I am reading Charles Nash's book on differential topology and QFT. In particular, I have question on the part calculating dimension of instanton moduli space. The question split into conceptual part and calculation part.

In the book p235-236, the dimension problem is recast into cohomology problem: \begin{equation} \dim \mathcal{M} = \dim \frac{{\ker d_A^ - }}{{{\mathop{\rm Im}\nolimits} {d_A}}} \end{equation}

and it is identified as the dimension of the second cohomology group of the complex \begin{equation} 0 \xrightarrow{\iota} \Gamma \left( {adP} \right)\xrightarrow{d_A} \Gamma \left( {{T^*}M \otimes adP} \right) \xrightarrow{d_A^- \equiv \pi^- d_A} \Gamma \left( {{\Lambda ^2}{T_-^*}M \otimes adP} \right) \xrightarrow{\pi^+} 0 \end{equation} with \begin{equation} {H^1} = \frac{{\ker {d_A}}}{{{\mathop{\rm Im}\nolimits} \iota }} = \ker {d_A},\;\;{H^2} = \frac{{\ker {\pi ^ - } \circ {d_A}}}{{{\mathop{\rm Im}\nolimits} {d_A}}},\;\;{H^3} = \frac{{\ker {\pi ^ + }}}{{{\mathop{\rm Im}\nolimits} {\pi ^ + } \circ {d_A}}} \end{equation}

Now we need to argue $h^3 = 0$ (for self-dual manifold with positive scalar curvature).

But my understanding is: ${\rm ker}\pi^+ = \Gamma \left( {\Lambda _ - ^2{T^*}M \otimes adP} \right)$ and the $H^3$ is just $\Gamma \left( {\Lambda _ - ^2{T^*}M \otimes adP} \right)/{\rm Im}d_A$. So $H^3$ tells us how many anti-self-dual forms there are which are **NOT** $d_A$-exct.

(1) **My feeling is that, among the many arbitrary anti-self-dual 2-forms, there must be some, right? Naively thinking, setting $A = 0$, there should be some anti-self-dual 2-form which is not even $d$-closed, let alone being $d$-exact.** So how should I understand when imposing curvature condition, all these (rather easy to exist) forms are forbidden to exist?

(2) The book compute the dimension $h^3$ by investigating the Laplacian \begin{equation} \Delta _A^{\left( 2 \right)} = d_A^ - {\left( {d_A^ - } \right)^\dagger } \end{equation} and the answer provided is \begin{equation} \Delta _A^{\left( 2 \right)} = \frac{1}{2}d_A^{\left( 1 \right)}{\left( {d_A^{\left( 1 \right)}} \right)^\dagger } + \frac{R}{6} - {W_ - } \end{equation} where $R$ is scalar curvature, $W_-$ is the anti-self-dual part of Weyl tensor.

I tried to do this in coordinate: \begin{equation} {\Delta _A}\omega \sim\frac{1}{2}\left( {{D_m}{D^k}{\omega _{kn}} - {D_n}{D^k}{\omega _{km}}} \right) - \frac{{\sqrt g }}{2}{\epsilon _{mn}}^{pq}{D_p}{D^k}{\omega _{kq}} \end{equation}

To produce curvature, I commute the $D^k$ with $D_m$ and $D_n$. There are a few terms, schematically

======================== OLD QUESTION ========================

(1) ${Ric_{m}}^l{\omega _{ln}}$

(2) ${R_{mknl}}{\omega ^{kl}}$

(3) $\pi^- d_A^\dagger d_A \omega$

(4) $D^kD_k\omega$

along with anti-symmetrization of $m$ and $n$, and anti-self-dual completion. However, non of the objects like scalar curvature and $d_A d_A^{\dagger}$ comes out.

**So I am wondering what has gone wrong? Or this is the right track and I should keep on expanding curvature into scalar, Ricci and Weyl tensor?**

======================================================

========================UPDATE========================

Expanding out curvature in terms of Ricci, Scalar, Weyl, the expression turns into basically of the form suggested by Nash with correct coefficient, except for the differential terms \begin{equation} = \frac{1}{2}\left[ {{D^k}\left( {{D_m}{\omega _{kn}} + {D_n}{\omega _{mk}}} \right) + {\rm{A.S.completion}}} \right] \end{equation} instead of $d_A d_A^\dagger$ in the book. It doesn't seem positive definite, which is a requirement for vanishing $h^2$: multiplying $\omega^{mn}$, integrated over $M$ and completing squares, the differential term seems to split into positive and negative term \begin{equation} \sim \int {{\omega ^{mn}}{D^k}\left( {{D_m}{\omega _{kn}} + {D_n}{\omega _{mk}} + {D_k}{\omega _{nm}}} \right) - \int {{D^k}{\omega ^{mn}}{D_k}{\omega _{mn}}} } +... \end{equation} where the 1st term is positive by partial integration, and ... denotes anti-self-dual completion.

So again, **Can some one points out what mistake I have made, or the differential term is secretly positive definite?**

Self-duality in four-dimensional Riemann geometry. It is very well written. $\endgroup$ – Liviu Nicolaescu Apr 19 '13 at 8:47