# 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.

The set-up mentioned in your question also appears in the paper "Differential characters and cohomology of the moduli of flat Connections" by Marco Castrillón López, Roberto Ferreiro Pérez. The desired computation of the curvature is in proposition 3 on page 6, and uses equivariant cohomology in the Cartan model. Hope this helps.

Recently, I had read this part in Donaldson and Kronheimer's book. Finally, I got some some reasonable view to understand them. I hope my opinion is correct, and helpful to you.

Because, the $$\Gamma$$-bundle $$p:\hat{Y}\to Y$$ is the quotient by $$\Gamma$$ action. So, I will look the tangent space like this $$T\hat{Y}\cong TY\oplus {\rm Lie}(\Gamma).$$ The following is my understanding picture for the relationship between the parallel transports of $$\hat{Y},\, Y$$ and the horizontal distribution $$H$$ (the subspace generated by parallel transport of $$\Gamma$$-bundle).

So if we restrict that $$\Phi\circ\theta(\hat{U}) = \hat{\nabla}_{\hat{U}}-p^*\nabla_{U}$$ on $$Y$$, then $$(\Phi\circ\theta)|_{Y}\in \Omega_Y^1({\rm Lie}(\Gamma))$$ is the connection of $$Y$$ which $${\rm Lie}(\Gamma)$$ value 1-form was determined by H. However the map $$\Phi$$ was defined as a linear map $$\Phi:{\rm Lie}(\Gamma) \to {\rm End}(\hat{E})$$, so $$\Phi$$ should be a inclusion map, or just an identity map.

Another difficulty is to understand the calculation result $$F(\nabla)(a,b) = \Theta_A(a,b)=-2G_A\{a,b\}|_x$$. It needs to use the definitions :

• $$a\in\mathscr{A}$$ and $$\displaystyle \hat{d}_a A = \lim_{t\to 0}\frac{\left( (A+ta)-A \right)}{t} = a$$.
• $$\theta_A(a)=-G_A(d_A^* a)$$
• if $$a\in T_A =\{ a\,|\, d_A^*a=0\}$$ then $$\theta_A(a)=0$$.

Thus, the curvature $$\Theta_A =\hat{d}\,\theta_A + \theta_A\wedge\theta_A$$ in the directions $$a,b\in T_A$$ will be $$\Theta_A (a,b) =\hat{d}\,\theta_A \, (a,b)$$ And $$\hat{d}_a\,\theta_A(b) = -(\hat{d}_a G_A)(d_A^*b) - G_A( (\hat{d}_a d_A^*) b) - G_A(d_A^* (\hat{d}_a b))$$ By definition the first term and last term ( because $$d_A^* (\hat{d}_a b)=d_A^* a=0$$) are zeros. And the middle term includes $$\begin{eqnarray*} (\hat{d}_a d_A^*)\,b &=& \star(\hat{d}_aA)\star b \\ &=& \star\, a \star b = \{ a,b \} \end{eqnarray*}$$