【スピン幾何】twisorスピノルと共形対称性

$$\begin{array}{rl}& e_i(g(\xi ,e_j))=\sum _i\alpha ⟨{\mathrm{\nabla }}_{e_i}\psi ,{\gamma }_j\psi ⟩+\sum _i\alpha ⟨\psi ,({\mathrm{\nabla }}_{e_i}{e}_j)\psi ⟩+\sum _i\alpha ⟨\psi ,{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi ⟩\\ \therefore & g({\mathrm{\nabla }}_{e_i}\xi ,e_j)=\sum _i\alpha ⟨{\mathrm{\nabla }}_{e_i}\psi ,{\gamma }_j\psi ⟩+\sum _i\alpha ⟨\psi ,{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi ⟩\\ & =\delta \sum _i\alpha ⟨{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi ,\psi ⟩+\sum _i\alpha ⟨{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi ,\psi {⟩}^{\ast }(\delta は\pm 1のどちらか)\\ & =\{\begin{array}{l}2\mathrm{Re}\sum _i\alpha ⟨{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi ,\psi ⟩(\delta =1)\\ -2\mathrm{Im}\sum _i\alpha ⟨{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi ,\psi ⟩(\delta =-1)\end{array}\end{array}$$
となる。以下$\delta =1$のときを議論する（$\delta =-1$でも同様）。$\psi$がtwistorスピノルなので、
$$\begin{array}{rl}& g({\mathrm{\nabla }}_{e_i}\xi ,e_j)+g({\mathrm{\nabla }}_{e_j}\xi ,e_i)=2\mathrm{Re}\sum _i\alpha ⟨{\gamma }_j{\mathrm{\nabla }}_{e_i}\psi +{\gamma }_i{\mathrm{\nabla }}_{e_j}\psi ,\psi ⟩\\ & =-\frac{2}{n}\mathrm{Re}\sum _i\alpha ⟨{\gamma }_j{\gamma }_{i}D\psi +{\gamma }_i{\gamma }_{j}D\psi ,\psi ⟩=\frac{4}{n}g(e_i,e_j)\mathrm{Re}\sum _i\alpha ⟨D\psi ,\psi ⟩\end{array}$$
が成り立ち、
$$\rho :=\frac{2}{n}\mathrm{Re}\sum _i\alpha ⟨D\psi ,\psi ⟩$$
とおくと
$$g({\mathrm{\nabla }}_{e_i}\xi ,e_j)+g({\mathrm{\nabla }}_{e_j}\xi ,e_i)=2\rho g(e_i,e_j)$$
を満たすから、$\xi$は共形Killingベクトルである。