# Formal adjoint of the exterior derivative on a manifold with boundary

Let $$(M,g)$$ a compact Riemmanian manifold with boundary. I want to define the formal adjoint of the exterior derivative $$\mathrm d\colon C^\infty(M)\to \Omega^1(M)$$.

If $$\partial M=\emptyset$$ and $$\xi$$ is $$1$$ differential form on $$M$$, I can define $$\mathrm d^\star \xi$$ from the formula $$\langle f,\mathrm d^\ast \xi\rangle =\langle\mathrm d f,\xi\rangle$$ imposing its validity for all $$f\in C^\infty(M)$$

If $$\partial M\ne \emptyset$$, Initially I thought substituting $$C^\infty(M)$$ with $$C^\infty_0(M)$$ (the space of scalar function with compact support contained in the interior of $$M$$) but I'm not sure that proceeding in this way $$\mathrm d^\ast\colon \Omega^1(M)\to C^\infty(M)$$ is well defined.

So, how I define $$\mathrm d^\ast\colon \Omega^1(M)\to C^\infty(M)$$ whene $$M$$ ad a non-void boundary?

The definition of the formal adjoint is usually given in both cases (compact manifold with empty or not empty boundary) by testing against $$C_0^\infty$$ sections (compactly supported). Of course in the case $$\partial M = \emptyset$$ it's the same as using $$C^\infty$$ sections.
The main difference is that now the formula $$\langle df, \omega\rangle_{L^2(M)} = \langle f, d^*\omega\rangle_{L^2(M)}$$ will hold only when $$\omega$$ (or $$f$$) vanish on $$\partial M$$. In general, if $$\omega$$ or $$f$$ do not vanish on the boundary there will be an extra term , more precisely $$\int_M \langle df, \omega\rangle \operatorname{vol}_M= \int_M \langle f, d^*\omega\rangle \operatorname{vol}_M+ \int_{\partial M}\langle \nu\wedge f, \omega\rangle\operatorname{vol}_{\partial M}$$ where $$\nu = \langle\cdot, \frac {\partial}{\partial n}\rangle \in \Omega^1(M)$$ is dual to the normal to the boundary.
The reason for this definition is functional Analysis. Indeed $$d$$ defines a closed, unbounded, densely defined operator $$L^2(M)\to L^2(\Lambda^1M)$$ ($$L^2$$-section), with a certain domain. Its Von-Neumann adjoint coincides (in its domain of definition) with $$d^*$$ (beware in general the Von-Neumann adjoint is defined on a subset of the domain of the natural extension of $$d^*$$ to $$L^2$$). What you gain is therefore a decomposition $$L^2(M) = \ker (d)\oplus \overline{Im(d^*)}$$ $$L^2(\Lambda^1M) = \overline{Im(d)}\oplus \ker(d^*)$$ where $$d, d^*$$ here are, with a little abuse of language, the extensions with the natural extended domain for $$d$$ and the domain given by the definition of Von-Neumann adjoint for $$d^*$$.
• Thanks for answering. Using $C_0^\infty(M)$, is $\mathrm d^\ast\omega$ well defined? If there were $\varphi\in \Gamma(E)$ verifying $\langle \mathrm d f,\omega\rangle=\langle f,\varphi\rangle$ forall $f\in C_0^\infty(M)$ I obtain $\langle f,\mathrm d^\ast f-\varphi\rangle=0$ forall $f\in C_0^\infty(M)$. From this I can't deduce $\mathrm d f=\varphi$ beacuse $C_0^\infty(M)$ is not dense in $L^2(M)$. Am i correct? May 8, 2022 at 13:52
• Is $C_0^\infty([0,1])$ dense in $L^2([0,1])$? May 8, 2022 at 14:07
• Yes and probably the following argument generalizes to my case. It's enough to prove the density of $C_0^\infty([0,1])$ in $C^\infty([0,1])$. Let $f\in C^\infty([0,1])$ and $\varepsilon>0$. I consider the compact subspace $K=\{ f : ||f||_2\ge \varepsilon/2 \}$ and I choose a smooth cut-off function $h\colon [0,1]\to \mathbb C$ with $\text{supp}h\subseteq (0,1)$ and $h|_K=1$. Then $hf\in C_0([0,1])$ and $||f-hf||_2=||f-hf||_{L^2([0,1]-K)}\le ||f||_{L^2([0,1]-K)}||1-h||\le\varepsilon$. May 8, 2022 at 15:33
• You read my mind! I post my question beacause I'm investigating the nature of the weak laplacian $\Delta\colon H^2(M)\to L^2(M)$. In order to define its action on $H^2(M)$, I was trying to extend $\mathrm d$ and $\mathrm d^\ast$. To extend $\mathrm d$, I thought I'd impose $\langle \mathrm d f,\varphi\rangle= f,\mathrm d^\ast \varphi\rangle$ for all $\varphi\in C_0^\infty(M)$: this define $\mathrm d f$ when $f\in L^2(M)$ is in the "maximal domain" of $\mathrm d$. Similarly, I thought I'd proceed for defining $\mathrm d^\ast \xi$ when $\xi\in L^2(\Lambda^1 M)$. May 8, 2022 at 16:31