I Vectors and isometries on a manifold

[davidge]

What do you think the general form for arbitrary rrr should be?

Since it has an r in one of its entries, I would say its form depends on the point in question.

 

[PeterDonis]

Since it has an r in one of its entries, I would say its form depends on the point in question.

The value of the components does, but the form does not. The general form is that the rotation maps $\begin{bmatrix} r \\ \theta \end{bmatrix}$ to $\begin{bmatrix} r \\ \theta + d \theta \end{bmatrix}$. If you work through the math, you will see that this is why the lower left component of the matrix I gave is $d\theta / r$.

 

[davidge]

The value of the components does, but the form does not

The general form is that the rotation maps [rθ][rθ]\begin{bmatrix} r \\ \theta \end{bmatrix} to [rθ+dθ]

Ok. But will this condition hold for $r = 0$? I've tried here and I found that at $r = 0$ we can't transform the second component from $\theta$ to $\theta + d\theta$. Also, as $r$ tends to zero the derivatives of the Killing Vector tends to infinity.

 

[PeterDonis]

will this condition hold for $r = 0$?

No, and that's ok, because the rotation leaves the origin $r = 0$ fixed, and the coordinate $\theta$ is singular there anyway, so "rotating" from $\theta$ to $\theta + d\theta$ is meaningless.

as rrr tends to zero the derivatives of the Killing Vector tends to infinity

Yes, that's true. So what? As I've said a number of times now, the derivatives of the KVF have nothing to do with generating rotations.

 

[PeterDonis]

Btw, if the rotation matrix being undefined at $r = 0$ bothers you, you can redo the analysis in Cartesian coordinates, which are nonsingular at the origin. In those coordinates you can show explicitly that the rotation matrix does nothing to the origin $x = 0$, $y = 0$.

 

[davidge]

So what? As I've said a number of times now, the derivatives of the KVF have nothing to do with generating rotations

Because I found that at least on $r = 0$ we cand find vectors $\xi^{\rho}$ that leave the point invariant and that obey Killing's equation:
$\xi^{r} = \xi_{r} = \xi^{\theta} = \xi_{\theta} = 0$, $\partial_{\theta} \xi^{r} = 0$, $\partial_{r} \xi^{\theta} = \frac{1}{r}$, and so

$V'^{\theta} = V^{\theta} + d{\theta} (\partial_{r} \xi^{\theta})V^{r} = \theta + d{\theta}$,
$V'^{r} = V^{r} + d{\theta} (\partial_{\theta} \xi^{r})V^{\theta} = V^{r} = r$.

These are not the Killing Vectors you wrote down in a previous post, because those were obtained from metric relations.

you can redo the analysis in Cartesian coordinates

I will

 

[PeterDonis]

I found that at least on $r = 0$ we cand find vectors $\xi^{\rho}$ that leave the point invariant and that obey Killing's equation

This doesn't make sense; $r = 0$ is just one point, and the vector $\xi$ vanishes at that point; but if you are only taking into account one point, derivatives are meaningless (you need at least a neighborhood). And at the point $r = 0$ it is also meaningless to talk about "rotating" $\theta$ to $\theta + d\theta$, since the coordinate $\theta$ is singular there.

Why are you persisting with trying to use derivatives of Killing vectors instead of Killing vectors themselves? That is not done anywhere in the literature that I'm aware of. If it's just your personal thing, then (a) it's not going to work, and (b) PF has rules about personal theories, which you should review.

 

[PeterDonis]

These are not the Killing Vectors you wrote down

Not quite, as you wrote it, because you used partial derivatives instead of covariant derivatives. But if you alter your definition to use covariant derivatives, you will see that, if we consider a neighborhood of $r = 0$ instead of just that point (so that derivatives are meaningful), and if you the components of $\xi$ at $r = 0$ plus the derivatives you gave there define a vector field $\xi$ on the neighborhood that is identical to the one I wrote down earlier.

(Also, your definition using partial derivatives is not well-defined at $r = 0$, since $1 / r$ is undefined there; so it doesn't work anyway. Whereas the definition in terms of covariant derivatives works fine at $r = 0$.)

 

[davidge]

Why are you persisting with trying to use derivatives of Killing vectors instead of Killing vectors themselves? That is not done anywhere in the literature

As I said before, I'm following what I learned from Weinberg's book and some two or three notes I found on web. (Also from a lecture I attended last month at the university.) Maybe I misunderstood what I've read?

at the point r=0r=0r = 0 it is also meaningless to talk about "rotating" θθ\theta to θ+dθθ+dθ\theta + d\theta, since the coordinate θθ\theta is singular there

I don't see why it is singular there.

you used partial derivatives instead of covariant derivatives

I did not show my complete derivation in post #56, but as I mentioned, I imposed the condition that partial derivatives of them are equal to the covariant ones.

