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--- lie_derivative [2023/06/05 13:02] – created spencer
+++ lie_derivative [2023/06/19 12:52] (current) – [In terms of a covariant derivative] spencer
@@ Line 15: / Line 15: @@
 ==== In terms of a covariant derivative ====
-Let $\alpha$ be any form on $M$ and $X$ any vector field; let $\nabla$ be a metric-compatible connection.
+Recall (p. 24 of DG notes) that $L_X = \nabla_X + e(\omega^i) e^*(\nabla_{e_j} X^\flat)$.
-We compute
+In the radial direction $\partial_r$ of geodesic coordinates,
+$$L_{\partial_r} = \nabla_{\partial_r} + e(\omega^i) e^*(\nabla_{e_j} dr)$$
+where, for example, $\partial_r = e_n$.
+Let $\omega^i$ be a coframe to the $e_i$s. The second fundamental form $h$ of a geodesic sphere is defined to be $h(X,Y) = g(\nabla_X \partial_r, Y)$.
+In coordinates, we have $h = h_{ij} e(\omega^i) e^*(\omega^j)$, where $h_{ij}$ satisfy $\nabla_{e_i} dr = \sum_{j=1}^{n-1} h_{ij} \omega^i$.
+We have
 \begin{align*}
-L_X (|\alpha|^2 vol) &= (X |\alpha|^2) vol + |\alpha|^2 L_X vol\tag{by axioms 1, 2} \\
+e(\omega^i) e^*(\nabla_{e_i} dr) &= e(\omega^i) e^*(h_{ij} \nabla_{e_i} \omega^j) \\
-&= \langle (2\nabla_X + \mathrm{div}(X)) \alpha, \alpha \rangle vol
+&= h_{ij} e(\omega^i) e^*(\omega^j) \\
+&= h,
 \end{align*}
-Thus we may express the Lie derivative in terms of the connection and divergence.
+where $h$ denotes the natural extension of the second fundamental form to operate on all forms, henceforth called $Q$.
-A typical choice of $X$ is a normal vector field to a hypersurface; the divergence of $X$ is then the trace of the second fundamental form of the embedding. For example, in Euclidean space it follows that
+Thus we have an expression for the Lie derivative in terms of the second fundamental form:
-$$\frac{1}{2} L_{\partial/\partial r} (|\alpha|^2 vol) = \langle (\nabla_{\partial/\partial r} + n/2)\alpha, \alpha\rangle vol.$$
+$$L_{\partial_r} = \nabla_{\partial_r} + Q.$$
-Also, writing $g$ for the metric then
+One could replace $e_1, \ldots, e_{n-1}$ with an (orthogonal) eigenbasis for the second fundamental form, say with eigenvalues $\lambda^1, \ldots, \lambda^{n-1}$; that is, so that $h_{ij} = \delta_{i}^j \lambda^j$ (here and unless otherwise stated in the sequel, no sum).
-$$2 \langle \nabla_X \alpha, \alpha \rangle = X |\alpha|^2 = L_X (g(\alpha, \alpha)) = (L_X g)(\alpha, \alpha) + 2g(L_X \alpha, \alpha).$$
+That is, in this basis, we have that $Q\omega^j = \lambda_j \omega^j$.
+Thus we have for $i < n$ that
+\begin{align*}
+L_{\partial_r} (\omega^i) &= \nabla_{\partial_r} \omega^i + \lambda^i \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
+$$L_{\partial_r} (\omega^i) = -\gamma^i_{nk} \omega^k + \lambda^i \omega^i.$$