymbol{l}_{2}= \bigl(0,-e^{x}\bigr). $$(4.13)
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For the subalgebra \(\langle \boldsymbol{e}_{1} \rangle \), \(H=\{ (0,r,0) \colon r \in \mathbb{R} \}\), and if \(\boldsymbol{g}=(t_{0},t_{1},t _{2})\) then \(\boldsymbol{g}H = \{(t_{0},t_{1}+re^{t_{0}},t_{2} ) \colon r \in \mathbb{R} \}\). For fixed \(t_{0}\), the points of \(\boldsymbol{g}H\) represent a line through \(\left ( {\scriptsize\begin{matrix}{} t_{1} \cr t_{2} \end{matrix}} \right ) \) also passing through \(\left ( {\scriptsize\begin{matrix}{} 0 \cr t_{2} \end{matrix}} \right ) \). So we identify \(\boldsymbol{g}H \equiv (t_{0},t_{2})\), setand choose$$ \boldsymbol{\pi } (t_{0},t_{1},t_{2})=(t_{0},t_{2}), $$(4.14)We calculate, with \(\boldsymbol{g}_{1} = (z_{0},z_{1},z_{2})\);$$ \boldsymbol{\sigma }(x,y) = (x,0,y). $$(4.15)Finally, the infinitesimal generators are:$$\begin{aligned} \boldsymbol{\lambda }(\boldsymbol{g}_{1} \boldsymbol{g}H) &= \boldsymbol{\pi }\bigl(\boldsymbol{g}_{1} \boldsymbol{\sigma }( \boldsymbol{g}H)\bigr)=\boldsymbol{\pi }\bigl( \boldsymbol{\psi } \bigl((z_{0},z_{1},z _{2}), (x,0,y)\bigr)\bigr) \\ &=\boldsymbol{\pi }(z_{0}+x,z_{1}+\tau _{1},z_{2}+\tau _{2}), \quad \text{where } \left ( \textstyle\begin{array}{c} \tau _{1} \\ \tau _{2} \end{array}\displaystyle \right ) =\left ( \textstyle\begin{array}{c@{\quad}c} e^{z_{0}}&0 \\ 0&e^{-z_{0}} \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{c} 0 \\ y \end{array}\displaystyle \right ) \\ &=\boldsymbol{\pi } \bigl(z_{0} + x, z_{1}, z_{2} + y e^{-z_{0}}\bigr) \\ &=\bigl(z_{0}+x, z_{2} + ye^{-z_{0}} \bigr). \end{aligned}$$(4.16)Since this is a 2-dimensional algebra of vector fields, we consider it no further.$$ \boldsymbol{l}_{0} (x,y) = (1,-y), \qquad \boldsymbol{l}_{1} (x,y) = (0,0), \qquad \boldsymbol{l}_{3} (x,y) = (0,1). $$(4.17)
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If \(\sigma = \left ( {\scriptsize\begin{matrix}{} 0&1 \cr -1&0 \end{matrix}} \right ) \), and (recalling the remark after (3.32)) we consider the subalgebra \(\langle \boldsymbol{e}_{0} \rangle \), then by the analogue of (4.1) we get \(H=\{ t \boldsymbol{e}_{0} \colon t \in \mathbb{R} \}\). Then \(\boldsymbol{g} H = \{ (z+t,x,y) \colon t \in \mathbb{R} \}\) if \(\boldsymbol{g}=(z,x,y)\) and we can take \(\boldsymbol{\pi }(z,x,y)=(x,y)\), \(\boldsymbol{\sigma }(x,y)=(0,x,y)\). With \(\boldsymbol{g}_{1} = (z_{1},x_{1},y_{1})\) one finds thatand calculates that the infinitesimal generators are:$$ \boldsymbol{\lambda }( \boldsymbol{g}_{1},\boldsymbol{g}H) = (x_{1} + x \cos z_{1} - y \sin z_{1}, y_{1} + x \sin z_{1} + y \cos z_{1} ), $$(4.18)$$ \boldsymbol{l}_{0} (x,y) = (-y,x), \qquad \boldsymbol{l}_{1} (x,y) = (1,0), \qquad \boldsymbol{l}_{2} (x,y) = (0,1). $$(4.19)
