For a surface with $K=0$ everywhere, show that the holonomy group reduces to the identity element.

Consider a connected surface $S$ embedded in $\Bbb R^3$ and let $\alpha$ be a closed path in $S$ connecting $p\in S$ back to itself. Now we define $P_{\alpha}$ as the effect of parallel transport on a vector as it moves along $\alpha$ from $p$ back to $p$. Now define $P_{\alpha}\circ P_{\beta} = P_{\alpha \circ \beta}$. Then I have already shown that $$H_p(S) = \{ P_{\alpha} :T_pS \to T_pS\ |\ \alpha\text{ is a path connecting }p\text{ to }p\}$$ is a group under the operation just defined, this group is called the holonomy group of $S$ at $p$. Also, for any other $q\in S$ we have that $$H_pS \cong H_qS.$$ This implies that we can talk about the holonomy group of a surface without referring to a specific point. Now consider a surface with Gaussian curvature $K=0$. Then the holonomy group reduces to the identity element. This last statement is what I would like to show. This is a question in Do Carmo's book on differential geometry.

In the case that $S$ is a region in the plane this is almost trivial. Any vector transported around a curve in the plane will always lie in the tangent space to the plane at each point on the curve. Therefore the effect of parallel transport will just return the vector to itself after completing the path. Now if we consider a general surface $S$ with $K=0$ this is not so obvious (at least not to me). We do know that locally this surface is isometric to a region in the plane.

I was thinking maybe $S$ can be found to be globally isometric to a region $U$ in the plane via some isomorphism $\phi$. Then any curve $\alpha$ in $S$ is mapped to this region. Then we consider the mapping $\varphi :H_pS \to H_{\phi (s)}U$ such that $\varphi(P_{\alpha}) = P_{\phi(\alpha)} = id$. The last equality coming from the fact that $\phi(\alpha)$ lies in the plane. Now if I could show that $\varphi$ has trivial kernel then it would be an injection and I would be done. Assuming the result of my question is true $\varphi$ should indeed be an isomorphism but I can't see why. Also, I have not used the fact that $K=0$ on $S$ except in claiming the existence of an isomorphism between $S$ and $U$. But I have not used the distance preserving property here. In contrast, between any surface realized as a graph over some region of the plane there exists a trivial isomorphism to that region of the plane, but its holonomy group need not be trivial.

So this route led me nowhere, however I thought it would be nice to give an argument that is in the style of group theory instead of differential geometry. But in the end it seems this may not work out. I would greatly appreciate any input and help. Thanks!

• I'm skeptical. What if $S$ is a cone (missing its vertex)? It's true if $S$ is simply connected, as you can apply the Gauss-Bonnet Theorem to relate $\int_D KdA$ to the holonomy around $\partial D$. Commented Jan 9, 2015 at 23:04
• I am intuitively inclined to think that on the cone also the holonomy group should be trivial. Maybe I am missing something. Thanks for your comment! Commented Jan 10, 2015 at 0:49
• No, it's not. Make a cone out of a pacman figure (or solve the differential equation). Commented Jan 10, 2015 at 0:51
• Do Carmo probably meant a simply-connected surface; otherwise, as Ted Says, the claim is false. Incidentally, a Moebius band would be another example. Commented Jan 10, 2015 at 15:03
• Thanks for the comments! So how would I work in the requirement that the surface has to be simply connected? Commented Jan 10, 2015 at 17:43

Hint: First show that if S is a simply-connected Riemannian surface of zero curvature, then S admits a local isometry $S\to R^2$. (This is where you use simple connectivity.) Next, use this to show that any two paths from p to q result in the same parallel transport $T_pS\to T_q S$.