Normal Bundle of Twistor lines I am reading the paper "hyperkaehler metrics and supersymmetry" by Hitchin etc.. Here is the link: Link . On page 555 it is said that the normal bundle of the twistor line is $T_{m} \times S^{2}$, where $T_{m}$ is the tangent space at $m \in M$. I dont quiet see this. How is this to be understood? In my opinion the normal bundle is $TZ|_{s(\mathbb{C}P^{1})} / T(\{m\} \times \mathbb{C}P^{1})$, where $s : \mathbb{C}P^{1} \rightarrow Z = M \times \mathbb{C}P^{1}, \lambda \mapsto (m,\lambda)$ is the twistor line. Hence locally (at point $(m,\lambda) \in s(\mathbb{C}P^{1})$ ) we get $T_{(m,\lambda)}Z / T(\{m\}) \oplus T_{\lambda}\mathbb{C}P^{1} = T_{m}M \oplus T_{\lambda}\mathbb{C}P^{1} / \{0\} \oplus T_{\lambda}\mathbb{C}P^{1} = T_{m}M$. I dont understand how the $S^{2} \cong \mathbb{C}P^{1}$ is appearing there? Does anyone have an idea?
thanks
monica
 A: Without wanting to teach Grandma to suck eggs, given a hyperkaehler manifold $M^{4n}$ with metric $g$, complex structures $I, J$ and $K$, and symplectic forms $\omega_I$ , $\omega_J$ and $\omega_K$ that are all compatible, in the sense
that $IJ = K$ and
\begin{align}
g(I\ast,\ast) &= g(\ast,\ast)\\
g(J\ast,\ast) &= g(\ast,\ast)\\
g(K\ast,\ast) &= g(\ast,\ast)
\end{align}we can define what is called the twistor space of $M$. The main observation that allows us to construct this space, is the following
$$(aI+bJ+cK)^2=-(a^2+b^2+c^2)$$
for real numbers $a, b$ and $c$.
In other words, if we have $a^2 + b^2 + c^2 = 1,$ then $aI + bJ + cK$ defines a new complex structure. Such $(a, b, c)$ form a sphere, which we can give a complex structure $I_0$ through the standard one-point compactification of the complex plane. The complex coordinate of this sphere will be $\zeta$. Then we are free to pick
$$(a, b, c) = \frac{1}{1 + |\zeta|^2}(1 − |\zeta|^2, 2\Re(\zeta), −2\Im(\zeta))$$
The twistor space is then $Z=M \times S^2$. Now onto the Normal Bundle.
Since $Z = M \times S^2$, there exists an obvious map $p : M \times S^2 \to S^2$. Actually, this projection extends to a holomorphic map $p : Z \to C\mathbb{P}^1$, which makes $Z$ into a holomorphic fiber bundle over $C\mathbb{P}^1$. 
The bundle admits a family of holomorphic sections, namely $\sigma_m : \zeta \to (m, \zeta)$, the standard embedding into $M × S^2$. The sections of the map $p$ are then the twistor lines, $P_m = im(σ_m)$.
With respect to this embedding, we can define the normal bundle $N$ of $P_m \subset M × S^2$. This bundle fits into the short exact sequence
$$0 \to TC\mathbb{P}^1\to T Z|_{P_m} \to N \to 0$$
Topologically, it is clear that $$T Z|P_m = N \oplus TC\mathbb{P}^1 = T_m M \oplus TC\mathbb{P}^1$$ with the normal bundle $N = T_m M$ over $P_m$ trivial