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For the subalgebra \(\langle \boldsymbol{e}_{1} \rangle \), we get \(H= \{ t \boldsymbol{e}_{1} \colon t \in \mathbb{R} \}\), and with \(\boldsymbol{g} = (z,x,y)\), \(\boldsymbol{g} H = \{ (z,x+t \cos z, y+t \sin z) \colon t \in \mathbb{R} \}\). For fixed \((z,x,y)\), \(\boldsymbol{g} H\) represents a line in the plane \(z=\) constant, through \((z,x,y)\), making an angle \(z\) with the \(x\)-axis. The point nearest the origin (i.e., nearest \((z,0,0)\)) is \((z, (y \cos z - x \sin z) (- \sin z, \cos z))\). Therefore we can identify \(\boldsymbol{g}H\) byand choose as section the point nearest the origin, so$$ \boldsymbol{ \pi } (\boldsymbol{g})=(z,y \cos z - x \sin z), $$(4.20)One calculates, in the manner of the previous bullet points, that with \(g_{1} = (z_{1},x_{1},y_{1})\),$$ \sigma (x,y) = (x, -y \sin x, y \cos x). $$(4.21)Finally the infinitesimal generators are:$$ \boldsymbol{\lambda }( \boldsymbol{g}_{1},\boldsymbol{g}H)= \bigl(z_{1} + z, y \cos z - x \sin z -x_{1} \sin (z_{1} + z) + y_{1} \cos (z_{1} + z)\bigr). $$(4.22)Note that the change of basis \(\boldsymbol{l}_{0} \to \boldsymbol{l} _{0}\), \(\boldsymbol{l}_{1} \to + \boldsymbol{l}_{2}\), \(\boldsymbol{l} _{2} \to - \boldsymbol{l}_{1}\) does not change the commutation relations, so the infinitesimal generators can also be written as$$ \boldsymbol{l}_{0} (x,y) = (1,0), \qquad \boldsymbol{l}_{1} (x,y) = (0,- \sin x), \qquad \boldsymbol{l}_{2} (x,y) = (0,\cos x). $$(4.23)$$ \boldsymbol{l}_{0} (x,y) = (1,0), \qquad \boldsymbol{l}_{1} (x,y) = (0, \cos x), \qquad \boldsymbol{l}_{2} (x,y) = (0,\sin x). $$(4.24)
4.2 Direct Construction of Canonical Vector Fields
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In the case that \(\sigma = \left ( {\scriptsize\begin{matrix}{} 1&0 \cr 0&-1 \end{matrix}} \right ) \), (4.35) givesfor some functions \(\alpha , \bar{\beta },\gamma ,\delta \) (the functions \(\beta (\cdot )\) and \(\bar{\beta }(\cdot )\) are unrelated). Now one of \(\bar{\beta } (0)\), \(\delta (0)\) is non-zero (since \(\dim \langle \boldsymbol{v}_{0}, \boldsymbol{v}_{1},\boldsymbol{v} _{2} \rangle =2\)). Suppose that \(\bar{\beta } (0) \neq 0\), without loss of generality. By change of coordinates \(x \to \tilde{x} = x + f(y)\), \(y \to \tilde{y} = g(y)\), where \(f' = - \alpha / \bar{\beta }\), \(g' = e^{-f(y)} / \bar{\beta }\), we get that \(\boldsymbol{v}_{1} = (0,e ^{-x})^{\top }\) and that \(\boldsymbol{v}_{2}\) has the functional form given in (4.35), in the new coordinates. Then (4.25)1 gives$$ \boldsymbol{v}_{1} (x,y)=\bigl(\alpha (y) e^{-x}, \bar{\beta } (y) e^{-x} \bigr)^{ \top }, \qquad \boldsymbol{v}_{2} (x,y)=\bigl( \gamma (y) e^{x}, \delta (y) e ^{x} \bigr)^{\top }, $$(4.36)which implies that \(\gamma ' (y) = 0\), \(- \gamma (y) = \delta ' (y)\). Thus \(\gamma (y)=0\), since if \(\gamma (0) \neq 0\), \(\boldsymbol{v} _{1}\) and \(\boldsymbol{v}_{2}\) are linearly independent at the origin. So \(\delta (y)\) is a constant, and we may presume that \(\delta (y) = -1\) (cf. (3.30)). Thus$$ \bigl[ \bigl(0,e^{-x}\bigr)^{\top }, \bigl(\gamma (y) e^{x}, \delta (y) e^{x} \bigr) ^{\top }\bigr] = \boldsymbol{0}, $$(4.37)and these are the vector fields given in (4.13).$$ \boldsymbol{v}_{0} (x,y) = (1,0)^{\top }, \qquad \boldsymbol{v}_{1} (x,y) = \bigl(0,e^{-x}\bigr)^{\top }, \qquad \boldsymbol{v}_{2} (x,y) = \bigl(0,-e^{x} \bigr)^{ \top }, $$(4.38)
