Suppose you have a torus and you carefully make a cylindrical cut down the center. Then you stretch out the outer half and glue together annular regions of the plane in the empty space. Now you have a torus that has a flat top and bottom
Is this still a $C^\infty$ manifold? Or did we mess it up and it's merely a topological manifold. I ask because the curvature is discontinuous (it is zero on the flat parts and non-zero on the curvy parts).
In a lower dimension, this question still bothers me: This is a stretched out circle (I'll include the parametric functions to make it painfully obvious, how I got this shape):
Can someone please tell me the nature of these manifolds? Also (bonus:) If we've lost the $C^\infty$-ness by this construction, how can we maintain it? What is a smoothing procedure for, say, taking the strecthed out circle above and making it $C^\infty$ by somehow fixing the connecting points?
Thanks!
This is a $C^1$ manifold that is not $C^2$. It is not hard to give it an intrinsic structure as a $C^∞$ manifold, but its embedding in 3-dimensional space will still only be $C^1$, as the curvature jumps suddenly from a positive constant to zero.
For the flattening out, here is one way to do it: Pick a $C^∞$ function $\psi$ so that $\psi(x)=0$ for $0\le x≤\frac12$ in a neigbourhood of $0$, and $\psi(x)=x-1$ for $x≥\frac32$, and consider the graph $$\bigl\{\,(x,\sqrt{1-\psi(x)^2})\colon 0\le x\le2\,\bigr\}.$$ This is smooth; it is flat for $0≤x≤\frac12$, and the part with $x≥\frac32$ is a circular arc. Now join this with its mirror images through the axis, and you have a smooth flattened curve. Rotate it around the $y$ axis to get a smooth flattened torus.