I'm confused about the number of Killing vectors in Schwarzschild metric I'm trying to perform a calculation to derive the Killing vectors of a spherically symmetric metric (so I use the Schwarzschild metric without loss of generality because the Birkhoff theorem tells me that it's the only static spherically symmetric metric that solves Einstein equations in vacuum).
I want to solve the equation:
\begin{equation}
\xi_{\alpha;\beta}+\xi_{\alpha;\beta}=0
\end{equation}
I know that the metric has symmetries with respect time and the three independent rotations around x,y,z axis, so I should expect four Killing vectors that are solution of this equation.
From an independent calculation I know that a generic stationary metric has a time-like killing vector:
\begin{equation}
\xi^{\alpha}=(1,0,0,0)
\end{equation}
and if i lower the index using Schwarzschild Metric (with +2 signature) I get:
\begin{equation}
\xi_{\alpha}=\left( -\left(1-\frac{R_{S}}{r}\right),0,0,0\right)
\end{equation}
Next I write down all the nine differential equations for the four component of the killing vectors, putting inside the nine non-vanishing connections of the metric. The system is not too difficult to solve, since though they are coupled, they are not all independent. In particular one on them holds:
\begin{equation}
\frac{\partial{\xi_{t}}}{\partial{t}}=\frac{R_{s}}{2}\left(1-\frac{R_{s}}{r}\right) \xi_{r}
\end{equation}
so $\xi_{r}$ must be zero because $\xi_{t}$ is independent from t. With similar considerations I also get $\xi_{\theta}=0$, so the only non vanishing components of a general killing vector are t and $\phi$. (which I could define in the beginning becuse these are the only two coordinates the metric doesn't depend on).
The equations left lead to solve the two coupled equations:
\begin{equation}
\frac{\partial^2{\xi_{\phi}}}{\partial{r}\partial{\theta}} =2\frac{cos(\theta)}{\sin(\theta)} \frac{\partial{{\xi}_{\phi}}}{\partial{r}}
\end{equation}
and
\begin{equation}
\frac{\partial^2{\xi_{\phi}}}{\partial{\theta}\partial{r}} =\frac{2}{r} \frac{\partial{{\xi}_{\phi}}}{\partial{\theta}}
\end{equation}
which leads to solve the equation:
\begin{equation}
\frac{cos(\theta)}{sin(\theta)}\frac{\partial{{\xi}_{\phi}}}{\partial{r}}=\frac{1}{r} \frac{\partial{{\xi}_{\phi}}}{\partial{\theta}}
\end{equation}
using separation of variables I find this solution:
\begin{equation}
{\xi}_{\phi}(r,\theta)= \left(A_{k}r^{k}+B_{k}\right) \left(C_{k}sin(k\theta)+D_{k}\right)
\end{equation}
My question is the following: is there a way to fix the constants A,B,C,D to get three independent killing vectors?
I don't know if my calculations are correct, but in my opinion there's no way  my solution can have three independet vectors with a single coordinate ($\phi$)"free". Hope someone can help.
 A: A brute force (and ugly) derivation of the Killing fields of Schwarzschild metric
The Schwarzschild metric is
\begin{equation}
  ds^2 = -\left(1-\frac{R_{\text{S}}}{r}\right) \text{d} t^2 + \left(1-\frac{R_{\text{S}}}{r}\right)^{-1} \text{d} r^2 + r^2 (\text{d} \theta^2 + \sin^2\theta \,\text{d} \phi^2).
 \end{equation}
The Christoffel symbols are
\begin{align*}
  \Gamma^t_{tr} &= \frac{R_{\text{S}}}{2r(r-R_{\text{S}})}, & \Gamma^r_{tt} &= \frac{R_{\text{S}}}{2r^3}(r-R_{\text{S}}), & \Gamma^r_{rr} &= \frac{-R_{\text{S}}}{2r(r-R_{\text{S}})}, \\
  \Gamma^\theta_{r\theta} &= \frac{1}{r}, & \Gamma^r_{\theta\theta} &= -(r-R_{\text{S}}), & \Gamma^\phi_{r\phi} &= \frac{1}{r}, \\
  \Gamma^r_{\phi\phi} &= -(r-R_{\text{S}}) \sin^2 \theta, & \Gamma^\theta_{\phi\phi} &= -\sin\theta \cos\theta, & \Gamma^\phi_{\theta\phi} &= \frac{\cos\theta}{\sin\theta}.
 \end{align*}
The Killing equations are
\begin{align}
   \tag{1a} &K_{t,t} - \Gamma^r_{tt} K_r = 0, \\
   \tag{1b} &K_{t,r} + K_{r,t} - 2 \Gamma^t_{tr} K_t = 0, \\
   \tag{1c} &K_{t,\theta} + K_{\theta,t} = 0, \\
   \tag{1d} &K_{t,\phi} + K_{\phi,t} = 0, \\
   \tag{1e} &K_{r,r} - \Gamma^r_{rr} K_r = 0, \\
   \tag{1f} &K_{r,\theta} + K_{\theta,r} - 2 \Gamma^\theta_{r\theta} K_\theta = 0, \\
   \tag{1g} &K_{r,\phi} + K_{\phi,r} - 2 \Gamma^\phi_{r\phi} K_\phi = 0, \\
   \tag{1h} &K_{\theta,\theta} - \Gamma^r_{\theta\theta} K_r = 0, \\
   \tag{1i} &K_{\theta,\phi} + K_{\phi,\theta} - 2 \Gamma^\phi_{\theta\phi} K_\phi = 0, \\
   \tag{1j} &K_{\phi,\phi} - \Gamma^r_{\phi\phi} K_r - \Gamma^\theta_{\phi\phi} K_\theta = 0.
