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the title is more or less self-explanatory.

Why is it not possible to embed the Real Projective Plane $\mathbb{R}P^2$ in to the euclidean space $\mathbb{R}^3$?

I stumbled upon the following example: Let $\mathbb{Z}_2$ act on $\mathbb{R}^3$ by reflections (f(x)=x,f(x)=-x). Then the quotient space $Q$ is a cone over $\mathbb{R}P^2$. $Q$ is not a manifold. How to prove this statement?

I tried the following: Suppose $0$ would have a neighbourhood which is homeomorphic to $\mathbb{R}^3$. Then by restricting this homeomorphism one could Embed the projective plane into $\mathbb{R}^3$. Which is not possible (by google). However I did not find out why.

Is there an algebraic invariant, which prohibits it and is easy to visualize?

Thomas Andrews
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  • wiki has a non-rigourious proof for that. – Seyhmus Güngören Feb 28 '15 at 14:50
  • Wait wait wait, you can! What's Boyes' Surface! – Alec Teal Feb 28 '15 at 15:18
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    @Alec: Boy's surface is immersed in $\mathbb {RP}^3$, not embedded . – Georges Elencwajg Feb 28 '15 at 15:32
  • @GeorgesElencwajg what's the difference (I am not saying they are the same, I just... it's a new area to me!) – Alec Teal Feb 28 '15 at 16:08
  • An embedding $f:X\to Y$ can be thought of as an "exact" copy of $X$ that happens to live in $Y$. An immersion $g:X\to Z$, is does not have to be injective, which is where the self-intersections come in. Like the Klien bottle, it is also non-orientable, so to immerse it in $\mathbb{R}^3$, it must pass through itself. – N. Owad Feb 28 '15 at 17:11

2 Answers2

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Every codimension one closed submanifold of $\mathbb R^3$ is orientable but $\mathbb{R}P^2$ is not orientable .

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Georges Elencwajg
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$\newcommand{\Reals}{\mathbf{R}}\newcommand{\Proj}{\mathbf{P}}$You have two questions:

  1. Why does $\Reals\Proj^{2}$ not embed in $\Reals^{3}$? (Your title question, which Georges has answered.)

  2. Why is $Q = \Reals^{3}/{\pm}$ not a manifold? (The motivating question in your post, which 1. answers indirectly.)

Here's an alternative take on 2.: The "link" of the vertex of $Q$ (i.e., the boundary of the image of a ball centered at the origin of $\Reals^{3}$) is $\Reals\Proj^{2}$ instead of $S^{2}$. If some neighborhood $U$ of the vertex of $Q$ were homeomorphic to $\Reals^{3}$, there would exist a closed ball $B$ centered at the origin of $\Reals^{3}$ whose image in $Q$ is (contractible and) contained in $U$. Following with a homeomorphism $\phi:U \to \Reals^{3}$ would give a contractible subset of $\Reals^{3}$ whose boundary is $\Reals\Proj^{2}$.

The same idea implies, for example, that a cone on a $2$-torus is not a manifold, even though a $2$-torus does embed in $\Reals^{3}$.

Andrew D. Hwang
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