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

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.

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 $$\langle d\alpha,\gamma d\alpha\rangle_{L^2} =\pm\varpi \langle d\alpha,\star d\alpha\rangle_{L^2}=\pm\varpi\int_V d\alpha\wedge d\alpha=\pm\varpi\int_V d(\alpha\wedge d\alpha)=0.$$