Whereas the definition in terms of covariant derivatives works fine at r=0r=0r = 0.

I don't think so. Consult your post on the derivation of the covariant derivatives to see that at $r = 0$, $\nabla_{r}\xi^{\theta}$ is undefined.

if you are only taking into account one point, derivatives are meaningless (you need at least a neighborhood)

I agree.

 

[PeterDonis]

Maybe I misunderstood what I've read?

I think you must be. Unfortunately I don't have Weinberg's book to check the relevant sections for myself.

I don't see why it is singular there.

The metric is $ds^2 = dr^2 + r^2 d\theta^2$. This metric has no inverse at $r = 0$ (its determinant is zero), because $g_{\theta \theta} = 0$ there. Saying that $\theta$ is singular at $r = 0$ is a (somewhat sloppy) shorthand for that.

I imposed the condition that partial derivatives of them are equal to the covariant ones.

You can't "impose" this condition; it is either satisfied by the coordinates you've chosen or it isn't. For polar coordinates, it isn't.

Consult your post on the derivation of the covariant derivatives to see that at $r = 0$, $\nabla_{r}\xi^{\theta}$ is undefined.

Yes, but that doesn't stop $\xi^\theta$ itself from being well-defined (and nonzero) at $r = 0$ by my definition; it's just $\xi^\theta = 1$, like it is everywhere else (and $\partial r \xi^\theta = 0$ everywhere, including $r = 0$). What vanishes according to my definition at $r = 0$ is the norm of $\xi$, i.e., $\sqrt{g_{\mu \nu} \xi^\mu \xi^\nu}$. From the metric you will see that this norm is just $r$. Whereas by your definition, you tried to make $\xi^\theta$ itself vanish at $r = 0$, which doesn't work.

 

[davidge]

You can't "impose" this condition

but, since $\nabla_{\mu}\xi^{\nu} = \partial_{\mu}\xi^{\nu} + \xi^{\sigma} \Gamma^{\nu}_{\sigma \mu}$, it follows that if I require $\xi^{\sigma} = 0$, then $\nabla_{\mu}\xi^{\nu} = \partial_{\mu}\xi^{\nu}$. Is it not so?

This metric has no inverse at r=0r=0r = 0

Oh yea. I forgot about it.

Yes, but that doesn't stop ξθξθ\xi^\theta itself from being well-defined (and nonzero) at r=0r=0r = 0 by my definition

The thing is that I'm using a different definition. What would my different definition imply on the action of the rotation?

 

[PeterDonis]

"A metric space is said to be isotropic about a given point $X$ if there exist infinitesimal isometries that leave the point $X$ fixed, so that $\xi ^{\lambda}(X) = 0$, and for which the first derivatives $\xi_{\lambda ; \ \nu}(X)$ take all possible values [...]. In particular, in N dimensions we can choose a set of N(N-1)/2 Killing vectors..."

"As an example of a maximally symmetric space, consider an N-dimensional flat space, with vanishing curvature tensor. [...]
We can choose a set of N(N+1)/2 Killing vectors as follows:

$$\xi_{\mu}^{(\nu)}(X) = \delta_{\mu}^{\nu}$$
$$\xi_{\mu}^{(\nu \lambda)}(X) = \delta_{\mu}^{\nu} x^{\lambda} - \delta_{\mu}^{\lambda} x^{\nu}$$

...

The N vectors $\xi_{\mu}^{(\nu)}(X)$ represent translations, whereas the N(N-1)/2 vectors $\xi_{\mu}^{(\nu \lambda)}(X)$ represent infinitesimal rotations [...]"

Let me try and unpack this. We have been discussing the 2-dimensional flat space, i.e., the plane. It has 3 Killing vectors total: two translations and one rotation. The two translations in Cartesian coordinates are just $\partial / \partial x$ and $\partial / \partial y$, i.e., the Cartesian basis vectors. In the notation of the above quote these would be $\xi_\mu^{(\nu)} (X)$. The rotation is the Killing vector we've been discussing; it is $\xi_{\mu}^{(\nu \lambda)}(X)$ in the notation of the quote above, and is $\partial / \partial \theta$ in polar coordinates, and its components in Cartesian coordinates I'll leave to you (but the second formula above should give a good hint).

