The subalgebras of the rank two symplectic Lie algebra

TL;DR

This paper completes the classification of all subalgebras of the rank 2 symplectic Lie algebra, including semisimple, Levi decomposable, and solvable subalgebras, providing a comprehensive understanding of its subalgebra structure.

Contribution

It provides the first complete classification of solvable subalgebras of rak{sp}(4,c), completing the classification of all subalgebras of this algebra.

Findings

Classified all solvable subalgebras of rak{sp}(4,c)

Completed the classification of subalgebras of rank 2 semisimple Lie algebras

Unified the understanding of subalgebra structures in rak{sp}(4,c)

Abstract

The semisimple subalgebras of the rank $2$ symplectic Lie algebra $\mathfrak{sp}(4,\mathbb{C})$ are well-known, and we recently classified its Levi decomposable subalgebras. In this article, we classify the solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$, up to inner automorphism. This completes the classification of the subalgebras of $\mathfrak{sp}(4,\mathbb{C})$. More broadly speaking, in completing the classification of the subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ we have completed the classification of the subalgebras of the rank $2$ semisimple Lie algebras.

Tables

Table 1. Table 1. Semisimple classes. Representatives with distinct values of a , b 𝑎 𝑏 a,b give classes which are inequivalent under the Adjoint action of B 𝐵 B . The third column indicates equivalences under the action of P 𝑃 P while the fourth indicates additional equivalences under the action of S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C}) . The last column gives the number of distinct eigenvalues. The pairs of B 𝐵 B -classes with the same 3 3 3 distinct eigenvalues are equivalent under P 𝑃 P . The pairs of B 𝐵 B -classes with the same 2 2 2 distinct eigenvalues are equivalent under S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C}) . Note that the regular elements are exactly those with 4 4 4 distinct eigenvalues.

Representative	Conditions	$P$	$Sp(4,ℂ)$	Eigenvalues
$T_{a,b}$	$a\ne 0$,	$(a,b)\sim (b,a)$	$(a,b)$	$4$
	$b\ne 0$		$\sim (-a,-b)$
	$a\ne \pm b$		$\sim (a,-b)$
$T_{a,0}$	$a\ne 0$	$(a,0)\sim (0,a)$	$(a,0)$	$3$
			$\sim (-a,0)$
$T_{0,a}$	$a\ne 0$	$(0,a)\sim (a,0)$	$(0,a)$	$3$
			$\sim (0,-a)$
$T_{a,-a}$	$a\ne 0$	$(a,-a)$	$(a,-a)$	$2$
		$\sim (-a,a)$	$\sim (a,a)$
$T_{a,a}$	$a\ne 0$		$(a,a)$	$2$
			$\sim (a,-a)$
			$\sim (-a,-a)$
$T_{0,0}$				$1$

Table 2. Table 2. The Nonsemisimple Classes. Distinct values of the parameter a 𝑎 a give representatives for the nonsemisimple classes up to equivalence under conjugation by B 𝐵 B . The third column lists equivalences under conjugation by P 𝑃 P , and the fourth lists additional equivalences under conjugation by S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C}) . The final column gives the block sizes of the Jordan normal form and the corresponding eigenvalues.

Representative	Conditions	$P$	$Sp(4,ℂ)$	JCF
$T_{a,0}+X_{\alpha }$	$a\ne 0$	$\begin{array}{c}\mathrm{conjugate}\\ \mathrm{to}\mathrm{below}\end{array}$	$\begin{array}{c}(a,0,-a,0)\\ \sim (-a,0,a,0)\end{array}$	$\begin{array}{c}(2,1,1)\\ 0,a,-a\end{array}$
$T_{0,a}+X_{\alpha +2\beta }$	$a\ne 0$	$\begin{array}{c}\mathrm{conjugate}\\ \mathrm{to}\mathrm{above}\end{array}$	$\begin{array}{c}(0,a,0,-a)\\ \sim (0,-a,0,a)\end{array}$	$\begin{array}{c}(2,1,1)\\ 0,a,-a\end{array}$
$T_{a,a}+X_{\beta }$	$a\ne 0$		$\begin{array}{c}(a,a)\\ \sim (-a,-a)\\ \mathrm{and}\mathrm{to}\mathrm{below}\end{array}$	$\begin{array}{c}(2,2)\\ a,-a\end{array}$
$T_{a,-a}+X_{\alpha +\beta }$	$a\ne 0$		$\begin{array}{c}(a,-a)\\ \sim (-a,a)\\ \mathrm{and}\mathrm{to}\mathrm{above}\end{array}$	$\begin{array}{c}(2,2)\\ a,-a\end{array}$
$X_{\beta }$		$\begin{array}{c}\\ \mathrm{all}\end{array}$		$\begin{array}{c}(2,2)\\ 0,0\end{array}$
$X_{\alpha +\beta }$		$\begin{array}{c}\mathrm{three}\end{array}$		$\begin{array}{c}(2,2)\\ 0,0\end{array}$
$X_{\alpha }+X_{\alpha +2\beta }$		$\begin{array}{c}\mathrm{conjugate}\end{array}$		$\begin{array}{c}(2,2)\\ 0,0\end{array}$
$X_{\alpha +2\beta }$		$\begin{array}{c}\mathrm{conjugate}\\ \mathrm{to}\mathrm{below}\end{array}$		$\begin{array}{c}(2,1,1)\\ 0,0,0\end{array}$
$X_{\alpha }$		$\begin{array}{c}\mathrm{conjugate}\\ \mathrm{to}\mathrm{above}\end{array}$		$\begin{array}{c}(2,1,1)\\ 0,0,0\end{array}$
$X_{\alpha }+X_{\beta }$				$\begin{array}{c}\left(4\right)\\ 0\end{array}$

Table 3. Table 3. One-dimensional subalgebras of 𝔰 ​ 𝔭 ​ ( 4 , ℂ ) 𝔰 𝔭 4 ℂ \mathfrak{sp}(4,\mathbb{C}) , up to conjugacy by S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C})

Representative	Conditions	Equivalences	Isomorphism Class	Isomorphism Class
			(de Graaf [dGr05])	(Šnobl and Winternitz [ŠW14])
Semisimple
$⟨T_{a,1}⟩$	$a\ne 0,\pm 1$	$⟨T_{a,1}⟩\sim ⟨T_{a^{\prime },1}⟩$	$J$	${𝔫}_{1,1}$
		if and only if $a^{\prime }=\pm a,\pm a^{-1}$
$⟨T_{1,0}⟩$			$J$	${𝔫}_{1,1}$
$⟨T_{1,1}⟩$			$J$	${𝔫}_{1,1}$
Nilpotent
$⟨X_{\alpha }⟩$			$J$	${𝔫}_{1,1}$
$⟨X_{\beta }⟩$			$J$	${𝔫}_{1,1}$
$⟨X_{\alpha }+X_{\beta }⟩$			$J$	${𝔫}_{1,1}$
Non-Trivial Jordan Decomposition
$⟨T_{1,0}+X_{\alpha }⟩$			$J$	${𝔫}_{1,1}$
$⟨T_{1,1}+X_{\beta }⟩$			$J$	${𝔫}_{1,1}$

Table 4. Table 4. Two-dimensional subalgebras of 𝔰 ​ 𝔭 ​ ( 4 , ℂ ) 𝔰 𝔭 4 ℂ \mathfrak{sp}(4,\mathbb{C}) , up to conjugacy by S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C})

Representative	Conditions	Equivalences	Isomorphism Class	Isomorphism Class
			(de Graaf [dGr05])	(Šnobl and Winternitz [ŠW14])
Semisimple
$𝔱$			$K^1$	2${𝔫}_{1,1}$
Containing both Semisimple and Nilpotent Elements
$⟨T_{3,1},X_{\alpha }+X_{\beta }⟩$			$K^2$	${𝔰}_{2,1}$
$⟨T_{a,1},X_{\alpha }⟩$	$a\ne 0,\pm 1$	$⟨T_{a,1},X_{\alpha }⟩\sim ⟨T_{-a,1},X_{\alpha }⟩$	$K^2$	${𝔰}_{2,1}$
$⟨T_{a,1},X_{\beta }⟩$	$a\ne 0,\pm 1$	$⟨T_{a,1},X_{\beta }⟩\sim ⟨T_{a^{-1},1},X_{\beta }⟩$	$K^2$	${𝔰}_{2,1}$
$⟨T_{1,0},X_{\alpha }⟩$			$K^1$	2${𝔫}_{1,1}$
$⟨T_{1,0},X_{\beta }⟩$			$K^2$	${𝔰}_{2,1}$
$⟨T_{1,0},X_{\alpha +2\beta }⟩$			$K^2$	${𝔰}_{2,1}$
$⟨T_{1,1},X_{\alpha }⟩$			$K^2$	${𝔰}_{2,1}$
$⟨T_{1,1},X_{\beta }⟩$			$K^1$	2${𝔫}_{1,1}$
$⟨T_{1,1},X_{\alpha +\beta }⟩$			$K^2$	${𝔰}_{2,1}$
Containing no Semisimple Elements, but not Nilpotent
$⟨T_{1,1}+X_{\beta },X_{\alpha +2\beta }⟩$			$K^2$	${𝔰}_{2,1}$
$⟨T_{1,0}+X_{\alpha },X_{\alpha +\beta }⟩$			$K^2$	${𝔰}_{2,1}$
$⟨T_{1,0}+X_{\alpha },X_{\alpha +2\beta }⟩$			$K^2$	${𝔰}_{2,1}$
Nilpotent
$⟨X_{\alpha },X_{\alpha +\beta }⟩$			$K^1$	2${𝔫}_{1,1}$
$⟨X_{\alpha },X_{\alpha +2\beta }⟩$			$K^1$	2${𝔫}_{1,1}$
$⟨X_{\beta }+X_{\alpha },X_{\alpha +2\beta }⟩$			$K^1$	2${𝔫}_{1,1}$

Table 5. Table 5. Three-dimensional solvable subalgebras of 𝔰 ​ 𝔭 ​ ( 4 , ℂ ) 𝔰 𝔭 4 ℂ \mathfrak{sp}(4,\mathbb{C}) , up to conjugacy by S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C})

Representative	Conditions	Equivalences	Isomorphism Class	Isomorphism Class
			(de Graaf [dGr05])	(Šnobl and Winternitz [ŠW14])
Containing a Cartan Subalgebra
$⟨𝔱,X_{\alpha }⟩$			$L_0^3$	${𝔫}_{1,1}\oplus {𝔰}_{2,1}$
$⟨𝔱,X_{\beta }⟩$			$L_0^3$	${𝔫}_{1,1}\oplus {𝔰}_{2,1}$
Containing a Regular Semisimple Element but not a Cartan Subalgebra
$⟨T_{a,1},X_{\alpha },X_{\alpha +\beta }⟩$	$a\ne 0,\pm 1,-3$		$L_{-2\left(a+1\right)/{\left(a+3\right)}^2}^3$	${𝔰}_{3,1,A=\frac{1+2\alpha +\sqrt{1+4\alpha }}{-2\alpha }}$
				where $\alpha =-2\left(a+1\right)/{\left(a+3\right)}^2$
$⟨T_{-3,1},X_{\alpha },X_{\alpha +\beta }⟩$			$L_1^4$	${𝔰}_{3,1,A=-1}$
$⟨T_{a,1},X_{\alpha },X_{\alpha +2\beta }⟩$	$a\ne 0,\pm 1$	$⟨T_{a,1},X_{\alpha },X_{\alpha +2\beta }⟩\sim ⟨T_{a^{-1},1},X_{\alpha },X_{\alpha +2\beta }⟩$	$L_{-a/{\left(a+1\right)}^2}^3$	${𝔰}_{3,1,A=\frac{1+2\alpha +\sqrt{1+4\alpha }}{-2\alpha }}$
				where $\alpha =-a/{\left(a+1\right)}^2$
$⟨T_{3,1},X_{\alpha }+X_{\beta },X_{\alpha +2\beta }⟩$			$L_{-3/16}^3$	${𝔰}_{3,1,A=1/3}$
Containing a Non-regular Semisimple Element but not a Cartan Subalgebra
$⟨T_{1,0},X_{\alpha },X_{\alpha +\beta }⟩$			$L_0^3$	${𝔫}_{1,1}\oplus {𝔰}_{2,1}$
$⟨T_{1,0},X_{\alpha },X_{\alpha +2\beta }⟩$			$L_0^3$	${𝔫}_{1,1}\oplus {𝔰}_{2,1}$
$⟨T_{1,0},X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$L_{-2/9}^3$	${𝔰}_{3,1,A=1/2}$
$⟨T_{1,-1},X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$L_0^3$	${𝔫}_{1,1}\oplus {𝔰}_{2,1}$
$⟨T_{1,-1},X_{\alpha },X_{\alpha +2\beta }⟩$			$L_1^4$	${𝔰}_{3,1,A=-1}$
$⟨T_{1,-1},X_{\beta },X_{\alpha +2\beta }⟩$			$L^2$	${𝔰}_{3,1,A=1}$
$⟨T_{1,1},X_{\alpha },X_{\alpha +2\beta }⟩$			$L^2$	${𝔰}_{3,1,A=1}$
Containing no Semisimple Elements but not Nilpotent
$⟨T_{1,1}+X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$L_{-1/4}^3$	${𝔰}_{3,2}$
$⟨T_{1,-1}+X_{\alpha +\beta },X_{\alpha },X_{\alpha +2\beta }⟩$			$L_1^4$	${𝔰}_{3,1,A=-1}$
$⟨T_{1,0}+X_{\alpha },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$L_{-2/9}^3$	${𝔰}_{3,1,A=1/2}$
Nilpotent
${𝔫}_{𝔭}$			$L^1$	$3{𝔫}_{1,1}$
$⟨X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$L_0^4$	${𝔫}_{3,1}$
$⟨X_{\alpha }+X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$L_0^4$	${𝔫}_{3,1}$

Table 6. Table 6. Four-dimensional solvable subalgebras of 𝔰 ​ 𝔭 ​ ( 4 , ℂ ) 𝔰 𝔭 4 ℂ \mathfrak{sp}(4,\mathbb{C}) , up to conjugacy by S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C})

Representative	Conditions	Equivalences	Isomorphism Class	Isomorphism Class
			(de Graaf [dGr05])	(Šnobl and Winternitz [ŠW14])
Containing a Cartan Subalgebra
$⟨𝔱,X_{\alpha },X_{\alpha +\beta }⟩$			$M^8$	${𝔰}_{4,12}$
$⟨𝔱,X_{\alpha },X_{\alpha +2\beta }⟩$			$M^8$	${𝔰}_{4,12}$
Containing a Regular Semisimple Element but not a Cartan Subalgebra
$⟨T_{a,1},{𝔫}_{𝔭}⟩$	$a\ne 0,\pm 1$	$⟨T_{a,1},{𝔫}_{𝔭}⟩\sim ⟨T_{a^{-1},1},{𝔫}_{𝔭}⟩$	$M_{A,B}^6$, where	see Subsection 10.3
		$\sim ⟨T_{-a,1},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$	$A=\frac{4a}{27{\left(a+1\right)}^2}$, $B=-\frac{2\left(a^2+4a+1\right)}{9{\left(a+1\right)}^2}$
$⟨T_{a,1},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$	$a\ne 0,\pm 1$	$⟨T_{a,1},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$	$M_{\left(1-a^2\right)/\left(4a^2\right)}^{13}$	${𝔰}_{4,8,A=\frac{1+2\alpha +\sqrt{1+4\alpha }}{-2\alpha }}$
		$\sim ⟨T_{-a,1},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$		where $\alpha =\left(1-a^2\right)/\left(4a^2\right)$
$⟨T_{3,1},X_{\alpha }+X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$M_{-2/9}^{13}$	${𝔰}_{4,8,A=1/2}$
Containing a Non-regular Semisimple Element but not a Cartan Subalgebra
$⟨T_{0,1},{𝔫}_{𝔭}⟩$			$M_{0,-2/9}^6$	${𝔫}_{1,1}\oplus {𝔰}_{3,1,A=1/2}$
$⟨T_{0,1},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$M_1^{14}$	${𝔰}_{4,6}$
$⟨T_{1,0},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$M^{12}$	${𝔰}_{4,8,A=1}$
$⟨T_{1,1},{𝔫}_{𝔭}⟩$			$M^2$	${𝔰}_{4,3,A=B=1}$
$⟨T_{1,-1},{𝔫}_{𝔭}⟩$			$M_{0,1}^7$	${𝔫}_{1,1}\oplus {𝔰}_{3,1,A=-1}$
$⟨T_{1,1},X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$M_0^{13}$	${𝔰}_{4,11}$
Containing no Semisimple Elements but not Nilpotent
$⟨T_{1,1}+X_{\beta },X_{\alpha },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$M_{1/27,-1/3}^6$	${𝔰}_{4,2}$
$⟨T_{1,0}+X_{\alpha },X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$			$M_{-1/4}^{13}$	${𝔰}_{4,10}$
Nilpotent Subalgebras
$𝔫$			$M_{0,0}^7$	${𝔫}_{4,1}$

Table 7. Table 7. Five- and six-dimensional solvable subalgebras of 𝔰 ​ 𝔭 ​ ( 4 , ℂ ) 𝔰 𝔭 4 ℂ \mathfrak{sp}(4,\mathbb{C}) , up to conjugacy by S ​ p ​ ( 4 , ℂ ) 𝑆 𝑝 4 ℂ Sp(4,\mathbb{C})

Dimension	Representative	Conditions	Isomorphism Class
			(Šnobl and Winternitz [ŠW14])
Containing a Cartan Subalgebra
$5$	$⟨𝔱,{𝔫}_{𝔭}⟩$		${𝔰}_{5,41,A=B=\frac{1}{2}}$
$5$	$⟨𝔱,X_{\beta },X_{\alpha +\beta },X_{\alpha +2\beta }⟩$		${𝔰}_{5,44}$
Containing a Regular Semisimple Element but not a Cartan Subalgebra
$5$	$⟨T_{a,1},𝔫⟩$	$a\ne 0,\pm 1$	${𝔰}_{5,35,A=\frac{2}{a-1}}$
Containing a Non-regular Semisimple Element but not a Cartan Subalgebra
$5$	$⟨T_{1,-1},𝔫⟩$		${𝔰}_{5,35,A=-1}$
$5$	$⟨T_{1,1},𝔫⟩$		${𝔰}_{5,37}$
$5$	$⟨T_{1,0},𝔫⟩$		${𝔰}_{5,36}$
$5$	$⟨T_{0,1},𝔫⟩$		${𝔰}_{5,33}$
$6$	$𝔟$		${𝔰}_{6,242}$

