Timeline for Lie algebra "generated" by matrix-valued curve?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 3, 2016 at 17:59 | comment | added | Benjamin | This is common in the physics literature and I've never understood why it is used. | |
| Apr 9, 2012 at 14:59 | history | edited | AlexArvanitakis | CC BY-SA 3.0 |
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| Apr 9, 2012 at 2:20 | vote | accept | AlexArvanitakis | ||
| Apr 9, 2012 at 1:45 | comment | added | Robert Bryant | But, if you take my $1$-variable solution, your $2$-variable solution is expressed as $U(t,t_0) = U(t)U(t_0)^{-1}$, so there's no need for a $2$-variable expression. | |
| Apr 9, 2012 at 1:43 | answer | added | Robert Bryant | timeline score: 8 | |
| Apr 9, 2012 at 1:19 | comment | added | AlexArvanitakis | The point of using $U(t,t_0)$ is that the matrices $U(t_1,t_2)$ possess the semigroup property $U(t,t_0)=U(t,t_1)U(t_1,t_0)$. | |
| Apr 9, 2012 at 1:15 | comment | added | Robert Bryant | Still, what is the point of having two arguments for $U$? | |
| Apr 9, 2012 at 1:14 | history | edited | AlexArvanitakis | CC BY-SA 3.0 |
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| Apr 9, 2012 at 1:14 | comment | added | AlexArvanitakis | Sorry, I was being sloppy. I meant $U(t_0,t_0)=\mathbf{1}_n$. Fixing... | |
| Apr 9, 2012 at 1:06 | comment | added | Robert Bryant | I don't understand your notation $U(t,t_0)$, and I'm sure you don't mean to have $U(t,t_0) = \mathbf{1}_n$, which doesn't make sense. Don't you just mean $U'(t) = A(t)U(t)$ with initial condition $U(t_0)=\mathbf{1}_n$? | |
| Apr 8, 2012 at 20:14 | history | edited | AlexArvanitakis | CC BY-SA 3.0 |
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| Apr 8, 2012 at 19:52 | history | asked | AlexArvanitakis | CC BY-SA 3.0 |