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--- lie_derivative [2023/06/19 12:46] – [In terms of a covariant derivative] spencer
+++ lie_derivative [2023/06/19 12:52] (current) – [In terms of a covariant derivative] spencer
@@ Line 35: / Line 35: @@
 That is, in this basis, we have that $Q\omega^j = \lambda_j \omega^j$.
-Thus we have
+Thus we have for $i < n$ that
 \begin{align*}
-L_{\partial_r} (a_i \omega^i) &= \nabla_{\partial_r} (a_i \omega^i) + Q(a_i \omega^i) \\
+L_{\partial_r} (\omega^i) &= \nabla_{\partial_r} \omega^i + \lambda^i \omega^i.
-&= \sum_{i=1}^{n-1} (\partial_r a_i + \lambda^i a_i) \omega^i + (\partial_r a_n) dr + \sum_{i=1}^n a_i \nabla_{\partial_r} \omega^i.
 \end{align*}
+Setting $\nabla_{e_i} e_j = \gamma^k_{ij} e_k$, then $\nabla_{e_i} (\omega^j (e_k)) = (\nabla_{e_i} \omega^j)(e_k) + \omega^j (\nabla_{e_i} e_k)$.
+Thus
+$$(\nabla_{e_i} \omega^j)(e_k) = -\omega^j( \gamma^\ell_{ik} e_\ell) = -\gamma^j_{ik}.$$
+Thus
+$$\nabla_{e_i} \omega^j = -\gamma^j_{ik} \omega^k.$$
+So
+$$L_{\partial_r} (\omega^i) = -\gamma^i_{nk} \omega^k + \lambda^i \omega^i.$$