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I have heard that the least dimension $m$ required for $\mathbb{R}P_2$ to be embedded in the Euclidean space is 4, thus I wanted to find an explicit formulae for it. I found two possible strategies, but is not sure that they'll work.

  1. Define $\phi([x_1,x_2,x_3])=(|x_1|,|x_2|,|x_3|,x_1x_2+x_2x_3+x_3x_1)$, where $[x_1,x_2,x_3]$ is the eq.class under quotient from $S^2$. I hope that would be an embedding, since the last is a symmetric polynomial which is equal for $(x_1,x_2,x_3)$ and $-(x_1,x_2,x_3)$.

  2. My second stragy is more geometric approach. Note that the projective plane deleted a circle is a Möbius band $M$. Thus if I could "paste" the boundary of a closed disk (via the fourth dimension) onto the boundary circle of a Möbius Band, then I'm done. But since I'm no good at "visualizing" four dimensions, I don't know exactly how to proceed.

My question is:

1) is 1. an embedding or not? (Or give another more elegant imbedding)

2) is there a way to realize my second approach? or is it hopeless?

3) Is there a more systematic way of doing such embeddings? (Frankly, If our world is in $\mathbb{R}_2$, I would probably not even be able to imagine how to imbed the torus into $\mathbb{R}_3$)

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    from http://en.wikipedia.org/wiki/Real_projective_plane#Embedding_into_4-dimensional_space $f(x,y,z)=(xy,xz,y^2-z^2,2yz)$, $f:S^2\to\mathbb{R}^4$, satisfies $f(-x,-y,-z)=f(x,y,z)$ so factors through the real projective plane.2011-12-16
  • 1
    Related: http://math.stackexchange.com/questions/40833/embedding-of-mathbbrp2-in-mathbbr42011-12-16

3 Answers 3