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ALMOST ABELIAN LIE GROUPS, SUBGROUPS AND QUOTIENTS

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Abstract

An almost Abelian Lie group is a non-Abelian Lie group with a codimension 1 Abelian normal subgroup. The majority of 3-dimensional real Lie groups are almost Abelian, and they appear in all parts of physics that deal with anisotropic media—cosmology, crystallography etc. In theoretical physics and differential geometry, almost Abelian Lie groups and their homogeneous spaces provide some of the simplest solvmanifolds on which a variety of geometric structures, such as symplectic, Kähler, spin etc., are currently studied in explicit terms. Recently, almost Abelian Lie algebras were classified and studied in details. However, a systematic investigation of almost Abelian Lie groups has not been carried out yet, and the present paper is devoted to an explicit description of properties of this wide and diverse class of groups. The subject of investigation are real almost Abelian Lie groups with their Lie group theoretical aspects, such as the exponential map, faithful matrix representations, discrete and connected subgroups, quotients and automorphisms. The emphasis is put on explicit description of all technical details.

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Funding

The work of Z. Avetisyan, M. Almora Rios, K. Berlow, I. Martin and K. Yang was supported by NSF REU grant DMS 185066. Z. Avetisyan acknowledges the support of the Ministry of Education and Science of Russia, Agreement No. 075-02-2021-1386. H. Zhang was supported by a UCSB CCS SURF grant. Authors G. Rakholia and Z. Zhao received support from the UCSB department of mathematics.

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Correspondence to Zhirayr Avetisyan.

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Dedicated to the memory of Prof. Nikolai Karapetiants on the occasion of his 80th birthday.

Appendix: : proof of Lemma 7

Appendix: : proof of Lemma 7

Let \(1<k\in \mathbb {N}\) and \((v_{1},t_{1}),\ldots ,(v_{k},t_{k})\in \mathbb {R}^{d}\times \mathbb {R}\) such that ti = nit0, \(n_{i}\in \mathbb {Z}\) for i = 1,…,k, where \(t_{0}\in \mathbb {R}\).

Lemma 7 There exists a change of generators \(A\in \text {GL}(\mathbb {Z},k)\) such that

$$\left(\begin{array}{cccc} u_{1} & u_{2} & {\ldots} & u_{k}\\ s_{1} & 0 & {\ldots} & 0 \end{array}\right)=\left(\begin{array}{cccc} v_{1} & v_{2} & {\ldots} & v_{k}\\ t_{1} & t_{2} & {\ldots} & t_{k} \end{array}\right) \cdot A.$$

Remark 12

Here, s1 = dt0, where \(d_{*}=\gcd (n_{1},\ldots ,n_{k})\).

Proof

The statement amounts to the existence of an \(A\in \text {GL}(\mathbb {Z},k)\) such that

$$(d_{*},0,\ldots,0)=(n_{1},n_{2},\ldots,n_{k})\cdot A.$$

Dividing both sides by d we reduce the problem to finding an \(A\in \text {GL}(\mathbb {Z},k)\) such that

$$(1,0,\ldots,0)=(\tilde{n}_{1},\tilde{n}_{2},\ldots,\tilde{n}_{k})\cdot A,$$
(20)

where \(\tilde {n}_{i}=n_{i}/d_{*}\) for i = 1,…,k and \(\gcd (\tilde {n}_{1},\ldots ,\tilde {n}_{k})=1\). Denote

$$d_{1}\doteq\gcd(\tilde{n}_{2},\tilde{n}_{3},\ldots,\tilde{n}_{k}),\quad d_{2}\doteq\gcd(\tilde{n}_{1},\tilde{n}_{3},\ldots,\tilde{n}_{k}),\quad\ldots,\quad d_{k}\doteq\gcd(\tilde{n}_{1},\tilde{n}_{2},\ldots,\tilde{n}_{k-1}),$$
$$m_{1}\doteq\frac{\tilde{n}_{1}}{d_{2}d_{3}{\ldots} d_{k}},\quad m_{2}\doteq\frac{\tilde{n}_{2}}{d_{1}d_{3}{\ldots} d_{k}},\quad\ldots,\quad m_{k}\doteq\frac{\tilde{n}_{k}}{d_{1}d_{2}{\ldots} d_{k-1}},$$

so that \(\tilde {n}_{i}=m_{i}d_{1}d_{2}{\ldots } d_{k}/d_{i}\) for i = 1,…,k and \(\gcd (m_{i},m_{j})=1\) for all ij.

