I Is a Manifold with a Boundary Considered a True Manifold?

[PeterDonis]

ok, which is the obvious parameterization to be used for the case $V=2x\partial_x$ ?

You should be able to work that out by analogy with what I said in post #179. I'll give you the first step: the components of $W$ for this case are $(2x, 0)$.

 

[cianfa72]

You should be able to work that out by analogy with what I said in post #179. I'll give you the first step: the components of $W$ for this case are $(2x, 0)$.

$(x,y)=(t^2,y_0)$ ?

 

[PeterDonis]

$(x,y)=(t^2,y_0)$ ?

No. Try writing out explicitly the equation for the $x$ component of $W$ that you need to integrate.

 

[cianfa72]

No. Try writing out explicitly the equation for the $x$ component of $W$ that you need to integrate.

$\dot x=2x \Rightarrow x=e^{2t} \Rightarrow (e^{2t},y_0)$

 

[PeterDonis]

$\dot x=2x \Rightarrow x=e^{2t} \Rightarrow (e^{2t},y_0)$

Yes.

 

[jbergman]

The components of $W$ are $(2y, 0)$. So an integral curve of $W$ will have a parameterization $(x, y) = (2y_0 t, y_0)$, where $y_0$ is a constant (the unchanging $y$ coordinate of the curve) and $t$ is the parameter.

This parameterization doesn't look correctly since for $t=0$ you want the mapping to be the identity mapping. I believe it should be $\theta_t(x,y) = (x+ 2yt, y)$.

 

[jbergman]

$\dot x=2x \Rightarrow x=e^{2t} \Rightarrow (e^{2t},y_0)$

This also doesn't look correct to me. I already gave the parameterization in the post where I computed the pullback.

We have two first order ODEs of the following form.
$$ \frac{d}{dt}\Bigr|_{t=0}\theta_t(x,y) = (2x, 0)$$
with initial conditions
$$\theta_t(x,y) = (x,y)$$
The solution to this is just
$$\theta_t(x,y) = (x + 2xt, y)$$

 

[DrGreg]

This also doesn't look correct to me. I already gave the parameterization in the post where I computed the pullback.

We have two first order ODEs of the following form.
$$ \frac{d}{dt}\Bigr|_{t=0}\theta_t(x,y) = (2x, 0)$$
with initial conditions
$$\theta_t(x,y) = (x,y)$$
The solution to this is just
$$\theta_t(x,y) = (x + 2xt, y)$$

Your solution isn't a one-parameter group i.e. $\theta_{s+t} \neq \theta_s \circ \theta_t$.

I'd like to change the notation slightly and write$$(x_t, y_t) = \theta_t(x_0, y_0).$$Then the differential equation to solve is$$
\frac{d}{dt}\theta_t(x_0,y_0) = V(\theta_t(x_0,y_0)) = V(x_t, y_t) = (2x_t, 0)
$$for all $t$, not just $t=0$, i.e.$$
\frac{dx_t}{dt} = 2x_t \quad ; \quad \frac{dy_t}{dt} = 0
$$which has solutions$$
x_t = x_0 \, e^{2t} \quad ; \quad y_t = y_0
$$i.e.$$
\theta_t(x_0, y_0) = (x_0 \, e^{2t}, y_0)
$$

 

[jbergman]

Your solution isn't a one-parameter group i.e. $\theta_{s+t} \neq \theta_s \circ \theta_t$.

I'd like to change the notation slightly and write$$(x_t, y_t) = \theta_t(x_0, y_0).$$Then the differential equation to solve is$$
\frac{d}{dt}\theta_t(x_0,y_0) = V(\theta_t(x_0,y_0)) = V(x_t, y_t) = (2x_t, 0)
$$for all $t$, not just $t=0$, i.e.$$
\frac{dx_t}{dt} = 2x_t \quad ; \quad \frac{dy_t}{dt} = 0
$$which has solutions$$
x_t = x_0 \, e^{2t} \quad ; \quad y_t = y_0
$$i.e.$$
\theta_t(x_0, y_0) = (x_0 \, e^{2t}, y_0)
$$

Yikes. Bumbled the differential equation. Thanks for the correction!

 

[PeterDonis]

This parameterization doesn't look correctly since for $t=0$ you want the mapping to be the identity mapping. I believe it should be $\theta_t(x,y) = (x+ 2yt, y)$.

Yes, you're right, there should be a constant term in $x$ to make the transformation general; what I wrote was really for the special case of $x = 0$ for $t = 0$.

 

[jbergman]

Yes, you're right, there should be a constant term in $x$ to make the transformation general; what I wrote was really for the special case of $x = 0$ for $t = 0$.

My solution was wrong as pointed out by Dr. Greg.

 

[PeterDonis]

My solution was wrong as pointed out by Dr. Greg.

@DrGreg was talking about the vector field $V = 2 x \partial_x$ that you originally brought up, not the vector field $W = 2 y \partial_x$ that I originally brought up. My post that you just responded to was about $W$. The differential equation for $W$ is $\dot{x} = 2 y$, which integrates to $x = 2yt + C$, where $C$ is a constant that obviously must be the starting value of $x$ so that for $t = 0$ the mapping is the identity.

