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--- integration_by_parts [2023/06/07 13:36] – [Integration by parts I ("Dot product by parts")] spencer
+++ integration_by_parts [2023/06/08 08:21] (current) – [Integration by Parts II (the formal adjoint of the exterior derivative)] spencer
@@ Line 1: / Line 1: @@
 ====== Integration by parts ======
-I forget how to do integration by parts on a manifold every day of the week. Here's a wiki page to remind me how it should work.
+I forget how to do integration by parts on a manifold every day of the week. Here's a wiki page to remind me how it should work. See also [[https://mathoverflow.net/questions/289064/boundary-terms-of-formal-adjoints-of-differential-operators/289070|this MO question]] for some discussion of boundary.
 ==== Preliminaries ====
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 Write $f = \langle a, b\rangle$ and consider $d(f \iota_X \mathrm{vol}) = Xf \mathrm{vol}$ (seen to be true locally and then extended globally).
 Thus
-$$\int_M X\langle a,b\rangle \mathrm{vol}_M = \int_{\partial M} f \iota_X \mathrm{vol}.$$
+$$\int_M X\langle a,b\rangle \mathrm{vol}_M = \int_{\partial M} f \iota_X \mathrm{vol}_M.$$
 Now like in the divergence theorem, the only part of $\iota_X$ that 'survives is the normal direction $\langle X, \nu\rangle \nu$, whence
 $$\int_M X\langle a,b\rangle \mathrm{vol}_M = \int_{\partial M} \langle a,b\rangle \langle X, \nu\rangle \mathrm{vol}_{\partial M}$$
 giving the integration by parts formula
-$$(\nabla_X, \alpha, \beta) = (\alpha, -\nabla_X \beta) + \int_M \langle \alpha, \beta \rangle \langle X, \nu\rangle \mathrm{vol}{\partial M}.$$
+$$(\nabla_X, \alpha, \beta) = (\alpha, -\nabla_X \beta) + \int_{\partial M} \langle \alpha, \beta \rangle \langle X, \nu\rangle \mathrm{vol}_{\partial M}.$$
+In particular, if any of $X, \alpha, \beta$ are compactly supported then we have the expected integration by parts formula $(\nabla_X, \alpha) = (\alpha, -\nabla_X \beta)$.
+==== Integration by Parts II (the formal adjoint of the exterior derivative) ====
+On a compact manifold, or on any manifold and operating only on the compactly supported sections, the above integration by parts formula gives that $(\nabla_X, \alpha) = (\alpha, -\nabla_X \beta)$ for all $k$-forms $\alpha, \beta$.
+In a local frame $e_1,\ldots,e_n$ with coframe $\omega^1,\ldots,\omega^n$ we have the formula for the exterior derivative: $d = e(\omega^i) \nabla_{e_i}$.
+Thus we may compute its formal adjoint: set $\delta = -e^*(\omega^i) \nabla_{e_i}$ and compute
+\begin{align*}
+(d \alpha, \beta) &= \left(e(\omega^i) \nabla_{e_i} \alpha, \beta \right) \\
+&= \left(\nabla_{e_i}  \alpha, e^*(\omega^i) \beta \right) \\
+&= \int_M e_i \langle \alpha, e^*(\omega^i) \beta \rangle - \langle \alpha, \nabla_{e_i} e^*(\omega^i) \rangle \mathrm{vol}_M \\
+&= \int_M L_{e_i} (\langle \alpha, e^*(\omega^i) \beta \rangle) - \langle \alpha, e^*(\omega^i) \nabla_{e_i}  \rangle + \langle \alpha, e^*(\nabla_{e_i} \omega^i) \mathrm{vol}_M \\
+&= \int_M L_{e_i} (\langle \alpha, e^*(\omega^i) \beta \rangle \mathrm{vol}_M) - \int_M \langle \alpha, e^*(\omega^i) \beta \rangle L_{e_i} \mathrm{vol}_M - \int_M \left(\langle \alpha, e^*(\omega^i) \nabla_{e_i} \beta \rangle + \gamma^k_{ii} \langle \alpha, e^*(\omega^k) \beta\rangle\right) \mathrm{vol}_M \\
+&= \int_{\partial M} e^*(\omega^i) (\langle \alpha, e^*(\omega^i) \beta \rangle \mathrm{vol}_M) + (\alpha, \delta \beta) - \int_M \langle\alpha, e^*(\omega^i) \beta\rangle (\mathrm{div}(e_i) + \gamma_{kk}^i) \mathrm{vol}_M.
+\end{align*}
+On the other hand, the divergence of $e_i$ is $\sum_k (e_i)_{k; k}$, and $(e_i)_{k; k} = \langle \nabla_{e_k} e_i, e_k\rangle = -\gamma^i_{kk}$ so the last term vanishes.
+As before, the only boundary term with contribution is the unit outward normal one, and we conclude that with $\nu$ the unit outward normal,
+$$(d\alpha, \beta) = (\alpha, \delta \beta) + \int_{\partial M} \langle \alpha, \iota_{\nu} \beta\rangle \mathrm{vol}_{\partial M}.$$
+Probably. Would have to check whether boundary terms arise in the first step where we move $e(\omega^i)$ to the other side too.
+Certainly in the compactly supported case $(d\alpha,\beta) = (\alpha, \delta \beta)$ holds just fine.
+Of course, we also have $\delta = d^* = (-1)^{|\alpha| + 1} \star^{-1} d \star$ for $\star$ the Hodge star operator.