We will define the auxiliary matrix \(B\in \text {GL}(\mathbb {Z},k)\) depending on whether k is even or odd. If k = 2r, then define numbers \(q_{1},\ldots ,q_{k}\in \mathbb {Z}\) such that, by Bézout’s identity, m2j− 1q2j− 1 + m2jq2j = 1 for j = 1,…,r. Then, B is the following matrix,

$$B=\left(\begin{array}{cccccccc} q_{1} & 0 & {\ldots} & 0 & -m_{2} & 0 & {\ldots} & 0\\ q_{2} & 0 & {\ldots} & 0 & m_{1} & 0 & {\ldots} & 0\\ 0 & q_{3} & {\ldots} & 0 & 0 & -m_{4} & {\ldots} & 0\\ 0 & q_{4} & {\ldots} & 0 & 0 & m_{3} & {\ldots} & 0\\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & \vdots\\ 0 & 0 & {\ldots} & q_{k-1} & 0 & 0 & {\ldots} & -m_{k}\\ 0 & 0 & {\ldots} & q_{k} & 0 & 0 & {\ldots} & m_{k-1} \end{array}\right).$$

It is easy to see that indeed, \(|\det B|=1\) and

$$(m_{1},\ldots,m_{r} | m_{r+1},\ldots,m_{k})\cdot B=(1,\ldots,1 | 0,\ldots,0).$$

If on the other hand k = 2r + 3, then we introduce the numbers \(q_{1},\ldots ,q_{k-3}\in \mathbb {Z}\) as before, m2j− 1q2j− 1 + m2jq2j = 1 for j = 1,…,r. Then, again powered by Bézout’s identity, we define integers \(q_{k-2},q_{k-1},q_{k},s_{k-2},s_{k}\in \mathbb {Z}\) such that mk− 2qk− 2 + mk− 1qk− 1 + mkqk = 1 and \(m_{k-2}s_{k_{2}}+m_{k}s_{k}=1\). Now, the matrix B is as follows,

$$B=\left(\begin{array}{ccccccccccc} q_{1} & 0 & {\ldots} & 0 & 0 & -m_{2} & 0 & {\ldots} & 0 & 0 & 0 \\ q_{2} & 0 & {\ldots} & 0 & 0 & m_{1} & 0 & {\ldots} & 0 & 0 & 0 \\ 0 & q_{3} & {\ldots} & 0 & 0 & 0 & -m_{4} & {\ldots} & 0 & 0 & 0 \\ 0 & q_{4} & {\ldots} & 0 & 0 & 0 & m_{3} & {\ldots} & 0 & 0 & 0 \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ 0 & 0 & {\ldots} & q_{k-4} & 0 & 0 & 0 & {\ldots} & -m_{k-4} & 0 & 0 \\ 0 & 0 & {\ldots} & q_{k-3} & 0 & 0 & 0 & {\ldots} & m_{k-3} & 0 & 0 \\ 0 & 0 & {\ldots} & 0 & q_{k-2} & 0 & 0 & {\ldots} & 0 & m_{k} & -m_{k-1}s_{k-2}\\ 0 & 0 & {\ldots} & 0 & q_{k-1} & 0 & 0 & {\ldots} & 0 & 0 & 1 \\ 0 & 0 & {\ldots} & 0 & q_{k} & 0 & 0 & {\ldots} & 0 & -m_{k-2} & -m_{k-1}s_{k} \end{array}\right).$$

Again, it can be observed that \(|\det B|=1\) and

$$(m_{1},\ldots,m_{r+1} | m_{r+2},\ldots,m_{k})\cdot B=(1,\ldots,1 | 0,\ldots,0).$$

For every \(l\in \mathbb {N}\) denote by \(C_{l}\in \text {GL}(\mathbb {Z},l)\), the matrix

$$C_{l}=\left(\begin{array}{ccccc} 1 & -1 & 0 & {\ldots} & 0\\ 0 & 1 & -1 & {\ldots} & 0\\ {\vdots} & {\vdots} & {\vdots} & {\vdots} & \vdots\\ 0 & 0 & 0 & {\ldots} & 1 \end{array}\right).$$

