Deformation-Obstruction Theory of YM Instantons In Donaldson-Kronhiemer Section 4.2.5. (local models of the moduli space of YM instantons) they first get local models of the moduli space $M$ inside the space of all connections modulo gauge $\mathcal{B}$ by taking $(F^+)^{-1}(0)/\Gamma_A$, where ($\Gamma_A$ is the isotropy group of the connection).
Now here comes my confusion. They then remark its differential at zero is $d^+_A$, but that $H^1$ of its deformation-obstruction complex $$\Omega^0(\mathfrak{g}_E) \xrightarrow{d_A} \Omega^1(\mathfrak{g}_E) \xrightarrow{d^+_A} \Omega^+(\mathfrak{g}_E)$$
is ker $\delta_A$, where $\delta_A = d^+_A \oplus d_A^\ast : \Omega^0 \to \Omega^1 \oplus \Omega^+$ and this shows we have finite-dimensional local models of $M$ cut out by the zeros of a $\Gamma_A$-equivariant map $$f : \text{ker } \delta_A \to \text{coker }d^+_A.$$
But why the extra operator $d^\ast_A$? Why should the space of infinitesimal deformations be ker $\delta_A$ rather than ker $d^+_A?$ After all, it is $d^+_A$ which is the differential of $\psi$, not $\delta_A$. Is the Zariski tangent space not defined as the kernel of the differential of the map $\psi$ which cuts out a local model?
 A: $\newcommand{\A}{\mathscr{A}}$  $\newcommand{\G}{\mathscr{G}}$   Denote by $\A$ the space of connections, by $\A_-$ the space of ASD connections and by $\G$ the gauge group. For ssimplicity I will not keep track of various Sobolev decorations.   The moduli space  $\newcommand{\M}{\mathscr{M}}$ $\M$  is defined as a set  by the equality
$$\M=\A_-/\G\cdot A. $$
Thus the tangent space to $\M$ at an ASD connection ought to be
$$T_A\M=T_A\A_-/T_1\G\cdot A.$$
Observe that $\newcommand{\gog}{\mathfrak{g}}$
$$ T_A\A_-= \ker  d_A^+,\;\; T_1\G\cdot A\cong\mathrm{Im}\;\big(d_A:\Omega^1(\gog_E)\to\Omega^1(\gog_E)\;\big). $$
Now you need to invoke a basic fact in Hodge theory.  If you have a  cochain complex
$$ 0\to V_0\stackrel{d_0}{\to} V_1\stackrel{d_1}{\to} V_2  \stackrel{d_2}{\to}\cdots  $$
then, 
$$ H^k(V_\bullet)\cong \ker\big(\; d_k\oplus d_{k-1}^*: V_k\to V_{k+1}\oplus V_{k-1}\;\big). $$
This happens under certain assumptions, e.g. , if the above complex is elliptic.
When applied to the deformation complex $\mathbf{C}$ of the ASD equation which is elliptic  we get
$$T_A\M=\ker d_A^+/ \mathrm{Im}\;(d_A)=H^1(\mathbf{C}) \cong \ker d_A^+ \oplus d_A^*. $$
A very  good  exercise I recommend solving is to verify that the  deformation complex $\mathbf{C}$ is indeed elliptic.