Equations

$$\begin{array}[]{llll}J&\text{The abelian Lie algebra of dimension $1$}\end{array}$$

$$\begin{array}[]{llll}K^{1}&\text{The abelian Lie algebra of dimension $2$}\\ K^{2}&[x_{1},x_{2}]=x_{1}\end{array}$$

$$\begin{array}[]{llll}L^{1}&\text{The abelian Lie algebra of dimension $3$}\\ L^{2}&[x_{3},x_{1}]=x_{1},[x_{3},x_{2}]=x_{2}\\ L^{3}_{A}&[x_{3},x_{1}]=x_{2},[x_{3},x_{2}]=Ax_{1}+x_{2}\\ L^{4}_{A}&[x_{3},x_{1}]=x_{2},[x_{3},x_{2}]=Ax_{1}\end{array}$$

$$\begin{array}[]{llll}M^{2}&[x_{4},x_{1}]=x_{1},[x_{4},x_{2}]=x_{2},[x_{4},x_{3}]=x_{3}\\ M^{6}_{A,B}&[x_{4},x_{1}]=x_{2},[x_{4},x_{2}]=z_{3},[x_{4},x_{3}]=Ax_{1}+Bx_{2}+x_{3}\\ M^{7}_{A,B}&[x_{4},x_{1}]=x_{2},[x_{4},x_{2}]=x_{3},[x_{4},x_{3}]=Ax_{1}+Bx_{2}\\ M^{8}&[x_{1},x_{2}]=x_{2},[x_{3},x_{4}]=x_{4}\\ M^{12}&[x_{4},x_{1}]=x_{1},[x_{4},x_{2}]=2x_{2},[x_{4},x_{3}]=x_{3},[x_{3},x_{1}]=x_{2}\\ M^{13}_{A}&[x_{4},x_{1}]=x_{1}+Ax_{3},[x_{4},x_{2}]=x_{2},[x_{4},x_{3}]=x_{1},[x_{3},x_{1}]=x_{2}\\ M^{14}_{A}&[x_{4},x_{1}]=Ax_{3},[x_{4},x_{3}]=x_{1},[x_{3},x_{1}]=x_{2},A\neq 0\end{array}$$

$$\begin{array}[]{llll}\mathfrak{n}_{1,1}&\text{The abelian Lie algebra of dimension $1$}\\ \end{array}$$

$$\begin{array}[]{llll}\mathfrak{s}_{2,1}&[e_{2},e_{1}]=e_{1}\end{array}$$

$$\begin{array}[]{llll}\mathfrak{n}_{3,1}&[e_{2},e_{3}]=e_{1}\\ \mathfrak{s}_{3,1}&[e_{3},e_{1}]=e_{1},[e_{3},e_{2}]=Ae_{2}\\ &0<|A|\leq 1,~{}\text{if}~{}|A|=1~{}\text{then}~{}\arg(A)\leq\pi\\ \mathfrak{s}_{3,2}&[e_{3},e_{1}]=e_{1},[e_{3},e_{2}]=e_{1}+e_{2}\end{array}$$

$$\begin{array}[]{llll}\mathfrak{n}_{4,1}&[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2}\\ \mathfrak{s}_{4,2}&[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=e_{1}+e_{2},[e_{4},e_{3}]=e_{2}+e_{3}\\ \mathfrak{s}_{4,3}&[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=Ae_{2},[e_{4},e_{3}]=Be_{3}\\ &0<|B|\leq|A|\leq 1,(A,B)\neq(-1,-1)\\ \mathfrak{s}_{4,6}&[e_{2},e_{3}]=e_{1},[e_{4},e_{2}]=e_{2},[e_{4},e_{3}]=-e_{3}\\ \mathfrak{s}_{4,8}&[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=(1+A)e_{1},[e_{4},e_{2}]=e_{2},[e_{4},e_{3}]=Ae_{3}\\ &0<|A|\leq 1,~{}\text{if}~{}|A|=1~{}\text{then}~{}\arg(A)<\pi\\ \mathfrak{s}_{4,10}&[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=2e_{1},[e_{4},e_{2}]=e_{2},[e_{4},e_{3}]=e_{2}+e_{3}\\ \mathfrak{s}_{4,11}&[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=e_{2}\\ \mathfrak{s}_{4,12}&[e_{3},e_{1}]=e_{1},[e_{3},e_{2}]=e_{2},[e_{4},e_{1}]=-e_{2},[e_{4},e_{2}]=e_{1}\end{array}$$

$$\begin{array}[]{llll}\mathfrak{s}_{5,33}&[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2},[e_{5},e_{2}]=-e_{2},\\ &[e_{5},e_{3}]=-2e_{3},[e_{5},e_{4}]=e_{4}\\ \mathfrak{s}_{5,35}&[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2},[e_{5},e_{1}]=(A+2)e_{1},\\ &[e_{5},e_{2}]=(A+1)e_{2},[e_{5},e_{3}]=Ae_{3},[e_{5},e_{4}]=e_{4},\\ &A\neq 0,-2\\ \mathfrak{s}_{5,36}&[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2},[e_{5},e_{1}]=2e_{1},\\ &[e_{5},e_{2}]=e_{2},[e_{5},e_{4}]=e_{4}\\ \mathfrak{s}_{5,37}&[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2},[e_{5},e_{1}]=e_{1},\\ &[e_{5},e_{2}]=e_{2},[e_{5},e_{3}]=e_{3}\\ \mathfrak{s}_{5,41}&[e_{4},e_{1}]=e_{1},[e_{4},e_{3}]=Ae_{3},\\ &[e_{5},e_{2}]=e_{2},[e_{5},e_{3}]=Be_{3},\\ &0<|B|\leq|A|\leq 1\\ \mathfrak{s}_{5,44}&[e_{2},e_{3}]=e_{1},[e_{4},e_{1}]=e_{1},[e_{4},e_{2}]=e_{2},\\ &[e_{5},e_{2}]=e_{2},[e_{5},e_{3}]=-e_{3}\end{array}$$

$$\begin{array}[]{llll}\mathfrak{s}_{6,242}&[e_{2},e_{4}]=e_{1},[e_{3},e_{4}]=e_{2},[e_{5},e_{1}]=2e_{1},\\ &[e_{5},e_{2}]=e_{2},[e_{5},e_{4}]=e_{4},[e_{6},e_{1}]=e_{1},\\ &[e_{6},e_{2}]=e_{2},[e_{6},e_{3}]=e_{3}\end{array}$$

$$J=\left(\begin{array}{cccc}0& 0& 1& 0\\ 0& 0& 0& 1\\ -1& 0& 0& 0\\ 0& -1& 0& 0\end{array}\right).$$

$$T_{a,b}=\mathrm{diag}(a,b,-a,-b).$$

$$X_{\alpha }=\left(\begin{array}{cccc}0& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),$$

$$X_{\alpha +\beta }=\left(\begin{array}{cccc}0& 0& 0& 1\\ 0& 0& 1& 0\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right),$$

$$\mathfrak{b}=\left(\begin{array}{cccc}\ast & \ast & \ast & \ast \\ 0& \ast & \ast & \ast \\ 0& 0& \ast & 0\\ 0& 0& \ast & \ast \end{array}\right)\cap \mathfrak{sp}(4,\mathbb{C}),$$

$$\mathfrak{n}=\left(\begin{array}{cccc}0& \ast & \ast & \ast \\ 0& 0& \ast & \ast \\ 0& 0& 0& 0\\ 0& 0& \ast & 0\end{array}\right)\cap \mathfrak{sp}(4,\mathbb{C})=\left\{\left(\begin{array}{cccc}0& t& a& b\\ 0& 0& b& d\\ 0& 0& 0& 0\\ 0& 0& -t& 0\end{array}\right)\right\},$$

$$\mathfrak{p}=\left(\begin{array}{cccc}\ast & \ast & \ast & \ast \\ \ast & \ast & \ast & \ast \\ 0& 0& \ast & \ast \\ 0& 0& \ast & \ast \end{array}\right)\cap \mathfrak{sp}(4,\mathbb{C}),$$

$${\mathfrak{n}}_{\mathfrak{p}}=\left(\begin{array}{cccc}0& 0& \ast & \ast \\ 0& 0& \ast & \ast \\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\cap \mathfrak{sp}(4,\mathbb{C})=\left\{\left(\begin{array}{cccc}0& 0& a& b\\ 0& 0& b& d\\ 0& 0& 0& 0\\ 0& 0& 0& 0\end{array}\right)\right\},$$

$$\begin{array}[]{llllllll}\langle T_{1,b}\rangle\cong\langle T_{1,b^{-1}}\rangle,b\neq 0,\pm 1;&\langle T_{1,0}\rangle,;&\langle T_{1,1}\rangle.\end{array}$$

$$\begin{array}[]{llllllll}\langle X_{\beta}\rangle;&\langle X_{\alpha}\rangle;&\langle X_{\alpha}+X_{\beta}\rangle.\end{array}$$

$$\begin{array}[]{llllllll}\langle T_{1,0}+X_{\alpha}\rangle;&\langle T_{1,1}+X_{\beta}\rangle.\end{array}$$

$$T_{1,1}+c{X}_{\beta }{T}_{1,-1}+c{X}_{\alpha +\beta }{T}_{1,0}+c{X}_{\alpha }{T}_{0,1}+c{X}_{\alpha +2\beta },$$

$$T_{1,1}+X_{\beta }{T}_{1,-1}+X_{\alpha +\beta }{T}_{1,0}+X_{\alpha }{T}_{0,1}+X_{\alpha +2\beta }.$$

$$W=\left(\begin{array}{cccc}0& 1& 0& 0\\ -1& 0& 0& 0\\ 0& 0& 0& 1\\ 0& 0& -1& 0\end{array}\right)\in Sp(4,\mathbb{C}).$$

$$W(T_{1,0}+X_{\alpha })W^{-1}=T_{0,1}+X_{\alpha +2\beta }.$$

$$A=\left(\begin{array}{cccc}1& 0& 0& 0\\ 0& 0& 0& -1\\ 0& 0& 1& 0\\ 0& 1& 0& 0\end{array}\right)\in Sp(4,\mathbb{C}).$$

$$A(T_{1,1}+X_{\beta })A^{-1}=T_{1,-1}+X_{\alpha +\beta }.$$

$${\mathfrak{s}}_Z=\{X\in \mathfrak{gl}(2,\mathbb{C})|X=X^t\mathrm{and}\mathrm{tr}(X{Z}^t)=0\}$$

$$\begin{array}[]{lllll}&{\rm tr}(XZ^{t})\\ =&{\rm tr}(gXZ^{t}g^{-1})={\rm tr}(gXg^{t}g^{-t}Z^{t}g^{-1})={\rm tr}\bigl{(}gXg^{t}(g^{-t}Zg^{-1})^{t}\bigr{)}.\end{array}$$

$$⟨X_{\alpha },X_{\alpha +\beta }⟩\mathrm{and}⟨X_{\alpha },X_{\alpha +2\beta }⟩.$$

$$\begin{array}[]{llll}\text{ad}(T^{\prime\prime})(X)&=&\sum_{i}\lambda_{i}N_{i},\\ \text{ad}(T^{\prime\prime})^{2}(X)&=&\sum_{i}\lambda_{i}^{2}N_{i},\\ \text{ad}(T^{\prime\prime})^{3}(X)&=&\sum_{i}\lambda_{i}^{3}N_{i},\\ \text{ad}(T^{\prime\prime})^{4}(X)&=&\sum_{i}\lambda_{i}^{4}N_{i},\end{array}$$

Full text

1. Introduction

Semisimple subalgebras of semisimple Lie algebras have been extensively studied [dGr11, Dyn52a, Dyn52b, LG72, Min06]. For instance, the semisimple subalgebras of the exceptional Lie algebras have been classified, up to inner automorphism [Min06]. As another important example, de Graaf [dGr11] classified the semisimple subalgebras of the simple Lie algebras of ranks $\leq 8$, up to linear equivalence, which is somewhat weaker than a classification up to inner automorphism.

Much less research has examined general subalgebras of semisimple Lie algebras. By Levi’s Theorem [[ŠW14], Chapter II, Section 2], a subalgebra of a complex semisimple Lie algebra is either semisimple, solvable, or a nontrivial semidirect sum of the first two. A subalgebra that is a nontrivial semidirect sum of a semisimple subalgebra with a solvable subalgebra is called a Levi decomposable subalgebra.

We have made considerable progress towards classifying both solvable and Levi decomposable subalgebras of semisimple Lie algebras. Most relevant among this work to the present paper is our classification of the solvable, and Levi decomposable subalgebras of the rank $2$, semisimple Lie algebras $\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$ [DR16a], and $\mathfrak{sl}(3,\mathbb{C})$ [DR16b]. Since the classifications of semisimple subalgebras of $\mathfrak{sl}(2,\mathbb{C})\oplus\mathfrak{sl}(2,\mathbb{C})$, and $\mathfrak{sl}(3,\mathbb{C})$ are well-known, our work completes the classification of subalgebras of these rank $2$, semisimple Lie algebras.

The aim of the current paper is to complete the classification of subalgebras of the rank $2$, symplectic Lie algebra $\mathfrak{sp}(4,\mathbb{C})$–the remaining rank $2$, classical, semisimple Lie algebra whose subalgebras have not been classified. The semisimple subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ are well-known [dGr11], and the authors recently classified its Levi decomposable subalgebras [DR15].

Hence, in this article, we classify the most difficult case: the solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$, up to inner automorphism (equivalently, up to conjugacy by the symplectic group $Sp(4,\mathbb{C})$). By Levi’s theorem, this completes the classification of the subalgebras of $\mathfrak{sp}(4,\mathbb{C})$.

In addition, Mayanskiy [May16] recently posted a classification of the subalgebras of the exceptional Lie algebra $G_{2}$. In light of the above mentioned work, this article completes of the classification of the subalgebras of the rank $2$ semisimple Lie algebras.

In addition to the intrinsic mathematical significance of classifications of subalgebras of Lie algebras (or classifications of subgroups of corresponding Lie groups), such classifications also have physical significance and mathematical application, some of which are listed below:

• •

If a system of differential equations is invariant under a Lie group, then its subgroups can be used to construct group invariant solutions [Olv86].

• •

Subgroups of the symmetry groups of nonlinear partial differential equations provide a method for performing symmetry reduction (reducing the number of independent variables) [ORWZ90, GHP84, DKLW86].

• •

A knowledge of the subgroup structure of a Lie group $G$ is needed if we are interested in considering all possible contractions of $G$ to other groups [PW77].

• •

Physical models–such as the vibron model, and the interacting boson model– use chains of subalgebras, and these subalgebras need to be explicitly described in application [IA87].

The article is organized as follows. In Section 2, we describe two partial classifications of solvable Lie algebras that will be used in our classification of solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$: The classification of de Graaf [dGr05], and that described by Šnobl and Winternitz in [ŠW14]. Section $3$ contains preliminary background on $\mathfrak{sp}(4,\mathbb{C})$, and in Section $4$ we classify the one-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$.

In Section $5$ we develop preliminary results that will be used in classifications of subalgebras of dimension greater than one. In Sections $6$, $7$, and $8$ we classify the two-, three-, and four-dimensional solvable subalgebras, respectively. Section $9$ contains the classification of the five-, and six-dimensional solvable subalgebras.

Finally, in Section 10, we identify our classification of solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with respect to the classification of solvable Lie algebras of de Graaf [dGr05], and that described by Šnobl and Winternitz in [ŠW14]. The complete classification of the subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ is summarized in Tables 3 to 7.

All Lie algebras and representations in this article are finite dimensional, and over the complex numbers, unless otherwise stated.

2. Solvable Lie algebras of small dimension

A full classification of solvable Lie algebras is not known and thought to be an impossible task. However, partial classifications of solvable Lie algebras do exist. Two such partial classifications are that of de Graaf [dGr05], and that described by Šnobl and Winternitz in [ŠW14]. The classification of solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ in this article will be described with respect to both of these classifications, and we consider both of these classifications in this section.

De Graaf classified the solvable Lie algebras in dimensions $\leq 4$ over a field $\mathbb{F}$ of any characteristic [dGr05]. In his classification, de Graaf does not distinguish between indecomposable and decomposable Lie algebras. We describe the classification in its entirety up to and including dimension $3$, and include only those four-dimensional solvable subalgebras which appear in this article:

[TABLE]

[TABLE]

[TABLE]

Note that $L^{3}_{A}\cong L^{3}_{B}$ if and only if $A=B$; and $L^{4}_{A}\cong L^{4}_{B}$ if and only if there is an $\alpha\in\mathbb{F}^{*}$ with $A=\alpha^{2}B$.

[TABLE]

Note that $M^{6}_{A,B}\cong M^{6}_{C,D}$ if and only if $A=C$ and $B=D$; $M^{7}_{A,B}\cong M^{7}_{C,D}$ if and only if there is an $\alpha\in\mathbb{F}^{*}$ with $A=\alpha^{3}C$ and $B=\alpha^{2}D$; $M^{13}_{A}\cong M^{13}_{B}$ if and only if $A=B$; and $M^{14}_{A}\cong M^{14}_{B}$ if and only if there is an $\alpha\in\mathbb{F}^{*}$ with $A=\alpha^{2}B$.

An alternate classification of solvable Lie algebras is presented by Šnobl and Winternitz in [ŠW14], which is up to and including dimension $6$ and includes only indecomposable Lie algebras. This classification is an amalgam of results from various sources (e.g., [Bia98, Kru54, Lie88, Mor58, PSWZ76, ST13, Tur88, Tur90]).

We present the classification from [ŠW14] in its entirety up to and including dimension $3$. We give a partial description of the classification in dimensions $4$, $5$ and $6$, including just those algebras which appear in this article:

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

[TABLE]

In the classification from [ŠW14], an algebra designated with $\mathfrak{n}$ is nilpotent, and one with $\mathfrak{s}$ is solvable, but not nilpotent. The first subscript indicates the dimension, and the second index is for enumeration. So, $\mathfrak{s}_{6,242}$ is the $242^{\text{nd}}$ six-dimensional, solvable, non-nilpotent Lie algebra in the classification.

3. Preliminaries

The symplectic algebra $\mathfrak{sp}(4,\mathbb{C})$ is the Lie algebra of $4\times 4$ complex matrices $X$ satisfying $JX^{t}J=X$, where $J$ is the $4\times 4$ matrix

[TABLE]

The corresponding Lie group is the symplectic group $Sp(4,\mathbb{C})$ given by $\{g\in GL(4,\mathbb{C})\mid gJg^{t}=J\}$.

Let $\mathfrak{t}$ be the diagonal Cartan subalgebra, and $T$ the corresponding Cartan subgroup. For $a,b\in\mathbb{C}$, define

[TABLE]

If we choose the positive root vectors to be

[TABLE]

then $X_{\alpha}$ and $X_{\alpha+2\beta}$ correspond to the long roots, while $X_{\beta}$ and $X_{\alpha+\beta}$ correspond to the short roots. The corresponding Borel subalgebra is

[TABLE]

and we let $B\subset Sp(4,\mathbb{C})$ be the corresponding subgroup. Let $\mathfrak{n}$ be its nilpotent radical

[TABLE]

and let $N\subset Sp(4,\mathbb{C})$ be the corresponding subgroup. Also, let $\mathfrak{p}$ be the maximal parabolic subalgebra

[TABLE]

and let $P\subset Sp(4,\mathbb{C})$ be the corresponding subgroup.

Let $\mathfrak{n}_{\mathfrak{p}}$ be the nilpotent radical of $\mathfrak{p}$:

[TABLE]

and let $N_{P}\subset Sp(4,\mathbb{C})$ be the corresponding subgroup.

