0
$\begingroup$

Let $M \subset \mathbb{R}^d$ be a smooth 2-manifold that is homeomorphic to a sphere or a connected sum or tori. Does there always exists two points $x,y \in M$ such that the normals $\angle(n_x, n_y) \geq 90^{\circ}$? It seems intuitively obvious but how does one formally prove this?

$\endgroup$
2
  • 1
    $\begingroup$ This looks like a DG homework assignment. $\endgroup$ Commented Nov 24, 2014 at 23:19
  • 1
    $\begingroup$ It's not, self study. $\endgroup$
    – user62013
    Commented Nov 24, 2014 at 23:55

1 Answer 1

1
$\begingroup$

Assuming $M\subset \Bbb R^3$ and that is how you are measuring angles. Consider the map for each $x\in M$, $\phi_x: \Bbb R \rightarrow \Bbb R^3$ which moves along the normal at $x$. By the tranversality theorem, $\phi_x \pitchfork M$ for almost every $x\in M$. For all $x$, $|\phi_x(\Bbb R)\cap M| \geq 2$, so now you can just look at an intersection point adjacent to the one at $x$ (with no intersection points in between).

$\endgroup$
1
  • $\begingroup$ Actually I intended for $M \subset \mathbb{R}^d$, codimension $\geq 1$, but I think pointing me toward the transversality theorem should be enough. Thanks! $\endgroup$
    – user62013
    Commented Nov 24, 2014 at 23:54

Not the answer you're looking for? Browse other questions tagged .