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I am a little confused as to how to compute generally the Lie algebra of a Lie group and viceversa, namely the Lie groups (up to diffeomorphism) having a certain Lie algebra.

The way I did this for classical groups such as $O(n)$ or $SL_n(\mathbb{R})$ was to express them as fibres over a regular value of a smooth map and then explicitly computing the tangent space at the identity as the kernel of the differential. This method however works only in very specific cases.

  1. How does one, for instance, compute the Lie algebra of the group $SO(2) \bar{\times} \mathbb{R}^4$ (by $\bar{\times}$ I mean the semi-direct product).
  2. Which connected Lie groups up to diffeomorphism have the following Lie algebra

$$\left\{\left(\begin{array}{ccc} x & y & w \\ z & -x & v \\ 0 & 0 & 0 \end{array} \right), \qquad x,y,z,v,w\in \mathbb{R}\right\}?$$

Thanks in advance for any help.

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For 1 you can try embedding the Lie group into $\text{GL}_n$ for some $n$ and computing which matrices exponentiate into it. For 2, every connected Lie group $H$ having a particular Lie algebra $\mathfrak{g}$ is covered by a unique simply connected Lie group $G$ having Lie algebra $\mathfrak{g}$, and in fact $H$ is obtained from $G$ by quotienting by a discrete subgroup of its center. So to find all $H$ it suffices to find $G$ (which may not be easy in general) then to describe the discrete subgroups of its center. –  Qiaochu Yuan Aug 2 '12 at 21:35
    
Two more notes: for 1 I think a semidirect product of Lie groups has Lie algebra the corresponding semidirect product, and for 2 to find $G$ it suffices to find some $H$ and take its universal cover (although this may not be easy to describe in general). –  Qiaochu Yuan Aug 2 '12 at 21:36

1 Answer 1

1 was answered by a comment to yr question.

As for 2, let $W=\{(a,b,c)\in \mathbb R^3| c=0\}$ and let $G$ be the set of elements $g\in GL_3(\mathbb R)$ such that (1) $g$ leaves $W$ invariant ($gW=W$), (2) the induced action on $W$ satisfies $det=1$, and (3) the induced action on $\mathbb R^3/W$ is trivial (the identity). It follows from this definition that $G$ is a subgroup of $GL_3(\mathbb R)$ and a simple calculation shows that $$G=\left\{\left(\begin{array}{ccc} a & b & e \\ c & d & f \\ 0 & 0 & 1 \end{array} \right) | \quad a,b,c,d,e,f\in \mathbb{R}, \quad ad-bc=1\right\}.$$ Now you can check that the Lie algebra of $G$ consists of the matrices you indicated.

Of course there are other answers, as the $G$ above is not simply connected.

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