# Is every flat manifold with boundary locally isometric to the Euclidean half-space?

Let $M$ be a smooth manifold with boundary, endowed with a smooth Riemannian metric $g$.

Suppose $g$ is flat, and let $p \in \partial M$.

Is there an open neighbourhood of $p$ which is isometric to an open subset of the standard half-space $\mathbb{H}^n=\{ (x_1,...,x_n)| x_n \ge 0\}$, endowed with the standard Euclidean metric?

One approach would to be to use the exponential map at $p$ (which is a local isometry in the case with no boundary). However, I am not sure where $\exp_p$ is defined in this case.

I guess its domain will be the subset of inward-pointing vectors in $T_pM$.

Edit:

As seen by the example $M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \}$, the suggetsion to use the exponential map does not work. The problem is that to cover the boundary, we need to conside geodesics with initial velocities in $T_p\partial M$. But, if we consider these geodesics in the ambient space $\mathbb{R}^2$, then they are standard straight lines (which are tangent to the unit circle), hence do not lie in $M$ for any positive time.

So, indeed, the boundary "cannot be straightened out".

I would say no, because you cannot assure that the boundary will be straight. Consider the disk $$M = \{ x \in \mathbb R^2 \,:\, |x| \leq 1 \}\,.$$ It is clearly flat, but because the only isometries in $\mathbb R^2$ are translations and rotations you cannot straighten out the boundary, not even locally.
• What you are saying is very interesting. I also thought on the problem of bounding the sizes of the "good" neighbourhoods, where $f$ is invertible. In fact, this was the main motivation which made me ask the question... This is relevant to another question I asked on isometric immersions between manifolds with boundary, as you may see here: math.stackexchange.com/questions/2171418/… if you are interested. (The relevant part to our discussion is after "Edit 2"). Mar 14, 2017 at 7:41