The following lemmas will be used below.

Lemma 3.1**.**

The nonzero eigenvalues of an element of $\mathfrak{sp}(4,\mathbb{C})$ occur in negative pairs (that is, its eigenvalues are of the form $a,b,-a,-b$). The nonzero generalized eigenvalues of an element of $\mathfrak{sp}(4,\mathbb{C})$ occur in negative pairs.

Proof.

Any element is conjugate to an element of $\mathfrak{b}$. The two diagonal blocks of such an element are negative transposes, so have the negatives of each other’s eigenvalues and generalized eigenvalues. ∎

Lemma 3.2**.**

Suppose $X\in\mathfrak{b}$ is semisimple. Then there is $b\in B$ such that $bXb^{-1}\in\mathfrak{t}$, i.e., any semisimple element of $\mathfrak{b}$ is conjugate to an element of $\mathfrak{t}$ by an element of $B$.

Moreover, if $X=T+N$, with $T\in\mathfrak{t}$, $N\in\mathfrak{n}$, then $bXb^{-1}=T$.

Proof.

Since $X\in\mathfrak{b}$ is semisimple, it is an element of a Cartan subalgebra of $\mathfrak{b}$. Since all Cartan subalgebras of $\mathfrak{b}$ are conjugate under $B$ [[Hum72], Theorem 16.2], the first assertion follows.

For the second assertion, let $b\in B$ such that $bXb^{-1}\in\mathfrak{t}$. Then $bXb^{-1}=b(T+N)b^{-1}$. Clearly $bNb^{-1}\in\mathfrak{n}$, and $bTb^{-1}=T+N^{\prime}$, with $N^{\prime}\in\mathfrak{n}$. We must have $N^{\prime}=-bNb^{-1}$ and $bXb^{-1}=T$. ∎

We summarize the classification of the conjugacy classes of $\mathfrak{sp}(4,\mathbb{C})$ in the following two tables, which distinguish conjugacy by $B$, $P$, and $G=Sp(4,\mathbb{C})$.

4. One-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$

Any solvable subalgebra of $\mathfrak{sp}(4,\mathbb{C})$ is contained in a Borel subalgebra and hence is conjugate to a subalgebra of $\mathfrak{b}$. Accordingly, we shall focus on solvable subalgebras of $\mathfrak{b}$.

In this section, we classify the one-dimensional (solvable) subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ by separating them into three cases: subalgebras with semisimple generators (Theorem 4.1), subalgebras with nilpotent generators (Theorem 4.2), and subalgebras with generators that have non-trivial Jordan decompositions (Theorem 4.3). The results are summarized in Table 3.

Theorem 4.1**.**

Every semisimple element of $\mathfrak{sp}(4,\mathbb{C})$ is conjugate to an element $T_{a,b}$ in $\mathfrak{t}$ (c.f., Eq. (12)).

A complete list of one-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with semisimple generators, up to conjugacy in $Sp(4,\mathbb{C})$, is

[TABLE]

If $a,b\neq 0$, $b\neq\pm a$, then the subalgebras $\langle T_{a,b}\rangle$, $\langle T_{a,0}\rangle$ and $\langle T_{a,a}\rangle$ are pairwise inequivalent; $\langle T_{a,0}\rangle=\langle T_{1,0}\rangle$; $\langle T_{a,a}\rangle=\langle T_{1,1}\rangle$, which is conjugate to $\langle T_{1,-1}\rangle$; and $\langle T_{a,b}\rangle$ is conjugate to $\langle T_{a,^{\prime}b^{\prime}}\rangle$ if and only if $\{a,b\}=\{\lambda a^{\prime},\pm\lambda b^{\prime}\}$, for some $\lambda\in\mathbb{C}^{\times}$.

Proof.

Every semisimple element $T$ of $\mathfrak{sp}(4,\mathbb{C})$ is conjugate to an element in $\mathfrak{t}$ [[CM93], Corollary 2.2.2], so we may assume $T\in\mathfrak{t}$.

Two elements in $\mathfrak{t}$ are $Sp(4,\mathbb{C})$-conjugate if and only if they are $W$-conjugate [[CM93], Theorem 2.2.4] , where $W$ is the Weyl group corresponding to $\mathfrak{t}$. The Weyl group $W$ of $\mathfrak{sp}(4,\mathbb{C})$ has generator $s_{\alpha}$ and $s_{\beta}$ such that $s_{\alpha}(T_{a,b})=T_{a,-b}$ and $s_{\beta}(T_{a,b})=T_{b,a}$. The result follows. ∎

Theorem 4.2**.**

A complete list of inequivalent, one-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with nilpotent generators, up to conjugacy in $Sp(4,\mathbb{C})$, is

[TABLE]

Proof.

There are precisely three nonzero nilpotent orbits of $\mathfrak{sp}(4,\mathbb{C})$ [[CM93], Theorem 5.1.3] with representatives $X_{\alpha},X_{\beta}$, and $X_{\alpha}+X_{\beta}$. ∎

Theorem 4.3**.**

A complete list of one-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with generators having a nontrivial Jordan decomposition, i.e., that are neither semisimple nor nilpotent, up to conjugacy in $Sp(4,\mathbb{C})$, is

[TABLE]

The subalgebras $\langle T_{a,0}+X_{\alpha}\rangle$ and $\langle T_{0,a}+X_{\alpha+2\beta}\rangle$ are conjugate to $\langle T_{1,0}+X_{\alpha}\rangle$, for any $a\neq 0$. The subalgebras $\langle T_{a,a}+X_{\beta}\rangle$ and $\langle T_{a,-a}+X_{\alpha+\beta}\rangle$ are conjugate to $\langle T_{1,1}+X_{\beta}\rangle$, for any $a\neq 0$.

Proof.

The generator of such a subalgebra of $\mathfrak{b}$ must have Jordan decomposition $X=T+N$, with $T\in\mathfrak{t}$, $N\in\mathfrak{n}$, and $T\neq 0$, $N\neq 0$. Moreover, $T$ cannot be regular or $X$ would be semisimple. After multiplying by a scalar, we can assume $T=T_{1,1},T_{1,-1},T_{1,0}$, or $T_{0,1}$. Since $N$ and $T$ must commute, the possibilities are

[TABLE]

for some $c\neq 0$. After conjugation by a suitable diagonal element, we can assume $c=1$, so the possibilities are

[TABLE]

Let

[TABLE]

Then

[TABLE]

Let

[TABLE]

Then

[TABLE]

Since $T_{1,0}+X_{\alpha}$ has rank $3$ and $T_{1,1}+X_{\beta}$ has rank $4$, the subalgebras $\langle T_{1,0}+X_{\alpha}\rangle$ and $\langle T_{1,1}+X_{\beta}\rangle$ are not equivalent.

The result follows. ∎

5. Preliminary results to be used in classification of higher dimensional subalgebras

5.1. Two-dimensional subalgebras of $\mathfrak{n}_{\mathfrak{p}}$

If $Z$ is a symmetric $2\times 2$ matrix, then

[TABLE]

is a two-dimensional subspace of the symmetric matrices, and every such subspace is of this form for some such $Z$. Note that

[TABLE]

Accordingly, conjugating $\begin{pmatrix}0&\mathfrak{s}_{Z}\\ 0&0\\ \end{pmatrix}$ by $\begin{pmatrix}g&0\\ 0&g^{-t}\\ \end{pmatrix}$ amounts to acting on $\mathfrak{s}_{Z}$ by $X\mapsto gXg^{t}$, which in turn amounts to acting on $Z$ by $Z\mapsto g^{-t}Zg^{-1}$.

Consequently, there are two conjugacy classes of two-dimensional subspaces of $\mathfrak{n}_{\mathfrak{p}}$, corresponding to $Z=\begin{pmatrix}1&0\\ 0&0\\ \end{pmatrix}$ and $Z=\begin{pmatrix}0&1\\ 1&0\\ \end{pmatrix}$, respectively. Since these matrices $Z$ correspond to inequivalent forms, we see that these two classes are inequivalent. We have just proved the following result.

Lemma 5.1**.**

There are two conjugacy classes of two-dimensional subalgebras of $\mathfrak{n}_{\mathfrak{p}}$, with representatives

[TABLE]

5.2. Semisimple elements in $\mathfrak{b}$

Lemma 5.2**.**

Suppose $\mathfrak{a}$ is a solvable subalgebra of $\mathfrak{b}$. If $\mathfrak{a}$ contains semisimple elements, then it is possible to find $b\in B$ so that the conjugate $\mathfrak{a}^{b}={\rm Ad}(b)\mathfrak{a}$ intersects $\mathfrak{t}$. Moreover, if $\mathfrak{a}$ contains a Cartan subalgebra, a two-dimensional algebra of commuting semisimple elements, then it is possible to find $b\in B$ so that the conjugate $\mathfrak{a}^{b}$ contains $\mathfrak{t}$.

Proof.

The first assertion follows from Lemma 3.2. Since all Cartan subalgebras of $\mathfrak{b}$ are conjugate under $B$ [[Hum72], Theorem 16.2], the second assertion follows. ∎

Lemma 5.3**.**

Suppose $\mathfrak{a}$ is a solvable subalgebra of $\mathfrak{b}$. Suppose $\mathfrak{a}$ contains $X=T+N$, $X^{\prime}=T^{\prime}+N^{\prime}$, with $T,T^{\prime}\in\mathfrak{t}$, $N,N^{\prime}\in\mathfrak{n}$. Suppose moreover that $T$ and $T^{\prime}$ are linearly independent.

Then it is possible to find $b\in B$ so that the conjugate $\mathfrak{a}^{b}={\rm Ad}(b)\mathfrak{a}$ contains $\mathfrak{t}$.

Proof.

It is possible to find a linear combination of $X$ and $X^{\prime}$ that can be written as $X^{\prime\prime}=T^{\prime\prime}+N^{\prime\prime}$, with $N^{\prime\prime}\in\mathfrak{n}$ and $T^{\prime\prime}\in\mathfrak{t}$ such that $\text{ad}(T^{\prime\prime})$ has distinct eigenvalues on $\mathfrak{n}$. By Lemma 3.2, we can perform a conjugation so that $X^{\prime\prime}=T^{\prime\prime}\in\mathfrak{t}$. Then if $X\in\mathfrak{t}$, we are done. Otherwise, $X^{\prime\prime}$ and $X$ are linearly independent. If $X=T+N$, with $N\neq 0$, we can write $N=\sum_{i}N_{i}$, where each $N_{i}$ is an eigenvector of $\text{ad}(T^{\prime\prime})$ with eigenvalue $\lambda_{i}$, and the $\lambda_{i}$ are distinct. Then $\mathfrak{a}$ contains

[TABLE]

and hence $\mathfrak{a}$ contains all the $N_{i}$.

From this we find that $\mathfrak{a}$ contains $T=X-\sum_{i}N_{i}$, and hence $\mathfrak{a}\supseteq\langle T,T^{\prime\prime}\rangle=\mathfrak{t}$. ∎

6. Two-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$

In this section, we classify the two-dimensional (solvable) subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ according to two cases: Subalgebras containing a semisimple element (see Theorems 6.2 and 6.7), and subalgebras not containing any semisimple elements (see Theorems 6.9 and 6.10). Again, without loss of generality, we assume that each solvable subalgebra is in the Borel subalgebra $\mathfrak{b}$. The results are summarized in Table 4.

6.1. Two-dimensional subalgebras containing a semisimple element

6.1.1. Regular Semisimple Elements

Suppose $\mathfrak{a}\subset\mathfrak{b}$ is a solvable subalgebra of dimension $2$. By Lemma 5.2, we can assume that if it contains semisimple elements, then $\mathfrak{a}$ contains elements of $\mathfrak{t}$, and that if it contains a Cartan subalgebra, then it contains and hence equals $\mathfrak{t}$. Suppose it contains a regular diagonal element $T_{a,b}=diag(a,b,-a,-b)$, i.e., one such that the restriction of ${\rm ad}(T_{a,b})$ to $\mathfrak{n}$ is nonsingular. It is easily seen that this amounts to $a,b\neq 0,a\neq\pm b$. Then we can assume $b=1$, $a\neq 0,\pm 1$.

It is easy to check that, for $a\neq 0,\pm 1$, the eigenvalues of ${\rm ad}(T_{a,1})$ restricted to $\mathfrak{n}$ are distinct, with the single exception of $a=3$: ${\rm ad}(T_{3,1})$ has the same eigenvalue for $X_{\alpha}$ and $X_{\beta}$. This means that for any $r,s\in\mathbb{C}$, not both zero, $\langle T_{3,1},rX_{\alpha}+sX_{\beta}\rangle$ is two-dimensional.

Lemma 6.1**.**

If $r,s$ are both nonzero, then there is a diagonal element in $G$ that conjugates $rX_{\alpha}+sX_{\beta}$ to $X_{\alpha}+X_{\beta}$; it also fixes $T_{3,1}$.

Proof.

If $u^{2}=r$, then the diagonal element ${\rm diag}(\frac{1}{su},\frac{1}{u},su,u)\in Sp(4,\mathbb{C})$ conjugates $rX_{\alpha}+sX_{\beta}$ to $X_{\alpha}+X_{\beta}$ and, being diagonal, fixes $T_{3,1}$. ∎

By Lemma 6.1, $\langle T_{3,1},rX_{\alpha}+sX_{\beta}\rangle\sim\langle T_{3,1},X_{\alpha}+X_{\beta}\rangle$, provided $r,s\neq 0$.

If $a\neq 3$, then ${\rm ad}(T_{a,1})$ has distinct eigenvalues on $\mathfrak{n}$, so the only two-dimensional subalgebras of $\mathfrak{b}$ containing $T_{a,1}$ are of the form $\langle T_{a,1},X_{\gamma}\rangle$, where $\gamma$ is one of $\alpha,\beta,\alpha+\beta,\alpha+2\beta$.

Let $W\in Sp(4,\mathbb{C})$ be as in Eq. (22). Then

[TABLE]

Accordingly, under the Adjoint action of $P$,

[TABLE]

The element $A$ of Eq. (24) conjugates $X_{\beta}$ into $X_{\alpha+\beta}$ and $T_{a,1}$ into $-T_{-a,1}$, so

[TABLE]

Similarly, if $J$ is the matrix of Eq. (11), then $AJ$ conjugates $T_{a,1}$ to $T_{-a,1}$. It fixes $X_{\alpha}$ and takes $X_{\beta}$, $X_{\alpha+\beta}$, and $X_{\alpha+2\beta}$ out of $\mathfrak{b}$. In particular, $\langle T_{a,1},X_{\alpha}\rangle\sim\langle T_{-a,1},X_{\alpha}\rangle$.

We summarize: any two-dimensional algebra containing a regular diagonal matrix but not $\mathfrak{t}$ is conjugate under $Sp(4,\mathbb{C})$ to $\langle T_{3,1},X_{\alpha}+X_{\beta}\rangle$ or to $\langle T_{a,1},X_{\alpha}\rangle$ or $\langle T_{a,1},X_{\beta}\rangle$, for some $a\in\mathbb{C}$, with $a\neq 0,\pm 1$, with the understanding that $\langle T_{a,1},X_{\alpha}\rangle\sim\langle T_{-a,1},X_{\alpha}\rangle$ and $\langle T_{a,1},X_{\beta}\rangle\sim\langle T_{a^{-1},1},X_{\beta}\rangle$.

Next we consider the possibility of other equivalences between pairs of these algebras. Suppose $X\in\mathfrak{n}$ is such that $\langle T_{a,1},X\rangle$ is two-dimensional, with $a\neq 0,\pm 1$. Note that every element of $\langle T_{a,1},X\rangle$ is of the form $cT_{a,1}+dX$, for some $c,d\in\mathbb{C}$. If $c\neq 0$, this matrix has distinct eigenvalues and hence is semisimple. However, if $c=0$, the matrix is nilpotent, and unless $d=0$, it has rank equal to the rank of $X$. Now ${\rm rank}(X_{\alpha})=1$, ${\rm rank}(X_{\beta})=2$, and ${\rm rank}(X_{\alpha}+X_{\beta})=3$, so $\langle T_{3,1},X_{\alpha}+X_{\beta}\rangle$ is not equivalent to $\langle T_{a,1},X_{\alpha}\rangle$ or to $\langle T_{a,1},X_{\beta}\rangle$, for any $a\neq 0,\pm 1$. Moreover, $\langle T_{a,1},X_{\alpha}\rangle$ cannot be equivalent to $\langle T_{b,1},X_{\beta}\rangle$, for any $a,b\neq 0,\pm 1$.

Suppose $\langle T_{a,1},X_{\gamma}\rangle$ is conjugate by $g\in Sp(4,\mathbb{C})$ to $\langle T_{b,1},X_{\gamma}\rangle$, for some $b\in\mathbb{C}$, $b\neq 0,\pm 1$, where $\gamma=\alpha$ or $\beta$.

The elements of $\langle T_{b,1},X_{\gamma}\rangle$ are all of the form $cT_{b,1}+dX_{\gamma}$, for $c,d\in\mathbb{C}$, and the semisimple ones are those for which $c\neq 0$.

For $z\in\mathbb{C}$, consider

[TABLE]

A simple calculation shows that the last term is zero, and, if $c\neq 0$, choosing $z=\frac{d}{c\gamma(T_{b,1})}$ shows that the semisimple element $cT_{b,1}+dX_{\gamma}\in\langle T_{b,1},X_{\gamma}\rangle$ is conjugate to $cT_{b,1}$ by an element of the form $(id+zX_{\gamma})$, which normalizes $\langle T_{b,1},X_{\gamma}\rangle$.

Since $gT_{a,1}g^{-1}$ is a semisimple element of $\langle T_{b,1},X_{\gamma}\rangle$ there must be $c,d$ with $c\neq 0$ so that the eigenvalues $\pm a,\pm 1$ of $T_{a,1}$ equal those of $cT_{b,1}$. The only possibilities are $c=\pm 1$ and $b=\pm a$ or $c=\pm a$ and $b=\pm\frac{1}{a}$.

This amounts to saying that $T_{a,1}$ can be conjugated to $\pm T_{a,1}$, $\pm T_{-a,1}$, $\pm T_{1,a}$, or $\pm T_{1,-a}$, and we have already seen that all of these are possible.

Moreover, the eigenvalue of $\text{ad}(T_{a,1})$ or $\text{ad}(T_{-a,1})$ corresponding to the eigenvector $X_{\alpha}$ is $\alpha(T_{a,1})=2$ or $\alpha(T_{-a,1})=2$, while the eigenvalue of $\text{ad}(T_{1,a})$ or $\text{ad}(T_{1,-a})$ is $2a\neq\pm 2$ or $-2a\neq\pm 2$, respectively. This shows that there are no equivalences between $\langle T_{a,1},X_{\alpha}\rangle$ and $\langle T_{1,\pm a},X_{\alpha}\rangle$.

