Is a Manifold with a Boundary Considered a True Manifold?

[cianfa72]

It means that the metric is unchanged along integral curves of the Killing vector field that corresponds to the symmetry. For example, in Euclidean 3-space, there is a 3-parameter group of Killing vector fields corresponding to rotational symmetry.

The claim about the metric unchanged applies in a neighborhood of each point along the integral curves of Killing vector field corresponding to the symmetry.

 

[PeterDonis]

The claim about the metric unchanged applies in a neighborhood of each point along the integral curves of Killing vector field corresponding to the symmetry.

No, it doesn't. It only applies on the integral curves themselves. The metric on two adjacent integral curves does not have to be the same.

Euclidean space is actually a bad example because it's flat everywhere. N-spheres with the standard metric are also bad examples because they have constant curvature. A better example would be, for example, Schwarzschild spacetime outside the horizon, where the metric is a function of $r$, so adjacent integral curves of the timelike KVF with slightly different values of $r$ have different metrics.

 

[cianfa72]

A better example would be, for example, Schwarzschild spacetime outside the horizon, where the metric is a function of $r$, so adjacent integral curves of the timelike KVF with slightly different values of $r$ have different metrics.

In this case the integral curves of the timelike KVF are the worldlines of hovering observers at fixed $(r,\theta,\phi)$. So the spacetime metric will be different for such two hovering observers at slightly different $r$ values (their $\theta, \phi$ coordinates are irrelevant).

 

[cianfa72]

Euclidean space is actually a bad example because it's flat everywhere.

So in Euclidean space any (smooth) curve joining two points is actually an integral curve of a KVF, I believe.

 

[PeterDonis]

In this case the integral curves of the timelike KVF are the worldlines of hovering observers at fixed $(r,\theta,\phi)$. So the spacetime metric will be different for such two hovering observers at slightly different $r$ values (their $\theta, \phi$ coordinates are irrelevant).

Yes, that's what I said.

 

[PeterDonis]

in Euclidean space any (smooth) curve joining two points is actually an integral curve of a KVF, I believe.

No. But you are illustrating why I said that Euclidean space is a bad example for understanding symmetries and KVFs.

Euclidean 3-space has a 3-parameter group of rotational KVFs and a 3-parameter group of translational KVFs. The rotational KVF integral curves are circles and the translational KVF integral curves are straight lines. The metric is of course constant along any integral curve of any of those KVFs; but the converse is not true, that any curve along which the metric is constant must be an integral curve of some KVF. But a manifold like Euclidean space, or indeed any manifold with constant curvature, can mislead one into thinking that the converse should be true, as it apparently has misled you.

 

[cianfa72]

The metric is of course constant along any integral curve of any of those KVFs; but the converse is not true, that any curve along which the metric is constant must be an integral curve of some KVF.

So which is actually the distinguishing feature of a KVF and its integral curves? Thanks.

 

[jbergman]

So which is actually the distinguishing feature of a KVF and its integral curves? Thanks.

I am new to KVF's but according to wikipedia the distinguishing feature is
$$\mathcal{L}_{X} g = 0$$
i.e., the Lie Derivative of the metric along the Killing Vector Field is 0.

An equivalent statement given in Lee's "Introduction to Riemannian Manifolds" is that the metric is invariant under the flow generated by the KVF $X$.
$$\theta_t^{*}g=g$$
So, the flow generated by a KVF gives us isometries which preserve the metric.

 

[DrGreg]

So which is actually the distinguishing feature of a KVF and its integral curves? Thanks.

It's not just the spacetime that possesses the corresponding symmetry. The integral curves possesses the same symmetry.

For example, in Euclidean space, a family of parallel straight lines possesses translational symmetry in the common direction of the lines. A family of coaxial circles possesses rotational symmetry about the common axis.

For a rigorous definition, you have to look at the definition of a KVF.

There's also another way to look at it. If you find a coordinate system that has the property that, along every integral curve of the vector field, $(N-1)$ of the $N$ coordinates are constant, and all $N^2$ components of the metric tensor are independent of the $N$th coordinate, then your vector field is a KVF.