Now, as far as isotropy is concerned, the only Killing vector involved is the rotation Killing vector (the translation Killing vectors have to do with homogeneity, which is a different symmetry property). So let's focus on that. We have seen that it satisfies $\xi = 0$ at the point $X$, i.e., at the origin (note: I am assuming that Weinberg refers to the norm being zero, not components--you have to read his notation very carefully). The part I'm not sure about is the thing about the first derivatives taking "all possible values". The derivatives in question are covariant derivatives, as shown by the semicolon in the quoted formula; but I don't understand the "all possible values", since the values of all the possible covariant derivatives are as given by the formulas I posted previously, there is no wiggle room for them to have "all possible" values.

I notice that you left out text after the "all possible values"; could you possibly post it or at least a further excerpt, for more context?

 

[PeterDonis]

if I require $\xi^{\sigma} = 0$

You can't "require" $\xi^\sigma = 0$ (or any other value). The Killing vector is what it is. You can't make it be whatever you want.

I'm using a different definition.

You can't help yourself to whatever definition you want. Killing vectors are a geometric property of the manifold; you don't pick a definition for them, you find out what they are by looking at the geometry.

 

[davidge]

d its components in Cartesian coordinates I'll leave to you

I will work this out

I am assuming that Weinberg refers to the norm being zero, not components--you have to read his notation very carefully

I don't think it's what he is meaning

The part I'm not sure about is the thing about the first derivatives taking "all possible values".

There are pics below from that part on the book

You can't "require" ξσ=0ξσ=0\xi^\sigma = 0 (or any other value). The Killing vector is what it is. You can't make it be whatever you want.

It seems he was trying to define them by his own choice.

 

[PeterDonis]

There are pics below from that part on the book

These are helpful. First a general comment: Weinberg's books tend to assume a lot of background knowledge, so they aren't always suitable as introductory texts. Also they tend to assume a physicist's level of rigor rather than a mathematicians, which means issues like coordinate singularities are glossed over if they don't affect the physics (or in this case the geometry).

Now some more specific comments:

(1) I didn't say you were confusing a rotation with a coordinate transformation; I said you seemed confused about which interpretation of rotations you were using, the interpretation as a mapping from vectors to vectors (or points to points), holding the coordinates constant; or the interpretation as a coordinate transformation holding the underlying manifold and its points and vectors constant. Weinberg is clearly using the second interpretation, which is perfectly valid, but if you're going to use it you have to use it properly. On this interpretation, the only thing that changes under a rotation is the coordinates: you can't think of moving points and vectors around, you have to think of moving the coordinate grid lines around while holding all points and vectors constant. (I actually prefer the other interpretation because to me it seems more natural to think of moving points and vectors around and keeping the coordinate grid lines fixed; but that's a matter of personal preference.)

(2) Weinberg is not defining the Killing vectors "by his own choice". He is defining Killing vectors the standard way; a vector field that satisfies Killing's equation (his equation 13.1.5) is a Killing vector field. (And he is saying, which is true, that any Killing vector field defines an isometry.) He is then saying that if you have a Killing vector field that satisfies some additional conditions, then the metric space on which it exists is said to be "isotropic". But you can't just handwave such a Killing vector field into existence; you have to check to see if one exists given the geometry of the metric space. There are metric spaces on which no such Killing vector field exists; such spaces are simply not isotropic.

(3) As far as his statements $\xi^\lambda(X) = 0$ and $\xi_{\lambda ; \ \nu}(X)$ taking all possible values, I'll defer comment on them until you've worked things out in Cartesian coordinates. I don't think polar coordinates will work for evaluating these statements since they are singular at the point $X$ (the origin $r = 0$).

 

[PeterDonis]

Can you show explicitly the form that $\nabla{_\theta} \xi^{\theta}$, $\nabla{_r} \xi^{\theta}$, $\nabla{_\theta} \xi^{r}$ and $\nabla{_r} \xi^{r}$ takes

I answered this as it was asked, but since we have mentioned Killing's equation, I should clarify that for that equation these are not quite the relevant covariant derivatives, because the index on $\xi$ is upper instead of lower. Killing's equation in polar coordinates (the only non-trivial component) is

$$
\nabla_r \xi_\theta + \nabla_\theta \xi_r = 0
$$

Lowering the index makes a difference, because $\xi_\theta = g_{\theta \theta} \xi^\theta = r^2$ depends on $r$ (whereas $\xi^\theta$ with the upper index does not). So Killing's equation becomes (giving only the nonzero terms)

$$
\partial_r \xi_\theta - 2 \Gamma^\theta{}_{r \theta} \xi_\theta = 2r - 2 \frac{1}{r} r^2 = 0
$$

(Note the minus sign in front of the $\Gamma$ term because we are now "correcting" a lower index instead of an upper index in the covariant derivative.)