 

[cianfa72]

So the explicit calculation in #175 becomes:
$$\theta_t^{*} = d\theta_t = diag(e^{2t}, 1)$$
Hence as expected:
$$(\mathscr{L}_V g)_p = \lim_{t \rightarrow 0} \frac{(e^{2t})^2dx^2 +dy^2 - dx^2 - dy^2}{t}= \lim_{t \rightarrow 0} \frac{(e^{4t} - 1)}{t} dx^2=4 dx^2$$
At the end of the day, both vector fields $V=2x\partial_x$ and $V=2y\partial_x$ are not KVFs for the Euclidean metric.

A point looking weird to me is that in the former case if we take two points on two neighboring integral curves (i.e. straight lines parallel to the $x$ axis) with the same curve parameter $t_0$ and flow them along their respective integral curves up to the same parameter $t_1$, the Euclidean distance between them does not change. Yet the Lie Derivative of the euclidean metric tensor $g$ along $V=2x\partial_x$ is not null.

 

[DrGreg]

A point looking weird to me is that in the former case if we take two points on two neighboring integral curves (i.e. straight lines parallel to the $x$ axis) with the same curve parameter $t_0$ and flow them along their respective integral curves up to the same parameter $t_1$, the Euclidean distance between them does not change. Yet the Lie Derivative of the euclidean metric tensor $g$ along $V=2x\partial_x$ is not null.

But take two neighbouring points on the same curve and "flow" them both by the same increase in parameter...

 

[cianfa72]

But take two neighbouring points on the same curve and "flow" them both by the same increase in parameter...

Ah ok, so what the Lie derivative of the tensor metric tells us applies even for the 'distance' of neighboring points on the same integral curve (both "flowed" by the same increase of the parameter), not just for points on neighboring integral curves in the congruence.

 

[jbergman]

Ah ok, so what the Lie derivative of the tensor metric tells us applies even for the 'distance' of neighboring points on the same integral curve (both "flowed" by the same increase of the parameter), not just for points on neighboring integral curves in the congruence.

Yes. That's the point. I think we are getting two hung up on the integral curves, here. A vector field induces a smooth maps of the manifold to itself via the flows.

The Lie derivative is the limit of the pullback of the metric along the integral curve that the point is on. But it uses the differential of the smooth maps associated with the flow to compute the pullback by pushing forward tangent vectors.

This differential encodes information about distance changing in all directions.

So, in effect the Lie Derivative is like a directional derivative. It gives the first order change in the metric when moving along the integral curve at a point. But that change can encode changes in distance in any direction.

It's like an ordinary directional derivative of a vector quantity. The direction just specifies the direction of the step to take. It says nothing of the direction in which the vector will vary along that step.

 

[jbergman]

To beat a dead horse, there are ways the Lie Derivative of the metric can be non-zero.

1. The induced flow of the vector field moves points in a direction that the metric varies on a manifold
2. The induced flow moves points in the same neighborhood at different speeds there by distorting the distance between them. This is reflected in the pushforward of the induced flow and leads to tangent vectors being stretched or squished.

 

[sysprog]

This is reflected in the pushforward of the induced flow and leads to tangent vectors being stretched or squished.

I'm not sure that the 'squished' negative vector properly emerges when we take the first derivative (velocity) of position wrt time.

 

[jbergman]

I'm not sure that the 'squished' negative vector properly emerges when we take the first derivative (velocity) of position wrt time.

I'm not sure I understand this comment. Are you talking about Minkowski space?

 

[cianfa72]

To beat a dead horse, there are ways the Lie Derivative of the metric can be non-zero.

1. The induced flow of the vector field moves points in a direction that the metric varies on a manifold
2. The induced flow moves points in the same neighborhood at different speeds there by distorting the distance between them. This is reflected in the pushforward of the induced flow and leads to tangent vectors being stretched or squished.

About 2. I take it as, even though the metric tensor did not change along the induced flow of the vector field, a non-zero Lie Derivative would imply that the induced flow moves points in a neighborhood such that the 'distance' between them changes when 'flowed' by the same increase of the integral curves parameter.

 

[jbergman]

About 2. I take it as, even though the metric tensor did not change along the induced flow of the vector field, a non-zero Lie Derivative would imply that the induced flow moves points in a neighborhood such that the 'distance' between them changes when 'flowed' by the same increase of the integral curves parameter.

In both cases that is true. But the distance can change because they have moved in a direction where the metric naturally changes like in the radial direction with the Schwarzschild metric. Or with a constant metric we move points differing distances under the same increase of the parameter.

Of course you can also have both.

 

[cianfa72]

Or with a constant metric we move points differing distances under the same increase of the parameter.

Constant in the sense that it does not change along the integral curves of the vector field, not necessarily that it must stay constant at each point in the manifold.

 

[sysprog]

I'm not sure I understand this comment. Are you talking about Minkowski space?

I was merely quibbling about the terms.