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In the case that \(\sigma = \left ( {\scriptsize\begin{matrix}{} 0&1 \cr -1&0 \end{matrix}} \right ) \), then (4.35) giveswhere \(R(x)\), \(D(y)\) are defined in the obvious way (new function \(\beta (\cdot )\)). So$$ \left ( \textstyle\begin{array}{c@{\quad}c} a&b \\ c&d \end{array}\displaystyle \right ) = \left ( \textstyle\begin{array}{c@{\quad}c} \cos x&-\sin x \\ \sin x&\cos x \end{array}\displaystyle \right ) D(y) = R(x) \left ( \textstyle\begin{array}{c@{\quad}c} \alpha (y)&\beta (y) \\ \gamma (y)&\delta (y) \end{array}\displaystyle \right ) , $$(4.39)At least one of \(\beta (0)\), \(\alpha (0)\) is non-zero at the origin, without loss of generality assume that \(\beta (0) \neq 0\). Let \(\tilde{x} = x + f(y)\), \(\tilde{y} = g(y)\) and let \((\tilde{a} ( \tilde{x}, \tilde{y} ), \tilde{b} (\tilde{x},\tilde{y}) )^{\top }\), \((\tilde{c}(\tilde{x},\tilde{y}), \tilde{d}( \tilde{x},\tilde{y}))^{ \top }\) be the components of \(\boldsymbol{v}_{1},\boldsymbol{v}_{2}\) in the new coordinates. Then$$\begin{aligned} \begin{aligned} \boldsymbol{v}_{1}(x,y) &= ( \alpha \cos x - \gamma \sin x, \beta \cos x - \delta \sin x)^{\top }, \\ \boldsymbol{v}_{2}(x,y) &= (\alpha \sin x + \gamma \cos x, \beta \sin x + \delta \cos x)^{\top }. \end{aligned} \end{aligned}$$(4.40)Now choose \(f\) so that the 11 component of the matrix displayed in the line above is zero, recalling \(\beta (0) \neq 0\)$$\begin{aligned} &\left ( \textstyle\begin{array}{c@{\quad}c} \tilde{a}&\tilde{c} \\ \tilde{b}&\tilde{d} \end{array}\displaystyle \right ) _{\tilde{x} = \tilde{x} (x,y), \tilde{y} = \tilde{y} (y)} \\ &\quad= \left ( \textstyle\begin{array}{c@{\quad}c} 1&f'(y) \\ 0&g'(y) \end{array}\displaystyle \right ) \left ( \textstyle\begin{array}{c@{\quad}c} \alpha (y)&\gamma (y) \\ \beta (y)&\delta (y) \end{array}\displaystyle \right ) R(-x), \\ &\quad= \left ( \textstyle\begin{array}{c@{\quad}c} \alpha + \beta f'&\gamma + \delta f' \\ \beta g'&\delta g' \end{array}\displaystyle \right ) R(f) R(-x-f) \\ &\quad=\left ( \textstyle\begin{array}{c@{\quad}c} (\alpha + \beta f') \cos f + (\gamma + \delta f') \sin f&-(\alpha + \beta f') \sin f + (\gamma + \delta f') \cos f \\ g'(\beta \cos f + \delta \sin f)&g'(-\beta \sin f + \delta \cos f) \end{array}\displaystyle \right ) \\ &\qquad R(-x-f). \end{aligned}$$(4.41)Knowing \(f=f(y)\), choose \(g\) so that the 21 component of the matrix is 1:$$ f' = - \frac{(\alpha \cos f + \gamma \sin f)}{\beta \cos f + \delta \sin f}, \quad f(0)=0. $$(4.42)Then from (4.40)$$ g' = \frac{1}{\beta \cos f + \delta \sin f}, \quad g(0)=0. $$(4.43)for some functions \(\xi , \mu \), and dropping the tilde this gives$$ \left ( \textstyle\begin{array}{c@{\quad}c} \tilde{a}&\tilde{c} \\ \tilde{b}&\tilde{d} \end{array}\displaystyle \right ) _{\tilde{x},\tilde{y}} = \left ( \textstyle\begin{array}{c@{\quad}c} 0&\xi ( \tilde{y} ) \\ 1&\mu (\tilde{y}) \end{array}\displaystyle \right ) R(-\tilde{x}), $$(4.44)Since \(\boldsymbol{v}_{1}\) and \(\boldsymbol{v}_{2}\) are linearly dependent at the origin, \(\xi (0)=0\). We calculate that \([ \boldsymbol{v}_{1}, \boldsymbol{v}_{2}](\cdot )\) has components \(-(\xi ' + \xi ^{2}, \mu ' + \xi \mu )\). Hence \(\xi ' + \xi ^{2} = 0\), \(\xi (0) = 0\), so \(\xi (y)=0\), \(\mu '(y)=0\). So \(\mu (y)=\mu _{0} \in \mathbb{R}\) and \(\boldsymbol{v}_{1} (x,y)=(0,\cos x - \mu _{0} \sin x)^{\top }\), \(\boldsymbol{v}_{2} = (0, \sin x + \mu _{0} \cos x)^{ \top }\). Now from (3.31) we may replace \(\boldsymbol{v}_{1}\) by \(\tilde{\boldsymbol{v}}_{1} = \alpha \boldsymbol{v}_{1} - \beta \boldsymbol{v}_{2}\), \(\boldsymbol{v}_{2}\) by \(\tilde{\boldsymbol{v}} _{2} = \beta \boldsymbol{v}_{1} + \alpha \boldsymbol{v}_{2}\), \(\alpha , \beta \in \mathbb{R}\), \(\alpha ^{2} + \beta ^{2} >0\) without changing the commutation relations. By putting \(\alpha = (1 + \mu _{0} ^{2})^{-1}\), \(\beta = - \mu _{0} \alpha \) we obtain$$\begin{aligned} \begin{aligned} \boldsymbol{v}_{1} (x,y) &= \bigl(- \xi (y) \sin x, \cos x - \mu (y) \sin x\bigr)^{\top }, \\ \boldsymbol{v}_{2} (x,y) &= \bigl( \xi (y) \cos x, \sin x + \mu (y) \cos x\bigr)^{\top }. \end{aligned} \end{aligned}$$(4.45)and so recover (4.24).$$ \boldsymbol{v}_{0} (x,y) = (1,0)^{\top }, \qquad \boldsymbol{v}_{1} (x,y) = (0,\cos x)^{\top }, \qquad \boldsymbol{v}_{2} (x,y) = (0,\sin x)^{\top }, $$(4.46)
5 Sets of Points Obtained by Discrete Flow Along the Primary Lattice Vector Fields
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if \((x,\boldsymbol{x}) \in D\), then \(x=n \varepsilon \) for some \(n \in \mathbb{Z}\),
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if \((0,\boldsymbol{x}),(0,\boldsymbol{y}) \in D\), then \((0, \boldsymbol{x})^{l} (0,\boldsymbol{y})^{m} = (0,l \boldsymbol{x} + m \boldsymbol{y}) \in D\), \(l,m \in \mathbb{Z}\),
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if \((x,\boldsymbol{x}) \in D\) then \(\boldsymbol{x} = \boldsymbol{\alpha } \mu \), where \(\boldsymbol{\alpha }\) is independent of \(\mu \).
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The proposition holds independently of the choice of \(\hat{\boldsymbol{e}}_{1} = \left ( {\scriptsize\begin{matrix}{} a \cr b \end{matrix}} \right ) \neq \boldsymbol{0}\).It follows thatwhere \(\varepsilon \) is such that (5.14) holds, so \(\pi (D)\) is a two-dimensional lattice, with integral basis \(\mu \hat{\boldsymbol{e}}_{1}\), \(e^{\sigma ^{\top } \varepsilon } (\mu \hat{\boldsymbol{e}}_{1})\), for this choice of isotopy group/projection/group action. Note that the fundamental cell of the lattice can be arbitrarily small.$$ \pi (D) = \bigl\{ l(\mu \hat{\boldsymbol{e}}_{1}) + me^{\sigma ^{\top } \varepsilon } (\mu \hat{\boldsymbol{e}}_{1}) \colon l,m \in \mathbb{Z} \bigr\} , $$(5.20)