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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?

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$\newcommand{\A}{\mathscr{A}}$ $\newcommand{\G}{\mathscr{G}}$ 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.

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  • $\begingroup$ Great! But I still do not see why the obstruction space should be coker $ d^+_A$ rather than coker $\delta_A$. $\endgroup$ – user100272 Nov 19 '17 at 21:29
  • $\begingroup$ The deformation complex has three terms. The obstruction space is $\ker (d_A^+)^* $\endgroup$ – Liviu Nicolaescu Nov 19 '17 at 23:41
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    $\begingroup$ Right, I suppose my question is why $C$ is the deformation-obstruction complex. I see that $H^1_A \simeq T_A \mathcal{M}$ by your argument. But I do not see why $H^2_A$ should, a priori, be ker$ (d_A^\ast)^\ast$. $\endgroup$ – user100272 Nov 20 '17 at 10:29
  • $\begingroup$ Use the Hodge theoretic fact I mentioned in my answer. The obstruction space is by definition $H^2(\mathbf{C})$. The deformation complex has only three terms and the last differential is $0$. There is no $d_k$ in the operator $d_k\oplus d_{k-1}^*$ whose kernel is $H^k(V_\bullet)$. $\endgroup$ – Liviu Nicolaescu Nov 20 '17 at 10:45
  • $\begingroup$ I understand the obstruction space is defined as $H^2(\textbf{C})$. What I do not see is why this particular complex $\textbf{C}$ is the deformation-obstruction complex (rather than some other complex). $\endgroup$ – user100272 Nov 20 '17 at 11:19

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