Do Isometries Preserve Covariant Derivatives?

In summary: Well, let's call it ##\overline{\mathcal{P}}##, so as not to confuse it with the original point ##\mathcal{P}##.
  • #1
Gene Naden
321
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O'Neill's Elementary Differential Geometry, in problem 3.4.5, asks the student to prove that isometries preserve covariant derivatives. Before solving the problem in general, I decided to work through the case where the isometry is a simple inversion: ##F(p)=-p##, using a couple of simple vector fields: ##V(p)=(x,y,z)## and ##W(p)=(x^2,0,0)##. I was surprised to find that the covariant derivative changes sign and is not preserved. Have I made an error? Here is the statement of the problem:

Let F be an isometry of ##R^3##.
For each vector field V let ##\bar{V}## be the vector field such that
##F^*(V(p))=\overline{V}(F(p))## for all p.
Show ##\overline{\nabla_{V} W}=\nabla_{\overline{V}}\overline{W}##

The covariant derivative of W is defined as ##\nabla_VW=W(p+tV)^\prime(0)##

The tangent map (derivative map) of F is defined as ##F^*(v_p)=\frac{dF(p+tv)}{dt}|_{t=0}##

Here is my work:

##\nabla_V W=\nabla_{(x,y,z)}W=\frac{d(W((x,y,z)+t(x,y,z))}{dt}|_{t=0}=\frac{d((x+tx)^2,0,0)}{dt}=(2x^2,0,0)##
##\overline{\nabla_VW}(-p)=F^*((2x^2,0,0)_{x,y,z})=\frac{dF((x,y,z)+t(2x^2,0,0)}{dt}|_{t=0}=
\frac{d(-x-2x^2t,0,0)}{dt}=(-2x^2,0,0)##

##q=-p##

##\overline{\nabla_VW}(q)=(-2x^2,0,0)##
##\overline{W}(-p)=F*((x^2,0,0)_{(x,y,z)}=\frac{dF((x,y,z)+t(x^2,0,0)}{dt}|_{t=0}=\frac{d(-x-tx^2,0,0)}{dt}=(-x^2,0,0)##
##\overline{W}(q)=(-x^2,0,0)##
##\overline{V}(-p)=F^*((x,y,z)_{x,y,z})=\frac{dF((x,y,z)+t(x,y,z))}{dt}|_{t=0}=\frac{d(-x-tx,-y-ty,-z-tz)}{dt}
=(-x,-y,-z)##
##\overline{V}(q)=(-x,-y,-z)##

##\nabla_{\overline{V}}\overline{W}(q)=\nabla_{(-x,-y,-z)} (-x^2,0,0)=\frac{d(-(x-tx)^2,0,0)}{dt}=(2x^2,0,0) \neq \overline{\nabla_{V} W}##
 
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  • #2
It seems to me that you have: (letting ##\mathcal{P}## be the point with coordinates ##(x,y,z)## and ##\overline{\mathcal{P}}## be ##(-x,-y,-z)##.

##\overline{W}(\overline{\mathcal{P}}) = - W(\mathcal{P})##
##\overline{W}(\overline{\mathcal{P}} + \lambda \overline{V}) = - W(\mathcal{P} + \lambda V)##
##\Delta \overline{W} = \overline{W}(\overline{\mathcal{P}} + \lambda \overline{V}) - \overline{W}(\overline{\mathcal{P}}) = - W(\mathcal{P} + \lambda V) - (- W(\mathcal{P})) = -\Delta W##

Roughly speaking, ##\nabla_V W = lim_{\lambda \rightarrow 0} \Delta W##.

(I say "roughly speaking", because you can't directly subtract vectors at different points in space. But you seem to be assuming a flat, Cartesian coordinate system).
 