 

[davidge]

Weinberg's books tend to assume a lot of background knowledge

they tend to assume a physicist's level of rigor rather than a mathematicians

I noticed it

I answered this as it was asked, but since we have mentioned Killing's equation, I should clarify that for that equation these are not quite the relevant covariant derivatives, because the index on ξξ\xi is upper instead of lower

No problem. I realized that that was for vectors, not co-vectors.

I didn't say you were confusing a rotation with a coordinate transformation

Weinberg is clearly using the second interpretation

Weinberg is not defining the Killing vectors "by his own choice"

That is ok

He is then saying that if you have a Killing vector field that satisfies some additional conditions

you can't just handwave such a Killing vector field into existence

I did not realize this fact before.

As far as his statements ξλ(X)=0ξλ(X)=0\xi^\lambda(X) = 0 and ξλ; ν(X)ξλ; ν(X)\xi_{\lambda ; \ \nu}(X) taking all possible values, I'll defer comment on them until you've worked things out in Cartesian coordinates

A tentative:

The two translational Killing vectors could be $\xi^{1}$ and $\xi^{2}$.
For example, let $x$ and $y$ be two points on $\mathbb{R}^2$, related by $y = x +\epsilon \xi$, for arbitrary $\epsilon \in \mathbb{R}$.
$\xi^{1}(x) = \xi^{2}(x) = 1, x \doteq (0,0) \Longrightarrow y \doteq (x^{1} + \epsilon\xi^{1},x^{2} + \epsilon\xi^{2}) = \epsilon(1,1)$.

Following your work here on the polar coordinates case,

if $V$ is a vector with components $\begin{bmatrix}1\\0\end{bmatrix}$ at $x \doteq (0,0)$, the action of the Killing vector is to change $\begin{bmatrix}1\\0\end{bmatrix}$ to $\begin{bmatrix}1\\0\end{bmatrix} + \epsilon\begin{bmatrix}1\\1\end{bmatrix}$, remembering that in 2-d, Cartesian coordinates, the Killing vectors are $\partial/\partial x^1$ and $\partial/\partial x^2$ for a point $x$ with coordinates $(x^1,x^2)$.

The problem here is that it seems we don't have N(N+1)/2 = 3 KV, but only 2. It's missing one, because the derivatives all vanish for this metric.

 

[PeterDonis]

let $x$ and $y$ be two points on $\mathbb{R}^2$, related by $y = x +\epsilon \xi$,

You still appear to be confused about interpretations. If you want to use Weinberg's interpretation, then translations are coordinate transformations just like rotations. (As far as I can tell, his discussion of isometries as coordinate transformations and Killing vectors as generating infinitesimal coordinate transformations is not limited to rotations.) So you should be saying: let $x$ and $y$ be the coordinates of a chosen point in $\mathbb{R}^2$ in two different charts, related by $y = x + \epsilon \xi$.

In this case, the two translation Killing vectors, in Cartesian coordinates, are $\partial / \partial x^1$ and $\partial / \partial x^2$ (I won't use $x$ and $y$ since you used them to label the coordinate 2-tuples as a whole). In column vector notation these are $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 0 \\ 1 \end{bmatrix}$.

ξ¹(x)=ξ²(x)=1

No. There are two Killing vectors, not two components of one Killing vector. See above.

The correct actions of the two translation Killing vectors, using $x = \begin{bmatrix} a \\ b \end{bmatrix}$ and $y = \begin{bmatrix} a' \\ b' \end{bmatrix}$ for the two coordinate 2-tuples in column vector notation, are:

$$
\begin{bmatrix} a' \\ b' \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} + \epsilon \begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} a + \epsilon \\ b \end{bmatrix}
$$

$$
\begin{bmatrix} a' \\ b' \end{bmatrix} = \begin{bmatrix} a \\ b \end{bmatrix} + \epsilon \begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} a \\ b + \epsilon \end{bmatrix}
$$

As you can see, these are just infinitesimal translations in one of the two basis directions.

The problem here is that it seems we don't have N(N+1)/2 = 3 KV, but only 2.

You've forgotten the rotation Killing vector. It's still there. Have you tried figuring out what it looks like in Cartesian coordinates?

 

[davidge]

You still appear to be confused about interpretations

So you should be saying

Yes. I'm sorry for non using the correct language in my last post. I get confused with English language itself.

There are two Killing vectors, not two components of one Killing vector

Thanks for clarifying this for me. I thought there were only one vector.

As you can see, these are just infinitesimal translations in one of the two basis directions.

