I Frobenius theorem applied to frame fields

  • #31
cianfa72 said:
Sorry, but the above Borel subgroup of ##\operatorname{SL}(2,\mathbb R)## is not itself a Lie group ?
It is of course a Lie group. It's a two-dimensional smooth manifold. A typical element looks like
$$
\begin{pmatrix}e^t&c\\0&e^{-t}\end{pmatrix}
$$
The toral element, the diagonal matrix is a flow through time, and the unipotent element makes it non-abelian. The group is
$$
\biggl\langle \begin{pmatrix}e^t&c\\0&e^{-t}\end{pmatrix} \biggr\rangle =\left\{\left.\begin{pmatrix}x&y\\0&z\end{pmatrix}\,\right|\,x\cdot z= 1\right\} .
$$
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
508
  • · Replies 73 ·
3
Replies
73
Views
7K
  • · Replies 20 ·
Replies
20
Views
4K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
22
Views
2K
  • · Replies 10 ·
Replies
10
Views
962
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 51 ·
2
Replies
51
Views
5K