Identify antipodal points on the horizontal circles of a paraboloid

Question

Assume we have a paraboloid M, the image of $f:[0,2\pi) \times [0,1]\to\mathbb R^3$ defined by $f(t,r)=(r\cos(t),r\sin(t),r^2)$. We then identify the antipodal points of the horizontal circles $S_r=\{t\in [0,2\pi):(r\cos(t),r\sin(t),r^2)\}$ . Is the quotient $\tilde {M}$ a smooth manifold? In particular, I am not sure if $\tilde {M}$ is locally $\mathbb R^2$ at $(0,0,0)$.

Answer 1

$\newcommand{\Reals}{\mathbf{R}}$In a word, "yes". To start with an analogy, consider the equivalence relation $z \sim -z$ on the complex plane. The squaring map $z \mapsto z^{2}$ may be viewed as a smooth map to the quotient, and the image, the complex plane, is a smooth manifold. (To transfer this to the paraboloid, view the paraboloid as the graph of $z \mapsto |z|^{2}$.)

The presumably confusing aspect of this example is, the map $$ (r\cos t, r\sin t, r^{2}) \mapsto (r\cos(2t), r\sin(2t), r^{2}) $$ is not smooth at the origin. The squaring map corresponds instead to $$ (r\cos t, r\sin t, r^{2}) \mapsto (r^{2}\cos(2t), r^{2}\sin(2t), r^{4}). $$

Separately, note that in the analogous situation of the graph of $f(x) = \|x\|^{2}$ (with $x$ denoting a point of $\Reals^{n}$, $n > 2$), the quotient space is not a manifold at the origin: In this case, the "link" of the origin is the $(n - 1)$-dimensional projective space, not the $(n - 1)$-dimensional sphere.