Yea

You've forgotten the rotation Killing vector. It's still there. Have you tried figuring out what it looks like in Cartesian coordinates?

Yes. I've tried, but all covariant derivatives vanish and I was expecting to find it from that derivatives. Also, in polar coordinates, the effect of using the rotation KV to a general vector seems to be the same as using the translation KV in your previous post. Is it due the difference between Cartesian and Polar coordinates?

 

[PeterDonis]

all covariant derivatives vanish

All covariant derivatives of what? All covariant derivatives of the basis vectors vanish, yes. But there is nothing that says a Killing vector has to be a coordinate basis vector.

If you're having trouble guessing an answer, just write down the most general possible vector in Cartesian coordinates and plug it into Killing's equation, and see what conditions its components have to satisfy for the vector to satisfy Killing's equation. In Cartesian coordinates this is actually very simple because all of the Christoffel symbols are zero, so covariant derivatives are just partial derivatives. Also there is only one component of Killing's equation that is not trivial in 2 dimensions.

 

[davidge]

Please correct me if I'm wrong in what follows.
Solving Killing's equation I found that its components should satisfy $\xi^{1} = - \xi^{2}$. Then I proceeded as follows:

In Cartesian coordinates, suppose we have a vector with components $\begin{bmatrix}1\\0\end{bmatrix}$. If we rotate it by 90°, its new components* should become $\begin{bmatrix}0\\1\end{bmatrix}$. It can be done through the action of a matrix operator on the original vector.

By trying combinations of elements of this matrix, knowing what the resultant vector will look, I found that the matrix should be
$\begin{bmatrix}a_{11}&\xi^1\\-\xi^1&a_{22}\end{bmatrix}$.

Now, I found what $a_{11}$ and $a_{22}$ are by operating on the vector $\begin{bmatrix}0\\1\end{bmatrix}$ which becomes $\begin{bmatrix}1\\0\end{bmatrix}$ after rotation.
The matrix is then $\begin{bmatrix}cos(\theta)&sin(\theta)\\-sin(\theta)&cos(\theta)\end{bmatrix}$ for a rotation through an angle $\theta$. The killing vector components are $sin(\theta)$ and $-sin(\theta)$.

* Here I'm considering a same vector rotated by an angle, so it has new components in its new "position". However, the interpretation you mentioned of seeing the rotation as a mapping of a vector $\begin{bmatrix}0\\1\end{bmatrix}$ into another vector $\begin{bmatrix}1\\0\end{bmatrix}$ makes more sense to me.

That is more like a heuristic derivation of the Killing vector components. If there is a more rigorous derivation please let me know.

I'm not sure the above is right .

 

[PeterDonis]

I found that its components should satisfy $\xi^{1} = - \xi^{2}$.

That doesn't look right. How are you deriving that requirement? The only nontrivial component of Killing's equation in 2 dimensions, in Cartesian coordinates (so covariant derivatives are just partial derivatives) is $\frac{\partial}{\partial x^1} \xi_2 + \frac{\partial}{\partial x^2} \xi_1 = 0$. I don't see how to get from that to what you wrote.

suppose we have a vector with components $\begin{bmatrix}1\\0\end{bmatrix}$ . If we rotate it by 90°, its new components* should become $\begin{bmatrix}0\\1\end{bmatrix}$ .

This is not an infinitesimal rotation; you should start with rotating $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ to something rotated by an infinitesimal angle $d \theta$ from that. That is what will have a direct relationship to the Killing vector.

By trying combinations of elements of this matrix, knowing what the resultant vector will look, I found that the matrix should be
$\begin{bmatrix}a_{11}&\xi^1\\-\xi^1&a_{22}\end{bmatrix}$ .

This isn't right either. See above comments.

If you want to guess what the infinitesimal matrix might look like, you could try expanding the components of the finite rotation matrix (which you got correct, see below) as Taylor series in $\theta$ (or $d \theta$ for an infinitesimal rotation) and dropping terms higher than first order. Also remember that, since we are using Cartesian coordinates, you will need to ultimately express everything in terms of $x^1$ and $x^2$ (or what would normally be called $x$ and $y$ if we used those names for the coordinates), so you will need to figure out how to express the rotation angle $\theta$ in terms of them (or $d\theta$ in terms of $dx$ and $dy$).

The matrix is then $\begin{bmatrix}cos(\theta)&sin(\theta)\\-sin(\theta)&cos(\theta)\end{bmatrix}$ for a rotation through an angle $\theta$.

