Coordinate representation of a diffeomorphism

  • #1
stevendaryl
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I'm trying to understand diffeomorphisms, and I thought I basically understood them, but when I tried to work out a problem I created for myself, I realized I didn't know how to answer it.

So let's consider a diffeomorphism generated by a vector field ##V##. If ##X## is a point on our manifold, then we can define a parametrized path ##Y_X(\lambda)## via:

##Y_X(0) = X##

##\frac{d}{d\lambda} Y_X(\lambda) = V|_{Y_X(\lambda)}##

So far, ##X## and ##Y## are points on the manifold, not coordinates, but now I'd like to switch to a coordinate system. I'll just use ##X^\alpha## and ##Y^\alpha## to mean the coordinates of points ##X## and ##Y##, and use ##V^\alpha## to mean the components of ##V## in the coordinate basis for our coordinate system.

The first uncertainty on my part is whether the following is true:

If we want to know the coordinates for the point ##Y_X(\lambda)##, I think it's just a matter of integration:

##Y^\alpha_X(\lambda) = Y^\alpha_X(0) + \int V^\alpha|_{Y_X(\lambda)} d\lambda##
## = X^\alpha + \int V^\alpha|_{Y_X(\lambda)} d\lambda##

Assuming that is correct, what I want to compute next is the dependence of ##Y^\alpha_X(\lambda)## on ##X##. At this point, my first inclination would be:

##\frac{\partial Y^\alpha_X(\lambda)}{\partial X^\beta} = \frac{\partial X^\alpha}{\partial X^\beta}+ \int \frac{\partial V^\alpha}{\partial Y^\mu}|_{Y_X(\lambda)} \frac{\partial Y^\mu_X(\lambda)}{\partial X^\beta} d\lambda##

My uncertainty is with the expression ##\frac{\partial V^\alpha}{\partial Y^\mu}##. Should that be ##\nabla_\mu V^\alpha## rather than the partial derivative? My feeling is that it shouldn't be, since the concept of diffeomorphism is independent of whether there is a connection, or not.
 

Answers and Replies

  • #2
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I would have written the integration as ##Y_X(\lambda) = Y_X(0) + \int_0^\lambda V(Y_X(\mu)) \,d\mu## locally around ##\lambda = 0##.

So the first question is, whether ##\left(\int_0^\lambda V(Y_X(\mu)) \,d\mu \right)^\alpha = \int_0^\lambda V^\alpha(Y_X(\mu)) \,d\mu \,## and next whether integration and (partial) differentiation can be switched. If both is true, then we get ##\dfrac{\partial V^\alpha}{\partial X^\beta} = \dfrac{\partial V^\alpha (Y_X(\mu))}{\partial X^\beta}## and we have to apply the chain rule on ##V^\alpha \circ Y_X##.
 
  • #3
martinbn
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The first uncertainty on my part is whether the following is true:

If we want to know the coordinates for the point ##Y_X(\lambda)##, I think it's just a matter of integration:

##Y^\alpha_X(\lambda) = Y^\alpha_X(0) + \int V^\alpha|_{Y_X(\lambda)} d\lambda##
## = X^\alpha + \int V^\alpha|_{Y_X(\lambda)} d\lambda##
If I understand correctly your question and notations, I think this is not true. It will not be just integration, you need to solve a system of differential equations. Take any example. Say ##\mathbb R^2## with the usual coordinates ##(x,y)##, let the vector field be ##V=x^2\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}##. Then the parametrized path you are looking for satisfies
##\frac{dx}{d\lambda}=x^2##
##\frac{dy}{d\lambda}=y##
Solving the equations is not just a matter of integration. Well in a way it is because I chose them simple to solve, but in general the right sides will depend on all coordinate function, they will not be decoupled as in this example. The dependence on the initial point will come from the initial conditions for the equations ##x(0)=x_0## and ##y(0)=y_0##.
 
  • #4
stevendaryl
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If I understand correctly your question and notations, I think this is not true. It will not be just integration, you need to solve a system of differential equations. Take any example. Say ##\mathbb R^2## with the usual coordinates ##(x,y)##, let the vector field be ##V=x^2\frac{\partial}{\partial x}+y\frac{\partial}{\partial y}##. Then the parametrized path you are looking for satisfies
##\frac{dx}{d\lambda}=x^2##
##\frac{dy}{d\lambda}=y##
Solving the equations is not just a matter of integration. Well in a way it is because I chose them simple to solve, but in general the right sides will depend on all coordinate function, they will not be decoupled as in this example. The dependence on the initial point will come from the initial conditions for the equations ##x(0)=x_0## and ##y(0)=y_0##.
Very good point. However, I would say it is true that ##x(\lambda) = x(0) + \int_0^{\lambda} V^x(x(s), y(s)) ds##, but unhelpful. My question is not about how to solve it, but about the meaning.
 
  • #5
martinbn
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I see, I have misunderstood you question. Yes, you are right, that you can write the equations in that form (as integral equations), but they are still equations, as the unknown appears on both sides. They are not solved.
 

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