Timeline for Commutator of closed subgroups

6 events

when toggle format	what		by	license	comment
Oct 26, 2023 at 23:01	history	edited	Tom Goodwillie	CC BY-SA 4.0	added 52 characters in body
Oct 26, 2023 at 23:00	comment	added	Tom Goodwillie		Oh, you're right. What I wrote was nonsense.
Oct 26, 2023 at 15:27	comment	added	David E Speyer		But $\exp(px) = \begin{bmatrix} e^p & 0 \\ 0 & e^{-p} \end{bmatrix}$ and $\exp(qy) = \begin{bmatrix} \cosh(q) & \sinh(q) \\ \sinh(q) & \cosh(q) \end{bmatrix}$. I compute that the commutator of these matrices is $\begin{bmatrix} \cosh^2(q) - e^{2p} \sinh^2(q) & (e^{2p}-1) \cosh(q) \sinh(q) \\ (e^{-2p}-1) \cosh(q) \sinh(q) & \cosh^2(q) - e^{-2p} \sinh^2(q) \end{bmatrix}$, which is not of the form $\begin{bmatrix} \cos & \sin \\ - \sin & \cos \end{bmatrix}$.
Oct 26, 2023 at 15:24	comment	added	David E Speyer		I don't think this works. You are assuming that $(\exp(\mathbb{R}x), \exp(\mathbb{R}y)) = \exp(\mathbb{R}[x,y])$. (Group commutator on the left, Lie algebra commutator on the right.) Take $x = \begin{bmatrix} 1&0 \\ 0&-1 \end{bmatrix}$ and $y = \begin{bmatrix} 0&1 \\ 1&0 \end{bmatrix}$. Then $[x,y] = \begin{bmatrix} 0 & 2 \\ -2 & 0 \end{bmatrix}$ so $\exp(\theta [x,y]) = \begin{bmatrix} \cos 2 \theta & \sin 2 \theta \\ - \sin 2 \theta & \cos 2 \theta \end{bmatrix}$. (continued)
Mar 6, 2012 at 20:36	vote	accept	W. Politarczyk
Mar 4, 2012 at 22:35	history	answered	Tom Goodwillie	CC BY-SA 3.0