This is correct, but for a finite rotation, not an infinitesimal one. (Actually, reversing the signs on the $\sin \theta$ components is also a valid solution; which one corresponds to rotating by $\theta$ and which by $- \theta$ depends on your convention for which direction, clockwise or counterclockwise, corresponds to increasing $\theta$.) Also, the way you are getting there is not valid; see above.

The killing vector components are $sin(\theta)$ and $sin(\theta)$.

This is not correct. See above.

 

[davidge]

How are you deriving that requirement?

covariant derivatives are just partial derivatives is ∂∂x1ξ2+∂∂x2ξ1=0

you should start with rotating [10][10]\begin{bmatrix} 1 \\ 0 \end{bmatrix} to something rotated by an infinitesimal angle dθdθd \theta from that

$V \doteq \begin{bmatrix}V^{1}\\V^{2}\end{bmatrix} = \begin{bmatrix}1\\0\end{bmatrix} \\ V'^{1} = V^{1} + d\theta \xi^{1}_{,2}V^{2} = V^{1} \\ V'^{2} = V^{2} + d\theta \xi^{2}_{,1}V^1 = d\theta$
where a prime denotes the transformed components and $V^{1} = 1, V^{2} = 0, \xi^{1}_{,2} = - d\theta \xi^{2}_{,1} = 1$.

Would this be a solution?

If you want to guess what the infinitesimal matrix might look like, you could try expanding the components of the finite rotation matrix

for the infinitesimal case, would it be enough to take the limit as $\theta$ tends to zero, in which case, $cos(\theta) \approx 1$ and $sin(\theta) \approx \theta$ (1), where we can replace $\theta$ in (1) with $d\theta$ because it's infinitesimal? (2)

you will need to figure out how to express the rotation angle θθ\theta in terms of them (or dθdθd\theta in terms of dxdxdx and dydydy).

Following (2), we would have $tan(\theta) \approx d\theta = dx^{2}/dx^{1}$.

 

[PeterDonis]

Would this be a solution?

I can't follow your notation. Are you trying to use derivatives of the Killing vector components again? Once more: isometries are generated by Killing vectors themselves, not their derivatives. If you are trying to use derivatives, you are doing it wrong.

for the infinitesimal case, would it be enough to take the limit as $\theta$ tends to zero

No, because in that limit, $cos \theta = 1$ and $\sin \theta = 0$, and that just gives you the identity matrix.

What you were actually doing to get $\cos \theta \approx 1$ and $\sin \theta \approx \theta$ was what I said: you expand the functions as Taylor series, so $\cos \theta = 1 - \frac{1}{2!} \theta^2 + \frac{1}{4!} \theta^4 - ...$ and $\sin \theta = \theta - \frac{1}{3!} \theta^3 + ...$, and then discard terms beyond first order in $\theta$.

So now you have the infinitesimal rotation matrix $\begin{bmatrix} 1 & - d\theta \\ d\theta & 1 \end{bmatrix}$ (where I have flipped the signs on the $d\theta$ terms from what you wrote down). What do you get when you operate on the vector $\begin{bmatrix} 1 \\ 0 \end{bmatrix}$ with this matrix? Or, more generally, what do you get when you operate on the vector $\begin{bmatrix} x^1 \\ x^2 \end{bmatrix}$?

 

[PeterDonis]

Would this be a solution?

To re-ground this in what you quoted from Weinberg's book, his equation for an infinitesimal coordinate transformation is

$$
x'^\mu = x^\mu + \epsilon \xi^\mu
$$

which we can rewrite in column vector notation as

$$
\begin{bmatrix} x'^1 \\ x'^2 \end{bmatrix} = \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} + d\theta \begin{bmatrix} \xi^1 \\ \xi^2 \end{bmatrix}
$$

where I have put $d\theta$ instead of $\epsilon$ because we are talking about rotations by an infinitesimal angle. But since we are talking about infinitesimal rotations, we must also have the equation $x' = R x$, which in matrix form is

$$
\begin{bmatrix} x'^1 \\ x'^2 \end{bmatrix} = \begin{bmatrix} R_{11} & R_{12} \\ R_{21} & R_{22} \end{bmatrix} \begin{bmatrix} x^1 \\ x^2 \end{bmatrix}
$$

The matrix $R$ is just the rotation matrix, whose infinitesimal form we know. This is what connects the two viewpoints (matrix and Killing vector).

 

[davidge]

in that limit, cosθ=1cosθ=1cos \theta = 1 and sinθ=0sin⁡θ=0\sin \theta = 0, and that just gives you the identity matrix.
What you were actually doing to get cosθ≈1cos⁡θ≈1\cos \theta \approx 1 and sinθ≈θsin⁡θ≈θ\sin \theta \approx \theta was what I said

Oh yea.