Similarly, the eigenvalue of $\text{ad}(T_{a,1})$ or $\text{ad}(-T_{1,a})$ corresponding to the eigenvector $X_{\beta}$ is $\beta(T_{a,1})=a-1$ or $\beta(-T_{1,a})=a-1$, while the eigenvalue of $\text{ad}(\pm T_{-a,1})$ or $\text{ad}(\pm T_{1,-a})$ is $\pm(-a-1)\neq a-1$ or $\pm(1+a)\neq a-1$, respectively. This shows that there are no equivalences between $\langle T_{a,1},X_{\beta}\rangle$ and $\langle T_{1,-a},X_{\beta}\rangle$ or $\langle T_{-a,1},X_{\beta}\rangle$.

We combine the above remarks.

Theorem 6.2**.**

Up to equivalence under the action of $Sp(4,\mathbb{C})$, a complete set of representatives for the two-dimensional subalgebras containing a regular semisimple element is

[TABLE]

with the understanding that $\langle T_{a,1},X_{\alpha}\rangle\sim\langle T_{-a,1},X_{\alpha}\rangle$ and $\langle T_{a,1},X_{\beta}\rangle\sim\langle T_{a^{-1},1},X_{\beta}\rangle$, but that otherwise these algebras are pairwise not equivalent under $Sp(4,\mathbb{C})$.

We also note that

[TABLE]

6.1.2. Algebras Containing Non-Regular Semisimple Elements

Any non-regular semisimple element of $\mathfrak{t}$ is a scalar multiple of $T_{0,1}$, $T_{1,0}$, $T_{1,1}$, or $T_{1,-1}$.

6.1.2.1 $T_{0,1}$

Suppose a two-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ contains $T_{0,1}$ but not $\mathfrak{t}$. Eigenvectors of $\text{ad}(T_{0,1})$ in $\mathfrak{n}$ are scalar multiples of the root vectors $X_{\alpha}$, $X_{\beta}$, $X_{\alpha+\beta}$, or $X_{\alpha+2\beta}$, which have the distinct eigenvalues $2,-1,1$, and [math], respectively. The corresponding algebras are $\langle T_{0,1},X_{\alpha}\rangle$, $\langle T_{0,1},X_{\beta}\rangle$, $\langle T_{0,1},X_{\alpha+\beta}\rangle$, and $\langle T_{0,1},X_{\alpha+2\beta}\rangle$.

If $A$ is the matrix defined in Eq. (24), then $A$ conjugates $T_{0,1}$ to $-T_{0,1}$, $X_{\beta}$ to $X_{\alpha+\beta}$, and $X_{\alpha+\beta}$ to $-X_{\beta}$. This shows that $\langle T_{0,1},X_{\beta}\rangle$ is equivalent to $\langle T_{0,1},X_{\alpha+\beta}\rangle$, so it suffices to consider $\langle T_{0,1},X_{\beta}\rangle$.

A comparison of the eigenvalues of $\text{ad}(T_{0,1})$ shows that $\langle T_{0,1},X_{\alpha}\rangle$ cannot be equivalent to $\langle T_{0,1},X_{\beta}\rangle$.

We observe that $\langle T_{0,1},X_{\alpha+2\beta}\rangle$ is abelian, so it is inequivalent to $\langle T_{0,1},X_{\alpha}\rangle$ and $\langle T_{0,1},X_{\beta}\rangle$.

We summarize in the following lemma.

Lemma 6.3**.**

The algebras $\langle T_{0,1},X_{\alpha}\rangle$, $\langle T_{0,1},X_{\beta}\rangle$, and $\langle T_{0,1},X_{\alpha+2\beta}\rangle$ are all inequivalent. Up to equivalence, they are representatives of all two-dimensional solvable subalgebras $\mathfrak{a}\subset\mathfrak{b}$ containing $T_{0,1}$ but not $\mathfrak{t}$.

Also, $\langle T_{0,1},X_{\beta}\rangle\sim\langle T_{0,1},X_{\alpha+\beta}\rangle$. Moreover, $\langle T_{0,1},X_{\alpha+2\beta}\rangle$ is abelian.

6.1.2.2 $T_{1,0}$

Suppose a two-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ contains $T_{1,0}$ but not $\mathfrak{t}$. Eigenvectors of $\text{ad}(T_{1,0})$ in $\mathfrak{b}$ are scalar multiples of $X_{\alpha}$ or $X_{\alpha+2\beta}$ or linear combinations of $X_{\beta}$ and $X_{\alpha+\beta}$, with eigenvalues $0,2,1$, respectively.

If $\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in SL(2,\mathbb{C})$, then the matrix

[TABLE]

centralizes $T_{1,0}$ and conjugates $X_{\beta}$ to

[TABLE]

an arbitrary nonzero linear combination of $X_{\beta}$ and $X_{\alpha+\beta}$.

We find that the algebras $\langle T_{1,0},cX_{\beta}+dX_{\alpha+\beta}\rangle$ for any $c,d\in\mathbb{C}$, not both zero are all pairwise equivalent. In particular, it suffices to consider $\langle T_{1,0},X_{\beta}\rangle$.

Since $\langle T_{1,0},X_{\alpha}\rangle$ is abelian, it is not equivalent to $\langle T_{1,0},X_{\beta}\rangle$ or to $\langle T_{1,0},X_{\alpha+2\beta}\rangle$. A comparison of the eigenvalues of $\text{ad}(T_{1,0})$ shows that $\langle T_{1,0},X_{\beta}\rangle$ cannot be equivalent to $\langle T_{1,0},X_{\alpha+2\beta}\rangle$.

We summarize in the following lemma.

Lemma 6.4**.**

The algebra $\langle T_{1,0},X_{\alpha}\rangle$ is abelian. The algebras $\langle T_{1,0},cX_{\beta}+dX_{\alpha+\beta}\rangle$ for any $c,d\in\mathbb{C}$, not both zero, are all equivalent. The algebras $\langle T_{1,0},X_{\alpha}\rangle$, $\langle T_{1,0},X_{\beta}\rangle$, and $\langle T_{1,0},X_{\alpha+2\beta}\rangle$ are all inequivalent. Up to equivalence, they are representatives of all two-dimensional solvable subalgebras $\mathfrak{a}\subset\mathfrak{b}$ containing $T_{1,0}$ but not $\mathfrak{t}$.

6.1.2.3 Equivalences between subalgebras containing $T_{1,0}$ and $T_{0,1}$

From Eq. (30), we see that $\langle T_{0,1},X_{\alpha}\rangle\sim\langle T_{1,0},X_{\alpha+2\beta}\rangle$ and $\langle T_{1,0},X_{\alpha}\rangle\sim\langle T_{0,1},X_{\alpha+2\beta}\rangle$. We also see that $\langle T_{1,0},X_{\beta}\rangle\sim\langle T_{1,0},X_{\alpha+\beta}\rangle\sim\langle T_{0,1},X_{\alpha+\beta}\rangle$.

Since $\langle T_{1,0},X_{\alpha}\rangle$ is abelian and $\langle T_{0,1},X_{\alpha}\rangle$ is not, they are inequivalent. A comparison of the eigenvalues of $\text{ad}(T_{1,0})$ and $\text{ad}(T_{0,1})$ shows that neither $\langle T_{1,0},X_{\beta}\rangle$ nor $\langle T_{0,1},X_{\beta}\rangle$ can be equivalent to $\langle T_{1,0},X_{\alpha}\rangle$, $\langle T_{1,0},X_{\alpha+2\beta}\rangle$, $\langle T_{0,1},X_{\alpha}\rangle$, or $\langle T_{0,1},X_{\alpha+2\beta}\rangle$.

We summarize in the following lemma.

Lemma 6.5**.**

A two-dimensional solvable algebra containing $T_{0,1}$ or $T_{1,0}$ is equivalent to one of the following:

[TABLE]

These algebras are all pairwise inequivalent. The algebra $\langle T_{1,0},X_{\alpha}\rangle$ is abelian.

6.1.2.4 $T_{1,1}$

Suppose a two-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ contains $T_{1,1}$ but not $\mathfrak{t}$. Eigenvectors of $\text{ad}(T_{1,1})$ in $\mathfrak{b}$ are scalar multiples of $X_{\beta}$ or linear combinations of $X_{\alpha}$, $X_{\alpha+\beta}$, and $X_{\alpha+2\beta}$. In other words, the $2$-eigenspace of $\text{ad}(T_{1,1})$ in $\mathfrak{b}$ is $\mathfrak{n}_{\mathfrak{p}}$.

Conjugating a matrix $T_{1,1}+N$, with $N\in\mathfrak{n}_{\mathfrak{p}}$, by a matrix of the block form $\begin{pmatrix}g&0\\ 0&g^{-t}\\ \end{pmatrix}$ fixes $T_{1,1}$ and takes $N$ to $gNg^{t}$, so by Lemma $4.3$, every subalgebra of the form $\langle T_{1,1},N\rangle$, with $0\neq N\in\mathfrak{n}_{\mathfrak{p}}$, is equivalent to either $\langle T_{1,1},X_{\alpha}\rangle$ or $\langle T_{1,1},X_{\alpha+\beta}\rangle$, and these two subalgebras are inequivalent because their nilpotent elements have different ranks.

Since these subalgebras are nonabelian and $\langle T_{1,1},X_{\beta}\rangle$ is abelian, the algebras $\langle T_{1,1},X_{\alpha}\rangle$, $\langle T_{1,1},X_{\beta}\rangle$, and $\langle T_{1,1},X_{\alpha+\beta}\rangle$ are inequivalent. Any semisimple element in any of these subalgebras has eigenvalues that are a multiple of the eigenvalues of $T_{1,1}$. Similarly, any semisimple element in any of the subalgebras mentioned in Lemma 6.5 has eigenvalues that are a multiple of the eigenvalues of $T_{1,0}$. So none of these subalgebras containing $T_{1,1}$ can be equivalent to any of the subalgebras mentioned in Lemma 6.5.

Finally, the $2$-eigenspace of $\text{ad}(T_{1,1})$ is $\mathfrak{n}_{\mathfrak{p}}$, and the equivalence class of $\langle T_{1,1},N\rangle$, for $N\in\mathfrak{n}_{\mathfrak{p}}$, is determined by ${\rm rank}(N)$. We summarize.

Lemma 6.6**.**

Every two-dimensional solvable subalgebra containing $T_{1,1}$ but not $\mathfrak{t}$ is equivalent to one of the inequivalent subalgebras $\langle T_{1,1},X_{\alpha}\rangle$, $\langle T_{1,1},X_{\beta}\rangle$, or $\langle T_{1,1},X_{\alpha+\beta}\rangle$. None of these subalgebras is equivalent to any of the subalgebras mentioned in Lemma 6.5 or in Theorem 6.2.

We also note that, for $a,b,c\in\mathbb{C}$, not all zero, $\langle T_{1,1},aX_{\alpha}+bX_{\alpha+\beta}+cX_{\alpha+2\beta}\rangle\sim\langle T_{1,1},X_{\alpha+\beta}\rangle$ unless $aX_{\alpha}+bX_{\alpha+\beta}+cX_{\alpha+2\beta}$ has rank $1$, in which case $\langle T_{1,1},aX_{\alpha}+bX_{\alpha+\beta}+cX_{\alpha+2\beta}\rangle\sim\langle T_{1,1},X_{\alpha}\rangle$. In particular, $\langle T_{1,1},X_{\alpha+2\beta}\rangle\sim\langle T_{1,1},X_{\alpha}\rangle$.

6.1.2.5 $T_{1,-1}$

The $Sp(4,\mathbb{C})$ matrix $A$ in Eq. (24) conjugates $T_{1,1}$ to $T_{1,-1}$.

It also takes $X_{\beta}$ to $-X_{\alpha+\beta}$ and vice versa, fixes $X_{\alpha+2\beta}$, and takes $X_{\alpha}$ out of $\mathfrak{b}$.

The $Sp(4,\mathbb{C})$ matrix $AJ$ conjugates $T_{1,1}$ to $-T_{1,-1}$. It also fixes $X_{\alpha}$ and takes $X_{\beta}$, $X_{\alpha+\beta}$, and $X_{\alpha+2\beta}$ out of $\mathfrak{b}$.

We conclude that

[TABLE]

Since $\langle T_{1,1},X_{\alpha+2\beta}\rangle\sim\langle T_{1,1},X_{\alpha}\rangle$, we conclude that $\langle T_{1,-1},X_{\alpha+2\beta}\rangle\sim\langle T_{1,-1},X_{\alpha}\rangle$. We also note that for $c\neq 0$, $\langle T_{1,-1},X_{\beta}+cX_{\alpha+2\beta}\rangle\sim\langle T_{1,-1},X_{\beta}\rangle$. We summarize.

Theorem 6.7**.**

Every two-dimensional solvable subalgebra containing $T_{1,1}$ or $T_{1,-1}$ but not $\mathfrak{t}$ is equivalent to one of the inequivalent subalgebras $\langle T_{1,1},X_{\alpha}\rangle$, $\langle T_{1,1},X_{\beta}\rangle$, or $\langle T_{1,1},X_{\alpha+\beta}\rangle$. None of these subalgebras is equivalent to any of the subalgebras mentioned in Lemma 6.5 or in Theorem 6.2.

Moreover, $\langle T_{1,-1},X_{\alpha+2\beta}\rangle\sim\langle T_{1,1},X_{\alpha}\rangle$, and, for any $c\in\mathbb{C}$, $\langle T_{1,-1},X_{\beta}$ $+cX_{\alpha+2\beta}\rangle\sim\langle T_{1,-1},X_{\beta}\rangle$.

6.2. Two-dimensional subalgebras not containing any semisimple elements

6.2.1. Two-dimensional algebras containing a nonsemisimple element that is not nilpotent

Suppose $\mathfrak{a}$ is a two-dimensional subalgebra of $\mathfrak{b}$ which does not contain any semisimple elements, but whose elements are not all nilpotent.

If $\mathfrak{a}=\langle X,X^{\prime}\rangle$, where $X=T+N$, $X^{\prime}=T^{\prime}+N^{\prime}$, with $T,T^{\prime}\in\mathfrak{t}$ and $N,N^{\prime}\in\mathfrak{n}$, then we can assume $T\neq 0$. If $T^{\prime}$ is not a scalar multiple of $T$, then $T$ and $T^{\prime}$ span $\mathfrak{t}$, and some linear combination of $X,X^{\prime}$ would have distinct eigenvalues and therefore would be semisimple. So $T^{\prime}$ must be a multiple of $T$, and $X^{\prime}$ can be replaced by $X^{\prime\prime}=N^{\prime\prime}\in\mathfrak{n}$. Since $\dim(\mathfrak{a})=2$, $[X,X^{\prime\prime}]$ must be a scalar multiple of $X^{\prime\prime}$.

Without loss of generality, we can assume that $T$ has the form of one of the first four classes under the action of $B$ in Table 2. In fact, we can even assume that the parameter $a$ in these classes is equal to $1$.

For example, it is clear that $X=T_{1,1}+X_{\beta}$ commutes with $X_{\beta}$, so $\langle T_{1,1}+X_{\beta},X_{\beta}\rangle=\langle T_{1,1},X_{\beta}\rangle$ is a commutative two-dimensional algebra containing a semisimple element. A simple calculation shows that the only other possibility for $X^{\prime\prime}=N^{\prime\prime}\in\mathfrak{n}$ satisfying $[X,X^{\prime\prime}]=vX^{\prime\prime}$, for some $v\in\mathbb{C}$, is for $X^{\prime\prime}$ to be a multiple of $X_{\alpha+2\beta}$. So $\langle T_{1,1}+X_{\beta},X_{\alpha+2\beta}\rangle$ is a noncommutative two-dimensional algebra.

Similarly, it is clear that $X=T_{1,-1}+X_{\alpha+\beta}$ commutes with $X_{\alpha+\beta}$, so $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha+\beta}\rangle=\langle T_{1,-1},X_{\alpha+\beta}\rangle$ is a commutative two-dimensional algebra containing a semisimple element. A simple calculation shows that the only other possibilities for $X^{\prime\prime}=N^{\prime\prime}\in\mathfrak{n}$ satisfying $[X,X^{\prime\prime}]=vX^{\prime\prime}$, for some $v\in\mathbb{C}$, are for $X^{\prime\prime}$ to be a multiple of $X_{\alpha}$ or of $X_{\alpha+2\beta}$. So $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha}\rangle$ and $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are noncommutative two-dimensional algebras. Note that they are conjugate by the element $W$ of Eq. (22).

We will use the following lemma.

Lemma 6.8**.**

The $Sp(4,\mathbb{C})$ matrix $A$ in Eq. (24) conjugates

[TABLE]

for any $\lambda\in\mathbb{C}$.

Proof.

Straightforward calculation. ∎

Note too that by Lemma 6.8 and an easy calculation, the $Sp(4,\mathbb{C})$ matrix $A$ in Eq. (24) conjugates $\langle T_{1,1}+X_{\beta},X_{\alpha+2\beta}\rangle$ to $\langle T_{1,-1}+X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$ $\sim$ $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha}\rangle$.

For $X=T_{1,0}+X_{\alpha}$, clearly $\langle T_{1,0}+X_{\alpha},X_{\alpha}\rangle=\langle T_{1,0},X_{\alpha}\rangle$ is a commutative two-dimensional algebra containing a semisimple element. A simple calculation shows that the only other possibilities for $X^{\prime\prime}=N^{\prime\prime}\in\mathfrak{n}$ satisfying $[X,X^{\prime\prime}]=vX^{\prime\prime}$, for some $v\in\mathbb{C}$, are for $X^{\prime\prime}$ to be a multiple of $X_{\alpha+\beta}$ or of $X_{\alpha+2\beta}$. So $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta}\rangle$ and $\langle T_{1,0}+X_{\alpha},X_{\alpha+2\beta}\rangle$, are noncommutative two-dimensional algebras. Note that they are not conjugate by any element of $Sp(4,\mathbb{C})$ because the elements which have eigenvalues $1,-1,0,0$, i.e., $\pm T_{1,0}+N$, for some $N\in\mathfrak{n}$, have eigenvalue $\pm 1$ on $X_{\alpha+\beta}$ and eigenvalue $\pm 2$ on $X_{\alpha+2\beta}$.

For $X=T_{0,1}+X_{\alpha+2\beta}$, clearly $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha+2\beta}\rangle=\langle T_{0,1},X_{\alpha+2\beta}\rangle$ is a commutative two-dimensional algebra containing a semisimple element. A simple calculation shows that the only other possibilities for $X^{\prime\prime}=N^{\prime\prime}\in\mathfrak{n}$ satisfying $[X,X^{\prime\prime}]=vX^{\prime\prime}$, for some $v\in\mathbb{C}$, are for $X^{\prime\prime}$ to be a multiple of $X_{\alpha}$, $X_{\beta}$, or $X_{\alpha+\beta}$. So $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha}\rangle$, $\langle T_{0,1}+X_{\alpha+2\beta},X_{\beta}\rangle$, and $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha+\beta}\rangle$ are noncommutative two-dimensional algebras. Note that the last two of these are conjugate by the element $A$ of Eq. (24).

