I Do Isometries Preserve Covariant Derivatives?

[Gene Naden]

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:

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

$q=-p$

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

$\nabla_{\overline{V}}\overline{W}(q)=\nabla_{(-x,-y,-z)} (-x^2,0,0)=\frac{d(-(x-tx)^2,0,0)}{dt}=(2x^2,0,0) \neq \overline{\nabla_{V} W}$

 

[stevendaryl]

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).

 

[Gene Naden]

So you agree with my result? I am about convinced that $\nabla_{\overline{V}}\overline{W}(q)\neq \overline{\nabla_{V} W}$

 

[stevendaryl]

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 $\overline{\nabla_V W} = \nabla_{\overline{V}} \overline{W} = -\nabla_V W$

 

[stevendaryl]

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:

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

 

[fresh_42]

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 $-tV\,?$

 

[stevendaryl]

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, $\overline{V} = -V$.

 

[stevendaryl]

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

 

[Gene Naden]

@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 $\overline{\nabla_V W}=-\nabla_V W$?

 

[stevendaryl]

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 $(-x,-y,-z)$.

 

[Gene Naden]

@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.

 

[stevendaryl]

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:

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

 

[Gene Naden]

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

 

[lavinia]

Have you solved the general problem?

 

[Gene Naden]

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