It can be easily seen that

$$(1,\ldots,1,0,\ldots,0)\cdot\left[C_{l}\oplus\mathbf{1}_{k-l}\right]=(1,0,\ldots,0),$$

where exactly l non-zero entries are on the left-hand side. Finally, we define the auxiliary matrix \(D\in \text {GL}(\mathbb {Z},k)\) by \(D=B\cdot \left [C_{r}\oplus \mathbf {1}_{r}\right ]\) or \(D=B\cdot \left [C_{r+1}\oplus \mathbf {1}_{r+2}\right ]\) depending on whether k = 2r or k = 2r + 3, respectively. From what we had above, it is clear that

$$(m_{1},m_{2},\ldots,m_{k})\cdot D=(1,0,\ldots,0).$$
(21)

This property of D (as the more general (20)) is remarkable. It means that the first column D∗1 is a Bézout tuple for (m1,…,mk), while the k − 1 other columns D∗2,…,Dk span the hyperplane orthogonal to (m1,…,mk). It is clear that any other Bézout tuple for (m1,…,mk) is of the form D ⋅ (1,λ2,…,λk) with \((\lambda _{2},\ldots ,\lambda _{k})\in \mathbb {Z}^{k-1}\), and replacing the first column D∗1 in D with any other such tuple will not violate (21).

Remember that \(\gcd (\tilde {n}_{1},\ldots ,\tilde {n}_{k})=1\), so that there exists a Bézout tuple \((p_{1},\ldots ,p_{k})\in \mathbb {Z}^{k}\) such that \(\tilde {n}_{1}p_{1}+\ldots +\tilde {n}_{k}p_{k}=1\). It follows that

$$m_{1}\frac{d_{1}{\ldots} d_{k}}{d_{1}}p_{1}+\ldots+m_{k}\frac{d_{1}{\ldots} d_{k}}{d_{k}}p_{k}=1,$$

that is, d1dk ⋅ (p1/d1,…,pk/dk) is a Bézout tuple for (m1,…,mk), and we can afford setting D∗1 = d1dk ⋅ (p1/d1,…,pk/dk) without changing (21) or \(\det D\).

The desired matrix A can be constructed as below,

$$A=\left(\begin{array}{cccc} p_{1} & D_{1,2}d_{1} & {\ldots} & D_{1,k}d_{1} \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ p_{k} & D_{k,2}d_{k} & {\ldots} & D_{k,k}d_{k} \end{array}\right).$$

We check that (20) is true. Indeed, \((\tilde {n}_{1},\ldots ,\tilde {n}_{k})\cdot (p_{1},\ldots ,p_{k})^{\top }=1\) by definition, whereas

$$(\tilde{n}_{1},\ldots,\tilde{n}_{k})\cdot(D_{1,j}d_{1},\ldots,D_{k,j}d_{k})^{\top}=d_{1}{\ldots} d_{k}\cdot(m_{1},\ldots,m_{k})\cdot(D_{1,j},\ldots,D_{k,j})=0,\quad j=2,\ldots,k$$

follows from (21). Finally,

$$\det A=\left|\mathrm{diag}(d_{1},\ldots,d_{k})\cdot\left(\begin{array}{cccc} p_{1}/d_{1} & D_{1,2} & {\ldots} & D_{1,k} \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ p_{k}/d_{1} & D_{k,2} & {\ldots} & D_{k,k} \end{array}\right)\right|=$$
$$\left|\begin{array}{cccc} d_{1}{\ldots} d_{k}p_{1}/d_{1} & D_{1,2} & {\ldots} & D_{1,k} \\ {\vdots} & {\vdots} & {\vdots} & {\vdots} \\ d_{1}{\ldots} d_{k}p_{k}/d_{1} & D_{k,2} & {\ldots} & D_{k,k} \end{array}\right|=\det D,$$

which proves that \(A\in \text {GL}(\mathbb {Z},k)\).

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Rios, M.A., Avetisyan, Z., Berlow, K. et al. ALMOST ABELIAN LIE GROUPS, SUBGROUPS AND QUOTIENTS. J Math Sci 266, 42–65 (2022). https://doi.org/10.1007/s10958-022-05872-2

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