However, $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha}\rangle$ and $\langle T_{0,1}+X_{\alpha+2\beta},X_{\beta}\rangle$ are not conjugate by any element of $Sp(4,\mathbb{C})$ because the elements which have eigenvalues $1,-1,0,0$, i.e., $\pm T_{0,1}+N$, for some $N\in\mathfrak{n}$, have eigenvalue $\pm 2$ on $X_{\alpha}$ and eigenvalue $\pm 1$ on $X_{\beta}$.

Moreover, the element $W$ of Eq. (22) conjugates $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha}\rangle$ to $\langle T_{1,0}+X_{\alpha},X_{\alpha+2\beta}\rangle$ and $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha+\beta}\rangle\sim\langle T_{0,1}+X_{\alpha+2\beta},X_{\beta}\rangle$ to $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta}\rangle$.

Finally, note that none of these two-dimensional algebras containing a singular matrix $T_{1,0}+X_{\alpha}$ or $T_{0,1}+X_{\alpha+2\beta}$, but no semisimple elements, can be conjugate to any of the algebras containing an invertible matrix $T_{1,1}+X_{\beta}$ or $T_{1,-1}+X_{\alpha+\beta}$.

Theorem 6.9**.**

A two-dimensional subalgebra of $\mathfrak{b}$ which does not contain any semisimple elements, but whose elements are not all nilpotent, must be equivalent to one of the following inequivalent algebras:

[TABLE]

6.2.2. Two-dimensional algebras whose elements are all nilpotent

Suppose $\mathfrak{a}\subseteq\mathfrak{g}$ is a two-dimensional algebra whose elements are all nilpotent. If $\mathfrak{a}\subset\mathfrak{n}_{\mathfrak{p}}$, then there are exactly two possible classes, as listed in Lemma 5.1:

[TABLE]

Otherwise, we can assume that $\mathfrak{a}$ is generated by $X=X_{\beta}+sX_{\alpha}+tX_{\alpha+\beta}+uX_{\alpha+2\beta}$ and $X^{\prime\prime}=rX_{\alpha}+vX_{\alpha+\beta}+wX_{\alpha+2\beta}$. A simple calculation shows that the only way that $[X,X^{\prime\prime}]$ can be a scalar multiple of $X^{\prime\prime}$ is for $X^{\prime\prime}$ to be a (nonzero) multiple of $X_{\alpha+2\beta}$, in which case $\mathfrak{a}$ is commutative.

Now $\langle X_{\beta}+sX_{\alpha}+tX_{\alpha+\beta}+uX_{\alpha+2\beta},$ $X_{\alpha+2\beta}\rangle\sim\langle X_{\beta}+sX_{\alpha}+tX_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$, and $Id+tX_{\alpha}$ conjugates the latter to $\langle X_{\beta}+sX_{\alpha},X_{\alpha+2\beta}\rangle$. Then, if $s\neq 0$ and $z^{2}=s$, ${\rm diag}(1/z,1/z,z,z)$ conjugates this into $\langle X_{\beta}+X_{\alpha},X_{\alpha+2\beta}\rangle$.

On the other hand, if $s=0$, then $\mathfrak{a}=\langle X_{\beta},X_{\alpha+2\beta}\rangle$. If $A$ and $W$ are the matrices in Eqs. (24) and (22), respectively, then $WA$ conjugates $\langle X_{\beta},X_{\alpha+2\beta}\rangle$ to $\langle X_{\alpha},X_{\alpha+\beta}\rangle$.

The algebra $\langle X_{\beta}+X_{\alpha},X_{\alpha+2\beta}\rangle$ contains elements of rank 3, so it is inequivalent to either of the algebras contained in $\mathfrak{n}_{\mathfrak{p}}$, whose elements have rank 2 or less.

Note that $\langle X_{\alpha},X_{\alpha+2\beta}\rangle$ contains two lines consisting of elements of rank $1$, while $\langle X_{\alpha},X_{\alpha+\beta}\rangle$ contains only one such line. Hence they are inequivalent. And all the elements in either of them have rank at most $2$, whereas $\langle X_{\beta}+X_{\alpha},X_{\alpha+2\beta}\rangle$ contains elements of rank 3. Accordingly, these three algebras are pairwise inequivalent.

We summarize.

Theorem 6.10**.**

Each two-dimensional subalgebra whose elements are all nilpotent is conjugate to exactly one of the following inequivalent abelian algebras:

[TABLE]

7. Three-dimensional solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$

In this section, we classify the three-dimensional solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$, dividing them into four cases: Three-dimensional subalgebras containing a Cartan subalgebra (see Lemma 7.1); three-dimensional subalgebras containing a semisimple element but not a Cartan subalgebra (see Theorems 7.2, 7.5 and 7.7); non-nilpotent three-dimensional solvable algebras containing no semisimple elements (see Theorem 7.8); and three-dimensional nilpotent subalgebras (see Theorem 7.9). Again, without loss of generality, we assume that each solvable subalgebra is in the Borel subalgebra $\mathfrak{b}$. The results are summarized in Table 5. We begin with a lemma.

Lemma 7.1**.**

Each three-dimensional solvable algebra containing a Cartan subalgebra is conjugate to exactly one of the following inequivalent subalgebras:

[TABLE]

Moreover, $\langle\mathfrak{t},X_{\alpha}\rangle\sim\langle\mathfrak{t},X_{\alpha+2\beta}\rangle$ and $\langle\mathfrak{t},X_{\beta}\rangle\sim\langle\mathfrak{t},X_{\alpha+\beta}\rangle$.

Proof.

We can assume that such an algebra is contained in $\mathfrak{b}$ and contains $\mathfrak{t}$. Then it must contain exactly one of the positive root vectors.

But the matrix $W$ of Eq. (22) conjugates $\langle\mathfrak{t},X_{\alpha}\rangle$ to $\langle\mathfrak{t},X_{\alpha+2\beta}\rangle$, and the matrix $A$ of Eq. (24) conjugates $\langle\mathfrak{t},X_{\beta}\rangle$ to $\langle\mathfrak{t},X_{\alpha+\beta}\rangle$.

Since $\langle\mathfrak{t},X_{\beta}\rangle$ contains nilpotent elements of rank $2$ while the nilpotent elements of $\langle\mathfrak{t},X_{\alpha}\rangle$ all have rank $1$, these two algebras are inequivalent. ∎

7.1. Three-dimensional subalgebras containing a semisimple element but not a Cartan subalgebra

7.1.1. Regular semisimple elements

Suppose $\mathfrak{a}$ is a three-dimensional solvable subalgebra of $\mathfrak{b}$ which contains a regular semisimple element but not $\mathfrak{t}$. Then we can assume it contains $T_{a,1}$, for some $a\neq 0,\pm 1$, but not $\mathfrak{t}$. By Lemma 5.3, we can assume $\mathfrak{a}=\langle T_{a,1},N,N^{\prime}\rangle$, with $N,N^{\prime}\in\mathfrak{n}$.

If $a\neq 3$, the only possibilities are for $N,N^{\prime}$ to be two commuting positive root vectors:

[TABLE]

If $a=3$, there is the additional possibility $\langle T_{3,1},X_{\alpha}+X_{\beta},X_{\alpha+2\beta}\rangle$. It is inequivalent to any of the others because it contains the nilpotent element $X_{\alpha}+X_{\beta}$, which has rank $3$, while all the others contain nilpotent elements of rank at most $2$.

The element $W$ of Eq. (22) conjugates $\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle$ to $\langle T_{a^{-1},1},$ $X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ and $\langle T_{a,1},X_{\alpha},X_{\alpha+2\beta}\rangle$ to $\langle T_{a^{-1},1},X_{\alpha},$ $X_{\alpha+2\beta}\rangle$. Also, the matrix $A$ of Eq. (24) conjugates $\langle T_{a,1},X_{\beta},X_{\alpha+2\beta}\rangle$ to $\langle T_{-a,1},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$. Accordingly, any three-dimensional solvable subalgebra containing a regular semisimple element but not all of $\mathfrak{t}$ must be equivalent to one of the following:

[TABLE]

However, we do know that for $a\neq 0,\pm 1$, $\langle T_{a,1},X_{\alpha},X_{\alpha+2\beta}\rangle$ $\sim$ $\langle T_{a^{{-1}},1},$ $X_{\alpha},X_{\alpha+2\beta}\rangle$. Otherwise, we claim that these algebras are inequivalent. Indeed, by consideration of the ranks of the nilpotent elements, we have seen that the algebra in the last line is inequivalent to any of the others. Moreover, those in the first line contain nilpotent elements of rank $2$ except for those on the single line spanned by $X_{\alpha}$, while those in the second line contain nilpotent elements of rank $2$ except for those on the two lines spanned by $X_{\alpha}$ and by $X_{{\alpha+2\beta}}$. None of the algebras on the first line can be equivalent to any of the algebras on the second, even for different values of $a\neq 0,\pm 1$.

So consider $\mathfrak{a}=\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle$, with $a\neq 0,\pm 1$. The only other elements of $\mathfrak{t}$ with the same eigenvalues as $T_{a,1}$ are $-T_{a,1}$, $\pm T_{-a,1}$, $\pm T_{1,\pm a}$. So we can assume that if $g\in Sp(4,\mathbb{C})$ conjugates $\mathfrak{a}$ to an algebra $\langle T_{b,1},X_{\alpha},X_{\alpha+\beta}\rangle$, other than $\mathfrak{a}$ itself, then it conjugates $T_{a,1}$ to one of these seven matrices. The vectors $X_{\alpha}$ and $X_{\alpha+\beta}$ are eigenvectors of $\text{ad}(T_{a,1})$ with eigenvalues $2$ and $a+1$, respectively. The element $g\in Sp(4,\mathbb{C})$ must conjugate eigenvectors to eigenvectors. But since they have different ranks, it must conjugate each to a multiple of itself.

The only one of the seven matrices above whose adjoint has eigenvalue $2$ on $X_{\alpha}$ is $T_{-a,1}$, and its eigenvalue on $X_{\alpha+\beta}$ is $1-a$, which cannot equal $1+a$, since $a\neq 0$. We conclude that the algebras $\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle$, with $a\neq 0,\pm 1$, are pairwise inequivalent.

Now consider $\mathfrak{a}=\langle T_{a,1},X_{\alpha},X_{\alpha+2\beta}\rangle$, with $a\neq 0,\pm 1$. The vectors $X_{\alpha}$ and $X_{\alpha+2\beta}\rangle$ are eigenvectors of $\text{ad}(T_{a,1})$ with eigenvalues $2,2a$, respectively. The only one of the seven conjugates of $T_{a,1}$ in $\mathfrak{t}$ with the same adjoint eigenvalues is $T_{1,a}$, corresponding to the equivalence we have already seen arising from conjugation by $W$.

Theorem 7.2**.**

Each three-dimensional solvable subalgebra which contains a regular semisimple element but not a Cartan subalgebra is conjugate to one of the following:

[TABLE]

Apart from the equivalences noted, the algebras listed above are pairwise inequivalent.

7.1.2. Singular semisimple elements

Suppose $\mathfrak{a}$ is a three-dimensional solvable subalgebra of $\mathfrak{b}$ which contains a singular semisimple element but not a Cartan subalgebra. We can assume $\mathfrak{a}$ contains one of $T_{1,1}$, $T_{1,-1}$, $T_{1,0}$, or $T_{0,1}$, but not $\mathfrak{t}$.

7.1.2.1 $T_{0,1}$

Suppose a three-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ contains $T_{0,1}$ but not $\mathfrak{t}$. The eigenvectors of $\text{ad}(T_{0,1})$ in $\mathfrak{n}$ are scalar multiples of the root vectors $X_{\alpha}$, $X_{\beta}$, $X_{\alpha+\beta}$, or $X_{\alpha+2\beta}$, which have the distinct eigenvalues $2,-1,1$, and [math], respectively. The only possible three-dimensional algebras are $\langle T_{0,1},X_{\alpha},X_{\alpha+\beta}\rangle$, $\langle T_{0,1},X_{\alpha},X_{\alpha+2\beta}\rangle$, $\langle T_{0,1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, and $\langle T_{0,1},X_{\beta},X_{\alpha+2\beta}\rangle$.

If $A$ is the matrix defined in Eq. (24), then $A$ conjugates $T_{0,1}$ to $-T_{0,1}$, $X_{\beta}$ to $X_{\alpha+\beta}$, $X_{\alpha+\beta}$ to $-X_{\beta}$, and $X_{\alpha+2\beta}$ to itself. This shows that $\langle T_{0,1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ is equivalent to $\langle T_{0,1},X_{\beta},X_{\alpha+2\beta}\rangle$.

A comparison of the eigenvalues of $\text{ad}(T_{0,1})$ shows that the three remaining algebras are inequivalent.

We summarize in the following lemma.

Lemma 7.3**.**

Every solvable three-dimensional subalgebra of $\mathfrak{b}$ which contains $T_{0,1}$ but not $\mathfrak{t}$ is equivalent to one of the following inequivalent algebras:

[TABLE]

Also, $\langle T_{0,1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim\langle T_{0,1},X_{\beta},X_{\alpha+2\beta}\rangle$.

7.1.2.2 $T_{1,0}$

Suppose a three-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ contains $T_{1,0}$ but not $\mathfrak{t}$. From the eigenvalues and eigenvectors of $\text{ad}(T_{1,0})$ in $\mathfrak{n}$ as described in §6.1.2.2, we find that the only possible three-dimensional algebras are $\langle T_{1,0},X_{\alpha},X_{\alpha+\beta}\rangle$, $\langle T_{1,0},X_{\alpha},X_{\alpha+2\beta}\rangle$, and $\langle T_{1,0},aX_{\beta}+bX_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

The matrix defined in Eq. (36) centralizes $T_{1,0}$ and conjugates $X_{\beta}$ to an arbitrary linear combination of $X_{\beta}$ and $X_{\alpha+\beta}$. This shows that $\langle T_{1,0},aX_{\beta}+bX_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim\langle T_{1,0},X_{\beta},X_{\alpha+2\beta}\rangle$, for any $a,b\in\mathbb{C}$, not both zero.

A comparison of the eigenvalues of $\text{ad}(T_{1,0})$ shows that there are no additional equivalences.

We summarize in the following lemma.

Lemma 7.4**.**

Every solvable three-dimensional subalgebra of $\mathfrak{b}$ which contains $T_{1,0}$ but not $\mathfrak{t}$ is equivalent to one of the following inequivalent algebras:

[TABLE]

Also, for any $a,b\in\mathbb{C}$, not both zero, $\langle T_{1,0},aX_{\beta}+bX_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim\langle T_{1,0},X_{\beta},X_{\alpha+2\beta}\rangle\sim\langle T_{1,0},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

Theorem 7.5**.**

Every solvable three-dimensional subalgebra of $\mathfrak{b}$ which contains a semisimple element which has zero as an eigenvalue but does not contain all of $\mathfrak{t}$ is equivalent to one of the following inequivalent algebras:

[TABLE]

Proof.

The element $W$ defined in Eq. (22) takes $T_{0,1}$ to $T_{1,0}$, $X_{\alpha+\beta}$ to $-X_{\alpha+\beta}$, $X_{\alpha}$ to $-X_{\alpha+2\beta}$, and $X_{\alpha+2\beta}$ to $-X_{\alpha}$. The result follows from Lemmas 7.3 and 7.4. ∎

7.1.2.3 $T_{1,-1}$

From the eigenvalues and eigenvectors of $\text{ad}(T_{1,-1})$ in $\mathfrak{n}$ as described in §6.1.2.5, we find that the three-dimensional subalgebras of $\mathfrak{b}$ which contain $T_{1,-1}$ but not $\mathfrak{t}$ are:

[TABLE]

The element $W\in Sp(4,\mathbb{C})$ conjugates $\langle T_{1,-1},X_{\alpha},X_{\alpha+\beta}\rangle$ to $\langle T_{1,-1},$ $X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, and comparison of the eigenvalues of $\text{ad}(T_{1,-1})$ shows that this is the only equivalence.

Lemma 7.6**.**

Any three-dimensional solvable subalgebra of $\mathfrak{b}$ which contains $T_{1,-1}$ but not $\mathfrak{t}$ is equivalent to one of the following inequivalent algebras:

[TABLE]

Moreover, $\langle T_{1,-1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim\langle T_{1,-1},X_{\alpha},X_{\alpha+\beta}\rangle$.

7.1.2.4 $T_{1,1}$

We recall the eigenvalues and eigenvectors of $\text{ad}(T_{1,1})$ in $\mathfrak{n}$ as described in §6.1.2.4. Because the $2$-eigenspace of $\text{ad}(T_{1,1})$ in $\mathfrak{n}$ is $\mathfrak{n}_{\mathfrak{p}}$, we find from Lemma 5.1 that any three-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ containing $T_{1,1}$ but not $\mathfrak{t}$ is conjugate to one of the following:

[TABLE]

The matrix $A$ defined in Eq. (24) conjugates $\langle T_{1,-1},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$ to $\langle T_{1,1},X_{\beta},X_{\alpha+2\beta}\rangle$ and $\langle T_{1,-1},X_{\beta},X_{\alpha+2\beta}\rangle$ to $\langle T_{1,1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

Since the eigenvalues of $\text{ad}(T_{1,-1})$ on $X_{\alpha}$ and $X_{\alpha+2\beta}$ are $-2$ and $2$, respectively, while those of $\text{ad}(T_{1,1})$ on $X_{\alpha}$ and $X_{\alpha+2\beta}$ are both $2$, we see that $\langle T_{1,1},X_{\alpha},X_{\alpha+2\beta}\rangle$ is not equivalent to $\langle T_{1,-1},X_{\alpha},X_{\alpha+2\beta}\rangle$. Since the eigenvalue of $\text{ad}(T_{1,1})$ on $X_{\beta}$ is [math], we also see that $\langle T_{1,1},X_{\alpha},X_{\alpha+2\beta}\rangle$ is not equivalent to $\langle T_{1,1},X_{\beta},X_{\alpha+2\beta}\rangle$.

Finally, the two-dimensional space spanned by $X_{\alpha+\beta}$ and $X_{\alpha+2\beta}$ consists of elements which are all of rank $2$, except for those on a single line, which are of rank $1$. However, the span of $X_{\alpha}$ and $X_{\alpha+2\beta}$ contains two lines of elements of rank $1$. This shows that $\langle T_{1,1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ is not equivalent to $\langle T_{1,1},X_{\alpha},X_{\alpha+2\beta}\rangle$.

We summarize:

Theorem 7.7**.**

Any three-dimensional solvable subalgebra of $\mathfrak{b}$ containing a semisimple element with eigenvalues $1,1,-1,-1$ but not containing $\mathfrak{t}$ is equivalent to one of the following inequivalent algebras:

[TABLE]

7.2. Non-nilpotent three-dimensional solvable algebras containing no semisimple elements

Any three-dimensional solvable algebra $\mathfrak{a}$ containing no semisimple elements but which does not consist entirely of nilpotent elements must contain an element conjugate to one of the first four matrices in Table 2. Possibly after a conjugation, we can assume that $\mathfrak{a}$ contains one of the following elements:

[TABLE]

Note that each of these elements is of the form $T+X$, where $0\neq T\in\mathfrak{t}$ is non-regular and $X$ is the only positive root vector that commutes with $T$. Note that if $N$ is any linear combination of the other positive root vectors, then $T+N$ is semisimple. By Lemma 5.3, $\mathfrak{a}$ has a basis consisting of one of the above elements and two elements of $\mathfrak{n}$ which are linear combinations of the positive root vectors that do not commute with it.

