# Solution to Killing equations

## Homework Statement

Find all Killing vector solutions of the metric
$$g_x{_x}=x^2, g_x{_y}=g_y{_x}=0, g_y{_y}=x$$
where $(x^a)=(x^0, x^1)=(x, y)$

## Homework Equations

Killing equations:
$$L_Xg_a{_b} = X^e\partial_eg_a{_b}+g_a{_d}\partial_bX^d+g_b{_d}{\partial}_aX^d = 0$$

## The Attempt at a Solution

$$L_Xg_x{_x} = X^a+x\partial_xX^a=0$$
$$L_Xg_x{_y}= L_Xg_y{_x}=x\partial_yX^a+\partial_xX^a=0$$
$$L_Xg_y{_y}=X^a+2x\partial_yX^a=0$$
In the back of the book, it says the solution is $\frac{\partial}{{\partial}y}$. I don't really know what they mean by that. I've always seen $\frac{\partial}{{\partial}y}$ as an operation that takes the partial derivative of something with respect to y, not a value. I thought perhaps it meant any function that depends only on y and not on x, but if I plug that into the equations above, I get:
$$f(y)\neq0$$
$$x\partial_yf\neq0$$
$$f(y)+2x\partial_yf\neq0$$
I'm pretty sure I've got the equations correct; I just don't know what they mean when they say the solution is $\frac{\partial}{{\partial}y}$.

The equations in the first post are wrong. The equations should be:$$L_Xg_x{_x}=2xX^x+2x^2\partial_xX^x=0$$$$L_Xg_x{_y}=L_Xg_y{_x}=x^2\partial_yX^x+x\partial_xX^y=0$$$$L_Xg_y{_y}=X^x+2x\partial_yX^y=0$$
The first equation restricts you to $X^x=\frac{1}{x}f(y)$, and anything I do with the second and third equations, I come up with $-\frac{1}{2x^3}f(y)=\partial_y\partial_yf(y)$. Obviously, there is no solution to this except $f(y)=0$, so I'm still finding the only possible solution to be $X^x=0, X^y=c$.

I found out that $\frac{\partial}{{\partial}y}$ indeed means any function of y. I'm not sure why they would offer that as a solution when there are 2 vectors I am solving for. Can anyone figure out why $\frac{\partial}{{\partial}y}$ is a solution in the back of the book?

Last edited:
Dick