# Curvature of tautological connections over the space of connections

My question is about computing the curvature of a quotient connection, specifically for the case of the quotient of the tautological connection of a universal bundle on the moduli space of connections.

The calculation I am trying to understand shows up in a few places, for instance in the book "Geometry of Four-Manifolds" by Donaldson and Kronheimer, in section 5.2.3. I am confused on the proof of Proposition 5.2.17, which I will recount below after introducing some notation.

Let $$E\to X$$ be a complex smooth bundle over a smooth four-manifold $$X$$ with hermitian structure $$h$$. (Donaldson and Kronheimer assume it is an $$SU(2)$$-bundle, but I think this can be generalized to the structure group $$U(n)$$.) Let $$\mathscr{A}$$ be the affine space of unitary connections on $$E$$, and let $$\mathscr{A}^*\subset \mathscr{A}$$ be the space of irreducible connections. Set $$\mathbb{E}:= \pi_2^*E$$ where $$\pi_2:\mathscr{A}^*\times X \to X$$ is the projection, and let $$\mathfrak{g}_\mathbb{E}$$ be the associated bundle of Lie algebras. Both $$\mathbb{E}$$ and $$\mathfrak{g}_\mathbb{E}$$ carry a tautological connection $$\hat\nabla$$, which restricts to a connection $$A$$ on $$E$$ on every slice $$\{A\}\times X\subset \mathscr{A}^*\times X$$, and which is trivial in the $$\mathscr{A}$$-directions.

Let $$\mathscr{G}$$ be the group of gauge transformations of $$E$$, and $$\mathscr{G}_0:= \mathscr{G}/U(1)$$ be the group of reduced gauge transformations. We then have an orbit space $$\mathscr{B}^*:= \mathscr{A}^*/\mathscr{G}_0$$.

Here is Proposition 5.2.17, which involves calculating the curvature of the quotient connection $$\nabla$$ in the quotient $$\mathfrak{g}_\mathbb{P}:=\mathfrak{g}_\mathbb{E}/\mathscr{G}_0\to \mathscr{B}^*\times X$$.

Proposition (5.2.17) Let $$\hat{\nabla}$$ be the tautological connection on $$\mathfrak g_{\mathbb{E}}$$, 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 $$([A], x)\in \mathscr{B}^*\times X$$ 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 $$\operatorname{Lie}(G)$$, with $$G=U(n)$$ being the gauge group.

I am specifically confused about how they apply equation (5.2.16) (which is marked as equation $$(*)$$ 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 calculate the map $$\Phi$$ below 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 $$g_{\mathbb{E}}$$, with :

• $$\Gamma=\mathscr{G}_0$$
• $$\hat{Y} = \mathscr{A}^*\times X$$
• $$H$$ is the connection on the $$\mathscr{G}_0$$-bundle $$p:\mathscr{A}^*\to \mathscr{B}^*$$ obtained from slice neighborhoods for the action of the gauge transformations
• $$\hat{E} = \mathfrak{g}_\mathbb{E}$$
• $$\hat{\nabla}$$ is the tautological connection on $$\mathfrak g_{\mathbb{E}}$$

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. I can see the answer they get looks like the restriction map $$\Omega^0(\mathfrak{g}_E)\to \operatorname{Lie}(G)_x$$, but how they deduce this is completely opaque to me.