7.2.1. $T_{1,1}+X_{\beta}$

If $\mathfrak{a}$ contains $T_{1,1}+X_{\beta}$, then it cannot contain an element of the form $X_{\alpha}+bX_{\alpha+\beta}+cX_{\alpha+2\beta}$, because any such algebra would contain all of $\mathfrak{n}_{\mathfrak{p}}$ and thus have dimension $4$. The only possible algebra is $\langle T_{1,1}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

7.2.2. $T_{1,-1}+X_{\alpha+\beta}$

If $\mathfrak{a}$ contains $T_{1,-1}+X_{\alpha+\beta}$, then it cannot contain an element of the form $aX_{\beta}+bX_{\alpha}+cX_{\alpha+2\beta}$, with $a,b\neq 0$, because any such algebra would contain all of $\mathfrak{n}$ and hence $T_{1,-1}$. The only possibilities are $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha},X_{\alpha+2\beta}\rangle$ and $\langle T_{1,-1}+X_{\alpha+\beta},X_{\beta},X_{\alpha+2\beta}\rangle$.

7.2.3. $T_{1,0}+X_{\alpha}$

If $\mathfrak{a}$ contains $T_{1,0}+X_{\alpha}$, then it cannot contain an element of the form $X_{\beta}+bX_{\alpha+\beta}+cX_{\alpha+2\beta}$, because any such algebra would contain all of $\mathfrak{n}_{\mathfrak{p}}$ and hence $T_{1,0}$. The only possibility is $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

7.2.4. $T_{0,1}+X_{\alpha+2\beta}$

Note that $X_{\alpha+2\beta}$ commutes with all the other positive root vectors, and they have distinct eigenvalues for $\text{ad}(T_{0,1})$ and hence for $\text{ad}(T_{0,1}+X_{\alpha+2\beta})$. So if $\mathfrak{a}$ contains $T_{0,1}+X_{\alpha+2\beta}$, then it has a basis consisting of $T_{0,1}+X_{\alpha+2\beta}$ and two positive root vectors other than $X_{\alpha+2\beta}$.

It cannot contain $X_{\alpha}$ and $X_{\beta}$ because they generate $\mathfrak{n}$. It cannot contain $X_{\beta}$ and $X_{\alpha+\beta}$ because then it would contain $X_{\alpha+2\beta}$ and hence $T_{0,1}$. The only possibility is $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha},X_{\alpha+\beta}\rangle$.

7.2.5. Equivalences

The algebras $\langle T_{1,1}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ and $\langle T_{1,-1}$ $+X_{\alpha+\beta},X_{\alpha},X_{\alpha+2\beta}\rangle$ cannot be equivalent, since in the former, the restriction of $\text{ad}(T_{1,1}+X_{\beta})$ to the nilpotent subspace has a two-dimensional generalized eigenspace with eigenvalue $2$, whereas in the latter, the restriction of $\text{ad}(T_{1,-1}+X_{\alpha+\beta})$ to the nilpotent subspace has distinct eigenvalues $2,-2$.

However, the element $W$ of Eq. (22) takes $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$ to $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha},X_{\alpha+\beta}\rangle$, showing that they are equivalent. The matrix $A$ defined in Eq. (24) conjugates $\langle T_{1,1}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ to $\langle T_{1,-1}+X_{\alpha+\beta},X_{\beta},X_{\alpha+2\beta}\rangle$.

Finally, we note that $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ consists entirely of singular elements, so it cannot be conjugate to either $\langle T_{1,1}+X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$ or $\langle T_{1,-1}$ $+X_{\alpha+\beta},X_{\alpha},X_{\alpha+2\beta}\rangle$, both of which contain invertible matrices.

Theorem 7.8**.**

Any three-dimensional solvable algebra $\mathfrak{a}$ containing no semisimple elements but which does not consist entirely of nilpotent elements must be equivalent to one of the following inequivalent subalgebras:

[TABLE]

7.3. Three-dimensional nilpotent subalgebras

Since $[X_{\beta},X_{\alpha}]=X_{\alpha+\beta}$ and $[X_{\beta},X_{\alpha+\beta}]=2X_{\alpha+2\beta}$, $X_{\beta}$ and $X_{\alpha}$ generate $\mathfrak{n}$ and $X_{\beta}$ and $X_{\alpha+\beta}$ generate $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$. In fact, any algebra containing an element of the form $X_{\beta}+N$, with $N\in\mathfrak{n}_{\mathfrak{p}}$, and an element of the form $X_{\alpha}+bX_{\alpha+\beta}+cX_{\alpha+2\beta}$ must contain $\mathfrak{n}$. For any $r,s\neq 0$, the algebra $\langle rX_{\beta}+sX_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ is conjugate by an element of the diagonal subgroup $T\subset Sp(4,\mathbb{C})$ to $\langle X_{\beta}+X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$. We summarize.

Theorem 7.9**.**

Any three-dimensional solvable subalgebra $\mathfrak{a}$ consisting entirely of nilpotent elements must be equivalent to one of the following inequivalent subalgebras:

[TABLE]

8. Four-dimensional solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$

In this section, we classify the four-dimensional solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ into five cases: Four-dimensional subalgebras containing a Cartan subalgebra (see Theorem 8.1); solvable subalgebras containing a regular semisimple element but not all of $\mathfrak{t}$ (see Theorem 8.2); solvable subalgebras containing a non-regular semisimple element but not all of $\mathfrak{t}$ (see Theorem 8.4); non-nilpotent solvable subalgebras containing no semisimple elements (see Theorem 8.5); and nilpotent subalgebras (see Theorem 8.6). Again, without loss of generality, we assume that each solvable subalgebra is in the Borel subalgebra $\mathfrak{b}$. The results are summarized in Table 6.

8.1. Four-dimensional Subalgebras Containing a Cartan Subalgebra

The four-dimensional subalgebras of $\mathfrak{b}$ that contain $\mathfrak{t}$ are

[TABLE]

The element $W$ of Eq. (22) conjugates $\langle\mathfrak{t},X_{\alpha},X_{\alpha+\beta}\rangle$ to $\langle\mathfrak{t},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$, and the element $A$ of Eq. (24) conjugates $\langle\mathfrak{t},X_{\beta},X_{\alpha+2\beta}\rangle$ to $\langle\mathfrak{t},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

Consider the eight elements $T_{\pm 2,\pm 1},T_{\pm 1,\pm 2}\in\mathfrak{t}$. They are all the elements of $\mathfrak{t}$ that have distinct eigenvalues $\pm 1,\pm 2$. Acting on $\mathfrak{n}$ via the adjoint representation, they all have the property that they act with even eigenvalues on the root vectors corresponding to long roots and with odd eigenvalues on the root vectors corresponding to short roots. This shows that $\langle\mathfrak{t},X_{\alpha},X_{\alpha+\beta}\rangle$ is not equivalent to $\langle\mathfrak{t},X_{\alpha},X_{\alpha+2\beta}\rangle$.

We summarize.

Theorem 8.1**.**

Any four-dimensional subalgebra of $\mathfrak{b}$ that contains $\mathfrak{t}$ is equivalent to one of the following inequivalent subalgebras.

[TABLE]

8.2. Four-dimensional solvable subalgebras containing a regular

semisimple element but not a Cartan subalgebra

Suppose $a\neq 0,\pm 1$. If $a\neq 3$, then it is easy to see that any four-dimensional subalgebra $\mathfrak{a}\subset\mathfrak{b}$ which contains $T_{a,1}$ but not $\mathfrak{t}$ must be one of the following:

[TABLE]

In the first case, $[\mathfrak{a},\mathfrak{a}]=\mathfrak{n}_{\mathfrak{p}}$, which is abelian, while in the second, $[\mathfrak{a},\mathfrak{a}]=\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, which is not. In particular, the two algebras are not equivalent.

We also observe that the matrix $W$ of Eq. (22) conjugates $\langle T_{a,1},\mathfrak{n}_{\mathfrak{p}}\rangle$ to $\langle T_{a^{-1},1},\mathfrak{n}_{\mathfrak{p}}\rangle$. As in §7.1.1, we argue that any other equivalence between pairs of such algebras must take $T_{a,1}$ to one of $\pm T_{a,1}$, $\pm T_{1,a}$, $\pm T_{-a,1}$, or $\pm T_{1,-a}$. The only possibilities with the right eigenvalues on $\mathfrak{n}_{\mathfrak{p}}$ are $T_{a,1}$ and $T_{1,a}$, so the above equivalence is the only one.

Similarly, the matrix $A$ of Eq. (24) conjugates $\langle T_{a,1},X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$ to $\langle T_{-a,1},X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$. In any equivalence between two algebras of this form, the only rank $1$ eigenvector $X_{\alpha+2\beta}$ must go to a multiple of itself, so its eigenvalue must be preserved, and the only possibilities are that $T_{a,1}$ must go to itself or $T_{a,-1}$. Again the above equivalence is the only one.

If $a=3$, then two analogous algebras occur, with the same equivalences, but for any $r,s\neq 0$, there is also the algebra $\langle T_{3,1},rX_{\alpha}+sX_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$. By Lemma 6.1, for any nonzero $r,s$, this algebra is equivalent to $\mathfrak{a}=\langle T_{3,1},X_{\alpha}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

This algebra $\mathfrak{a}$ is not equivalent to $\langle T_{a,1},\mathfrak{n}_{\mathfrak{p}}\rangle$, for any $a\neq 0,\pm 1$, because $[\mathfrak{a},\mathfrak{a}]=\langle X_{\alpha}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, which is not abelian. Moreover, it is not equivalent to $\langle T_{a,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, for any $a\neq 0,\pm 1$, because the corresponding commutator subalgebras are $\langle X_{\alpha}+X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$ and $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, whose elements are generically of rank $3$ and rank $2$, respectively.

We summarize.

Theorem 8.2**.**

*Any four-dimensional subalgebra of $\mathfrak{b}$ that contains a regular semisimple element but not a Cartan subalgebra is equivalent to one of the following subalgebras, which are inequivalent apart from the equivalences noted. *

[TABLE]

where $a\neq 0,\pm 1$.

8.3. Four-dimensional solvable subalgebras containing a non-regular

semisimple element but not a Cartan subalgebra

As was observed in §6.1.2, the non-regular elements in $\mathfrak{t}$ are multiples of $T_{1,0}$, $T_{0,1}$, $T_{1,1}$, and $T_{1,-1}$.

8.3.1. $T_{0,1}$

Since the positive root vectors have different eigenvalues for $\text{ad}(T_{0,1})$, the only possibilities are algebras spanned by three of them together with $T_{0,1}$. The only possibilities are $\langle T_{0,1},\mathfrak{n}_{\mathfrak{p}}\rangle$ and $\langle T_{0,1},X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$. Since the eigenvalues of $\text{ad}(T_{0,1})$ on $\mathfrak{n}_{\mathfrak{p}}$ are $0,1,2$ while those on $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are $-1,1,0$, we see they are inequivalent.

8.3.2. $T_{1,0}$

The eigenvalues of the positive root vectors for $\text{ad}(T_{1,0})$ are $1$ for $X_{\beta}$ and for $X_{\alpha+\beta}$, [math] for $X_{\alpha}$, and $2$ for $X_{\alpha+2\beta}$.

It is not difficult to check that the only possibilities are $\langle T_{1,0},\mathfrak{n}_{\mathfrak{p}}\rangle$ and $\langle T_{1,0},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$. Since the eigenvalues of $\text{ad}(T_{1,0})$ on $\mathfrak{n}_{\mathfrak{p}}$ are $0,1,2$ while those on $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are $1,1,2$, we see they are inequivalent.

The matrix $W$ of Eq. (22) conjugates $\langle T_{0,1},\mathfrak{n}_{\mathfrak{p}}\rangle$ to $\langle T_{1,0},\mathfrak{n}_{\mathfrak{p}}\rangle$, but $\langle T_{0,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ and $\langle T_{1,0},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are inequivalent because in the former, $\text{ad}(T_{0,1})$ is singular on the nilpotent subalgebra while in the latter, $\text{ad}(T_{1,0})$ is nonsingular on the nilpotent subalgebra.

We summarize.

Lemma 8.3**.**

A four-dimensional solvable algebra containing $T_{0,1}$ or $T_{1,0}$ but not a Cartan subalgebra is equivalent to one of the following:

[TABLE]

These algebras are all pairwise inequivalent.

8.3.3. $T_{1,1}$

The eigenvalues of the positive root vectors for $\text{ad}(T_{1,1})$ are [math] for $X_{\beta}$ and $2$ for $X_{\alpha}$, $X_{\alpha+\beta}$, and $X_{\alpha+2\beta}$.

It is not difficult to check that the only possibilities are $\langle T_{1,1},\mathfrak{n}_{\mathfrak{p}}\rangle$ and $\langle T_{1,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$. Since the eigenvalues of $\text{ad}(T_{1,1})$ on $\mathfrak{n}_{\mathfrak{p}}$ are $2,2,2$ while those on $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are $0,2,2$, we see they are inequivalent.

8.3.4. $T_{1,-1}$

The eigenvalues of the positive root vectors for $\text{ad}(T_{1,-1})$ are $2$ for $X_{\beta}$ and $X_{\alpha+2\beta}$, $-2$ for $X_{\alpha}$, and [math] for $X_{\alpha+\beta}$. It is not difficult to check that the only possibilities are $\langle T_{1,-1},\mathfrak{n}_{\mathfrak{p}}\rangle$ and $\langle T_{1,-1},X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$. Since the eigenvalues of $\text{ad}(T_{1,-1})$ on $\mathfrak{n}_{\mathfrak{p}}$ are $-2,0,2$ while those on $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are $2,0,2$, we see they are inequivalent.

The matrix $A$ of Eq. (24) conjugates $\langle T_{1,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ to $\langle T_{1,-1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$. Consideration of the eigenvalues of $\text{ad}(T_{1,-1})$ on $\mathfrak{n}_{\mathfrak{p}}$ shows that $\langle T_{1,1},\mathfrak{n}_{\mathfrak{p}}\rangle$ and $\langle T_{1,-1},\mathfrak{n}_{\mathfrak{p}}\rangle$ are not equivalent.

We summarize.

Theorem 8.4**.**

A four-dimensional solvable algebra containing a non-regular semisimple element but not a Cartan subalgebra is equivalent to one of the following:

[TABLE]

These algebras are all pairwise inequivalent.

8.4. Non-nilpotent four-dimensional solvable algebras containing no semisimple elements

As in §7.2, we see that, possibly after a conjugation, any four-dimensional solvable algebra $\mathfrak{a}$ containing no semisimple elements but which does not consist entirely of nilpotent elements must contain one of the following elements:

[TABLE]

By Lemma 5.3 and the argument given at the beginning of §§7.2, $\mathfrak{a}$ has a basis consisting of one of the above elements and three elements of $\mathfrak{n}$ which are linear combinations of the positive root vectors that do not commute with it. However, since $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha},X_{\beta},X_{\alpha+2\beta}\rangle$ and $\langle T_{0,1}+X_{\alpha+2\beta},X_{\alpha},X_{\beta},X_{\alpha+\beta}\rangle$ both contain $X_{\alpha}$ and $X_{\beta}$, they both contain all of $\mathfrak{n}$, and hence the former contains the semisimple element $T_{1,-1}$ and the latter contains $T_{0,1}$.

The remaining two algebras $\langle T_{1,1}+X_{\beta},X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ and $\langle T_{1,0}+X_{\alpha},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$ are clearly inequivalent, because the former contains invertible elements while the latter does not.

Theorem 8.5**.**

Any four-dimensional solvable algebra $\mathfrak{a}$ containing no semisimple elements but which does not consist entirely of nilpotent elements must be conjugate to one of the following inequivalent algebras:

[TABLE]

8.5. Four-dimensional nilpotent subalgebras

Theorem 8.6**.**

The only four-dimensional nilpotent subalgebra of $\mathfrak{b}$ is $\mathfrak{n}$.

9. Five- and six-dimensional solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$

9.1. Five-dimensional solvable subalgebras

A five-dimensional solvable subalgebra $\mathfrak{a}\subset\mathfrak{b}$ must contain some semisimple elements. By Lemma 5.2, we can assume it contains $\mathfrak{t}$ if it contains a Cartan subalgebra and $0\neq T\in\mathfrak{t}$ otherwise.

If $\mathfrak{a}\subset\mathfrak{b}$ contains $\mathfrak{t}$, then it must be spanned by $\mathfrak{t}$ and three of the root vectors $X_{\alpha}$, $X_{\beta}$, $X_{\alpha+\beta}$, and $X_{\alpha+2\beta}$. As above, the possibilities are $\mathfrak{a}=\mathfrak{t}+\mathfrak{n}_{\mathfrak{p}}$ and $\mathfrak{a}=\langle\mathfrak{t},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$.

If $\mathfrak{a}\subset\mathfrak{b}$ does not contain $\mathfrak{t}$, then it must contain $0\neq T\in\mathfrak{t}$. We can find a basis for $\mathfrak{a}$ consisting of $T$ and four elements of $\mathfrak{n}$. But these elements span $\mathfrak{n}$, and $\mathfrak{a}=\langle T,\mathfrak{n}\rangle$.

Now it is easy to check that the normalizer of $\mathfrak{n}$ in $Sp(4,\mathbb{C})$ is $B$. So if Ad$(g)$ takes $\langle T,\mathfrak{n}\rangle$ to $\langle T^{\prime},\mathfrak{n}\rangle$, for some $T,T^{\prime}\in\mathfrak{t}$, then $g$ must preserve the nilpotent radical $\mathfrak{n}$ and therefore must be in $B$. Using Lemma 5.2, this means that $T$ and $T^{\prime}$ must be conjugate under $B$. Of course, for any nonzero $r\in\mathbb{C}$, we have that $\langle T,\mathfrak{n}\rangle=\langle rT,\mathfrak{n}\rangle$. So the subalgebra $\langle T,\mathfrak{n}\rangle$ is determined by the nonzero scalar multiples of the $B$-conjugacy class of $T$.

Using the list in Table 1, we find the following list of representatives: $T_{1,a}$, $a\neq 0,\pm 1$; $T_{1,0}$, $T_{0,1}$, $T_{1,1}$, and $T_{1,-1}$.

We summarize.

Theorem 9.1**.**

Up to equivalence, every five-dimensional solvable subalgebra is one of the following. In §§10.4, we shall show that they are in fact pairwise nonisomorphic.

[TABLE]

9.2. Six-dimensional solvable subalgebras

The only six-dimensional subalgebra of $\mathfrak{b}$ is $\mathfrak{b}$ itself.

10. Isomorphisms

10.1. Dimension $2$

We first identify the two-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with respect de Graaf’s classification [dGr05]. For each algebra of de Graaf’s classification that appears, we then identify which algebra it is isomorphic to in the classification described by Šnobl and Winternitz in [ŠW14].

