Diffeomorphisms and active transformations

The important point is that the domain of definition of the Lie derivative must be an interval containing 0 and all sufficiently small numbers.
  • #1
"Don't panic!"
601
8
I've been reading Sean Carroll's notes on General Relativity, http://arxiv.org/pdf/gr-qc/9712019.pdf . I've got to chapter 5 (page 133) and am reading the section on diffeomorphisms in which Sean relates diffeomorphisms to active transformations. When he says this does he mean that one defines a diffeomorphism [itex]\phi : M\rightarrow M[/itex] that maps a region on the manifold to another region on the manifold (i.e. in essence it changes the location we are considering on the manifold), such that one evaluates the points in the new region (mapped to by [itex]\phi[/itex]) with respect to a coordinate system defined in the original region that we were located, via a pullback? If this is the case, then would it be correct to say that diffeomorphism invariance of a quantity means that when mapped to a different point on the manifold it has the same form as it had at the original point, such that, when evaluated at the new point and then "pulled back" to the original point it has the same value as the original quantity evaluated at the original point? (sorry, I realize that I may not have worded this particularly well).

Also, he then goes on to say that this suggests that we can define another kind of derivative operator that allows us to categorise the rates of change in tensors as they change under such diffeomorphisms. He says that for that "a single discrete diffeomorphism is insufficient ; we require a one-parameter family of diffeomorphisms, [itex]\phi_{t}[/itex]", parametrised by some real parameter, [itex]t\in\mathbb{R}[/itex]. Is this because one discrete diffeomorphism would only enable us to evaluate the derivative of a tensor for only a few points (on an individual basis), analogously to when one first defines the derivative of a function at a point [itex]x_{0}\in\mathbb{R}[/itex] as $$ \lim_{h\rightarrow 0}\frac{f(x_{0}+h)-f(x_{0})}{h}$$? To generalise this notion we note that if [itex]f[/itex] is differentiable at every point in some region then we can define a function [itex]f' : \mathbb{R}\rightarrow\mathbb{R}[/itex] that maps each point in that region to the derivative of [itex]f[/itex] at that point, i.e. [itex]x\mapsto f'(x)[/itex]. We define this function such that
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
Given this we can then define an operator [itex]\frac{d}{dx}\left(\right)[/itex]; this requires a set of differentiable functions such that [itex]\frac{d}{dx}\left(\right)[/itex] maps each differentiable function [itex]f[/itex] to its corresponding derivative function [itex]f'[/itex], i.e. $$\frac{d}{dx}\left(f\right)=f'$$ Is this the point, that the domain of a derivative operator is the set of differentiable functions and not the set of real numbers and thus, going back to differential geometry, in order to define a derivative operator using diffeomorphisms we require a family of diffeomorphisms - one discrete diffeomorphism is not enough?!

Finally, when he introduces this family of diffeomorphisms he says that "if we consider what happens to the point [itex]p\in M[/itex] under the entire family [itex]\phi_{t}[/itex], it is clear that it describes a curve in [itex]M[/itex]". Is this because, if we consider an individual diffeomorphism (for fixed [itex]t[/itex]) then each point [itex]p[/itex] in the domain of the diffeomorphism is mapped to a single discrete point. However if we fix the point [itex]p[/itex] and allow the diffeomorphism to vary (i.e. we vary the values of [itex]t[/itex]) then this describes a curve in [itex]M[/itex] as [itex]p[/itex] will be mapped to different points in a continuous fashion as [itex]\phi_{t}[/itex] varies?

Sorry for the long-windedness of this post, I will really appreciate any feedback though!
 
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  • #2
"Don't panic!" said:
"if we consider what happens to the point [itex]p\in M[/itex] under the entire family [itex]\phi_{t}[/itex], it is clear that it describes a curve in [itex]M[/itex]".
The map ##t\mapsto\phi_t(p)## is from a subset of ##\mathbb R## to a subset of ##M##. That makes it a curve.

"Don't panic!" said:
...another kind of derivative operator...
He says that for that "a single discrete diffeomorphism is insufficient ; we require a one-parameter family of diffeomorphisms, [itex]\phi_{t}[/itex]", parametrised by some real parameter, [itex]t\in\mathbb{R}[/itex].
The Lie derivative of a vector field Y in the direction of the vector field X, at the point p, is
$$(\mathcal L_XY)_p=\lim_{t\to 0}\frac{(\phi_{-t})_* X_{\phi_t(p)}-X_p}{t},$$ where ##\phi_t(p)## is defined so that for each ##p##, ##X_p## is the tangent vector at ##p## of the curve ##t\mapsto\phi_t(p)##. The numerator must be defined for enough values of t for the limit to make sense. To be more precise, the map ##t\mapsto (\phi_{-t})_* X_{\phi_t(p)}## must be differentiable at 0.
 
  • #3
Fredrik said:
The map tϕt(p)t\mapsto\phi_t(p) is from a subset of R\mathbb R to a subset of MM. That makes it a curve.

Would it be correct to say that it describes a curve because of the reason I gave:

"Don't panic!" said:
if we fix the point pp and allow the diffeomorphism to vary (i.e. we vary the values of tt) then this describes a curve in MM as pp will be mapped to different points in a continuous fashion as ϕt\phi_{t} varies?

Fredrik said:
The numerator must be defined for enough values of t for the limit to make sense. To be more precise, the map t↦(ϕt)∗Xϕt(p)t\mapsto (\phi_{-t})_* X_{\phi_t(p)} must be differentiable at 0.

So is it that we need [itex](\phi_{-t})_{\ast}[/itex] to vary in a continuous manner, induced by [itex]t[/itex] varying in a continuous manner such that we can legitimately compare nearby values such that the limit (given above in your quote) exists?
 
  • #4
I like to think in terms of functions and their domains, not in terms of variables that "vary" in a manner that can be described either as continuous or not continuous. But yes, your way of describing these things is reasonable.
 
  • #5
Ah, ok. I agree your way of thinking is better, but I sometimes struggle to understand it conceptually doing it that way. Thanks for your help.
 

Related to Diffeomorphisms and active transformations

1. What is a diffeomorphism?

A diffeomorphism is a smooth, invertible transformation between two differentiable manifolds. It preserves the differentiable structure and enables us to map points from one manifold to another in a smooth and continuous way.

2. How are diffeomorphisms related to active transformations?

Diffeomorphisms are a type of active transformation, which means they actively change the coordinates of points in a space. They are used in physics and mathematics to describe how objects move and change in a space.

3. What is the importance of diffeomorphisms in physics?

In physics, diffeomorphisms are used to describe the symmetries of physical systems. They play a crucial role in theories such as general relativity, where the laws of physics must be invariant under coordinate transformations.

4. How are diffeomorphisms different from other types of transformations?

Diffeomorphisms are different from other types of transformations, such as rigid transformations or passive transformations, because they actively change the coordinates of points in a space. This means that the shape and structure of the space itself can be transformed, rather than just the coordinates of objects within the space.

5. Can you give an example of a diffeomorphism in real life?

A common example of a diffeomorphism in real life is a map projection. A map projection is a transformation of the Earth's curved surface onto a flat map, while preserving distances and angles between points. This is a diffeomorphism because it is a smooth, invertible transformation between two differentiable surfaces.

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