I Finding Killing Vectors for $$ds^2 = dr^2 + r^2d\theta^2$$

[davidge]

I have tried to find the three Killing vectors for the metric $$ds^2 = dr^2 + r^2d \theta^2$$ that is, the Euclidean metric of $\mathbb{R}^2$ written in polar coordinates. I found these to be

$$\bigg(\text{first}\bigg) \ \ \xi_r = \text{Cos} \theta \\
\xi_\theta = -\text{rSin} \theta \\

\bigg(\text{second}\bigg) \ \ \xi_r = \text{Sin} \theta \\
{\xi_\theta = \text{rCos} \theta} \\

\bigg(\text{third}\bigg) \ \ \xi_r = 0 \\
\xi_\theta = \text{r²}$$ As I have found solutions only for 3d on web, I would like to know whether these are correct or not.

 

[Orodruin]

Why don't you check whether or not they satisfy the Killing equations?

 

[davidge]

Why don't you check whether or not they satisfy the Killing equations?

I did
And they do satisfy the Killing equation.

 

[Orodruin]

I did
And they do satisfy the Killing equation.

And thus they are Killing vector fields ...

 

[davidge]

And thus they are Killing vector fields ...

 

[davidge]

What bothers me is that in 2d we should have only two independent vectors. So I should be able to get one of those three above by a linear combination of the other two, but when I do that, I get non constant coefficients multiplying them.

 

[Orodruin]

These are vector fields, not vectors.

 

[davidge]

These are vector fields, not vectors.

So they are'nt vectors? Can you say a bit more on this please

 

[Orodruin]

There is no such thing as a "Killing vector". The Killing equation is a differential equation and as such describes vector fields, ie, assignments of one vector to each point in the space.

 

[davidge]

assignments of one vector to each point in the space

For instance, what could be such one vector?

 

[Orodruin]

You wrote down several vector fields (in coordinate basis) in the firs post.

 

[davidge]

You wrote down several vector fields (in coordinate basis) in the firs post.

But you say they aren't vectors. I asked for an example of assigment of a vector by a vector field

 

[Orodruin]

Your field is an assignment of a (dual) vector to every point in space!

For example, for $\theta = 0$ (and arbitrary r) your first field takes the value $\xi = dr$, where $dr$ is the coordinate basis dual vector.

 

[davidge]

Your field is an assignment of a (dual) vector to every point in space!

For example, for $\theta = 0$ (and arbitrary r) your first field takes the value $\xi = dr$, where $dr$ is the coordinate basis dual vector.

I got it. Thanks.

 

[davidge]

Is it correct to say that they are three independent vector fields? In the sense that one cannot be expressed as a multiple of another one.

Also, if we evaluate any of them at a particular point $(r, \theta)$ do they form three linearly dependent vectors?