I am trying to understand a certain passage in the book "Geometry of Four-Manifolds" by Donaldson and Kronheimer (specifically, a computation in section 5.2.3). I am confused on the proof of Proposition 5.2.17, which states:

Proposition (5.2.17) Let $\hat{\nabla}$ be the tautological connection on the $SO(3)$-bundle $\mathfrak g_{\pi_2^*E}$, and $\nabla$ the quotient of this connection on the quotient bundle $\mathfrak g_{\mathbb{P}}\to \mathscr{B}^*\times X$. The three components of the curvature of $\nabla$ at a point $([A], x)$ are given by

  1. $F(\nabla)(u,v) = F(A)(u,v)$
  2. $F(\nabla)(a,v) = \langle a, v\rangle$
  3. $F(\nabla)(a,b) = - 2 G_A\{a,b\}|_x$.

Here, $u,v\in T_x X$, $a,b\in \Omega^1(\mathfrak g_E)$ satisfying $d_A^*a=d_A^*b=0$; $G_A$ is the Green's operator for the Laplacian $d_A^*d_A$ on $\Omega^0(\mathfrak g_E)$; and $\{,\}$ is the natural pairing formed from a metric on $X$ and the Lie bracket on $\operatorname{Lie}(G)$, $G$ being the gauge group.

Here, $\mathscr{B}^*$ is the space of irreducible connections on a bundle $E\to X$ modulo the action of the group of gauge transformations.

I am specifically confused about how they apply equation (5.2.16) (which is marked as $(*)$ below) to deduce this curvature, mainly because I do not understand how they figure out what $\Phi$ "does" in the case of the qoutient of the tautological connection. So, my question is,

How do they determine what $\Phi$ is in order to apply equation $(*)$ below to deduce this Proposition?

Here are the relevant details from this passage in the book. Suppose a Lie group $\Gamma$ acts freely and properly on a manifold $\hat{Y}$. Also assume we have a bundle $\hat{E}\to \hat{Y}$ and an action of $\Gamma$ on $\hat{E}$ that is linear on fibers and that covers the group action on $\hat{Y}$. Let $Y := \hat{Y}/\Gamma$ and $E:= \hat{E}/\Gamma$.

We now suppose we are given two things:

  1. A connection $\hat{\nabla}$ in $\hat{E}$ invariant under $\Gamma$.
  2. A connection $H$ in the $\Gamma$-bundle $p:\hat{Y}\to Y$.

One then gets a quotient connection $\nabla$ in $E$ from this. Then, in order to compute its curvature, introduce the 1-form $B \in \Omega_{ \hat Y }^1 \otimes \operatorname{End}(\hat{E})$ given by $$ B:= \hat{\nabla} - p^* \nabla.$$ Then, because $B$ vanishes on $H$-horizontal vectors, we can write $B$ as $\Phi \circ \theta$, where $\theta$ is the connection 1-form for $H$ and $\Phi: \operatorname{Lie}(\Gamma) \to \operatorname{End}(\hat{E})$ is a linear map. One can then compute that $$(*)\quad F(\nabla)(U,V) = F(\hat{\nabla})(\hat{U},\hat{V}) - \Phi\circ \Theta(U,V)$$ where $U,V\in T_y Y$ and $\hat{U},\hat{V}$ are horizontal lifts to $T\hat{Y}$.

They apply this to the $SO(3)$-bundle of Lie algebras $g_{\pi_2^*E}$ obtained from the pullback of $E$ along $\pi_2:\mathscr{A}^*\times X \to X$ and $\mathscr A^*$ is the space of irreducible connections on $E$; where $\Gamma$ the group of gauge transformations modulo $\pm 1$; with $\hat{Y} = \mathscr{A}^*\times X$; with $H$ being the connection on $\mathscr{A}^*$ obtained from slice neighborhoods for the action of the gauge transformations; and $\hat{\nabla}$ is the tautological connection on $\pi_2^*(E)$ (or, rather, the induced one on $\mathfrak g_{\pi_2^* E}$).

They use the results that for $H$, the connection form $\theta$ and curvature form $\Theta$ are $\theta_A(a) = -G_A d_A^* a$ and $\Theta_A(a,b) = -2G_A{a,b}$, which I am fine with. What is really bothering me is how they figure what $\Phi$ (or $B$ for that matter) is, since the pullback of the quotient of the tautological connection $p^* \nabla$ gives me a serious headache. Really, any advice or other resources for this calculation would be appreciated.

Edit: I have attempted to try and calculate it from basic principles. Let's call the Let's say we are at a point $(A, x)$ of $\mathscr{A}^*\times X$. The mapping $\gamma:\operatorname{Lie}(G)\to T_\nabla \mathscr{A}^* \times T_x X$ is given by taking a section $\xi \in \Omega^0(\mathfrak{g}_E)$ and sending it to $$\gamma(\xi) = (d_A\xi, 0)\in \Omega^1(\mathfrak{g}_E)\times T_x X = T_\nabla \mathscr{A}^*\times T_xX.$$ The value of $\Phi(\xi)$ at $x$ is the endomorphism on $\mathfrak{g}_{\pi_2^*E}$ determined by $B(\gamma(\xi))$. Then we proceed to determine how $\hat{\nabla}_{\gamma(\xi)}$ and $(p^*\nabla)_{\gamma(\xi)}$ perform on sections of $\mathfrak{g}_{\pi_2^*E}$.

I think $p^*\nabla$ is easiest to calculate: since a pullback connection is uniquely determined by the general formula $$ (p^*\nabla)_v(p^*s) = p^*(\nabla_{p_*v} s), \quad v\in T\hat{Y}, s:Y\to E, $$ we see that $p^*(\nabla)_{\gamma(\xi)}$ is identically $0$ on all sections because $p_*(\gamma (\xi))=0$, which is true because $\gamma$ maps into the vertical subspace. However, because $\hat\nabla$ is tautological, it is trivial in the $\mathscr{A}^*$ directions, so I also think $\hat\nabla_{\gamma (\xi)}=0$. This leads me to think $\Phi$ is identically $0$, which seems suspect.

  • $\begingroup$ I am currently trying to digest the same passage. I have no answer, but a question about your definition of $B$: it seems $p^* \nabla$ is a connection in $p^* E$ while $\hat{\nabla}$ is a connection in $\hat{E}$. So, they are connections on different bundles. How can you subtract them? $\endgroup$
    – user505117
    Mar 24, 2021 at 13:41
  • 1
    $\begingroup$ @user505117 Great question. I hadn't ever thought of it, but I imagine there's a canonical identification between $\hat{E}$ and $p^*E$ since $p$ is such a nice map. Perhaps one could show that the diagram with $\hat{E}, \hat{Y}, E, $ and $Y$ is a fiber square. $\endgroup$
    – Andrew
    Apr 2, 2021 at 21:37
  • $\begingroup$ I think the canonical identification of $\hat{E}$ and $p^* E$ is given as follows: for $y \in \hat{Y}$, map $v \in \hat{E}_y$ to $[v] \in E_{[x]}$ and then to $[v] \in p^*E_x$. Both maps are canonical and well-defined and that defines a map $\hat{E} \rightarrow p^*E$. $\endgroup$
    – user505117
    May 6, 2021 at 15:30


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