  • #3
So you agree with my result? I am about convinced that ##\nabla_{\overline{V}}\overline{W}(q)\neq \overline{\nabla_{V} W}##
 
  • #4
Gene Naden said:
So you agree with my result? I am about convinced that ##\nabla_{\overline{V}}\overline{W}(q)\neq \overline{\nabla_{V} W}##

No, I came up with ##\overline{\nabla_V W} = \nabla_{\overline{V}} \overline{W} = -\nabla_V W##
 
  • #5
As I said, when the connection coefficients are all zero, we can write:

##\nabla_{\overline{V}} \overline{W}|_{\overline{\mathcal{P}_0}} = \frac{d}{dt} \overline{W}(\overline{\mathcal{P}_0} + t \overline{V})##

But the way that the transformation is defined, ##\overline{W}(\overline{\mathcal{P}}) = -W(\mathcal{P})##. So we have:

##\nabla_{\overline{V}} \overline{W}|_{\overline{\mathcal{P}_0}} = -\frac{d}{dt} W(\mathcal{P}_0 + t V) = - \nabla_V W|_{\mathcal{P}_0}##
 
  • #6
stevendaryl said:
As I said, when the connection coefficients are all zero, we can write:

##\nabla_{\overline{V}} \overline{W}|_{\overline{\mathcal{P}_0}} = \frac{d}{dt} \overline{W}(\overline{\mathcal{P}_0} + t \overline{V})##

But the way that the transformation is defined, ##\overline{W}(\overline{\mathcal{P}}) = -W(\mathcal{P})##. So we have:

##\nabla_{\overline{V}} \overline{W}|_{\overline{\mathcal{P}_0}} = -\frac{d}{dt} W(\mathcal{P}_0 + t V) = - \nabla_V W|_{\mathcal{P}_0}##
Why doesn't ##V## change its sign, from ##+tV## to ##-tV\,?##
 
  • #7
fresh_42 said:
Why doesn't ##V## change its sign, from ##+tV## to ##-tV\,?##

It did. The key fact is that we're mapping both vectors and positions. So if ##\mathcal{P}## is a location, then ##\overline{\mathcal{P}}## is the mapped location. The relationship between ##\overline{W}## and ##W## is this:

##\overline{W}(\overline{\mathcal{P}}) = - W(\mathcal{P})##

So ##\overline{W}(\overline{\mathcal{P}}+ t \overline{V}) = - W(\mathcal{P}+ tV)##

On the left-hand side, ##\overline{V} = -V##.
 
  • #8
diffeo.jpg


This is how I see it: ##\nabla_\overline{V} \overline{W} = lim_{t \rightarrow 0} \overline{\Delta W}##, which is just the negative of ##\Delta W##.
 

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  • #9
@stevendaryl wrote ##\overline{\nabla_V W}=\nabla_{\overline{v}}\overline{W}=-\nabla_V W##

I get ##\nabla_{\overline{v}}\overline{W}=-\nabla_V W## but how did you get ##\overline{\nabla_V W}=\nabla_{\overline{v}}\overline{W}## or ##\overline{\nabla_V W}=-\nabla_V W##?
 
  • #10
That's the definition of the mapping: For any vector field ##X## and any point ##\mathcal{P}##,

##\overline{X}(\overline{\mathcal{P}}) = - X(\mathcal{P})##

(Where if ##\mathcal{P}## has coordinates ##(x,y,z)##, then ##\overline{\mathcal{P}}## has coordinates ##(-x,-y,-z)##.

In your original derivation, the only mistake I see was the last line.

What you want to evaluate is ##\nabla_{\overline{V}} \overline{W}|_\overline{\mathcal{P}}##. Looking at just the x-component, we have:

##(\nabla_{\overline{V}} \overline{W})^x|_\overline{\mathcal{P}} = \frac{d}{dt} - (-x - tx)^2##

not ##\frac{d}{dt} - (x-tx)^2##. You're evaluating the covariant derivative at the point with coordinates ##(-x,-y,-z)##.
 
  • #11
@stevendaryl, I see the error you pointed out in my last step. It has caused me to reexamine the entire problem.

I have defined F so that ##F^*:v_p\rightarrow (-v)_{\overline{p}}## where the coodinates of ##\overline{p}## are the negatives of the coordinates of p. And two vector fields ##V(p)=(x,y,z)## and ##W(p)=(x^2,0,0)##. But I don't understand the basic premise of the problem: "For each vector field V let ##\bar{V}## be the vector field such that ##F^*(V(p))=\overline{V}(F(p))## for all p." What does this mean?

I tried to use this statement to construct ##\overline{W}## and ##\overline{V}##, but frankly I didn't know what I was doing.
 
