# Extension of smooth maps at a cusp

There is a short remark in deCarmos "Riemannian Geometry" (p. 67)

and I wonder about the condition that the vertex angles must be $\neq \pi$. If $s_1$ and $s_2$ are two differentiable maps on an open set containing $A$ and if $p$ is a cusp point in $\partial A$, then how is it possible that $ds_1|_p \neq ds_2|p$?

Taking the first open set $U$ in the above quote, what about a sequence $x_n$ in $U$ sucht that $x_n \rightarrow p$ and then by continuity $ds_1|_p = ds_2|p$? Where does the condition about the angle $\neq \pi$ enter the discussion?

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How are two segments meeting with a vertex angle of $\pi$ different from a single line segment? (I.e., vertices of angle $\pi$ are completely superfluous.) – Eric Towers Jun 6 '14 at 8:10
Don't confuse it with an angle of $0$ (or $2\pi$), it is a cusp at this vertices. – Blah Jun 6 '14 at 8:15
I don't... I confuse it with a straight line because it is a straight line. In particular, I can take an analytic curve and announce that every point on it is a vertex with angle $\pi$. This is not a profitable announcement. – Eric Towers Jun 6 '14 at 8:17
Hm, the angle is defined via the metric as usual, so an angle of $\pi$ means opposite direction, between the left and right velocity vectors of our piecewise smooth curve at a vertex. – Blah Jun 6 '14 at 8:22
Are you certain they are not taken as limits of directions approaching the vertex along each curve? – Eric Towers Jun 6 '14 at 8:25

The question is whether the values of $s$ on $A$ determine $Ds_p$ for all $p \in A$. Knowing $Ds_p$ is equivalent to knowing $Ds_p (v)$ for two linearly independent $v$. Knowing $Ds_p (v)$ is equivalent to knowing $s(p + t_n v)$ for some sequence $0<t_n \to 0$.
If $p \in U$ then there is no issue - we can take an open ball around $p$ contained in $A$. If $p\in \partial A$ is not a vertex, then exactly half the vectors $v \in T_p M$ have rays $p + t v$ that stay in $A$ for some positive time, and thus the values of $s$ on $A$ easily determine $Ds_p$.
When $p$ is a vertex with turning angle $\theta$, there is a cone of vectors with angle $\pi - \theta$ with rays staying in $A$ for a bit, so as long as $\theta < \pi$ there are two linearly independent such vectors.
The issue when $\theta = \pi$ is that the values of $s$ on $A$ only determine $Ds_p (v)$ for a single direction $v$ - in every other direction there is a short time for which the rays are disjoint from $A$, and thus the behaviour of $Ds_p$ in these directions is not determined by $s|_A$. I don't have time to work out a specific example but it shouldn't be too hard.
I believe your argument using sequences rules out this possibility if you assume that $s$ is continuously differentiable.