Finding killing vector fields of specific spacetime

[Aemmel]

Homework Statement

Find explicit expressions for a complete set of Killing vector fields for the space $ds^2 = -(dudv+dvdu)+a^2(u)dx^2+b^2(u)dy^2$ (with coordinates $\{u,v,x,y\}$) where $a$ and $b$ are unspecified functions of $u$. (Carroll Exercise 3.13)

Relevant Equations

Hints from the exercise: there are five Killing vectors in all and all of them have a vanishing $u$ component $K^u$

I have been at this exercise for the past two days now, and I finally decided to get some help. I am learning General Relativity using Carrolls Spacetime and Geometry on my own, so I can't really ask a tutor or something. I think I have a solution, but I am really unsure about it and I found 6 Killing Vectors instead of 5. Maybe someone could take a look at it and tell me where I went wrong or if my answer is actually correct and I am not seeing it.

Straight up we can see 3 Killing vectors: <br /> K^{(1)\mu} = \left(0, 1, 0, 0\right) ,\<br /> K^{(2)\mu} = \left(0, 0, 1, 0\right) ,\<br /> K^{(3)\mu} = \left(0, 0, 0, 1\right)<br /> which we will also find again later on.
From the Killing equation \nabla_{(\mu} K_{\nu)} = g_{\nu\sigma} \nabla_\mu K^\sigma + g_{\mu\sigma} \nabla_\nu K^\sigma = 0 (I tried doing it with lower indices on K but got completely stuck. Raising the index seems to make it a bit easier) we can find the Killing vectors. Maybe there's a clever trick I am not seeing, but I can't find it. So I tried to just write out all the equations and go from there.
Writing them all out and using K^u = 0 we get
<br /> (I) \ \ \ \ \ \partial_u K^v = 0 \\<br /> (II) \ \ \ \ \ \partial_x K^x = 0 \\<br /> (III) \ \ \ \ \ \partial_y K^y = 0 \\<br /> (IV) \ \ \ \ \ \partial_v K^v = 0 \\<br /> (V) \ \ \ \ \ a^2 \partial_u K^x - \partial_x K^v = 0 \\<br /> (VI) \ \ \ \ \ b^2 \partial_u K^y - \partial_y K^v = 0 \\<br /> (VII) \ \ \ \ \ \partial_v K^x = 0 \\<br /> (VIII) \ \ \ \ \ \partial_v K^y = 0 \\<br /> (IX) \ \ \ \ \ b^2 \partial_x K^y + a^2 \partial_y K^x = 0<br />

From these it's directly clear that K^v=K^v(x,y) (from I, IV), K^x=K^x(u,y) (from II, VII) and K^y=K^y(u,x) (from III, VIII). Using this and differentiating V, VI, IX we find {\partial_x}^2 K^v = 0, {\partial_y}^2 K^v = 0, {\partial_x}^2 K^y = 0 and {\partial_y}^2 K^x = 0. Integrating these we find
<br /> K^v(x,y) = V_1 + V_2 x + V_3 y + V_4 xy \\<br /> K^x(u,y) = X_1(u) y + X_2(u) \\<br /> K^y(u,x) = Y_1(u) x + Y_2(u)<br />
where the V_i are constant with regard to any coordinate.
Plugging these into V and VI, rearranging with regard to x or y and comparing coefficients we can find that
<br /> \partial_u X_1 \equiv X_1&#039; = \frac{V_4}{a^2} \\<br /> X_2&#039; = \frac{V_2}{a^2} \\<br /> Y_1&#039; = \frac{V_4}{b^2} \\<br /> Y_2&#039; = \frac{V_3}{a^2}<br />
Now, using X_2&#039;=\frac{V_2}{V_4}X_1&#039; we can write X_2=\frac{V_2}{V_4}X_1 + C_x (where C_x is a constant regarding any coordinate) and a smiliar expression for Y_2. So that we get
<br /> K^x = X_1 y + \frac{V_2}{V_4} X_1 + C_x \\<br /> K^y = Y_1 y + \frac{V_3}{V_4} Y_1 + C_y<br />
Putting this into V again and using the known derivative for X_1 we find
<br /> X_1(u) = V_4 \frac{b}{a(ab&#039;-a&#039;b)}<br />
which is where I actually find the first problem. This would imply that the relation a = b a&#039;&#039; / b&#039;&#039; has to hold (because we know the derivative of X_1). But in the problem it was mentioned they are arbitrary. So I thought my problem had to be there, but I just can't see why my derivation up to this point is wrong (or the spacetime in general implies that the relation has to hold, otherwise those Killing vectors wouldn't exist. That would also be possible). Let's continue with this result for now.
Plugging our K^x and K^y into IX we get Y_1 = -\frac{a^2}{b^2}X_1 Thus arriving at
<br /> K^u = 0 \\<br /> K^v = V_1 + V_2 x + V_3 y + V_4 xy \\<br /> K^x = V_4 \frac{b}{a(ab&#039;-a&#039;b)} y + V_2 \frac{b}{a(ab&#039;-a&#039;b)} + C_x \\<br /> K^y = - V_4 \frac{a}{b(ab&#039;-a&#039;b)} x - V_3 \frac{a}{b(ab&#039;-a&#039;b)} + C_y<br />
which finally gives us 6 Killing vectors, by setting one of the 6 constants to 1 and the rest to 0
<br /> K^{(1)\mu} = \left(0, 1, 0, 0\right) \\<br /> K^{(2)\mu} = \left(0, 0, 1, 0\right) \\<br /> K^{(3)\mu} = \left(0, 0, 0, 1\right) \\<br /> K^{(4)\mu} = \left(0, x, \frac{b}{a(ab&#039;-ba&#039;)}, 0\right) \\<br /> K^{(5)\mu} = \left(0, y, 0, -\frac{a}{b(ab&#039;-a&#039;b)}\right) \\<br /> K^{(6)\mu} = \left(0, xy, \frac{b}{a(ab&#039;-ba&#039;)} y, -\frac{a}{b(ab&#039;-a&#039;b)} x\right) <br />

