How to find Killing vector field? I am studying about Killing vector from an online source as follows:

I couldn't obtain the six equations (50 - 55) from equation (49).
Also, I couldn't find $\partial^2_{x} \xi^{y}=0$ as mentioned (by applying $\partial_{x}$ to equation (52) and use (51)).
May someone show the detail steps to solve the six Killing vectors. Thank you so much.
 A: In the usual three dimensional Euclidean space in Cartesian coordinates, one has
$g_{\mu \sigma} = \delta_{\mu \sigma}$.
This is a constant, so its partial derivatives are zero.
This leads to the equalities
$g_{\mu \sigma} \partial_{\rho}\xi^{\mu} = \partial_{\rho}\xi^{\sigma}$,
$g_{\rho \nu}\partial_{\sigma} \xi^{\nu}= \partial_{\sigma}\xi^{\rho}$,
and
$\partial_{\mu}g_{\rho \sigma}= 0$.
So that (49) becomes, in this specific context
$$
\forall \rho, \sigma \in \{x,y,z\}, \quad \partial_{\rho}\xi^{\sigma} + \partial_{\sigma}\xi^{\rho} = 0.
$$
Setting $\rho = \sigma = x$ gives (twice) $(50)$.
Setting $\rho = x$, $\sigma = y$ gives $(51)$.
Setting $\rho = x$, $\sigma = z$ gives $(52)$.
Setting $\rho = \sigma = y$ gives (twice) $(53)$.
Setting $\rho = z$, $\sigma = y$ gives $(54)$.
Finally, setting $\rho= \sigma = z$ gives (twice) $(55)$.
Regarding the last claim: there is a typo in the document.
You should use $(50)$ (and not $(51)$) to obtain $\partial_x^2 \xi^y = 0$, after having applied $\partial_x$ to $(51)$ (and not $(52)$).
Indeed, partial derivatives commute, so that $\partial_x (\partial_y \xi^x) = \partial_y (\partial_x \xi^x) = 0$ by $(50)$.

Edit.
Let $\xi$ be a solution.
Then as we have seen, we have
$$
\partial_x\xi^x = \partial_y\xi^y = \partial_z\xi^z = 0
$$
and
$$
\partial^2_x\xi^y = \partial^2_x \xi^z
= \partial^2_y\xi^x = \partial^2_y\xi^z
= \partial^2_z\xi^x = \partial^2_z\xi^y
=0.
$$
This means that:

*

*$\xi^x$ is independent of $x$, and is an affine function of $y$ and $z$,

*$\xi^y$ is independent of $y$, and is an affine function of $x$ and $z$,

*$\xi^z$ is independent of $z$, and is an affine function of $x$ and $y$.

Thus, there are nine constants $a^x,b^x,c^x,a^y,b^y,c^y,a^z,b^z$ and $c^z$ such that
\begin{align}
\xi^x &= a^x\phantom{x} + b^x y + c^x z, \\
\xi^y &= a^y x + b^y \phantom{y} + c^y z, \\
\xi^z &= a^z x + b^z y + c^z \phantom{z}.
\end{align}
But equations $(51)$, $(52)$ and $(54)$ now give
$$
a^y = -b^x, \quad a^z = -c^x, \quad \text{and} \quad b^z = -c^y.
$$
Therefore, there exist six constants $c_1,c_2,\ldots,c_6$ such that
$$
\begin{align}
\xi^x &= \phantom{-} c_1\phantom{x}  -c_4 y + c_6 z, \\
\xi^y &= \phantom{-} c_4 x + c_2 \phantom{y} - c_5 z, \\
\xi^z &= -c_6 x + c_5 y + c_3 \phantom{z}.
\end{align}
$$
The set of Killing vector fields is thus contained in the six dimensional space spanned by the vector fields $\{\xi_1,\ldots,\xi_6\}$, where $\xi_j$ is chosen such that $c_j = 1$ and $c_k = 0$ if $k\neq j$.
You can check that indeed, these vector fields are Killing, so that we have found all of them.
This is geometrically clear what these vector fields represent:
$\xi^1,\xi^2$ and $\xi^3$ generate constant speed translations along the $x$-axis, $y$-axis and $z$-axis respectively, while $\xi^4$, $\xi^5$ and $\xi^6$ generate constant speed rotations about the $z$-axis, $x$-axis and $y$-axis respectively.
Note that this was expectable: we know that the isometries of the affine space $\Bbb R^3$ are precisely the rotations-translations.
So we could have guessed in the first place that these six linearly-independent vector fields are Killing by just checking that they satisfy the Killing field equation.
Since the space of Killing vector fields is of dimension at most $6(=\frac{3(3+1)}{2})$, this would have completed the proof, without solving any PDE!
