A question on moduli space of Hitchin's equations

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

$d_1dot{psi}=(d_{A}dot{psi},[Phi, dot{psi}])$

$d_2(dot{A},dot{Phi})=(d_Adot{A}+[dot{Phi},Phi^*]+[Phi,dot{Phi}^*], d^{primeprime}_{A}dot{Phi}+[dot{A}^{0,1},Phi])$

Then, he claims that $(d^*_2+d_1)(psi_1,psi_2)=0$ if and only if



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 $d^*$.

MathOverflow Asked on February 5, 2021

1 Answers

One Answer

The operator $$ d_2^*+d_1colon Omega^0(M,ad Potimesmathbb C)oplusOmega^0(M,ad Potimes mathbb C)toOmega^{0,1}(M,ad Potimes mathbb C)oplusOmega^{1,0}(M,ad Potimes mathbb C)$$ is just the sum of $d_1$ and $d_2^*,$ and it suffices to describe these two operators using the identification $$Omega^{0,1}(M,ad Potimesmathbb C)=Omega^1(M,ad P).$$ We decompose $psi_1inOmega^0(M,ad Potimesmathbb C)$ into real and imaginary parts $$psi_1=omega+i eta$$ for $omega,etain Omega^0(M,ad P),$ and let $d_1$ act on the real part ($omega$) and $d_2^*$ on the imaginary part ($eta$).

Then, we have $$d_1(psi_1,psi_2)=((d_Aomega)^{0,1},[Phi,omega])=(d_A''omega,[Phi,omega])$$ by definition. The dual of the operator $$d_A''colonOmega^{1,0}(M,ad Potimes mathbb C)to Omega^2(M,ad Potimes mathbb C); ; psi_2mapsto d_A''psi_2$$ is by Serre duality the operator $$d_A''colon Omega^0(M,ad Potimes mathbb C)to Omega^{0,1}(M,ad Potimes mathbb C);; psi_2mapsto d_A''psi_2,$$ and using the hermitian metric and the fact that $d_A$ is unitary the adjoint operator gets identified with $$d_A'colon Omega^0(M,ad Potimes mathbb C)to Omega^{1,0}(M,ad Potimes mathbb C);; psi_2mapsto d_A'psi_2.$$ The adjoint of the operator $$phiinOmega^2(M,ad P)mapsto ([phi,Phi^*]+[Phi,phi^*])^{0,1}inOmega^{0,1}(M,ad P)$$ becomes $$psi_2mapsto -[psi_2,Phi^*]$$ ( taking the $i-$ part into account).

It remains to describe the operator $d_2^*$ acting on the imaginary part $ieta$ of $psi_1.$ As above, this becomes $$d_2^*((ieta))=(id_A''eta,i[Phi,eta]).$$ Putting together the pieces proves the claim.

Answered by Sebastian on February 5, 2021

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