I'm learning gauge theoretic topics about 4-manifolds, and I get stuck when I try to calculate the dimension of ASD moduli space.
For an oriented closed 4-manifold $M$, we first fix a riemannian metric $g$ hence we get a hodge star operator $*$ w.r.t. $g$ and we define $F^+_A=P^+d_A d_A$, here $P^+$ means projection to the eigenvector space with positive eigen value(view $*$ the linear map in $H^2(M)$), we call a connection is ASD if $F^+A=0$. So to calculate the dimension of moduli space, it suffices to calculate the dimension that {$a\in\Omega(adP)|d_A^*a=0$ and $F_{A+a}^+=0$}, so I consider the elliptic complex $0\rightarrow \Omega^0(adP)\rightarrow \Omega^1(adP)\rightarrow \Omega^2(adP)\rightarrow 0 $. The second map is $d_A$ and the third map is $P^+d_A$, where $A$ is a irreducible ASD connection.
To calculate the dimension, we only need to calculate the euler characteristic of the complex above, since $A$ is irreducible and $ker d_A$ consist of elements in $\Omega(adP)$ that is covariant constant, so $ker d_A\neq 0$ equivalent to the existence of non trivial gauge transformation satisfy $g•A=A$. So by the Irrduciblity $H^0=Ker d_A=0$.
However, I have no idea how to calculate $-dim H^1+dim H^2$, since $H^1 =ker (d^*_A+d^+_A)$, if $H^2=0$, which also means $Coker (d^*_A+d^+_A)=0$, so I can use index theorem to calculate the dimension of $H^1$. However the index theorem of arbitrary elliptic operator seems too complicate to calculate, but how to calculate this dimension?
Thanks for your answers or comments!