Do Isometries Preserve Covariant Derivatives?

  • #1
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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}##
 

Answers and Replies

  • #2
stevendaryl
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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
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So you agree with my result? I am about convinced that ##\nabla_{\overline{V}}\overline{W}(q)\neq \overline{\nabla_{V} W}##
 
  • #4
stevendaryl
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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
stevendaryl
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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
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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
stevendaryl
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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
stevendaryl
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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
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@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
stevendaryl
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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
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@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
stevendaryl
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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
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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
 
Last edited:
  • #14
lavinia
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Have you solved the general problem?
 
  • #15
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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
 

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