In the case of Schwarzschild coordinates, all metric components are independent of both $t$ and $\phi$ (but not $r$ and $\theta$), so that gives you two different families of integral curves corresponding to time translation symmetry (stationary spacetime) and rotational symmetry.

 

[jbergman]

A coordinate chart is a labeling of points in the space by n-tuples of real numbers, with certain requirements about continuity, etc. A coordinate transformation is just a change of labeling.

A symmetry of a space corresponds to a Killing vector field, which is an invariant property that is independent of any choice of coordinates.No. It means that the metric is unchanged along integral curves of the Killing vector field that corresponds to the symmetry. For example, in Euclidean 3-space, there is a 3-parameter group of Killing vector fields corresponding to rotational symmetry. Integral curves of any particular one of these Killing vector fields correspond to circles centered on a particular axis of rotation. Moving along the flow of such integral curves is not a coordinate transformation.

Thank you for this informative answer. I wasn't familiar with KVFs before this, but your point about distinguishing between coordinate transformations and invariant properties of the manifold was helpful.

That said, I might quibble a bit with your answer after looking into this. As I understand it, the KVFs only give us continuous symmetries. So something like a reflection in Euclidean space would not be included in them and that if we want to talk about all symmetries of a space then we would be talking about all isometries of that space.

 

[cianfa72]

I am new to KVF's but according to wikipedia the distinguishing feature is
$$\mathcal{L}_{X} g = 0$$
i.e., the Lie Derivative of the metric along the Killing Vector Field is 0.

So, in Euclidean space if we pick a generic curve (that is not the integral curve of any KVF) the Lie Derivative of the metric tensor $g$ along the curve's tangent vector field $X$ does not vanish, namely $\mathcal{L}_{X} g \neq 0$. Yet the metric tensor $g$ does not change along that curve.

Edit: since the curve's tangent vector is not defined off the curve, I believe we need a congruence filling the Euclidean space such that one of its curves is actually the given curve joining the points. Then we can evaluate the Lie derivative of the metric tensor $g$ along this vector field $X$.

 

[jbergman]

So, in Euclidean space if we pick a generic curve (that is not the integral curve of any KVF) the Lie Derivative of the metric tensor $g$ along the curve's tangent vector field $X$ does not vanish, namely $\mathcal{L}_{X} g \neq 0$. Yet the metric tensor $g$ does not change along that curve.

I don't think that's true. Think of a dilation, i.e. a vector field that points outward but gets larger the further out you go. The flow associated with that changes the distance between points and the metric. You have to flow all the points of the manifold the same flow parameter time.

 

[PAllen]

Just a quick point about the special case of a constant scalar curvature manifold (at every point, sectional curvature defined by a pair of vectors is the same for every pair, and also the same for every point). In this case, every tangent direction at every point is a killing direction. It is for this reason that any curve acts like it is an integral curve of a kvf.

 

[jbergman]

Just a quick point about the special case of a constant scalar curvature manifold (at every point, sectional curvature defined by a pair of vectors is the same for every pair, and also the same for every point). In this case, every tangent direction at every point is a killing direction. It is for this reason that any curve acts like it is an integral curve of a kvf.

Right, but you can construct vector fields that are not KVFs and the corresponding flow will not preserve the metric or have 0 Lie Derivative.

 

[cianfa72]

Just a quick point about the special case of a constant scalar curvature manifold. In this case, every tangent direction at every point is a killing direction. It is for this reason that any curve acts like it is an integral curve of a kvf.

So in this special case (i.e. constant scalar curvature manifold) the Lie Derivative of the metric tensor $g$ along every tangent vector at every point is always null (strictly speaking we need a vector field $X$ along which calculate the Lie derivative: any smooth vector field will do the job).

 

[jbergman]

So in this special case (i.e. constant scalar curvature manifold) the Lie Derivative of the metric tensor $g$ along every tangent vector at every point is always null (strictly speaking we need a vector field $X$ along which calculate the Lie derivative: any smooth vector field will do the job).

