【Spin幾何】曲率作用素とLichnerowiczの公式

(1)
\begin{align} 4\sum_ie_i\cdot R(e_i,X)(\psi)&=\sum_{i,k,l}g(R(e_i,X)e_k,e_l)e_ie_ke_l\psi\\ &=\sum_{i,l,k:{\rm differ}}g(R(e_i,X)e_k,e_l)e_ie_ke_l\psi\ \ ({\rm vanish\ since\ 1st\ Jacobi\ id})\\ &+\sum_{i,l:{\rm differ}}g(R(e_i,X)e_i,e_l)e_ie_ie_l\psi\\ &+\sum_{i,k:{\rm differ}}g(R(e_i,X)e_k,e_i)e_ie_ke_i\psi\\ &=2Ric(X,e_k)e_k\psi=2Ric(X)\cdot\psi \end{align}

(2)
$$ \sum_je_j\cdot Ric(e_j)\cdot\psi=2\sum_{i,j}e_je_i R(e_i,e_j)(\psi)=\frac{1}{2}\sum_{i,j,k,l}g(R(e_i,e_j)e_k,e_l)e_je_ie_ke_l $$
($ i\ne j,k\ne l$のものしか考えなくてよい)
$$ =-\frac{1}{2}\sum_{i,j,l}g(R(e_i,e_j)e_j,e_l)e_ie_je_je_l-\frac{1}{2}\sum_{i,l}\sum_{j\ne k}g(R(e_i,e_j)e_k,e_l)e_ie_je_ke_l $$
($ k=j$と$ k\ne j$で別けた)
$$ =\frac{1}{2}\sum_{i,l}Ric(e_i,e_l)e_ie_l-\frac{1}{2}\sum_{i,j,l}g(R(e_i,e_j)e_i,e_l)e_ie_je_ie_l-\frac{1}{2}\sum_l\sum_{i,j,k:{\rm differ}}g(R(E_i,E_j)e_k,e_l)e_ie_je_ke_l $$
(第二項目を$ k=i$と$ k\ne i$で別けた）
$$ =-\frac{1}{2}R-\frac{1}{2}\sum_{i,j}g(R(e_i,e_j)e_i,e_j)e_ie_je_ie_j-\frac{1}{2}\sum_{i,j,l:{\rm differ}}g(R(e_i,e_j)e_i,e_l)e_ie_je_ie_l $$
(第二項目を$ l=j$と$ l\ne j$で別けた）
$$ =-\frac{1}{2}R-\frac{1}{2}R=-R $$

公式１(1)より
\begin{align} &\frac{1}{2}Ric(X)\cdot\psi=\sum_ie_i\cdot R(e_i,X)(\psi)\\ &=\sum_ie_i(\nabla_{e_i}\nabla_X\psi-\nabla_X\nabla_{e_i}\psi-\nabla_{[e_i,X]}\psi)\\ &=D(\nabla_X\psi)-\nabla_X(D\psi)+\nabla_Xe_i\cdot\nabla_{e_i}\psi-e_i\nabla_{[e_i,X]}\psi\\ &=D(\nabla_X\psi)-\nabla_X(D\psi)+\nabla_Xe_i\cdot\nabla_{e_i}\psi-e_i\nabla_{\nabla_{e_i}X}\psi+e_i\nabla_{\nabla_Xe_i}\psi\\ &=D(\nabla_X\psi)-\nabla_X(D\psi)+\omega^j_{~i}(X)e_j\cdot\nabla_{e_i}\psi-e_i\nabla_{\nabla_{e_i}X}\psi+e_i\nabla_{\nabla_Xe_i}\psi\\ &=D(\nabla_X\psi)-\nabla_X(D\psi)-e_j\cdot\nabla_{\omega^i_{~j}(X)e_i}\psi-e_i\nabla_{\nabla_{e_i}X}\psi+e_i\nabla_{\nabla_Xe_i}\psi\\ &=D(\nabla_X\psi)-\nabla_X(D\psi)-e_j\cdot\nabla_{\nabla_Xe_j}\psi-e_i\nabla_{\nabla_{e_i}X}\psi+e_i\nabla_{\nabla_Xe_i}\psi\\ &=D(\nabla_X\psi)-\nabla_X(D\psi)-e_i\nabla_{\nabla_{e_i}X}\psi \end{align}
である。($\sum$を書くの面倒なので省略)

1/2Ricci公式に$ e_i$をかけて$ X=e_i$として縮約すると
\begin{align} \sum_ie_i\frac{1}{2}Ric(e_i)\cdot\psi&=\sum_ie_iD(\nabla_{e_i}\psi)-\sum_ie_i\nabla_{e_i}(D\psi)-\sum_{i,j}e_ie_j\cdot\nabla_{\nabla_{e_j}e_i}\psi\\ \frac{1}{2}R\psi&=\sum_{i,j}e_ie_j\nabla_{e_j}\nabla_{e_i}\psi-D^2\psi-\sum_{i,j}e_ie_j\nabla_{\nabla_{e_j}e_i}\psi\\ &=\sum_i(-\nabla_{e_i}\nabla_{e_i}\psi+\nabla_{\nabla_{e_i}e_i}\psi)-D^2\psi+\sum_{i\ne j}e_ie_j(\nabla_{e_j}\nabla_{e_i}\psi-\nabla_{\nabla_{e_j}e_i}\psi) \end{align}
となる。ここで
\begin{align} &\sum_{i\ne j}e_ie_j(\nabla_{e_j}\nabla_{e_i}\psi-\nabla_{\nabla_{e_j}e_i}\psi)=\frac{1}{2}\sum_{i\ne j}e_ie_jR(e_j,e_i)\psi\\ &=\frac{1}{2}\sum_{i, j}e_ie_jR(e_j,e_i)\psi=\frac{1}{4}\sum_ie_iRic(e_i)\psi=-\frac{1}{4}R\psi \end{align}
であるから、
\begin{align} -\frac{1}{2}R\psi&=\Delta\psi-D^2\psi-\frac{1}{4}R\psi\\ \therefore\ D^2\psi&=\Delta\psi+\frac{1}{4}R\psi \end{align}
である。