2
$\begingroup$

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?

Thanks in advance

$\endgroup$

1 Answer 1

2
$\begingroup$

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 $d^*$.

$\endgroup$
4
  • $\begingroup$ 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? $\endgroup$ May 8, 2022 at 13:52
  • $\begingroup$ Is $C_0^\infty([0,1])$ dense in $L^2([0,1])$? $\endgroup$ May 8, 2022 at 14:07
  • $\begingroup$ 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 $||f-hf||_2=||f-hf||_{L^2([0,1]-K)}\le ||f||_{L^2([0,1]-K)}||1-h||\le\varepsilon$. $\endgroup$ May 8, 2022 at 15:33
  • $\begingroup$ 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 $\xi\in L^2(\Lambda^1 M)$. $\endgroup$ May 8, 2022 at 16:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .