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?



Deformation-Obstruction Theory of YM Instantons
$\newcommand{\A}{\mathscr{A}}$ $\newcommand{\G}{\mathscr{G}}$ 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*.


