Green's operator, differential forms

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?

• Not sure if it is proved in the book, but $(\Delta|_{{H^p}^\perp})^{-1}$ is bounded. – user99914 Dec 9 '14 at 9:11
• I haven't found it there, but $\Delta(E^p) = H^{p \perp}$. Could that imply boundedness? – Sasha Dec 9 '14 at 9:17
• I don't think so, you need some PDE estimate..... – user99914 Dec 9 '14 at 9:19
• Is that an exercise in the book? – user99914 Dec 9 '14 at 9:22
• 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) – 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)

• That's very nice. Thank you a lot! – Sasha Dec 9 '14 at 10:01
• I have one question, though. Why do we restrict $G$ to $(H^p)^\perp$? – Sasha Dec 9 '14 at 10:10
• @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. – user99914 Dec 9 '14 at 10:16
• So if we don't restrict $G$, we need to change $(\Delta |_{(H^p)^\perp})^{-1} (\alpha - H \alpha)$ ? – Sasha Dec 9 '14 at 10:22
• Yes, that can be done. – user99914 Dec 9 '14 at 10:30