I would like to know what the simply-connected Lie groups of dimension $2$ are. It is well-known that for every Lie algebra, there is exactly one simply-connected Lie group having it as its Lie algebra. I know that there are two two-dimensional Lie algebras, namely the abelian Lie algebra and another Lie algebra with basis ${x, y}$ such that $[x, y]=x$. The simply-connected Lie group corresponding to the abelian Lie algebra is $\mathbb{R}^2$. Can you tell me, what the Lie group corresponding to the other Lie algebra is?
What are the simply-connected two-dimensional Lie groups?
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3 Answers
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The group $\textrm{Aff}(\Bbb R) \cong \Bbb R \rtimes \Bbb R^*$ of affine transformations of the real line (under composition) has dimension $2$ and is nonabelian, so by your classification its Lie algebra is the nonabelian algebra you mention (for an appropriate choice of basis). The above isomorphism identifies the usual composition of affine functions $t \mapsto a t + b$ and $t \mapsto a' t + b'$ with the operation $$(b, a) \cdot (b', a') := (a b' + b, a a').$$
One can realize $\textrm{Aff}(\Bbb R)$ as the matrix group $$\left\{\pmatrix{a&b\\0&1} : a, b \in \Bbb R; a \neq 0 \right\} < \textrm{GL}(2, \Bbb R)$$ via $t \mapsto a t + b \leftrightarrow \pmatrix{a&b\\0&1} .$
The induced Lie algebra is $$\left\{\pmatrix{a&b\\0&0} : a, b \in \Bbb R\right\} < \frak{gl}(2, \Bbb R),$$ and that the basis $(x, y)$ given by $$x = \pmatrix{0&1\\0&0}, \quad y = \pmatrix{-1&0\\0&0}$$ satisfies the given relation $[x, y] = x$.
Finally, we can just as well realize this group on the set $\Bbb R^2$ with a nonstandard multiplication $\ast$: Conjugating the group operation $\cdot$ given above by the map $\Bbb R^2 \to \Bbb R \rtimes \Bbb R^+$ given by $(r, s) \mapsto (r, e^s)$ gives the rule $$(c, d) \ast (c', d') = (c + e^d c', d + d').$$
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Here is perhaps a more explicit answer. Since $G$ is simply connected, it suffices to look at the Lie algebra. Let $G$ be a Lie group with a two-dimensional Lie algebra $\mathfrak{g}$ spanned by vectors $X,A.$ If all Lie brackets of $\mathfrak{g}$ are trivial then $\mathfrak{g}$ is just a commutative Lie algebra and there is really nothing else to say. Suppose next that $\mathfrak{g}$ is not commutative. Next, let us consider the adjoint representation of $\mathfrak{g}$ with respect to the ordered basis $\mathcal{B}=\left( X,A\right) .$ Then the matrix representation of $ad\left( X\right) $ with respect to this fixed basis is given by $$ \left[ ad\left( X\right) \right] _{\mathcal{B}}=\left[ \begin{array} [c]{cc}% 0 & a\\ 0 & b \end{array} \right] $$ Similarly, the matrix representation of $ad\left( A\right) $ with respect to our fixed basis is given by $$ \left[ ad\left( A\right) \right] _{\mathcal{B}}=\left[ \begin{array} [c]{cc}% c & 0\\ d & 0 \end{array} \right] . $$ Next, $\left[ ad\left( A\right) \right]_{\mathcal{B}}$ has for eigenvalues $c,0.$ Note that $c$ cannot be zero. Otherwise, we could then find a basis for the Lie algebra such that the endomorphism $adA$ is zero; contradicting the fact that $\mathfrak{g}$ is not commutative. Assuming next that $c$ is not zero, then $c^{-1}ad\left( A\right) $ has for eigenvalue $1,0.$ As such, $c^{-1}ad\left( A\right) $ is diagonalizable and its Jordan canonical form is $$ \left[ \begin{array} [c]{cc}% 1 & 0\\ 0 & 0 \end{array} \right] . $$ Consequently, there exists a basis $\left( Y,B\right) $ for the Lie algebra $\mathfrak{g}$ such that the only non-trivial Lie brackets are given by $\left[ B,Y\right] =Y.$ This completes the classification of all two-dimensional Lie algebras.
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It is the group of matrrices $\{ \pmatrix{ a& b\cr 0& 1}, a>0\}$