Let $E$ be a Euclidean vector bundle over the unit ball centered at the origin $B^n(0)$. Let $\nabla$ and $\nabla'$ be two metric connections such that the curvatures coincide globally, i.e. $F\_\nabla\equiv F\_{\nabla'}$. Is it true that there exists a Gauge transformation mapping one connection into the other? (maybe up to considering a smaller ball/analytic connections?)


If I am not mistaken, if we knew that the holonomy groups of the two connections coincide, then the result should follow from the Reduction Theorem (Theorem 7.1 Kobayashi--Nomizu Foundations of differential geometry). However we only know that the Lie algebras coincide (by Ambrose--Singer Theorem). Can we reconstruct the actual groups locally?



Does the curvature locally determine the connection?
The answer is 'not always'. Here is a simple case where you cannot recover the connection up to gauge transformation from the curvature: Let $n=2$, let the rank of $E$ be $m$, and, since $E$ is trivial over $B^2(0)$, we might as well take it to be $E = B^2(0)\times\mathbb{R}^m$, i.e., trivialized. Then $F$ is just a $2$-form with values in $\mathfrak{so}(m)$, and a connection $\alpha$ is a $1$-form on $B^2(0)$ with values in $\mathfrak{so}(m)$. If $F$ is the curvature of $\alpha$, we have $\mathrm{d}\alpha+\alpha\wedge\alpha = F$. If $F = f\,\mathrm{d}x^1\wedge\mathrm{d}x^2$ and $\alpha = a\_1\,\mathrm{d}x^1+a\_2\,\mathrm{d}x^2$, then this becomes the simple partial differential equation
$$
\frac{\partial a\_2}{\partial x^1}-\frac{\partial a\_1}{\partial x^2}
+[a\_1,a\_2] = f.
$$
If you specify $a\_1$ (for example), this becomes a linear first order partial differential equation for $a\_2$, so there are many solutions, no matter what $F$ is. Essentially, you get to choose an arbitrary function $a\_1$ on $B^2(0)$ with values in $\mathfrak{so}(m)$ and then you get to choose $a\_2(0,x^2)$ arbitrarily. So $F$ certainly does not determine $\alpha$.


Well, what about up to gauge transformation? Suppose that $\beta$ be another connection with curvature $F$. Then $\mathrm{d}\beta+\beta\wedge\beta = F$. If $\beta$ were gauge equivalent to $\alpha$, then there would exist a map $g:B^2(0)\to\mathrm{SO}(m)$ such that $\beta = g^{-1}\mathrm{d}g + g^{-1}\alpha g$. This would imply that $F = g^{-1}Fg$, so $f = g^{-1} f g$.


If $f\equiv0$, this is no condition, but if $f$ is 'generic' in the sense that, at each point $p\in B^2(0)$, the centralizer of $f(p)\in\mathfrak{so}(m)$ is a maximal torus in $\mathrm{SO}(m)$, then $g(p)$ would have to take values in that maximal torus. When $m>2$, a maximal torus is a proper subgroup of $\mathrm{SO}(m)$, so the only possible gauge transformations that could work would be very restricted, too restricted to be able to account for the arbitrariness in $\alpha$.


To fix ideas, take $m=3$ and assume that $f\equiv f(0)$ is a nonzero constant function on $B^2(0)$. Then $\beta$ could be gauge equivalent to $\alpha$ only if the gauging function $g$ took values in the maximal torus (a circle) consisting of the elements of $\mathrm{SO}(3)$ that commute with $f(0)$. That forces $g$ to take values in a circle, so it depends on only one function of two variables, but $\alpha$ depends on three functions of two variables, so the vast majority of the solutions to $\mathrm{d}\alpha + \alpha\wedge\alpha = F$ will not be gauge equivalent.


In fact, one can find a solution $\alpha\_1$ that has holonomy in the maximal torus and a solution $\alpha\_2$ whose holonomy is all of $\mathrm{SO}(3)$: Let
$$
F = \begin{pmatrix}0&\mathrm{d}x\_1\wedge\mathrm{d}x\_2&0\\
-\mathrm{d}x\_1\wedge\mathrm{d}x\_2&0&0\\
0&0&0
\end{pmatrix}
$$
while
$$
\alpha\_1 = \begin{pmatrix}0&x\_1\,\mathrm{d}x\_2&0\\
-x\_1\,\mathrm{d}x\_2&0&0\\
0&0&0
\end{pmatrix}
\quad\text{and}\quad
\alpha\_2 = \begin{pmatrix}
0&0&\phantom{-}\mathrm{d}x\_1\\
0&0&-\mathrm{d}x\_2\\
-\mathrm{d}x\_1&\mathrm{d}x\_2&0
\end{pmatrix}.
$$


In dimension $n=3$, generically, there will be many connections $\alpha$ with a specified curvature $F$, and, when $m>2$, the generic $F$ will have the property that two connections with curvature $F$ will be gauge equivalent only when they are equal.


Meanwhile in dimensions $n >3$, for the generic $F$, the equation $\mathrm{d}\alpha + \alpha\wedge\alpha = F$ will completely determine $\alpha$ algebraically by the Bianchi identity $\mathrm{d}F = F\wedge\alpha-\alpha\wedge F$. Thus, a 'generic' curvature determines its connection, and not just up to gauge equivalence.


