2

Is it possible to embed projective plane in 4-space? If not what is the reason and what is the smallest singularity set?

user113715
  • 1,101
  • It greatly saddens me that the StackExchange Rules Police felt the need to shut down a perfectly reasonable question like this one — in fact one that could have had some very interesting and useful answers — in the service of godknowswhat. – Dan Asimov Mar 08 '23 at 00:14

1 Answers1

4

Let $f:\mathbb{S}^2→\mathbb{R}^4$ be defined by $$f(x,y,z)=(yz,xz,xy,ax^2+by^2+cz^2)$$ this gives an embedding of $\mathbb{RP}^2$ in $\mathbb{R}^4$.

We have $f(x,y,z)=f(−x,−y,−z)$, so we get a map $F:\mathbb{RP}^2→\mathbb{R}^4$. The differential $dF$ has rank 2, so this is an immersion and $f$ is injective. Then have to embedding.

jimbo
  • 2,156
  • In $\mathbb{R}^3$ the projective space are self intersects. – jimbo Jan 15 '15 at 14:18
  • Ok, if the image is self intersection. How many singular points and what is (are) their type(s) – user113715 Jan 15 '15 at 16:17
  • A cross-cap is obtenided by gluing a disc to its boundary is a model of the real projective plane $\mathbb{RP}^2$. With an interval of self-intersection, and two points where this model is not an immersion of $\mathbb{RP}^2$. – jimbo Jan 15 '15 at 16:46
  • I think that (a, b, c) must be pairwise distinct. One of them may be zero. – Loic Jan 27 '23 at 21:34