  • #12
Gene Naden said:
But I don't understand the basic premise of the problem: "For each vector field V let ##\bar{V}## be the vector field such that ##F^*(V(p))=\overline{V}(F(p))## for all p." What does this mean?

We have two different mappings:
  1. ##F(p)## maps a point ##p## to another point, ##p'##
  2. ##F^*(V)## maps a vector ##V## at point ##p## to another vector, ##V'## at point ##p'##.
Note that for curved spaces, there is no such thing as just a "vector". Vectors are associated with particular points on the space (or manifold). So if ##V## is a vector defined at point ##p##, the mapping ##F^*(V)## does not give a new vector at the point ##p##. It gives a new vector, ##V'##, defined at the point ##F(p)##.

We can combine these two mappings to give a mapping from vector fields to vector fields. A vector field assigns a vector to each point in the manifold. So assume that we have a vector field ##V(p)## and we want to use those mappings to define a new vector field, ##\overline{V}(p')##. We compute ##\overline{V}(p')## in this way:
  1. Given ##p'##, we compute the point ##p## such that ##F(p) = p'##. If ##F## has an inverse, then ##p = F^{-1}(p')##.
  2. Now that we have ##p##, we can compute ##V(p)##. This is a vector defined at point ##p##.
  3. We can then apply the function ##F^*(V(p))## to get a new vector, ##V'## defined at point ##F(p)##.
  4. We define the value of ##\overline{V}## at point ##p'## to be the vector ##V'##.
So this means that:

##\overline{V}(p') = F^*(V(p))##

Or since ##p' = F(p)##, we can write:

##\overline{V}(F(p)) = F^*(V(p))##
 
  • #13
So with ##p^\prime=F(p)##, I get ##\overline{V}(p^\prime)=(-p_x,-p_y,-p_z)=(p^{\prime}_x,p^{\prime}_y,p^{\prime}_z)## and ##\overline{W}(p^\prime)=(-p_{x}^2,0,0)=(-p^{\prime2}_x,0,0)##

##F^*(\nabla_VW(p))=(-2p_x^2,0,0)=(-2p_x^{\prime 2},0,0)##
##\overline{\nabla_VW(p^\prime)}=(-2p_x^{\prime 2},0,0)##
##\nabla_{\overline{V}}\overline{W}=\frac{d}{dt}((-(p^\prime_x+t\overline{V}_x)^2,0,0)=(-2p^\prime_x\overline{V}_x,0,0)=(-2p^{\prime 2}_x,0,0)##

So I think I understand the meaning of ##\overline{W}##, etc. Thanks @stevendaryl
 
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  • #14
Have you solved the general problem?
 
  • #15
With the aid of the textbook:
##F^*(\nabla_V W)=F^*(W(p+tV)^\prime(0)## Definition of covariant derivative
##(p+tV)=TC(p+tV)=C(p+tV)+a=F(p)+tC(V)## General property of isometry
##F^*(W(p+tV)^\prime(0)=(\overline{W}(F(p)+tC(v))^\prime(0)## Given in the problem statement
##F^*(v)=C(v)## Theorem 2.1, first edition
##(\overline{W}(F(p)+tC(v))^\prime(0)=\nabla_{F^*(v)}\overline{W}## defintion of covariant derivative
 

FAQ: Do Isometries Preserve Covariant Derivatives?

What is an isometry?

An isometry is a transformation that preserves distances between points. In other words, it is a type of transformation that does not change the shape or size of an object.

What is a covariant derivative?

A covariant derivative is a mathematical tool used to measure how a geometric object changes as it moves along a curved space. It takes into account the changes in both the object's magnitude and direction.

Do isometries preserve covariant derivatives in all cases?

No, isometries do not always preserve covariant derivatives. They only preserve covariant derivatives in spaces that are flat or have a constant curvature.

How do isometries affect covariant derivatives?

Isometries do not affect covariant derivatives in spaces that are flat or have a constant curvature. However, in curved spaces, isometries can change the values of covariant derivatives.

What implications does the preservation of covariant derivatives by isometries have in science?

The preservation of covariant derivatives by isometries has important implications in the study of curved spaces, such as in general relativity. It allows for the development of mathematical tools to accurately describe and understand the behavior of objects in these spaces.

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