I don't think this is correct. Given any tangent vector you can construct a KVF that agrees with the tangent vector at that point, but the converse is not true that any vector field with matching tangent vector at that point has 0 Lie Derivative, see my example of the dilation of the plane.

 

[PeterDonis]

So which is actually the distinguishing feature of a KVF and its integral curves?

Killing's equation. This is the sort of question you should be looking up for yourself.

 

[PeterDonis]

So, in Euclidean space if we pick a generic curve (that is not the integral curve of any KVF) the Lie Derivative of the metric tensor $g$ along the curve's tangent vector field $X$ does not vanish, namely $\mathcal{L}_{X} g \neq 0$. Yet the metric tensor $g$ does not change along that curve.

Correct. I strongly suggest that you do the math instead of relying on your intuition.

 

[PeterDonis]

Given any tangent vector you can construct a KVF that agrees with the tangent vector at that point, but the converse is not true that any vector field with matching tangent vector at that point has 0 Lie Derivative

Yes, exactly. A tangent vector at a point is not sufficient by itself to define a global curve (and hence a global vector field that has that curve as an integral curve). There are an infinite number of curves passing through a given point with a given tangent vector at that point.

 

[PeterDonis]

the Lie Derivative of the metric tensor along every tangent vector

Does not exist. You don't take Lie derivatives along vectors, you take them along curves.

Again, I strongly suggest that you do the math instead of relying on your intuition.

 

[cianfa72]

Does not exist. You don't take Lie derivatives along vectors, you take them along curves.

Sorry, to take Lie derivatives we need not just a curve but a family of them (i.e. a congruence).

In other words the vector field has to be defined in a neighborhood of each point along the curve (i.e. off the curve).

 

[PeterDonis]

to take Lie derivatives we need not just a curve but a family of them

In other words, you don't take Lie derivatives along vectors, you take them along curves, as I said. To do it, the curves you take them along need to be part of a congruence. Yes, that's true. And your earlier statement about taking Lie derivatives along vectors is still wrong.

If you are that familiar with Lie derivatives, you should be able to do the math yourself regarding all the things you are asking about in this discussion, instead of using your intuition to make wrong claims that we then have to correct.

 

[cianfa72]

Maybe I misinterpreted post #118 from @PAllen that in case of constant scalar curvature manifold every tangent direction at every point is a killing direction.

 

[PeterDonis]

Maybe I misinterpreted post #118 from @PAllen that in case of constant scalar curvature manifold every tangent direction at every point is a killing direction.

Perhaps you did. What he meant is that there an integral curve of some KVF passing through every point in every direction. But, as has been pointed out, the converse is not true: not every curve in every direction passing through a given point is an integral curve of some KVF. A direction (i.e., vector in the tangent space) at a point is not sufficient by itself to define a curve; for every tangent vector at a given point there are an infinite number of curves through that point that have that tangent vector, and only some of those curves will be integral curves of a KVF.

Once more: you need to take the time to actually do the math for this instead of using your intuition.

 

[cianfa72]

What he meant is that there an integral curve of some KVF passing through every point in every direction.

Sorry, but the above statement is always true or is true only in case of constant scalar curvature ?

 

[Ibix]

Sorry, but the above statement is always true

Clearly not: consider that there is no KVF pointing radially in Schwarzschild spacetime.

 

[PeterDonis]

the above statement is always true or is true only in case of constant scalar curvature ?

The latter, since that was the context of the original statement by @PAllen that started this subthread.

 

[ergospherical]

Well, this thread got quite badly side-tracked somehow.

 

[PAllen]

A simple example of a congruence in Minkowski spacetime that does not define a (timelike) kvf is the Milne congruence. This is despite the fact that each curve of the congruence is a timelike geodesic. What is true is that each curve of this congruence is a component integral curve of a different timelike kvf.

 

[jbergman]

Sorry, but the above statement is always true or is true only in case of constant scalar curvature ?

Personally, I think it's better to think about Isometries and flows than to think about the Killing vector fields or at least at first.

As explained on wikipedia, an isometry is a smooth mapping of a manifold of to itself that preserves the notion of distance.

