In "Foundations of Differential Manifolds and Lie Groups" by Frank Warner on page 225 there is defined Green's operator: $G: E^p(M) \rightarrow (H^p)^{\perp}$ by setting $G(\alpha)$ to equal the unique solution of $\Delta \omega = \alpha - H(\alpha)$ in $(H^p)^{\perp}$.


Here $E^p(M)$ is the set of all differential forms on the manifold $M$, $H^p$ is the set of all harmonic forms on $M$ and $H(\alpha)$ is the harmonic part of $\alpha$.

We define $*: \Lambda ^k M \rightarrow \Lambda^{m-k}M, \ \ \dim M =m$ in such a way that $*$ is $\mathbb{R}$-linear, $*(e_1 \wedge ... \wedge e_p) = e_{p+1} \wedge ... \wedge e_m$ for $e_1, ..., e_m$ - well oriented orthonormal basis and

$*(e_{i_1} \wedge ... \wedge e_{i_p}) = \pm e_{j_1} \wedge ... \wedge e_{j_{m-p}}$ depending on whether $e_{i_1}, ... ,e_{i_p}, e_{j_1} , ... , e_{j_{m-p}}$ 's orientation is the same or opposite to $e_1, ..., e_m$.

Then we define scalar product: on $E^p(M)$ $M$ is compact and oriented

$\langle\alpha, \beta\rangle = \int_M \alpha \wedge * \beta$

Now, I want to prove that $G$ is a bounded and self adjoint operator.

I think that we can write $G(\alpha) = (\Delta|_{H^{p \perp}})^{-1}(\alpha - H(\alpha))$ and I know that $\langle\Delta \alpha, \beta\rangle = \langle\alpha, \Delta \beta\rangle$, but it doesn't help me.

Could you help me with that?

  • $\begingroup$ Not sure if it is proved in the book, but $(\Delta|_{{H^p}^\perp})^{-1}$ is bounded. $\endgroup$ – user99914 Dec 9 '14 at 9:11
  • $\begingroup$ I haven't found it there, but $\Delta(E^p) = H^{p \perp}$. Could that imply boundedness? $\endgroup$ – Sasha Dec 9 '14 at 9:17
  • $\begingroup$ I don't think so, you need some PDE estimate..... $\endgroup$ – user99914 Dec 9 '14 at 9:19
  • $\begingroup$ Is that an exercise in the book? $\endgroup$ – user99914 Dec 9 '14 at 9:22
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    $\begingroup$ Now both $E^p$ and $(H^p)^\perp$ are given the $L^2$ products, so you want $||G(\alpha)|| \leq C ||\alpha||$ for some $C$, which is roughly the same as $||\omega || \leq C ||\Delta w ||$. Is that kinds of theorems proved in the book? (Sorry I really haven't read the book I do not know what is inside the book) $\endgroup$ – user99914 Dec 9 '14 at 9:36

Let $\alpha \in E^p$, then we can write $\alpha = H(\alpha) + (\alpha - H(\alpha))$. As $G(\alpha) \in (H^p)^\perp$ satisfies $\Delta G(\alpha) = \alpha - H(\alpha)$, we have

$$ ||G(\alpha)|| \leq C ||\Delta G(\alpha)|| = C ||\alpha - H(\alpha)|| \leq C||\alpha||,$$

thus $G$ is bounded. To show that it is self-adjoint, you need to restrict $G$ to $(H^p)^\perp$. Let $\alpha, \beta \in (H^p)^\perp$, $H(\alpha) = 0$ and by $\langle \alpha, \Delta \beta\rangle = \langle \Delta \alpha, \beta\rangle$ and changing $\alpha, \beta$ to $(\Delta |_{(H^p)^\perp})^{-1} \alpha$, $(\Delta |_{(H^p)^\perp})^{-1} \beta$ respectively, we have

$$\langle G(\alpha) , \beta\rangle = \langle \alpha, G(\beta)\rangle. $$

Thus $G$ is self adjoint. (Actually $G$ is also a compact operator)

  • $\begingroup$ That's very nice. Thank you a lot! $\endgroup$ – Sasha Dec 9 '14 at 10:01
  • $\begingroup$ I have one question, though. Why do we restrict $G$ to $(H^p)^\perp$? $\endgroup$ – Sasha Dec 9 '14 at 10:10
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    $\begingroup$ @Sasha: Actually for that formula we do not need to restrict to $(H^p)^\perp$, but it is more common that we only talk about a self adjoint operator if $G$ has the same domain and image. $\endgroup$ – user99914 Dec 9 '14 at 10:16
  • $\begingroup$ So if we don't restrict $G$, we need to change $(\Delta |_{(H^p)^\perp})^{-1} (\alpha - H \alpha)$ ? $\endgroup$ – Sasha Dec 9 '14 at 10:22
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    $\begingroup$ Yes, that can be done. $\endgroup$ – user99914 Dec 9 '14 at 10:30

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