Some time ago I wondered if there could be a smooth simple closed curve in the plane without a double normal — a chord of the curve that is perpendicular at both ends. This is a fun puzzle which I won't spoil for anyone who hasn't seen it. But today I wondered about the same question for a smooth surface in R^3 topologically equivalent to the 2-sphere S^2. Now for many smooth S^2 surfaces M — like convex ones — following the internal normal N(p) at point p of M until it first hits the surface, say at point Q(p) — the map Q : M —> M taking p to Q(p) is a continuous. If so, then having obtained Q(p) from p we now follow the internal normal N(Q(p)) at Q(p) until it first intersects the tangent plane T_p(M) of M at p. Let this point be f(p). Assuming no double normal, we know f has no fixed points: for all p in M, f(p) ≠ p. But then f(p) - p provides a vector field on S^2 continuous in p, known to be impossible (the hairy ball theorem), contradiction. But if that map Q above is *not* continuous, this argument doesn't work. So, Question: Does there exist a smooth S^2 in R^3 without a double normal? —Dan