A question on moduli space of Hitchin's equations

Question

I am reading Hitchin's Self-Duality paper. 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 $d^*$.

Answer 1

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.