Why are there only six linearly independent global conformal Killing vector fields on the two-sphere with the round metric?

Question

The title pretty much sums it up. For a 2-manifold with metric $\gamma_{AB}$, a vector field $Y^A$ is said to be a conformal Killing vector field if $\nabla_A Y_B + \nabla_B Y_A = \nabla_C Y^C \gamma_{AB}$. In the case of a two-sphere with the round metric, this means $\partial_{\bar{z}} Y^z = \partial_z Y^{\bar{z}} = 0$, where $z$ is the stereographic coordinate. The solutions to $Y^z$ is any holomorphic function of $z$, and for $Y^{\bar{z}}$ we have any holomorphic function of $\bar{z}$. Yet, I've seen references (such as arXiv: 1703.05448 [hep-th]) mention that there are only six linearly independent global conformal Killing vector fields: those with $Y^z = 1, z, z^2, i, iz, iz^2$ (and I suppose $Y^{\bar{z}}$ is given by the conjugate?). My question is then: why are these the only global conformal Killing vector fields? Why isn't, for example, $z^3$ a global conformal Killing vector field as well?

Answer 1

In the context of the Riemann sphere, $\Bbb S^2$, we can think of the standard coordinate $z$ as defining a coordinate chart $z : \Bbb S^2 \setminus \{N\} \to \Bbb C$ via stereographic projection with respect to the North Pole, $N$. In this language, we're interested in which conformal Killing fields of $\Bbb S^2 \setminus \{N\}$ (i.e., holomorphic vector fields on $\Bbb C$) can be extended continuously to $\Bbb S^2$.

To analyze the behavior of the vector fields at $N$, define a complementary coordinate $\zeta : \Bbb S^2 \setminus \{S\} \to \Bbb C$ by stereographic projection with respect to the South Pole, $S$. The transition map between the coordinates identifies $\zeta = \frac{1}{z}$, and the North Pole has coordinate $\zeta = 0$. In particular, in the $\zeta$ coordinate we have $$z^k \partial_z = -\zeta^{2 - k} \partial_\zeta .$$ So, the vector field $z^k \partial_z$ on $\Bbb S^2 \setminus \{N\}$, $k \in \Bbb Z_{\geq 0}$, can only be extended smoothly to $N$ ($\zeta = 0$), hence to all of $\Bbb S^2$, if $k \in \{0, 1, 2\}$.

The conformal Killing fields of a metric form a Lie algebra under the usual Lie bracket of vector fields, and we can identify that Lie algebra readily. The Riemann sphere (more explicitly, the $2$-sphere endowed with the conformal structure determined by the round metric) can be identified with the space of isotropic rays in Lorzentian space, $\Bbb R^{1, 3}$, and the faithful action of the Lorenztian group, $\operatorname{SO}(1, 3)$, preserves the conformal structure that space induces on $\Bbb S^2$. Differentiating the action thus identifies the $6$-dimensional Lie algebra $\mathfrak{so}(1, 3)$ with a Lie algebra of conformal Killing fields on $\Bbb S^2$. Just as well we can think of the Riemann sphere as the complex projective line $\Bbb C P^1$, which enjoys a group action by $\operatorname{PSL}(2, \Bbb C)$ preserving the conformal structure, and these two pictures are identified by the exceptional (real) Lie group isomorphism $\operatorname{PSL}(2, \Bbb C) \cong \operatorname{SO}^+(1, 3)$, or just as well the (real) Lie algebra isomorphism $\operatorname{sl}(2, \Bbb C) \cong \mathfrak{so}(1, 3)$.