I was trying to understand the following fundamental elliptic complex in Hodge theory (Here $M$ is a compact oriented four dimensional Riemannian manifold):
$$0\rightarrow \Omega^0(M) \stackrel{d}{\longrightarrow} \Omega^1(M) \stackrel{d^+}{\longrightarrow} \Omega^2_+(M)\rightarrow 0$$
How do we find the first cohomology group? i.e. decide $\omega \in \Omega^1(M)$ s.t. $d^+\omega=0 $ and $\omega$ is $L^2$-orthogonal to the image of $d$.
The answer is the harmonic 1-forms but I don't see why. Being $L^2$-orthogonal to image of $d$ means $\omega$ lies in the direct sum of harmonic 1-forms and image of $d^*$ by Hodge theory. but I don't see how the other condition served to eliminate the possibility of lying in image of $d^*$.
By Hodge decomposition $$ \Omega^2(M)=\operatorname{im}(d\colon\Omega^1(M)\to\Omega^2(M))\oplus\operatorname{im}(d^*\colon\Omega^3(M)\to\Omega^2(M))\oplus\ker(\Delta\colon\Omega^2(M)\to\Omega^2(M)). $$ But $$ \operatorname{im}(d^*\colon\Omega^3(M)\to\Omega^2(M))=\operatorname{im}(*d*\colon\Omega^3(M)\to\Omega^2(M))=\operatorname{im}(*d\colon\Omega^1(M)\to\Omega^2(M)),$$ so for $\alpha\in\ker d^+$ we must have $$ 0=\lVert d^+\alpha\rVert_{L^2}^2 =\frac14\lVert d\alpha+*d\alpha\rVert_{L^2}^2 =\frac14(\lVert d\alpha\rVert_{L^2}^2+\lVert *d\alpha\rVert_{L^2}^2) $$ i.e., we must have $d\alpha=0$. In other words, $\ker(d^+\colon\Omega^1(M)\to\Omega^2_+(M))=\ker(d\colon\Omega^1(M)\to\Omega^2(M))$ and so the cohomology $$ \frac{\ker(d^+\colon\Omega^1(M)\to\Omega^2_+(M))}{\operatorname{im}(d\colon\Omega^0(M)\to\Omega^1(M))}=\frac{\ker(d\colon\Omega^1(M)\to\Omega^2(M))}{\operatorname{im}(d\colon\Omega^0(M)\to\Omega^1(M))}=H^1(M). $$