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  <pbrock@nwu.edu> 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