= Solution
For the adjoint real bundle $\operatorname{ad}E$, the <ASD deformation complex> is
$$
\boxed{0\longrightarrow\Omega^0(\operatorname{ad}E)
\xrightarrow{\ d_A\ }\Omega^1(\operatorname{ad}E)
\xrightarrow{\ d_A^+\ }\Omega^{2,+}(\operatorname{ad}E)
\longrightarrow0,}
$$
where $d_A^+=P_+d_A$ and $P_+=(1+*)/2$. On an adjoint section $\xi$, the <vector-bundle curvature> identity gives $d_A^2\xi=[F_A,\xi]$. Since $P_+$ acts only on the two-form part and $F_A^+=0$,
$$
d_A^+d_A\xi=P_+[F_A,\xi]=[F_A^+,\xi]=0.
$$
This proves it is a complex. Its three <cohomology> spaces are the infinitesimal stabilizers $H_A^0$, infinitesimal deformations modulo gauge $H_A^1$, and obstructions $H_A^2$.
The associated gauge-fixed <elliptic differential operator> is
$$
D_A=d_A^*\oplus d_A^+:\Omega^1(\operatorname{ad}E)
\longrightarrow\Omega^0(\operatorname{ad}E)\oplus\Omega^{2,+}(\operatorname{ad}E).
$$
The <ASD deformation index> formula is
$$
\begin{aligned}
\operatorname{ind}D_A
&=\dim H_A^1-\dim H_A^0-\dim H_A^2\\
&=-2\langle p_1(\operatorname{ad}E),[X]\rangle
-\frac32\bigl(\chi(X)+\sigma(X)\bigr)\\
&=8e(E)-3(1-b_1(X)+b^+(X)).
\end{aligned}
$$
Here the <Atiyah-Singer index theorem> is used for the formula, with $p_1(\operatorname{ad}E)=-4c_2(E)$ and $e(E)=c_2(E)[X]$. For the simply connected negative-definite base in this question, $b_1=b^+=0$, so \b[$\operatorname{ind}D_A=8e(E)-3$]. The alternating Euler characteristic of the three-term complex is the negative of this operator index; stating which index is being used avoids a sign ambiguity.
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