# Why $\mathbb{RP}^2$ can not be embedded to $\mathbb{R}^3$?

Is there any answer of this question around basic theory of differentiable manifolds?

• en.wikipedia.org/wiki/Real_projective_plane#Examples gives an outline of a proof using the Jordan separation theorem. Nov 20, 2014 at 19:15
• Georges Elencwjag has deleted a much better answer, just using his simple proof that every embedded hypersurface $S$ is the zero locus of a single equation with nontrivial differential on $S$, which I'll leave here until such time as he should undelete. math.stackexchange.com/questions/879334/… Nov 20, 2014 at 22:48
• Dear @Kevin: your extremely kind words (for which I thank you) have made me decide to undelete my answer... Nov 20, 2014 at 23:01
• @GeorgesElencwajg glad to hear it! As you say, it's a bit embarrassing that the argument is not obvious to one who's studied some differential topology, let alone that it's not well known. Nov 20, 2014 at 23:56

Every closed smooth hypersurface $X\subset \mathbb R^n$ (=submanifold of dimension $n-1$), compact or not, has an equation, i.e. $X=f^{-1}(0)$ for some smooth $f:\mathbb R^n\to \mathbb R$ satisfying $df(x)\neq0$ for all $x\in X$.
In particular $X$ is orientable so that $\mathbb P^2(\mathbb R)$, which is not orientable, cannot be embedded into $\mathbb R^3$

(Non-) bibliography
The fact that every closed hypersurface in $\mathbb R^n$ has an equation is unfortunately not proved nor even mentioned in any book I know on differential manifolds or differential geometry.
It is however rather elementary and depends on methods developed 60 years ago in complex analysis : I wrote a proof here. (See also there, where only orientability of hypersurfaces is proved.)

• I guess the hypersurface has to be complete?
– user99914
Nov 20, 2014 at 23:41
• @John: Yes, for me a hypersurface is closed (else I'd speak of a locally closed hypersurface), but in order to prevent any misunderstanding I have edited the answer and added the word "closed" explicitly to the word "hypersurface" . Thanks for your comment. Nov 21, 2014 at 0:29

The Stiefel-Whitney class of $\Bbb RP^2$ is $(1+a)^3=a^2+a+1 \in H^*(\Bbb RP^2,\Bbb Z/2)$ where $a$ is the unique nonzero element in $H^1$. The inverse of this element in $H^*(\Bbb RP^2,\Bbb Z/2)$ is $1+a$. As the top Stiefel-Whitney class of the normal bundle of an embedding into $\Bbb R^n$ vanishes (c.f. Milnor-Stasheff Cor. 11.4) $1+a$ is not the Stiefel-Whitney class of the normal bundle to an embedding into $\Bbb R^3$, hence there is no embedding $\Bbb RP^2\rightarrow \Bbb R^3$. The theory of characteristic classes gives numerous obstructions to embeddings, immersions, cobordisms and other similar geometric constructions. A place to start is the Milnor-Stasheff book Characteristic Classes cited in this answer.

The above argument easily generalizes to show $\Bbb RP^{2^n}$ does not embed in $\Bbb R^{2^{n+1}-1}$, hence the inequality in the strong Whitney embedding theorem is sharp for dimensions arbitrarily large.

• Wonderful! I wanted to write that exact same argument but I got stuck because I couldn't see why a priori the first Stiefel-Whitney class of the normal bundle should vanish. Thanks for teaching me that great general vanishing result in Milnor-Stasheff, PVAL . Nov 21, 2014 at 0:26

The simplest solution is via Alexander duality, which shows immediately that every surface in $S^3$, in particular in $\mathbb{R}^3$, is orientable.

• Note that the separation theorem mentioned in the comments above is a special case of Alexander duality (which is much easier to prove than the full theorem).
– user98602
Nov 20, 2014 at 20:05
• The theorem Wiki cites is for a topological sphere. It seems like a big generalization to get to any surface-any idea whether it's actually still easier? Nov 20, 2014 at 22:35