Hodge theory, conformal manifolds and Fredholm modules-understanding the proof of one Lemma

Question

I would like to understand the proof of Lemma 1, page 339 in this book. Very briefly, the context is as follows: we have even dimensional oriented conformal manifold with the Hodge star operator chosen and we would like to construct a Fredholm module based on this data. In order to do so, we need an auxiliary operator, denoted by $S$ which has the property that the graph of $S$ coincides with the image of exterior derivative. There are several points which are not clear for me in the argument:

Q1. Why any differential form in $\mathcal{H}_0^+$ which is orthogonal to harmonic forms may be written as $\frac{1+\gamma}{2} d\alpha$ for some $\alpha$?

Hodge theory ensures us that such $\omega$ lies in the image of $d+d^*$ and the claim should somehow follow from the fact that $d^*=-*d*$ but I don't see how since there is no obvious commutation relation between the grading $\gamma$ and $d$.

Q2. I don't see why $\|\frac{1+\gamma}{2} d \alpha \|_2 =\| \frac{1-\gamma}{2}d\alpha \|_2$.

In fact after simple computation this is equivalent to the fact that $\langle d \alpha, \gamma d\alpha \rangle = -\langle \gamma d \alpha, d \alpha \rangle$. But how to get this minus sign? I thought that $\gamma$ should be self adjoint.

Q3. Even if we have this equality, does it really imply that $d\alpha$ is determined uniquely?

Leaving this specific context for a moment, we can imagine two vectors in $\mathbb{R}^2$ say $\xi_1=(1,1)$ and $\xi_2=(-1,1)$ and a projection $P(x,y):=(0,y)$ Then $P\xi_1=P\xi_2$ and $\| P \xi_1 \|= \| (I-P)\xi_1 \|$ but $(I-P)\xi_1 \neq (I-P)\xi_2$. In our context the role of $P$ is played by $\frac{1-\gamma}{2}$ (which indeed is a projection).

I suspect that this uniqueness is needed in order to define $S$ in such a way that $S (\frac{1-\gamma}{2} d\alpha):=\frac{1+\gamma}{2} d \alpha$. But even so I still don't see why the graph of $S$ should be equal to the image of $d$ (image or closure of image?) since I can't get rid of the harmonic part in the expression $\omega+S \omega$ for $\omega \in \mathcal{H}_0^{-}$ (which is mapped to 0 by $S$).

I realize that these are technical questions and can be asked separetly, nevertheless I think that the whole context may be here important therefore I decded to ask all my questions in one post-I hope that this will be fine.

Answer 1

Here is the answer to the questions. (Recall $V$ is a closed oriented conformal manifold of dimension $2n$). According to the Hodge-deRham decomposition, any $\omega\in L^2(V,\Lambda^nT^*V) $can be written $$\omega=h+d\alpha+d^*\beta$$ where $h\in L^2(V,\Lambda^nT^*V)$ is harmonic, $\alpha\in W^{1,2}(V,\Lambda^{n-1}T^*V)$ and $\beta\in W^{1,2}(V,\Lambda^{n+1}T^*V)$, moreover $\alpha$ (resp. $\beta$) is unique provided it is weakly coexact (resp. weakly exact). In fact, there is a unique $\eta\in W^{2,2}(V,\Lambda^{n}T^*V)$, orthogonal to the space of harmonic form such that: $$\omega=h+(dd^*+d^*d)\eta.$$ Now if $\gamma\omega=\omega$ and $\omega$ is orthogonal to the space of harmonic form then $$\omega=d\alpha+d^*\beta \ \mathrm{and}\ \omega=\gamma d\alpha+\gamma d^*\beta=d^*\tilde\alpha+d\tilde \beta$$ where (for some $\varpi,\varpi'\in \{1,i\}$) $\tilde\alpha=\pm\varpi \gamma\alpha$ and $\tilde\beta=\pm\varpi' \gamma\beta$, by uniqueness we get $$d\alpha=d\tilde \beta=\gamma d^*\beta \ \mathrm{and}\ d^*\beta=\gamma d\alpha=d^*\tilde\alpha$$ so that $$\omega=(1+\gamma)d\alpha=\frac{1+\gamma}{2} d\alpha.$$

For the next question, we have $$\langle d\alpha,\gamma d\alpha\rangle_{L^2} =\pm\varpi \langle d\alpha,\star d\alpha\rangle_{L^2}=\pm\varpi\int_V d\alpha\wedge d\alpha=\pm\varpi\int_V d(\alpha\wedge d\alpha)=0.$$