The classification of two-dimensional algebras amounts to whether they are abelian or not. Of course, $\mathfrak{t}$ is abelian, so $\mathfrak{t}\cong K^{1}$.

Since $[T_{3,1},X_{\alpha}+X_{\beta}]=2(X_{\alpha}+X_{\beta})$, $\langle T_{3,1},X_{\alpha}+X_{\beta}\rangle\cong K^{2}$.

Since $[T_{a,1},X_{\alpha}]=2X_{\alpha}$, $\langle T_{a,1},X_{\alpha}\rangle\cong K^{2}$.

Since $[T_{a,1},X_{\beta}]=(a-1)X_{\beta}$, $\langle T_{a,1},X_{\beta}\rangle\cong K^{2}$, provided $a\neq 1$.

Since $[T_{1,0},X_{\alpha}]=0$, $\langle T_{1,0},X_{\alpha}\rangle\cong K^{1}$.

Since $[T_{1,0},X_{\beta}]=X_{\beta}$, $\langle T_{1,0},X_{\beta}\rangle\cong K^{2}$.

Since $[T_{1,0},X_{\alpha+2\beta}]=2X_{\alpha+2\beta}$, $\langle T_{1,0},X_{\alpha+2\beta}\rangle\cong K^{2}$.

Since $[T_{1,1},X_{\alpha}]=2X_{\alpha}$, $\langle T_{1,1},X_{\alpha}\rangle\cong K^{2}$.

Since $[T_{1,1},X_{\beta}]=0$, $\langle T_{1,1},X_{\beta}\rangle\cong K^{1}$.

Since $[T_{1,1},X_{\alpha+\beta}]=2X_{\alpha+\beta}$, $\langle T_{1,1},X_{\alpha+\beta}\rangle\cong K^{2}$.

Since $[T_{1,1}+X_{\beta},X_{\alpha+2\beta}]=2X_{\alpha+2\beta}$, $\langle T_{1,1}+X_{\beta},X_{\alpha+2\beta}\rangle\cong K^{2}$.

Since $[T_{1,0}+X_{\alpha},X_{\alpha+\beta}]=X_{\alpha+\beta}$, $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta}\rangle\cong K^{2}$.

Since $[T_{1,0}+X_{\alpha},X_{\alpha+2\beta}]=2X_{\alpha+2\beta}$, $\langle T_{1,0}+X_{\alpha},X_{\alpha+2\beta}\rangle\cong K^{2}$.

Since $[X_{\alpha},X_{\alpha+\beta}]=0$, $\langle X_{\alpha},X_{\alpha+\beta}\rangle\cong K^{1}$.

Since $[X_{\alpha},X_{\alpha+2\beta}]=0$, $\langle X_{\alpha},X_{\alpha+2\beta}\rangle\cong K^{1}$.

Since $[X_{\beta}+X_{\alpha},X_{\alpha+2\beta}]=0$, $\langle X_{\beta}+X_{\alpha},X_{\alpha+2\beta}\rangle\cong K^{1}$.

The (only) nonabelian two-dimensional solvable algebra is $K^{2}$ in de Graaf’s classification, given by $[x_{1},x_{2}]=x_{2}$ and $\mathfrak{s}_{2,1}$ in Šnobl and Winternitz’s classification, which is given by $[e_{2},e_{1}]=e_{1}$. Clearly they are isomorphic by

[TABLE]

10.2. Dimension $3$

We first identify the three-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with respect de Graaf’s classification [dGr05]. For each algebra of de Graaf’s classification that appears, we then identify which algebra it is isomorphic to in the classification described by Šnobl and Winternitz in [ŠW14].

To see that $\langle\mathfrak{t},X_{\alpha}\rangle\sim L^{3}_{0}$, let

[TABLE]

and observe that these assignments respect the structure of $L^{3}_{0}$, whose only nonzero brackets are $[x_{3},x_{1}]=x_{2}$; $[x_{3},x_{2}]=x_{2}$.

To see that $\langle\mathfrak{t},X_{\beta}\rangle\sim L^{3}_{0}$, let

[TABLE]

and observe that these elements also respect the structure of $L^{3}_{0}$.

Lemma 10.1**.**

Consider the algebra $\mathfrak{a}$ with basis $\{T,A,B\}$ and relations $[A,B]=0$, $[T,A]=2A$ and $[T,B]=rB$, for some $r\in\mathbb{C}$. If $r\neq-2$, let $b=-\frac{2r}{(r+2)^{2}}$.

Then

[TABLE]

Proof.

Fix $k\in\mathbb{C}$, $k\neq 0$.

If we let $x_{1}=A+B$, $x_{2}=2kA+rkB$, and $x_{3}=kT$, then it is easy to check that $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=4k^{2}A+r^{2}k^{2}B$, $[x_{1},x_{2}]=0$.

If we want these elements to satisfy the additional condition $[x_{3},x_{2}]=bx_{1}+x_{2}$, for some $0\neq b\in\mathbb{C}$, this forces

[TABLE]

From these equations, it is easy to find that, provided $r\neq\pm 2$,

[TABLE]

So given $r\neq\pm 2$, and $k$ and $b$ defined as in Eq. (74), the algebra $\mathfrak{a}$ is given by the relations

[TABLE]

These are the relations for the algebra $L^{3}_{b}$.

If $r=2$, let $x_{1}=A$, $x_{2}=B$, $x_{3}=\frac{1}{2}T$. Then

[TABLE]

and these are the defining relations for the algebra $L^{2}$.

If $r=-2$, let $x_{1}=A+B$, $x_{2}=A-B$, $x_{3}=\frac{1}{2}T$. Then

[TABLE]

and these are the defining relations for the algebra $L^{4}_{1}$. ∎

We apply the remarks about the above algebra $\mathfrak{a}$ to various examples.

Fix $a\neq 0,\pm 1$, and consider the algebra $\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle$. Letting $T=T_{a,1}$, $A=X_{\alpha}$, $B=X_{\alpha+\beta}$, we find that $\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle$ is an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=a+1$. Letting $b=-\frac{2r}{(r+2)^{2}}=-2\frac{a+1}{(a+3)^{2}}$, the lemma tells us (c.f., Eq. (70)) that $\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle\sim L^{3}_{b}$, provided $r\neq\pm 2$, i.e., $a\neq-3$ (note that $a=1$ is excluded by hypothesis). It also tells us (c.f., Eq. (72)) that $\langle T_{a,1},X_{\alpha},X_{\alpha+\beta}\rangle\sim L^{4}_{1}$, if $a=-3$.

Now fix $a\neq 0,\pm 1$, and consider the algebra $\langle T_{a,1},X_{\alpha},X_{\alpha+2\beta}\rangle$. Letting $T=T_{a,1}$, $A=X_{\alpha}$, $B=X_{\alpha+2\beta}$, we find that $\langle T_{a,1},X_{\alpha},X_{\alpha+2\beta}\rangle$ is an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=2a$. Letting $b=-\frac{2r}{(r+2)^{2}}=-\frac{a}{(a+1)^{2}}$, we find that $\langle T_{a,1},X_{\alpha},X_{\alpha+2\beta}\rangle\cong L^{3}_{b}$.

Letting $T=T_{3,1}$, $A=X_{\alpha}+X_{\beta}$, $B=X_{\alpha+2\beta}$, we find an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=6$. We conclude that $\langle T_{3,1},X_{\alpha}+X_{\beta},X_{\alpha+2\beta}\rangle\cong L^{3}_{-3/16}$.

Next, letting $x_{1}=X_{\alpha}+X_{\alpha+\beta}$, $x_{2}=X_{\alpha+\beta}$, $x_{3}=T_{1,0}$, we find that $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=x_{2}$, $[x_{1},x_{2}]=0$, which are the defining relations for $L^{3}_{0}$. So $\langle T_{1,0},X_{\alpha},X_{\alpha+\beta}\rangle\cong L^{3}_{0}$.

Similarly, with $x_{1}=X_{\alpha}+X_{\alpha+2\beta}$, $x_{2}=X_{\alpha+2\beta}$, $x_{3}=\frac{1}{2}T_{1,0}$, we find that $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=x_{2}$, $[x_{1},x_{2}]=0$, which are the defining relations for $L^{3}_{0}$. Accordingly, $\langle T_{1,0},X_{\alpha},X_{\alpha+2\beta}\rangle\cong L^{3}_{0}$.

Letting $T=T_{1,0}$, $A=X_{\alpha+2\beta}$, $B=X_{\alpha+\beta}$, we find an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=1$. We conclude that $\langle T_{1,0},X_{\alpha+2\beta},X_{\alpha+\beta}\rangle\cong L^{3}_{-2/9}$.

Next, letting $x_{1}=X_{\alpha+\beta}+X_{\alpha+2\beta}$, $x_{2}=X_{\alpha+2\beta}$, $x_{3}=\frac{1}{2}T_{1,-1}$, we find that $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=x_{2}$, $[x_{1},x_{2}]=0$, which are the defining relations for $L^{3}_{0}$. So $\langle T_{1,-1},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\cong L^{3}_{0}$.

Consider the algebra $\langle T_{1,-1},X_{\alpha},X_{\alpha+2\beta}\rangle$. Letting $T=T_{1,-1}$, $A=X_{\alpha+2\beta}$, $B=X_{\alpha}$, we find that $\langle T_{1,-1},X_{\alpha},X_{\alpha+2\beta}\rangle$ is an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=-2$. From Eq. (72), we find that $\langle T_{1,-1},X_{\alpha},X_{\alpha+2\beta}\rangle\cong L^{4}_{1}$.

Consider the algebra $\langle T_{1,-1},X_{\beta},X_{\alpha+2\beta}\rangle$. Letting $T=T_{1,-1}$, $A=X_{\beta}$, $B=X_{\alpha+2\beta}$, we find that $\langle T_{1,-1},X_{\beta},X_{\alpha+2\beta}\rangle$ is an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=2$. From Eq. (71), we find that $\langle T_{1,-1},X_{\beta},X_{\alpha+2\beta}\rangle\cong L^{2}$.

Similarly, consider $\langle T_{1,1},X_{\alpha},X_{\alpha+2\beta}\rangle$. Letting $T=T_{1,1}$, $A=X_{\alpha}$, $B=X_{\alpha+2\beta}$, we find that $\langle T_{1,1},X_{\alpha},X_{\alpha+2\beta}\rangle$ is an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=2$. From Eq. (71), we find that $\langle T_{1,1},X_{\alpha},X_{\alpha+2\beta}\rangle\cong L^{2}$.

If we let $x_{1}=2(X_{\alpha+\beta}-X_{\alpha+2\beta})$, $x_{2}=X_{\alpha+\beta}$, and $x_{3}=\frac{1}{4}(T_{1,1}+X_{\beta})$, then we find that $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=-\frac{1}{4}x_{1}+x_{2}$, $[x_{1},x_{2}]=0$. So $\langle T_{1,1}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\cong L^{3}_{-1/4}$.

Consider the algebra $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha},X_{\alpha+2\beta}\rangle$. Letting $T=T_{1,-1}+X_{\alpha+\beta}$, $A=X_{\alpha+2\beta}$, $B=X_{\alpha}$, we find that $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha},X_{\alpha+2\beta}\rangle$ is an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=-2$. From Eq. (72), we find that $\langle T_{1,-1}+X_{\alpha+\beta},X_{\alpha},X_{\alpha+2\beta}\rangle\cong L^{4}_{1}$.

Letting $T=T_{1,0}+X_{\alpha}$, $A=X_{\alpha+2\beta}$, $B=X_{\alpha+\beta}$, we find an example of the algebra $\mathfrak{a}$ of Lemma 10.1, with $r=1$. We conclude that $\langle T_{1,0}+X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\cong L^{3}_{-2/9}$.

To see that $\langle X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim L^{4}_{0}$, let

[TABLE]

and observe that these elements generate an algebra isomorphic to $L^{4}_{0}$, whose only nonzero bracket is $[x_{3},x_{1}]=x_{2}$.

To see that $\langle X_{\alpha}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim L^{4}_{0}$, let

[TABLE]

and observe that these elements generate an algebra isomorphic to $L^{4}_{0}$.

The algebra $L^{1}$ in de Graaf’s classification is the abelian algebra. The algebra $L^{2}$ in de Graaf’s classification is given by $[x_{3},x_{1}]=x_{1}$, $[x_{3},x_{2}]=x_{2}$. It is isomorphic to $\mathfrak{s}_{3,1}$ in Šnobl and Winternitz’s classification, with the parameter having the value $A=1$. The algebra $\mathfrak{s}_{3,1}$ is given by $[e_{3},e_{1}]=e_{1}$, $[e_{3},e_{2}]=Ae_{2}$, subject to the following conditions:

[TABLE]

The isomorphism is just $x_{i}\longleftrightarrow e_{i}$, for $i=1,2,3$.

The algebra $L^{3}_{0}$ in de Graaf’s classification is isomorphic to $\mathfrak{n}_{1,1}\oplus\mathfrak{s}_{2,1}$ in Šnobl and Winternitz’s classification, with the parameter having the value $A=0$. The isomorphism is given by

[TABLE]

The algebra $L^{3}_{-1/4}$ in de Graaf’s classification is isomorphic to $\mathfrak{s}_{3,2}$ in Šnobl and Winternitz’s classification, which is given by $[e_{3},e_{1}]=e_{1}$, $[e_{3},e_{2}]=e_{1}+e_{2}$. The correspondence is given by

[TABLE]

The algebra $L^{4}_{0}$ in de Graaf’s classification is given by $[x_{3},x_{1}]=x_{2}$. It is isomorphic to $\mathfrak{n}_{3,1}$ in Šnobl and Winternitz’s classification, which is given by $[e_{2},e_{3}]=e_{1}$. The isomorphism is

[TABLE]

The algebra $L^{4}_{1}$ in de Graaf’s classification is given by $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=x_{1}$. It is isomorphic to $\mathfrak{s}_{3,1}$ in Šnobl and Winternitz’s classification, with the parameter having the value $A=-1$. The isomorphism is

[TABLE]

In de Graaf’s classification, $L^{3}_{\alpha}$ is given by $[x_{3},x_{1}]=x_{2}$, $[x_{3},x_{2}]=\alpha x_{1}+x_{2}$.

Lemma 10.2**.**

Suppose $\alpha\neq 0,-\frac{1}{4}$. Then there is one choice of the square root $\sqrt{1+4\alpha}$ for which the complex number

[TABLE]

satifies the conditions of Eq. (80).

Proof.

Fix one value of the square root and let $\lambda^{+}=\frac{1+\sqrt{1+4\alpha}}{2}$, $\lambda^{-}=\frac{1-\sqrt{1+4\alpha}}{2}$. Then $\lambda^{+}\lambda^{-}=-\alpha$, and either $\frac{\lambda^{+}}{\lambda^{-}}$ or $\frac{\lambda^{-}}{\lambda^{+}}$ satisfies the conditions of Eq. (80). If the former, the initial choice of square root is correct. If the latter, then make the other choice of square root. Either way, then $\lambda=\frac{\lambda^{+}}{\lambda^{-}}$ satisfies the conditions. ∎

If $\alpha\neq 0,-\frac{1}{4}$, then $L^{3}_{\alpha}\cong\mathfrak{s}_{3,1,A=\lambda}$, with parameter

[TABLE]

as described in Lemma 10.2.

The isomorphism is given by

[TABLE]

where $\lambda^{+}$ and $\lambda^{-}$ are as in the proof of Lemma 10.2. Note that $\lambda^{+}+\lambda^{-}=1$.

10.3. Dimension $4$

We first identify the four-dimensional subalgebras of $\mathfrak{sp}(4,\mathbb{C})$ with respect de Graaf’s classification [dGr05]. For each algebra of de Graaf’s classification that appears, we then identify which algebra it is isomorphic to in the classification described by Šnobl and Winternitz in [ŠW14].

To see that $\langle\mathfrak{t},X_{\alpha},X_{\alpha+\beta}\rangle\sim M^{8}$, let

[TABLE]

and observe that these elements generate an algebra isomorphic to $M^{8}$, whose only nonzero brackets are $[x_{1},x_{2}]=x_{2}$ and $[x_{3},x_{4}]=x_{4}$.

To see that $\langle\mathfrak{t},X_{\alpha},X_{\alpha+2\beta}\rangle\sim M^{8}$, let

[TABLE]

and observe that these elements generate an algebra isomorphic to $M^{8}$.

In $\langle T_{a,1},\mathfrak{n}_{\mathfrak{p}}\rangle$, $a\neq 0\pm 1$, let

[TABLE]

Then $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$, and $[x_{4},x_{3}]=\frac{2}{3(a+1)}X_{\alpha}+\frac{1}{3}X_{\alpha+\beta}+\frac{2a}{3(a+1)}X_{\alpha+2\beta}$.

It is not difficult to check that $[x_{4},x_{3}]=Ax_{1}+Bx_{2}+x_{3}$, where $A=\frac{4a}{27(a+1)^{2}}$ and $B=-\frac{2(a^{2}+4a+1)}{9(a+1)^{2}}$. These are the nonzero defining relations for $M^{6}_{A,B}$, and we find that $\langle T_{a,1},\mathfrak{n}_{\mathfrak{p}}\rangle\sim M^{6}_{A,B}=$

$M^{6}_{4a/(27(a+1)^{2}),-2(a^{2}+4a+1)/(9(a+1)^{2})}$.

In $\langle T_{a,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, with $a\neq 0\pm 1$, let $x_{1}=X_{\beta}+X_{\alpha+\beta}$, $x_{2}=\frac{8a}{a^{2}-1}X_{\alpha+2\beta}$, $x_{3}=\frac{2a}{a-1}X_{\beta}+\frac{2a}{a+1}X_{\alpha+\beta}$, $x_{4}=\frac{1}{2a}T_{a,1}$. Then it is easy to verify that the only nonzero relations are $[x_{4},x_{1}]=x_{1}+\frac{1-a^{2}}{4a^{2}}x_{3}$, $[x_{4},x_{2}]=x_{2}$, $[x_{4},x_{3}]=x_{1}$, $[x_{3},x_{1}]=x_{2}$. These are the defining relations for $M^{13}_{(1-a^{2})/(4a^{2})}$, so $\langle T_{a,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{13}_{(1-a^{2})/(4a^{2})}$.

In $\langle T_{3,1},X_{\alpha}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, let $x_{1}=X_{\alpha}+X_{\beta}+2X_{\alpha+\beta}$, $x_{2}=6X_{\alpha+2\beta}$, $x_{3}=3(X_{\alpha}+X_{\beta})+3X_{\alpha+\beta}$, $x_{4}=\frac{1}{6}T_{3,1}$.

Then it is easy to verify that the only nonzero relations are $[x_{4},x_{1}]=x_{1}-\frac{2}{9}x_{3}$, $[x_{4},x_{2}]=x_{2}$, $[x_{4},x_{3}]=x_{1}$, $[x_{3},x_{1}]=x_{2}$. These are the defining relations for $M^{13}_{-2/9}$, so $\langle T_{3,1},X_{\alpha}+X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{13}_{-2/9}$.