So now you have the infinitesimal rotation matrix [1−dθdθ1][1−dθdθ1]\begin{bmatrix} 1 & - d\theta \\ d\theta & 1 \end{bmatrix} (where I have flipped the signs on the dθdθd\theta terms from what you wrote down). What do you get when you operate on the vector [10][10]\begin{bmatrix} 1 \\ 0 \end{bmatrix} with this matrix? Or, more generally, what do you get when you operate on the vector [x1x2][x1x2]\begin{bmatrix} x^1 \\ x^2 \end{bmatrix}?

[x′1x′2]=[x1x2]+dθ[ξ1ξ2][x′1x′2]=[x1x2]+dθ[ξ1ξ2]​

\begin{bmatrix} x'^1 \\ x'^2 \end{bmatrix} = \begin{bmatrix} x^1 \\ x^2 \end{bmatrix} + d\theta \begin{bmatrix} \xi^1 \\ \xi^2 \end{bmatrix}

where I have put dθdθd\theta instead of ϵϵ\epsilon because we are talking about rotations by an infinitesimal angle. But since we are talking about infinitesimal rotations, we must also have the equation x′=Rxx′=Rxx' = R x, which in matrix form is

[x′1x′2]=[R11R12R21R22][x1x2][x′1x′2]=[R11R12R21R22][x1x2]​

\begin{bmatrix} x'^1 \\ x'^2 \end{bmatrix} = \begin{bmatrix} R_{11} & R_{12} \\ R_{21} & R_{22} \end{bmatrix} \begin{bmatrix} x^1 \\ x^2 \end{bmatrix}

Then the Killing vector components are $\xi^{1} = -x^2$ and $\xi^{2} = x^{1}$?

The matrix RRR is just the rotation matrix, whose infinitesimal form we know. This is what connects the two viewpoints (matrix and Killing vector).

I see now

 

[PeterDonis]

Then the Killing vector components are $\xi^{1} = -x^2$ and $\xi^{2} = x^{1}$?

Yes, you've got it.

 

[davidge]

you've got it.

Finally . Thank you for clarifying so many things through this thread.

 

[davidge]

There is only one problem. Weinberg says if we are to perform a rotation at a point $x$, the Killing vector components must vanish there, so that the point is left invariant, because any other point $x'^{\mu}$ would be equal to $x'^{\mu} = x^{\mu} + \epsilon \xi^{\mu}(x)$ in his notation, and since $\xi^{\mu}(x) = 0$, so $x'^{\mu} = x^{\mu}$.

So would this condition be necessary only if we want the metric to be the same? (i.e. an isotropy) And else, the rotation would not keep the point fixed, as you've shown?

 

[PeterDonis]

Weinberg says if we are to perform a rotation at a point $x$, the Killing vector components must vanish there

Yes, and the Killing vector we found satisfies that: the vector $\begin{bmatrix} - x^2 \\ x^1 \end{bmatrix}$ vanishes at the origin, since there $x^1 = x^2 = 0$.

would this condition be necessary only if we want the metric to be the same?

I'm not sure what you mean. The coordinate transformation generated by any Killing vector (i.e., any isometry) leaves the metric the same everywhere: the metric $g_{\mu \nu}$ takes the same form in the new coordinates as in the original ones. But a rotation only leaves the coordinates of point $X$ invariant; it changes the coordinates of all other points. (Note that translations--the other isometries Weinberg discusses--do not leave the coordinates of any points invariant.)

 

[PeterDonis]

One loose end that still remains is the statement of Weinberg's that for a rotation, the first derivatives of the Killing vector at $X$ take "all possible values". I think I understand what he's trying to say, although it seems to me to be a rather confusing way to say it.

If I'm right, the 2-d rotation case actually doesn't illustrate this very well, because there is only one Killing vector and only one first derivative of it at $X$. So we should try considering a case with multiple Killing vectors; the obvious case is 3-d rotations. Here there is a 3-parameter group of rotation Killing vectors, which in Cartesian coordinates have the form you would expect from our analysis of the 2-d case; they are:

$$
\xi = \begin{bmatrix} -x^2 \\ x^1 \end{bmatrix}
$$

$$
\upsilon = \begin{bmatrix} -x^3 \\ x^2 \end{bmatrix}
$$

$$
\zeta = \begin{bmatrix} -x^1 \\ x^3 \end{bmatrix}
$$

Weinberg has a qualifier after the "all possible values" statement, that this is "subject to the antisymmetry condition". What he means, I think, is simply that each Killing vector must satisfy Killing's equation, which constrains the partial derivatives of its components. Because of the antisymmetry, in $N$ dimensions Killing's equation will have $N(N-1)/2$ non-trivial components. For the 2-d case, that means just one component; for the 3-d case, there are three, which I have written in terms of the Killing vectors that satisfy them, as written above:

