I am reading Hitchin's Self-Duality [paper](http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.557.2243&rep=rep1&type=pdf). In section 5 (page 85), he is trying to prove that $Dim H^1=12(g-1)$. In doing so, he defines an operator $d^\*\_2+d\_1$, where $d^\*\_2$ and $d\_1$ are given by


$d\_1\dot{\psi}=(d\_{A}\dot{\psi},[\Phi, \dot{\psi}])$


$d\_2(\dot{A},\dot{\Phi})=(d\_A\dot{A}+[\dot{\Phi},\Phi^\*]+[\Phi,\dot{\Phi}^\*], d^{\prime\prime}\_{A}\dot{\Phi}+[\dot{A}^{0,1},\Phi])$


Then, he claims that $(d^\*\_2+d\_1)(\psi\_1,\psi\_2)=0$ if and only if


$d^{\prime\prime}\_{A}\psi\_1
+[\Phi^\*,\psi\_2]=0$


$d^{\prime}\_{A}\psi\_2
+[\Phi,\psi\_1]=0$


I am not able to derive this fact, and I spent quiet a lot of time on this, but unfortunately was not able to prove it. He says that he obtains this by calculating the explicit form of adjoint of $d\_2$. I was not able to perform this calculation. I am new to the subject, and I would really appreciate any help, or ideas on how to prove this. Thanks!


P.S. I know that a co-differential is defined by $ d^{\*}=(−1)^{n(k-1)+1}\*d\*:\Omega^{k}\to \Omega^{k-1}$ where $\*$ in the definition is Hodge star, and this is the adjoint of exterior derivative with respect to $L^2$ norm. But how would one apply hodge operator in this setting, or even use $L^2$ to get $d^\*$.



A question on moduli space of Hitchin's equations
The operator
$$ d\_2^\*+d\_1\colon \Omega^0(M,ad P\otimes\mathbb C)\oplus\Omega^0(M,ad P\otimes \mathbb C)\to\Omega^{0,1}(M,ad P\otimes \mathbb C)\oplus\Omega^{1,0}(M,ad P\otimes \mathbb C)$$
is just the sum of $d\_1$ and $d\_2^\*,$ and it suffices to describe these two operators using the identification $$\Omega^{0,1}(M,ad P\otimes\mathbb C)=\Omega^1(M,ad P).$$
We decompose $\psi\_1\in\Omega^0(M,ad P\otimes\mathbb C)$ into real and imaginary parts $$\psi\_1=\omega+i \eta$$ for $\omega,\eta\in \Omega^0(M,ad P),$ and let $d\_1$ act on the real part ($\omega$) and $d\_2^\*$ on the imaginary part ($\eta$).


Then, we have $$d\_1(\psi\_1,\psi\_2)=((d\_A\omega)^{0,1},[\Phi,\omega])=(d\_A''\omega,[\Phi,\omega])$$
by definition. The dual of the operator $$d\_A''\colon\Omega^{1,0}(M,ad P\otimes \mathbb C)\to \Omega^2(M,ad P\otimes \mathbb C); \; \psi\_2\mapsto d\_A''\psi\_2$$ is by Serre duality the operator $$d\_A''\colon \Omega^0(M,ad P\otimes \mathbb C)\to \Omega^{0,1}(M,ad P\otimes \mathbb C);\; \psi\_2\mapsto d\_A''\psi\_2,$$ and using the hermitian metric and the fact that $d\_A$ is unitary the adjoint operator gets identified with
$$d\_A'\colon \Omega^0(M,ad P\otimes \mathbb C)\to \Omega^{1,0}(M,ad P\otimes \mathbb C);\; \psi\_2\mapsto d\_A'\psi\_2.$$
The adjoint of the operator
$$\phi\in\Omega^2(M,ad P)\mapsto ([\phi,\Phi^\*]+[\Phi,\phi^\*])^{0,1}\in\Omega^{0,1}(M,ad P)$$ becomes
$$\psi\_2\mapsto -[\psi\_2,\Phi^\*]$$ ( taking the $i-$ part into account).


It remains to describe the operator $d\_2^\*$ acting on the imaginary part $i\eta$ of $\psi\_1.$ As above, this becomes $$d\_2^\*((i\eta))=(id\_A''\eta,i[\Phi,\eta]).$$ Putting together the pieces proves the claim.


