
#1
Jan2711, 06:21 AM

P: 12

hi there,
how can i show that the lie brocket is not connection? Many thanx =))) 



#2
Jan2711, 07:09 AM

Emeritus
Sci Advisor
PF Gold
P: 9,014

Can you think of a property that a connection has, that the Lie bracket for vector fields doesn't? The definition of "connection" doesn't list that many properties, so you can check them one at a time. For example, a connection is ℝlinear in both variables, and so is the Lie bracket for vector fields. That doesn't help, so you need to keep checking until you find a property that this Lie bracket doesn't have.




#3
Feb211, 04:08 PM

P: 12

Thanx Fredrik
could you please help me to find the property that is not satisfied? :( 



#4
Feb211, 04:26 PM

Emeritus
Sci Advisor
PF Gold
P: 9,014

The lie brocket
Yes, but I don't want to do your work for you. If you post the properties that you know a connection must have, and your attempts to determine if a Lie bracket has those properties too, I will tell you if you're doing something wrong. If you get stuck, show me where, and I'll try to help you get past that point.




#5
Feb311, 01:00 PM

P: 12

Hello !
Actually I check all the properties of the connection and I can not find out any different and I give up = any way, lets ignore it :\ I have another question =) If we want to prove that V(t) is a geodesic of 6(u,v) OK? So, I try to prove that by the fact " V(t) is geodesic iff V'' =0 " and also I try to do that by calculate V'^2 = constant. but both of them did not work =( is there any different way to prove that V(t) is geodesic ?? please your advice Thanx =) 



#6
Feb311, 01:49 PM

Emeritus
Sci Advisor
PF Gold
P: 9,014

[tex]\nabla_{X}(fY)(p)[/tex] [tex][X,fY]_p[/tex] Click the quote button, and you'll see how I did the LaTeX. If you try it, you need to keep in mind that there's a bug that makes the wrong images appear. The only workaround is to refresh and resend after each preview. I don't understand what you're asking. what is V(t)? What do you mean by 6(u,v)? What manifold are you talking about, and what metric/connection are you using? 



#7
Feb311, 03:44 PM

P: 12

If [tex]f(x)[/tex] is a positive function and
[tex]/sigma(u,v)[/tex] [tex]/eq(f(u)cos(v),f(u)sin (v),u)[/tex] then [tex]/gamma(t)[/tex] = [tex]/sigma(u(t),c)[/tex] is a geodesic where c is constant between 0 and [tex]/pi[/tex] that was the question and I tried to calculate the second derivative of /sigma but that did not work and we still have u in the first derivative which means it is not constant and thank you Fredrik =) 



#8
Feb311, 03:46 PM

P: 12

f is a real function



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