Timeline for standard form of antisymmetric matrix
Current License: CC BY-SA 3.0
3 events
| when toggle format | what | by | license | comment | |
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| Aug 16, 2016 at 1:19 | comment | added | Robert Bryant | @Zionnn: Well, for each continuous path $\theta_t$ in $A_d$ there certainly exists a continuous path $a_t$ satisfying $\pi(a_t) = \theta_t$. Whether you can choose $a_t$ depends on what you mean by 'choose'. For any noncontinuous map $T:A_d\to \mathrm{GL}(2d, \mathbb{R})$ that satisfies $\pi\circ T = \mathrm{id}$, there will exist a continuous path $\theta_t$ in $A_d$ such that $a_t = T(\theta_t)$ is not continuous. | |
| Aug 15, 2016 at 18:10 | comment | added | zionnn | Thanks for excellent answer. Is this still possible? if i have a continuous path of skew symmetric matrices $\theta_t \in A_d$ say indexed by closed interval, can I choose a continuous path $a_t \in GL(2d, \mathbb{R})$ such that $T(\theta_t ) = a_t$ for all $t$? | |
| Aug 14, 2016 at 18:56 | history | answered | Robert Bryant | CC BY-SA 3.0 |