Now there are two problems with my solution.
First: there are 6 vector fields instead of 5. K^{(6)} could be crafted with K^{(6)\mu} = yK^{(4)\mu} + xK^{(5)\mu} - xyK^{(1)\mu} but that would go against the knowledge I have of "linear independence". I had the idea that it might be different for vector fields (meaning non constant coefficients are fine) but that would mean there are only 3 independent Killing vectors in \mathbb{R}^3 which is not the case (as shown in Carrol 3.8).
Second: for K^{(4)} to K^{(6)} to be Killing vector fields, the equation a=ba&#039;&#039;/b&#039;&#039; has to hold, which, again, I don't see why that would be the case for arbitrary a, b.

 

[JD_PM]

I did not check your work carefully but I'd like to share what's the recipe I use when I want to get the Killing vectors out of a given metric describing a spacetime.

Let's work with the 2D anti-de Sitter metric as an example

$$(ds)^2 = -cosh^2 (\eta) d \tau^2 + d \eta^2$$

You should be able to follow the same logic to get your problem done.

Here we get one Killing vector by just having a look at the metric; as its components are independent of $\tau$ we get $K_0=\partial_{\tau}$

1- Get all non-zero Christoffel symbols out of the given metric

For the $AdS_2$ we find there are only two (remember that the Christoffel symbols are symmetric under interchange of lower indeces):

$$\Gamma^{\tau}_{\tau \eta} =\tanh(\eta), \qquad \Gamma^{\eta}_{\tau \tau} = \cosh (\eta) \sinh (\eta)$$

2 - Have clear what's the most general equation for the Killing vectors you're looking for

In this case, the spacetime coordinates look like $x^{\mu} = (\eta, \tau)$, so we expect a Killing vector equation of the form

$$K = f(\eta, \tau) \partial_{\mu} + g(\eta, \tau) \partial_{\tau} \ \ \ \ (1)$$

Based on it, we expect the components of the killing vector equation to be

$$K_{\eta} = f(\eta, \tau), \qquad K_{\tau} = -\cosh^2 (\eta) g(\eta, \tau)$$

3- Determine how many independent components your Killing vector has and get their corresponding equations

$\nabla_{(\mu}K_{\nu)}=\partial_{(\mu}K_{\nu)}-\Gamma_{\mu \nu}^{\rho}K_{\rho}$ has three independent components in this case: $(2) (\eta, \eta), (3) (\tau, \tau), (4) (\eta, \tau)$. Explicitly we get these three equations:

$$\partial_{\eta} K_{\eta} = 0 \ \ \ \ (2)$$

$$\partial_{\tau} K_{\tau} - \Gamma_{\tau \tau}^{\eta} K_{\eta}= 0 \ \ \ \ (3)$$

$$\partial_{\tau} K_{\eta} + \partial_{\eta} K_{\tau} - 2\Gamma_{\tau \eta}^{\tau} K_{\tau}= 0 \ \ \ \ (4)$$

Plugging the components of the Killing vector and and the Christoffel symbols into $(2), (3), (4)$ we get

$$\partial_{\eta} f(\eta, \tau) = 0 \ \ \ \ (5)$$

$$\partial_{\tau} g(\eta, \tau) + \tanh(\eta) f(\eta, \tau) = 0 \ \ \ \ (6)$$

$$\partial_{\tau} f(\eta, \tau) - \cosh^2 (\eta) \partial_{\eta} g(\eta, \tau) = 0 \ \ \ \ (7)$$

4- Solve such equations

To solve $(5)$ we set $f(\eta, \tau)=f(\tau)$

Then plug $f(\tau)$ into $(6)$ and get

$$\partial_{\tau} g(\eta, \tau)=-\tanh\eta f(\tau)$$

Which can be solved by setting

$$g(\eta, \tau)=\tanh(\eta)h(\tau), \qquad h'(\tau)=-f(\tau) \ \ \ \ (*)$$

We now can go to equation $(7)$ and get

$$f'(\tau) - \cosh^2 (\eta) \partial_{\eta} (\tanh (\eta) h(\tau)) = f'(\tau)-h(\tau)$$

As $h'(\tau) = -f(\tau)$ then $h''(\tau) = -f'(\tau)$

Thus equation $(7)$ ends up having the form

$$h''(\tau)=-h(\tau)$$

Which is a linear homogeneous second order differential equation whose solutions are

$$h_1(\tau)=e^{i\tau}, \qquad h_2(\tau)=e^{-i\tau}$$

Plug these solutions into $(*)$ and get

$$f_1(\tau) = -ie^{i\tau}, \qquad g_1(\eta, \tau)=\tanh(\eta)e^{i\tau}$$

$$f_2(\tau) = ie^{-i\tau}, \qquad g_2(\eta, \tau)=\tanh(\eta)e^{-i\tau}$$

5- Plug the functions you've got into (1)

In this case we get two Killing vectors

$$K_1 = e^{i\tau}(-i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$

$$K_2 = e^{-i\tau}(i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$

6- Conclude

Thus we get 3 Killing vectors out of the $AdS_2$ metric

$$K_0=\partial_{\tau}$$

$$K_1 = e^{i\tau}(-i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$

$$K_2 = e^{-i\tau}(i\partial_{\eta} + \tanh (\eta) \partial_{\tau})$$