The lie brocket

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  • #1
wii
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hi there,

how can i show that the lie brocket is not connection?

Many thanx =)))
 

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  • #2
Fredrik
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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 ℝ-linear 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.
 
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  • #3
wii
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Thanx Fredrik

could you please help me to find the property that is not satisfied?
:(
 
  • #4
Fredrik
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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
wii
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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
Fredrik
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Actually I check all the properties of the connection and I can not find out any different and I give up =|
Can you evaluate the following expressions, where X and Y are arbitrary vector fields, and f is an arbitrary smooth function?

[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 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
The policy around here is that people who ask for help with textbook-style problems have to show a complete statement of the problem, the definitions they're using, and their work so far, so that we can give hints that will help them move past the point where they are stuck.

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
wii
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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 :confused:

and thank you Fredrik =)
 
  • #8
wii
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f is a real function
 
  • #9
Fredrik
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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 :confused:

and thank you Fredrik =)
In LaTeX, you need to use \ instead of /. For example, \sigma instead of /sigma. I also recommend \cos instead of cos.

I don't have time to look at your problem today. Maybe someone else does.
 

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