Torus Killing vectors

Solution 1:

Look at your "final differential equation". Take derivative w.r.t. $\phi$. The RHS vanishes. The LHS becomes

$$ f''(\phi) + (1 - \frac{b}{a} \sin\theta ) f(\phi) = 0.$$

Taking now the $\theta$ derivative, you find $$ 0 - \frac{b}{a} \cos(\theta) f(\phi) = 0 $$

This equation is required to hold everywhere, and specifically when $\theta = 0$, which requires $f \equiv 0$. (In the case where $b = 0$ you no longer have a torus, and you do have other solutions!)

Plugging this into (1) you find that $\xi_\phi = h(\theta)$. And plugging this into (2) you find that

$$ h' = \frac{2a \cos\theta}{b + a\sin\theta} h $$

which is a separable ODE which you easily solve to find

$$ h(\theta) = C (b+a\sin\theta)^2 $$

So in conclusion:

the only Killing vector field of the 2D torus is the obvious one, corresponding to $\partial_\phi$.