From: rusin@vesuvius.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Immersion of Manifolds Date: 17 Dec 1998 06:44:07 GMT In article <366F83C6.A18F1532@nwu.edu>, paul wrote: >Let f: RP2 -> R3 be a function from the real projective plane to R3 >defined by >f(x,y,z) = (yz,xz,xy) > >Show that f fails to be an immersion at 6 points. A map f is an immersion on an open set if f'(p) is an injection at all points p in that open set. Your problem is a bit convoluted only because of the notation: you describe a map f : R^3 -> R^3, which you presumably wish to restrict to the sphere S^2 in R^3, and then view as a map on RP^2 (which is possible since f(-p) = f(p) ). We use f to describe all three of these maps into R^3. Since S^2 -> RP^2 is simply a covering, it suffices to show f is an immersion at all but 6 antipodal pairs of points of S^2. Since S^2 is a submanifold of R^3, it suffices to show that ker(f'(p)) \intersect T_p = {0} for all p in S^2 except those 12 points. Well, this isn't so hard. f'(x,y,z) is given in matrix form as [ 0 z y ] [ z 0 x ] , [ y x 0 ] a matrix whose characteristic polynomial is X^3 - X - 2xyz when x^2+y^2+z^2=1. In particular, the rank is at least 2 and f'(p) only has a nontrivial kernel if x, y, or z is zero. Now if, say, x=0, then the kernel of f'(p) is spanned by the column vector [0,-y,z]^t. No nozero multiple of this is in the tangent space T_p, in general, since T_p is the set of vectors perpendicular to [x,y,z]^t = [0,y,z]^t. So in order to find a vector lying in both ker(f'(p)) and T_p, we need -y^2 + z^2=0, i.e. y = +- z. Of course then x^2+y^2+z^2=1 forces y^2 = z^2 = 1/2. In this way, we find the only points p at which f fails to be an immersion on RP^2 are the six points [0, c, c], [0, c, -c] [c, 0, c], [c, 0, -c] [c, c, 0], [c, -c, 0] where c = sqrt(1/2). I trust the assignment due date has passed. dave