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?



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?

  • $\begingroup$ Perhaps the best place to look is the original source, namely the paper by Atiyah, Hitchin, Singer Self-duality in four-dimensional Riemann geometry. It is very well written. $\endgroup$ – Liviu Nicolaescu Apr 19 '13 at 8:47
  • $\begingroup$ @Liviu Nicolaescu: Thanks, I definitely will look at the paper, though maybe I should first try to complete the brute force expansion of curvature tensors. $\endgroup$ – Lelouch Apr 19 '13 at 14:22

Finally have time to update. After applying the projection operator, actually the $\pi^- d_A^* d_A$ merge with the original operator, and hence only positive-definite objets are left on the RHS, proving the statement form the book.

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