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I have found that affine subgroups is one of the two-dimensional subgroups of SL(2,R). Up to isomorphism is it the ultimate two dimensional subgroup? I have taken the two dimensional Lie algebras and see up to isomorphism there are two such cases, abelian and non-abelian. Non-abelian case gives us the affine group. Abelian case give us diagonal matrices but since trace zero in sl(2,R) it becomes one-dimensional.

José Carlos Santos
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Sandipan Dutta
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  • NB that the Lie group $G := \Bbb R \rtimes \Bbb R_+$ of oriented affine subgroups is (connected and) simply connected, so your classification of $2$-dimensional real Lie algebras any nonabelian $2$-dimensional real Lie group is (isomorphic to) a quotient of $G$ by a discrete, normal subgroup, but the only such subgroup is the trivial one, so (up to isomorphism) $G$ is the only nonabelian $2$-dimensional real Lie group. – Travis Willse Aug 13 '19 at 16:18
  • Thanks...But is the result only for connected subgroup..i.e connected two dimensional subgroup is isomorphic to Aff(1)..but what about disconnected two-dim subgr. of SL(2,R)..is there any ...so?..see this ans....https://math.stackexchange.com/questions/3257971/classification-of-two-dimensional-lie-groups?rq=1 – Sandipan Dutta Aug 14 '19 at 06:19
  • You're right, I should have said "connected". As for $\operatorname{SL(2, \Bbb R)}$ there are infinitely many disconnected $2$-dimensional subgroups, including $$\left{\pmatrix{a&b\&\frac{1}{a}} : a \in \Bbb R^*, b \in \Bbb R\right} .$$ I would guess, but am not sure, that all of the other such subgroups are conjugate to this one. – Travis Willse Aug 14 '19 at 15:37

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The Lie algebra $\mathfrak{sl}(2,\mathbb R)$ has no $2$-dimensional Abelian Lie subalgebras. Therefore, there are no Abelian $2$-dimensional Lie subgroups of $\mathrm{SL}(2,\mathbb R)$.

In order to prove that $\mathfrak{sl}(2,\mathbb R)$ has no $2$-dimensional Abelian Lie subalgebras. let$$H=\begin{bmatrix}1&0\\0&-1\end{bmatrix},\ X=\begin{bmatrix}0&1\\0&0\end{bmatrix},\text{ and }Y=\begin{bmatrix}0&0\\1&0\end{bmatrix}.$$Then $\{H,X,Y\}$ is a basis of $\mathfrak{sl}(2,\mathbb R)$ and we have:

  • $[H,X]=2X$,
  • $[H,Y]=-2Y$,
  • $[X,Y]=H$.

If $a_1H+a_2X+a_3Y,b_1H+b_2X+b_3Y\in\mathfrak{sl}(2,\mathbb R)$, then\begin{multline}[a_1H+a_2X+a_3Y,b_1H+b_2X+b_3Y]=0\iff\\\iff2(a_1b_2-a_2b_1)X-2(a_1b_3-a_3b_1)Y+(a_2b_3-a_3b_2)H=0\iff\\\iff\left\{\begin{array}{l}a_1b_2-a_2b_1=0\\a_1b_3-a_3b_1=0\\a_2b_3-a_3b_2=0.\end{array}\right.\end{multline}But this system is equivalent to $(a_1,a_2,a_3)\times(b_1,b_2,b_3)=0$ and this equality occurs if and only if one of the vectors $(a_1,a_2,a_3)$ and $(b_1,b_2,b_3)$ is a multiple of the other one. This proves that, if $X_1,X_2\in\mathfrak{sl}(2,\mathbb R)$ are such that $[X_1,X_2]=0$, then $\dim\operatorname{span}\bigl(\{X_1,X_2\}\bigr)\leqslant1$.

Max Demirdilek
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José Carlos Santos
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