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--- exterior_derivative [2022/09/05 05:20] – spencer
+++ exterior_derivative [2022/09/05 08:39] (current) – [Adjoint of the exterior derivative] spencer
@@ Line 24: / Line 24: @@
 Suppose that $M$ is closed and equipped with a Riemannian metric $g$, and that $E$ is equipped with a compatible fibre metric $h$.
-Then there is an induced metric on each fibre of each bundle $\Lambda^k T^*M \otimes E$, and by integrating there is an induced $L^2$ metric on each $\Omega^k(E)$.
 Let $p \in M$ be any point, and $(e_i)_{i=1}^n$ be an orthonormal frame of $TM$ in a neighbourhood about $p$.
 Let $(\omega^i)_{i=1}^n$ be the dual frame of $T^*M$ about $p$.
 **Claim:** The adjoint to the exterior derivative is given by
-\[ \mathrm{d}^* = -\omega(e_i)^* \nabla_{e_i}. \]
+\[ \mathrm{d}^* = -e(\omega^i)^* \nabla_{e_i}. \]
-**Proof:** Take $\sigma \in \Omega^{k+1}(E)$ and $\tau \in \Omega^k(E)$.
+**Proof 1:** We use Stokes's theorem.
-We compute:
+First, note that
+\[ e_i \langle e(\omega^i)\phi, \psi \rangle = \langle \nabla_{e_i} e(\omega^i)\phi, \psi\rangle + \langle e(\omega^i)\phi, \nabla_{e_i} \psi \rangle,\]
+and moreover that
+\[ \nabla_{e_i} e(\omega^i) \phi = e(\omega^i) \nabla_{e_i} \phi + e(\nabla_{e_i} \omega^i) \phi = = e(\omega^i) \nabla_{e_i} \phi + e(\nabla_{e_i} e_i) \phi. \]
+One therefore computes:
 \begin{align*}
-(\mathrm{d}\tau, \sigma)_{L^2} &= (e(\omega^i) \nabla_{e_i} \tau, \sigma)_{L^2} \\
+\langle \mathrm{d}\phi, \psi\rangle_{L^2} &= \int_M \langle \mathrm{d} \phi, \psi \rangle \mathrm{vol} \\
-&= (\nabla_{e_i} \tau, e(\omega^i)^* \sigma)_{L^2}
+&= \int_M \langle e(\omega^i) \nabla_{e_i} \phi, \psi \rangle \mathrm{vol} \\
+&= \int_M \left(e_i \langle e(\omega^i), \psi\rangle - \langle e(\nabla_{e_i} e_i) \phi, \psi\rangle - \langle e(\omega^i)\phi, \nabla_{e_i} \psi\rangle\right) \mathrm{vol}.
 \end{align*}
-On the other hand,
+The last term is the one we want to survive; so we hope to show that
-\begin{align*}
+\[ \left(e_i \langle e(\omega^i), \psi\rangle - \langle e(\nabla_{e_i} e_i\rangle\right)\mathrm{vol} \]
-(\tau, -\omega(e_i)^* \nabla_{e_i} \sigma)_{L^2} &= -(e(\omega^i)  \tau, \nabla_{e_i} \sigma)_{L^2} \\
+is exact.
-\end{align*}
-Now recall that by metric compatibility,
-\begin{align*}
-\nabla_{e_i} \langle e(\omega^i)\tau, \sigma\rangle  &= \langle \nabla_{e_i} e(\omega^i)\tau, \sigma\rangle + \langle e(\omega^i)\tau, \nabla_{e_i}\sigma\rangle \\
-&= \langle (\nabla_{e_i} \omega^i) \wedge \tau + e(\omega^i) \nabla_{e_i} \tau, \sigma\rangle + \langle e(\omega^i)\tau, \nabla_{e_i}\sigma\rangle.
-\end{align*}
-The left-hand side is a divergence; integrating (using round brackets everywhere for the $L^2$ norm, over a compact neighbourhood and multiplying by a suitable bump function to remove boundary terms), we have on a neighbourhood about $p$ that
-\begin{align*}
-( e(\omega^i)\tau, \nabla_{e_i}\sigma) &= -((\nabla_{e_i} \omega^i) \wedge \tau,\sigma) - ( e(\omega^i) \nabla_{e_i} \tau, \sigma) \\
-&= -( (\nabla_{e_i} \omega^i) \wedge \tau,\sigma)- ( \nabla_{e_i} \tau, e(\omega^i)^* \sigma),
-\end{align*}