In $\langle T_{0,1},\mathfrak{n}_{\mathfrak{p}}\rangle$, let $x_{1}=\frac{9}{4}X_{\alpha}+9X_{\alpha+\beta}+X_{\alpha+2\beta}$, $x_{2}=\frac{3}{2}X_{\alpha}+3X_{\alpha+\beta}$, $x_{3}=X_{\alpha}+X_{\alpha+\beta}$, $x_{4}=\frac{1}{3}T_{0,1}$.

Then $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$, and $[x_{4},x_{3}]=\frac{2}{3}X_{\alpha}+\frac{1}{3}X_{\alpha+\beta}$.

It is not difficult to check that $[x_{4},x_{3}]=-\frac{2}{9}x_{2}+x_{3}$. These are the nonzero defining relations for $M^{6}_{0,-2/9}$, so $\langle T_{0,1},\mathfrak{n}_{\mathfrak{p}}\rangle\sim M^{6}_{0,-2/9}$.

In $\langle T_{0,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, let $x_{1}=-X_{\beta}+X_{\alpha+\beta}$, $x_{2}=4X_{\alpha+2\beta}$, $x_{3}=X_{\beta}+X_{\alpha+\beta}$, $x_{4}=T_{0,1}$. Then $[x_{4},x_{1}]=x_{3}$, $[x_{4},x_{3}]=x_{1}$, and $[x_{3},x_{1}]=x_{2}$. These are the nonzero relations for $M^{14}_{1}$, so we find that $\langle T_{0,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{14}_{1}$.

In $\langle T_{1,0},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, let $x_{1}=X_{\alpha+\beta}$, $x_{2}=X_{\alpha+2\beta}$, $x_{3}=\frac{1}{2}X_{\beta}$, $x_{4}=T_{1,0}$. Then $[x_{4},x_{1}]=x_{1}$, $[x_{4},x_{2}]=2x_{2}$, $[x_{4},x_{3}]=x_{3}$, and $[x_{3},x_{1}]=x_{2}$. These are the nonzero relations for $M^{12}$, so we find that $\langle T_{1,0},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{12}$.

In $\langle T_{1,1},\mathfrak{n}_{\mathfrak{p}}\rangle$, let $x_{1}=X_{\alpha}$, $x_{2}=X_{\alpha+\beta}$, $x_{3}=X_{\alpha+2\beta}$, $x_{4}=\frac{1}{2}T_{1,1}$. Then the nonzero relations are $[x_{4},x_{1}]=x_{1}$, $[x_{4},x_{2}]=x_{2}$, $[x_{4},x_{3}]=x_{3}$, showing that $\langle T_{1,1},\mathfrak{n}_{\mathfrak{p}}\rangle\sim M^{2}$.

In $\langle T_{1,-1},\mathfrak{n}_{\mathfrak{p}}\rangle$ let $x_{1}=X_{\alpha}+X_{\alpha+\beta}+X_{\alpha+2\beta}$, $x_{2}=X_{\alpha}-X_{\alpha+2\beta}$, $x_{3}=X_{\alpha}+X_{\alpha+2\beta}$, $x_{4}=\frac{1}{2}T_{1,-1}$.

Then $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$, and $[x_{4},x_{3}]=X_{\alpha}-X_{\alpha+2\beta}=x_{2}$. These are the nonzero defining relations for $M^{7}_{0,1}$, so $\langle T_{1,-1},\mathfrak{n}_{\mathfrak{p}}\rangle\sim M^{7}_{0,1}$.

In $\langle T_{1,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, let $x_{1}=X_{\alpha+\beta}$, $x_{2}=X_{\alpha+2\beta}$, $x_{3}=\frac{1}{2}X_{\beta}+X_{\alpha+\beta}$, $x_{4}=\frac{1}{2}T_{1,1}$. Then $[x_{4},x_{1}]=x_{1}$, $[x_{4},x_{2}]=x_{2}$, $[x_{4},x_{3}]=x_{1}$, and $[x_{3},x_{1}]=x_{2}$, which are the nonzero relations for $M^{13}_{0}$. We conclude that $\langle T_{1,1},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{13}_{0}$.

In $\langle T_{1,1}+X_{\beta},X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, let $x_{1}=54X_{\alpha}$, $x_{2}=18X_{\alpha}+9X_{\alpha+\beta}$, $x_{3}=6X_{\alpha}+6X_{\alpha+\beta}+3X_{\alpha+2\beta}$, $x_{4}=\frac{1}{6}(T_{1,1}+X_{\beta})$. Then $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$, and $[x_{4},x_{3}]=2X_{\alpha}+3X_{\alpha+\beta}+3X_{\alpha+2\beta}$, which equals $\frac{1}{27}x_{1}-\frac{1}{3}x_{2}+x_{3}$. These are the only nonzero relations, so we find that $\langle T_{1,1}+X_{\beta},X_{\alpha},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{6}_{1/27,-1/3}$.

In $\langle T_{1,0}+X_{\alpha},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle$, let $x_{1}=-\frac{1}{2}X_{\beta}+\frac{1}{2}X_{\alpha+\beta}$, $x_{2}=-X_{\alpha+2\beta}$, $x_{3}=-X_{\beta}$, $x_{4}=\frac{1}{2}(T_{1,0}+X_{\alpha})$. Then $[x_{4},x_{1}]=x_{1}-\frac{1}{4}x_{3}$, $[x_{4},x_{2}]=x_{2}$, and $[x_{4},x_{3}]=x_{1}$, and $[x_{3},x_{1}]=x_{2}$. These are the nonzero relations for $M^{13}_{-1/4}$. We conclude that $\langle T_{1,0}+X_{\alpha},X_{\beta},X_{\alpha+\beta},X_{\alpha+2\beta}\rangle\sim M^{13}_{-1/4}$.

In $\mathfrak{n}$, let $x_{1}=X_{\alpha}$, $x_{2}=X_{\alpha+\beta}$, $x_{3}=2X_{\alpha+2\beta}$, $x_{4}=X_{\beta}$. Then $[x_{4},x_{1}]=x_{2}$ and $[x_{4},x_{2}]=x_{3}$ are the only nonzero relations, and we find that $\mathfrak{n}\sim M^{7}_{0,0}$.

The algebra $M^{2}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{1}$, $[x_{4},x_{2}]=x_{2}$, $[x_{4},x_{3}]=x_{3}$.

It is isomorphic to $\mathfrak{s}_{4,3}$ with parameters $A=B=1$ in Šnobl and Winternitz’s classification, which is given by $[e_{4},e_{1}]=e_{1}$, $[e_{4},e_{2}]=e_{2}$, $[e_{4},e_{3}]=e_{3}$. The isomorphism is $x_{i}\longleftrightarrow e_{i}$, for $i=1,\dots,4$.

The algebra $M^{8}$ in de Graaf’s classification is given by $[x_{1},x_{2}]=x_{2}$, $[x_{3},x_{4}]=x_{4}$. It is isomorphic to $K^{2}\oplus K^{2}$, so it is isomorphic to $\mathfrak{s}_{2,1}\oplus\mathfrak{s}_{2,1}$ in Šnobl and Winternitz’s classification, which is also isomorphic to $\mathfrak{s}_{4,12}$ over $\mathbb{C}$.

The algebra $M^{6}_{\alpha,\beta}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$, $[x_{4},x_{3}]=\alpha x_{1}+\beta x_{2}+x_{3}$.

This description of $M^{6}_{\alpha,\beta}$ shows that its nilradical $\mathfrak{n}$ is abelian, with basis $\{x_{1},x_{2},x_{3}\}$. The action of $x_{4}$ on $\mathfrak{n}$ is given, relative to this basis, by the matrix

[TABLE]

This is a companion matrix, with characteristic polynomial $\lambda^{3}-\lambda^{2}-\alpha\lambda-\beta$.

Since the roots add up to $1$, the only way there could be a single root of multiplicity $3$ would be for it to be $\frac{1}{3}$. This happens with $\beta=\frac{1}{27}$, $\alpha=-\frac{1}{3}$, and it is not difficult to see that in this case, the Jordan Canonical Form of $C$ has a single block. In particular, ${\rm ad}(3x_{4})$ has an eigenvector $u_{1}\in\mathfrak{n}$ with eigenvalue $1$, and there are vectors $v,w\in\mathfrak{n}$ so that $[3x_{4},v]=u_{1}+v$, $[3x_{4},w]=v+w$.

This algebra is isomorphic to $\mathfrak{s}_{4,2}$, which is given by $[e_{4},e_{1}]=e_{1}$, $[e_{4},e_{2}]=e_{1}+e_{2}$, $[e_{4},e_{3}]=e_{2}+e_{3}$. The isomorphism is given by

[TABLE]

Depending on $\alpha$ and $\beta$, the matrix $C$ may have three distinct eigenvalues, two distinct eigenvalues with one of them associated to a $2\times 2$ Jordan block, or the single eigenvalue $\lambda=\frac{1}{3}$, as discussed above.

Suppose $\beta\neq 0$, so that the eigenvalues are all nonzero.

If there are three distinct (nonzero) eigenvalues, we can assume they satisfy $|r^{\prime}|\geq|s^{\prime}|\geq|t^{\prime}|>0$. If $|r^{\prime}|>|s^{\prime}|$, then $\frac{1}{r^{\prime}}x_{4}$ has eigenvalues $1,s,t$, with $1>|s|\geq|t|>0$. If $|r^{\prime}|=|s^{\prime}|>|t^{\prime}|$, then $\frac{1}{r^{\prime}}x_{4}$ has eigenvalues $1,s,t$, with $1=|s|\geq|t|>0$ and we can assume $0<{\rm arg}(s)<2\pi$. If $|r^{\prime}|=|s^{\prime}|=|t^{\prime}|$, then $\frac{1}{r^{\prime}}x_{4}$ has eigenvalues $1,s,t$, with $1=|s|=|t|$ and we can assume $0<{\rm arg}(s)<{\rm arg}(t)<2\pi$. In each of these cases, let us write $u_{1},u_{s},u_{t}\in\mathfrak{n}$ for the corresponding eigenvectors.

In each of these cases, $M^{6}_{\alpha,\beta}\cong\mathfrak{s}_{4,3}$, with parameters $A=s$, $B=t$. This algebra is given by $[e_{4},e_{1}]=e_{1}$, $[e_{4},e_{2}]=Ae_{2}$, $[e_{4},e_{3}]=Be_{3}$. The isomorphism is given by

[TABLE]

If there are two distinct (nonzero) eigenvalues, we can assume that $r^{\prime}$ is associated to a $2\times 2$ Jordan block and $s^{\prime}$ is associated to a $1\times 1$ Jordan block. Then $\frac{1}{r^{\prime}}x_{4}$ has an eigenvector $u_{s}$ with eigenvalue $s=\frac{s^{\prime}}{r^{\prime}}$, an eigenvector $u_{1}$ with eigenvalue $1$, and a vector $v$ so that $[x_{4},v]=u_{1}+v$. In this case, $M^{6}_{\alpha,\beta}\cong\mathfrak{s}_{4,4}$, with parameter $A=s$.

This algebra is given by $[e_{4},e_{1}]=e_{1}$, $[e_{4},e_{2}]=e_{1}+e_{2}$, $[e_{4},e_{3}]=se_{3}$. The isomorphism is given by

[TABLE]

If $\beta=0$, $\alpha\neq 0,-\frac{1}{4}$, then $M^{6}_{0,\alpha}$ is isomorphic to $\mathfrak{n}_{1,1}\oplus\mathfrak{s}_{3,1}$ in Šnobl and Winternitz’s classification, with parameter

[TABLE]

With the choice of square root described in Lemma 10.2, the isomorphism is given by

[TABLE]

If $\beta=0$, $\alpha=-\frac{1}{4}$, then $M^{6}_{0,-1/4}$ is isomorphic to $\mathfrak{n}_{1,1}\oplus\mathfrak{s}_{3,2}$ in Šnobl and Winternitz’s classification. The isomorphism is given by

[TABLE]

We give some examples.

Suppose $\beta=1$, $\alpha=-1$. The algebra $M^{6}_{1,-1}$ is isomorphic to $\mathfrak{s}_{4,3}$. The isomorphism is given by

[TABLE]

Now suppose $\beta=-1$, $\alpha=1$. The algebra $M^{6}_{-1,1}$ is isomorphic to $\mathfrak{s}_{4,4}$. The isomorphism is given by

[TABLE]

The algebra $M^{13}_{\alpha}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{1}+\alpha x_{3}$, $[x_{4},x_{2}]=x_{2}$, $[x_{4},x_{3}]=x_{1}$, $[x_{3},x_{1}]=x_{2}$.

First consider the case in which $\alpha=0$. Then $M^{13}_{0}$ is isomorphic to $\mathfrak{s}_{4,11}$ in Šnobl and Winternitz’s classification, which is given by $[e_{4},e_{1}]=e_{1}$, $[e_{4},e_{2}]=e_{2}$, $[e_{2},e_{3}]=e_{1}$.

The isomorphism is

[TABLE]

Now suppose $\alpha=-\frac{1}{4}$. Then $M^{13}_{-1/4}$ is isomorphic to $\mathfrak{s}_{4,10}$ in Šnobl and Winternitz’s classification, which is given by $[e_{4},e_{1}]=2e_{1}$, $[e_{4},e_{2}]=e_{2}$, $[e_{4},e_{3}]=e_{2}+e_{3}$, $[e_{2},e_{3}]=e_{1}$.

The isomorphism is

[TABLE]

Now suppose $\alpha\neq 0,-\frac{1}{4}$. Then $M^{13}_{\alpha}$ is isomorphic to $\mathfrak{s}_{4,8}$, with parameter $\lambda=\frac{1+2\alpha+\sqrt{1+4\alpha}}{-2\alpha}$ as described in Lemma 10.2. Note that, with choice of square root specified in that lemma and the notation given in the proof of that lemma, $(1+\lambda)\lambda^{+}=\lambda$ and $(1+\lambda)\lambda^{-}=1$. The isomorphism is

[TABLE]

The algebra $M^{14}_{1}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{3}$, $[x_{4},x_{3}]=x_{1}$, $[x_{3},x_{1}]=x_{2}$.

It is isomorphic to $\mathfrak{s}_{4,6}$ in Šnobl and Winternitz’s classification, which is given by $[e_{4},e_{2}]=e_{2}$, $[e_{4},e_{3}]=-e_{3}$, $[e_{2},e_{3}]=e_{1}$.

The isomorphism is

[TABLE]

The algebra $M^{12}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{1}$, $[x_{4},x_{2}]=2x_{2}$, $[x_{4},x_{3}]=x_{3}$, $[x_{3},x_{1}]=x_{2}$.

It is isomorphic to $\mathfrak{s}_{4,8}$, with parameter $A=1$, in Šnobl and Winternitz’s classification, which is given by $[e_{4},e_{1}]=2e_{1}$, $[e_{4},e_{2}]=e_{2}$, $[e_{4},e_{3}]=e_{3}$, $[e_{2},e_{3}]=e_{1}$.

The isomorphism is

[TABLE]

The algebra $M^{7}_{0,1}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$, $[x_{4},x_{3}]=x_{2}$.

It is isomorphic to $\mathfrak{n}_{1,1}\oplus\mathfrak{s}_{3,1}$, with parameter $A=-1$ in Šnobl and Winternitz’s classification. The isomorphism is

[TABLE]

The algebra $M^{7}_{0,0}$ in de Graaf’s classification is given by $[x_{4},x_{1}]=x_{2}$, $[x_{4},x_{2}]=x_{3}$.

It is isomorphic to $\mathfrak{n}_{4,1}$ in Šnobl and Winternitz’s classification, which is given by $[e_{4},e_{2}]=-e_{1}$, $[e_{4},e_{3}]=-e_{2}$.

The isomorphism is

[TABLE]

10.4. Dimension $5$

Since the classification of de Graaf [dGr05] does not include solvable algebras of dimension greater than $4$, we will only identify the five-dimensional solvable subalgebras with respect to Šnobl and Winternitz’s classification [ŠW14].

The following isomorphism establishes $\mathfrak{s}_{5,41,A=B=\frac{1}{2}}\cong\langle\mathfrak{t},\mathfrak{n}_{\mathfrak{p}}\rangle$:

[TABLE]

The following isomorphism establishes $\mathfrak{s}_{5,44}\cong\langle\mathfrak{t},X_{\beta},X_{\alpha+\beta},$ $X_{\alpha+2\beta}\rangle$:

[TABLE]

The following isomorphism establishes $\mathfrak{s}_{5,35,A=\frac{2}{a-1}}\cong\langle T_{a,1},\mathfrak{n}\rangle$:

[TABLE]

The following isomorphism establishes $\mathfrak{s}_{5,35,A=-1}\cong\langle T_{1,-1},\mathfrak{n}\rangle$:

[TABLE]

The following isomorphism establishes $\mathfrak{s}_{5,37}\cong\langle T_{1,1},\mathfrak{n}\rangle$:

[TABLE]

The following isomorphism establishes $\mathfrak{s}_{5,36}\cong\langle T_{1,0},\mathfrak{n}\rangle$:

[TABLE]

The following isomorphism establishes $\mathfrak{s}_{5,33}\cong\langle T_{0,1},\mathfrak{n}\rangle$:

[TABLE]

10.5. Dimension $6$

The only solvable subalgebra of $\mathfrak{b}$ of dimension $6$ is $\mathfrak{b}$ itself. Referencing the classification in [ŠW14], the following isomorphism establishes $\mathfrak{s}_{6,242}\cong\mathfrak{b}$:

[TABLE]

11. Conclusions

The semisimple subalgebras of the rank $2$ symplectic Lie algebra $\mathfrak{sp}(4,\mathbb{C})$ are well-known [dGr09, dGr11], and we recently classified its Levi decomposable subalgebras [DR15]. In this article, we classified the solvable subalgebras of $\mathfrak{sp}(4,\mathbb{C})$, up to inner automorphism (equivalently, up to conjugacy by the symplectic group $Sp(4,\mathbb{C})$). By Levi’s theorem, this completes the classification of the subalgebras of $\mathfrak{sp}(4,\mathbb{C})$.

We summarized the classification in Tables 3 to 7. The classification is given with respect to the partial classification of solvable Lie algebras of de Graaf [dGr05], and that described by Šnobl and Winternitz in [ŠW14].

We have already classified the subalgebras of the special orthogonal algebra $\mathfrak{so}(4,\mathbb{C})$ [DR16a], and the subalgebras of the special linear algebra $\mathfrak{sl}(3,\mathbb{C})$ [DR16b]. And, Mayanskiy [May16] recently posted a classification of the subalgebras of the exceptional Lie algebra $G_{2}$. Hence, with this article, the classification of the subalgebras of the rank $2$ semisimple Lie algebras is complete.

Acknowledgements

The work of A.D. is partially supported by a research grant from the Professional Staff Congress/City University of New York (Grant No. TRADA-47-36). The work of J.R. is partially supported by the Natural Sciences and Engineering Research Council (Grant No. 3166-09). The authors would also like to thank the referee for his/her valuable comments.