# Coordinate representation of a diffeomorphism

• A
• stevendaryl
You can solve them locally, for small enough ##\lambda##, and here the integral equations will hold, but in general you cannot solve them globally.
stevendaryl
Staff Emeritus
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.

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##.

stevendaryl said:
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##.

martinbn said:
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.

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.

## 1. What is a diffeomorphism?

A diffeomorphism is a mathematical concept that describes a smooth and invertible transformation between two different manifolds. In simpler terms, it is a function that maps points from one space to another in a smooth and continuous manner.

## 2. What is the significance of coordinate representation in a diffeomorphism?

The coordinate representation of a diffeomorphism allows us to describe the transformation using a set of coordinates, making it easier to visualize and analyze the changes in the manifold. It also helps in studying the properties of the transformation and its effects on the underlying space.

## 3. How is a diffeomorphism different from a homeomorphism?

While both diffeomorphism and homeomorphism describe transformations between manifolds, the former is a more restrictive concept. Diffeomorphisms must also preserve the smoothness of the transformation, while homeomorphisms only require continuity. In other words, a diffeomorphism is a smooth homeomorphism.

## 4. Can a diffeomorphism be represented by a single coordinate system?

No, a diffeomorphism requires at least two coordinate systems to be represented. This is because a diffeomorphism maps points from one manifold to another, which may have different coordinate systems. The coordinate representation takes into account both the initial and final coordinate systems.

## 5. How is a diffeomorphism useful in science?

Diffeomorphisms have various applications in fields such as physics, engineering, and computer science. They are used to model and study natural phenomena, such as fluid dynamics and the behavior of particles in space. In engineering, diffeomorphisms are used to design efficient and optimal structures. In computer science, they are used in image processing, shape recognition, and data compression.

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