$$
\nabla_1 \xi_2 + \nabla_2 \xi_1 = 0
$$

$$
\nabla_2 \xi_3 + \nabla_3 \xi_2 = 0
$$

$$
\nabla_3 \xi_1 + \nabla_1 \xi_3 = 0
$$

So basically, I think that Weinberg's "all possible values" just means that, at the point $X$, where all the rotation Killing vectors vanish, we have $N(N-1)/2$ choices of distinct pairs of coordinate indexes that we can choose in in order to define the rotation, since each non-trivial component of Killing's equation involves a pair of coordinate indexes. To put it another way, any rotation picks out a plane containing the point $X$; but in $N$ dimensions there are $N(N-1)/2$ possible planes (more precisely, that number of mutually orthogonal planes), and each plane has its own Killing vector (more precisely, its own linearly independent Killing vector).

 

[davidge]

Are you trying to use derivatives of the Killing vector components again?

Now that we found what the Killing vector components are, we see that, for example:
$x'^{1} = x^{1} + d\theta \xi^{1}_{,2}x^{2}$, but $\xi^{1}_{,2} = -1$ and so $x'^{1} = x^{1} - d\theta x^{2}$, the same result we obtain in considering $x'^{1} = x^{1} + d\theta \xi^{1}$, because, as we know, $\xi^{1} = -x^{2}$. That is why I was insisting in using derivatives.
Now I think I understand the Weinberg's statement that the components should be zero but the derivatives should not. In our case this happens at the origin (0,0).

the Killing vector we found satisfies that: the vector [−x2x1][−x2x1]\begin{bmatrix} - x^2 \\ x^1 \end{bmatrix} vanishes at the origin

a rotation only leaves the coordinates of point XXX invariant; it changes the coordinates of all other points. (Note that translations--the other isometries Weinberg discusses--do not leave the coordinates of any points invariant.)

Indeed this is what you have been shown me

One loose end that still remains is the statement of Weinberg's that for a rotation, the first derivatives of the Killing vector at XXX take "all possible values". I think I understand what he's trying to say, although it seems to me to be a rather confusing way to say it.

we should try considering a case with multiple Killing vectors; the obvious case is 3-d rotations. Here there is a 3-parameter group of rotation Killing vectors, which in Cartesian coordinates have the form you would expect from our analysis of the 2-d case;

So basically, I think that Weinberg's "all possible values" just means that, at the point XXX, where all the rotation Killing vectors vanish, we have N(N−1)/2N(N−1)/2N(N-1)/2 choices of distinct pairs of coordinate indexes that we can choose in in order to define the rotation

Thank you for interpreting and explaining the meaning of that statement.

 

[PeterDonis]

we see that, for example:
$x'^{1} = x^{1} + d\theta \xi^{1}_{,2}x^{2}$,

No. No. No.

The correct equation is $x'^1 = x^1 + d\theta \xi^1$. That is the equation Weinberg wrote down. That is the equation that is always correct. And in this case it gives $x'^1 = x^1 - d\theta x^2$. (And $x'^2 = x^2 + d\theta x^1$.)

Your equation, even if it happens to look the same numerically by coincidence in this particular case, is not correct. I don't understand why you keep on trying to use it.

Now I think I understand the Weinberg's statement that the components should be zero but the derivatives should not

No, you don't. The derivatives should not be zero at the origin because if the derivatives were zero, the Killing vector itself would vanish everywhere. It has nothing whatever to do with your wrong equation.

If you look at the translation Killing vectors, btw, you will see that their derivatives vanish everywhere. But the Killing vectors themselves do not. So the "vanish at the origin but nonzero derivatives" requirement only applies to rotations, not translations.

If you ask why those two cases (rotations and translations) are the only two possibilities Weinberg considers, that's because they're the only two non-trivial possibilities that Killing's equation admits.

 

[davidge]

No. No. No.

Ok

The derivatives should not be zero at the origin because if the derivatives were zero, the Killing vector itself would vanish everywhere

If you look at the translation Killing vectors, btw, you will see that their derivatives vanish everywhere. But the Killing vectors themselves do not. So the "vanish at the origin but nonzero derivatives" requirement only applies to rotations, not translations.

I got it.

If you ask why those two cases (rotations and translations) are the only two possibilities Weinberg considers, that's because they're the only two non-trivial possibilities that Killing's equation admits.

I see