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--- exterior_derivative [2022/08/31 19:21] – created spencer
+++ exterior_derivative [2022/09/05 08:39] (current) – [Adjoint of the exterior derivative] spencer
@@ Line 5: / Line 5: @@
 \[ \Omega^k(E) = \Gamma(\Lambda^k T^* M \otimes E) = \Omega^k(M) \otimes \Gamma(E). \]
 (Note the inconsistency in notation: $M$ is a rank zero vector bundle over $M$, but $\Omega^k(M)$ does **not** mean the $M$-valued $k$-forms on $M$; it means the standard $k$-forms, and one really ought to write $\Omega^k(\mathbb{R} \times M)$. But this abuse is standard.)
+The last equality is not obvious. See [[section_isomorphisms|section isomorphisms]] for more information.
+The connection $\nabla$ can be thought of as a map $\Omega^0(E) \to \Omega^1(E)$.
+The //exterior derivative associated to $\nabla$//, written often as $\mathrm{d}$ or $\mathrm{d}^\nabla$ is a map $\Omega^k(E) \to \Omega^{k+1}(E)$ for each non-negative integer $k$ characterized by the following properties:
+  - Restricted to $\Omega^0(E)$, one has $\mathrm{d}^\nabla = \nabla$
+  - For a $k$-form $\omega$ and a section $\sigma$ of $E$, a Leibniz rule holds:
+\[ \mathrm{d}^\nabla (\omega \otimes \sigma) = \mathrm{d}\omega \otimes \sigma + (-1)^k \omega \otimes \mathrm{d}^\nabla \sigma. \]
+In this Leibniz rule, $\mathrm{d}\omega$ is the usual exterior derivative on forms.
+===== Formula for the exterior derivative in terms of the connection =====
+In a local orthonormal frame $(e_i)_{i=1}^n$ for $TM$, the exterior derivative takes the following form:
+\[ \mathrm{d}\alpha = e^i \wedge \nabla_{e_i} \alpha. \]
+The crux of the proof is that $\nabla$ is torsion-free; one can first argue that this identity holds in a coordinate frame (so, for the Levi-Civita connection, $\nabla_{e_i} \nabla_{e_j} = -\nabla_{e_j} \nabla_{e_i}$) and then extend by $C^\infty(M)$-linearity as appropriate.
+===== Adjoint of the exterior derivative ======
+Suppose that $M$ is closed and equipped with a Riemannian metric $g$, and that $E$ is equipped with a compatible fibre metric $h$.
+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
+\[ \mathrm{d}^* = -e(\omega^i)^* \nabla_{e_i}. \]
+**Proof 1:** We use Stokes's theorem.
+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*}
+\langle \mathrm{d}\phi, \psi\rangle_{L^2} &= \int_M \langle \mathrm{d} \phi, \psi \rangle \mathrm{vol} \\
+&= \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*}
+The last term is the one we want to survive; so we hope to show that
+\[ \left(e_i \langle e(\omega^i), \psi\rangle - \langle e(\nabla_{e_i} e_i\rangle\right)\mathrm{vol} \]
+is exact.