# Computing curvature of the quotient of the tautological connection

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.

• 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? Mar 24, 2021 at 13:41
• @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. Apr 2, 2021 at 21:37
• 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 $\hat{E} \rightarrow p^*E$. May 6, 2021 at 15:30