A flow is a family of diffeomorphisms parameterized by some parameter like $t$ that very continuously with that parameter and whose derivatives with respect to $t$ at each point are tangent vectors and very smoothly and hence are associated with a vector field. In the case that family of diffeomorphisms associated with a flow are Isometries then the generating vector field is a Killing vector field.

The isometries are easy to understand intuitively for Euclidean Space or a sphere. For Euclidean Space you just have rotations, translations and reflections. Rotations and Translations can be parameterized to define flows. The generators of these are the KVFs. Now there are many other diffeomorphisms of $\mathbb{R}^n$ that aren't isometries. The vector fields associated with these are not KVFs.

 

[PeterDonis]

As explained on wikipedia, an isometry is a smooth mapping of a manifold of to itself that preserves the notion of distance.

It might be worth expanding on this a bit.

An isometry is a mapping $M$ of a manifold to a manifold (which might be the same one or might not) such that, for any pair of points $p$, $q$, taken by the mapping to points $M(p)$, $M(q)$, the distance from $p$ to $q$ is the same as the distance from $M(p)$ to $M(q)$. So it's not that the metric itself stays the same: the distances between mapped points have to stay the same.

So, for example, in Minkowski spacetime, translations, rotations, and boosts are all isometries of the manifold into itself. But the Milne transformation, the mapping corresponding to the Milne congruence described by @PAllen in post #134, is not; distances between corresponding points are changed by the transformation. And that's true even though the Minkowski metric itself stays the same along each curve in the congruence.

 

[cianfa72]

So it's not that the metric itself stays the same: the distances between mapped points have to stay the same.

ok, now it makes sense to me.

distances between corresponding points are changed by the transformation. And that's true even though the Minkowski metric itself stays the same along each curve in the congruence.

ok got it.

Just to give an example that helps intuition, I drew the following picture for the Euclidean plane:

Take the triangle ABC, the blue and red arrows represent two of the three sides. Consider its flow along the dotted red curves (i.e. along the congruence filling the Euclidean plane). Each curve in the congruence is parametrized by the its euclidean length $s$.

We can see that even if the metric tensor $g$ does not change along the congruence's curves, the shape of the triangle along the flow for the same parameter $s$ actually changes. Hence the curves in the congruence are not integral curves of any KVF.

 

[PAllen]

I definitely agree it is more intuitive to approach kvf's starting from a congruence, which defines a vector field, and then require that any two points connected by a geodesic, when 'flowed' along the congruence by a fixed interval along congruence lines, results in a geodesic of the same length as before. Then showing that this property being true for any pair of points and any flow amount, is equivalent to more abstract definitions applied to the vector field itself. Note, of course, that none of this language involves coordinates (though building coordinates adapted to a killing flow ensures that certain features are present in the expression of the metric in those coordinates). Also note that none of this says anything about diffeomorphisms or coordinate changes.

However, I am quite confused by what any of this has to do with isometries. My understanding has been that for any diffeomorphism, when a pullback is applied to the metric, the result is an isometry. In particular, if you consider the coordinate transform from standard Minkowski coordinates to Milne coordinates, with the associated Milne metric, then for any pair of points connected by a geodesic in standard Minkowski coordinates, the result after transform, is still a geodesic of the same length when using the Milne metric.

 

[cianfa72]

However, I am quite confused by what any of this has to do with isometries. My understanding has been that for any diffeomorphism, when a pullback is applied to the metric, the result is an isometry.

Yes, the point is that an isometry is a special diffeomorphism such that the pullback of the metric tensor along it results in the same metric tensor defined at the point where the pullback is done (i.e. the Lie derivative of the metric tensor along the flow defined by the isometry is null).

 

[PAllen]

Yes, the point is that an isometry is a special diffeomorphism such that the pullback of the metric tensor along it results in the same metric tensor defined at the point where the pullback is done.

Yes, but any diffeomorphism has an associated pullback, and in GR, this is the only sense in which diffeomorphisms are applied. Also, this doesn't answer in any way, what this has to do with kvfs. That is claim I don't understand.