  \end{align}
From ($1$e) we have
\begin{equation}\tag{2}
  K_r = T(t,\theta,\phi) \left( \frac{r}{r-R_{\text{S}}} \right)^{1/2}.
 \end{equation}
Differentiating ($1$b) with respect to $t$ and substituting ($1$a), ($1$e) and ($2$) into the result, we obtain
\begin{equation}\tag{3}
  \left( \frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} \right) T(t,\theta,\phi) + \frac{\partial^2}{\partial t^2} T(t,\theta,\phi) = 0.
 \end{equation}
Since
\begin{equation}
  \frac{\partial}{\partial r} \Gamma^r_{tt} + 3 \Gamma^r_{tt} \Gamma^r_{rr} = \frac{R_{\text{S}}}{r^3}\left( -1 + \frac{3R_{\text{S}}}{4r} \right)  
 \end{equation}
is a function of $r$ only, and is not identically zero, ($3$) holds only when $T(t,\theta,\phi) \equiv 0$.  Then we have $K_r \equiv 0$.  So we can simplify the Killing equations to
\begin{align}
   \tag{4a} &K_{t,t} = 0, \\
   \tag{4b} &K_{t,r} = 2 \Gamma^t_{tr} K_t, \\
   \tag{4c} &K_{t,\theta} + K_{\theta,t} = 0, \\
   \tag{4d} &K_{t,\phi} + K_{\phi,t} = 0, \\
   \tag{4e} &K_{\theta,r} = 2 \Gamma^\theta_{r\theta} K_\theta, \\
   \tag{4f} &K_{\phi,r} = 2 \Gamma^\phi_{r\phi} K_\phi, \\
   \tag{4g} &K_{\theta,\theta} = 0, \\
   \tag{4h} &K_{\theta,\phi} + K_{\phi,\theta} = 2 \Gamma^\phi_{\theta\phi} K_\phi, \\
   \tag{4i} &K_{\phi,\phi} = \Gamma^\theta_{\phi\phi} K_\theta.   
  \end{align}
From ($4$a), ($4$b), ($4$e), ($4$f) and ($4$g) we have
\begin{align}
  K_t &= A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right), \\
  K_\theta &= B(t,\phi) \,r^2, \\
  K_\phi &= C(t,\theta,\phi) \,r^2.
 \end{align}
Substituting these results into ($4$c) and ($4$d), we obtain
\begin{align}
  \frac{\partial}{\partial\theta} A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right) + \frac{\partial}{\partial t} B(t,\phi) \,r^2 = 0, \\
  \frac{\partial}{\partial\phi} A(\theta,\phi) \left( 1 - \frac{R_{\text{S}}}{r} \right) + \frac{\partial}{\partial t} C(t,\theta,\phi) \,r^2 = 0.
 \end{align}
These equations hold only when $A$ is a constant and $B$ and $C$ are independent of $t$.  So we have
\begin{align}
   K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\
   K_\theta &= B(\phi) \,r^2, \\
   K_\phi &= C(\theta,\phi) \,r^2.
  \end{align}
Substituting $K_\theta$ and $K_\phi$ into ($4$h) and ($4$i), we obtain
\begin{align}
  \frac{\partial B}{\partial\phi} + \frac{\partial C}{\partial\theta} &= 2 \frac{\cos\theta}{\sin\theta} \,C, \\
  \frac{\partial C}{\partial\phi} &= -\sin\theta \cos\theta \,B.
 \end{align}
We can easily solve these PDEs to get
\begin{align}
  B(\phi) &= -D \sin\phi + E \cos\phi, \\
  C(\theta,\phi) &= -\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta,
 \end{align}
where $D,E,F$ are constants.  In summary, we have
\begin{align}
   K_t &= A \left( 1 - \frac{R_{\text{S}}}{r} \right), \\
   K_r &= 0, \\
   K_\theta &= (-D \sin\phi + E \cos\phi) \, r^2, \\
   K_\phi &= [-\sin\theta \cos\theta (D \cos\phi + E \sin\phi) + F \sin^2\theta] \, r^2.
  \end{align}
The general solution of the Killing equations of the Schwarzschild metric will be
\begin{equation}
  \begin{split}
   K &= g^{\mu\nu} K_\mu \partial_\nu \\
   &= g^{tt} K_t \frac{\partial}{\partial t} + g^{\theta\theta} K_\theta \frac{\partial}{\partial \theta} + g^{\phi\phi} K_\phi \frac{\partial}{\partial \phi} \\
   &= - A \frac{\partial}{\partial t} +(-D \sin\phi + E \cos\phi) \frac{\partial}{\partial \theta} + [ - \cot\theta (D \cos\phi + E \sin\phi) + F ] \frac{\partial}{\partial \phi} \\
   &= - A L_{(0)} + D L_{(1)} + E L_{(2)} + F L_{(3)},
  \end{split}
 \end{equation}
where
\begin{align}
   L_{(0)} &= \frac{\partial}{\partial t}, \\
   L_{(1)} &= -\sin\phi \frac{\partial}{\partial \theta} - \cot\theta \cos\phi \frac{\partial}{\partial \phi}, \\
   L_{(2)} &= \cos\phi \frac{\partial}{\partial \theta} - \cot\theta \sin\phi \frac{\partial}{\partial \phi}, \\
   L_{(3)} &= \frac{\partial}{\partial \phi}
  \end{align}
form a basis of the Lie algebra of the Killing fields of Schwarzschild metric.  $L_{(0)}$ is a timelike Killing vector field that is orthogonal to a foliation of spacelike hypersurfaces, representing a static spacetime.  $L_{(1)}, L_{(2)}, L_{(3)}$ are Killing fields of a 2-sphere, representing a spacetime with a SO(3) symmetry.
