I was wondering whether the real projective space $\Bbb{R}P^n$ embeds topologically into $\Bbb{R}^{n+1}$ for odd $n$.

It certainly doesn't for even $n$ because of Alexander duality. Also it doesn't embed smoothly for any $n$. I will prove the two above statements in my algebraic topological class, but I couldn't find anything in the literature for topological embeddings in the odd case.

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    $\begingroup$ Just to clarify: you're not assuming any nice behavior, eg locally flat, for your embedding. Right? $\endgroup$ – Danny Ruberman Jan 28 '18 at 17:50
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    $\begingroup$ If you restrict to locally flat embeddings, it looks like you can rule out $\mathbb{RP}^{4k-1} \hookrightarrow \mathbb{R}^{4k}$. $\endgroup$ – Marco Golla Jan 28 '18 at 18:48
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    $\begingroup$ The arguments that obstruct a smooth embedding will also obstruct the existence of a locally flat embedding. Put differently, by topological embedding I mean an injective continuous map (that is a homeomorphism onto its image). $\endgroup$ – Stefan Friedl Jan 28 '18 at 19:51
  • $\begingroup$ Possible duplicate of Is it true that all real projective space $RP^n$ can not be smoothly embedded in $R^{n+1}$ for n >1 $\endgroup$ – Moishe Kohan Jan 29 '18 at 0:15
  • $\begingroup$ It is essentially a duplicate of an earlier question which was about smooth embeddings but the proof did not use smoothness. $\endgroup$ – Moishe Kohan Jan 29 '18 at 0:15

The first page of W.~Massey's paper On the imbeddability of the real projective spaces in Euclidean space states that $\mathbb RP^n$ with $n>1$ cannot be imbedded topologically in $\mathbb R^{n+1}$ because its mod $2$ cohomology algebra does not satisfy a certain condition given by R. Thom.

The reference points to Theorem V.15 (p.180) of Thom's paper Espaces fibrés en sphères et carrés de Steenrod, Annales scientifiques de l'École Normale Supérieure, Série 3 : Volume 69 (1952) , p. 109-182.

EDIT: Actually, a proof can be found in Is it true that all real projective space $RP^n$ can not be smoothly embedded in $R^{n+1}$ for n >1.

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  • $\begingroup$ The paper by Thom seems to answer my question. It's slightly unfortunate that Thom doesn't explain what he means by "plonge", but since he distinguishes between "plonge" and "plonge differentiablement" it seems like he means a topological embedding. $\endgroup$ – Stefan Friedl Jan 29 '18 at 0:31
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    $\begingroup$ The reference to the other mathoverflow discussion is nice. But I don't see how it applies in the topological case. To make the Meyer-Vietoris argument work you need some type of regularity, isn't it? (So I am a little confused by the comment in that discussion that the argument is precisely the one that Thom gave). $\endgroup$ – Stefan Friedl Jan 29 '18 at 0:36
  • $\begingroup$ @StefanFriedl: Mayer-Vietoris sequence works for excisive triads. The solution is to use a fancier cohomology theory, e.g. if $A$, $B$, are closed subsets of the compact Hausdorff space $X$, then the triad $(X,A,B)$ is excisive for the Alexander-Spanier cohomology, see Theorem 8.14, Chapter II, section 8 in a book by W.Massey [Homology and cohomology theory, 1978]. In our case $X=S^{2n}$ and $A$, $B$ are closures of the components of $S^{2n}\setminus\mathbb RP^{2n-1}$. The linked argument is a formal consequence of the Mayer-Vietoris so it applies. $\endgroup$ – Igor Belegradek Jan 29 '18 at 1:58
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    $\begingroup$ The argument will also need the Alexander duality for the Alexander-Spanier cohomology $H^*$, which can be found in the same Massey's book. Thus $H^{2n}(A)=0=H^{2n}(B)$ with mod $2$ coefficients. $\endgroup$ – Igor Belegradek Jan 29 '18 at 2:24
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    $\begingroup$ I just noticed that Thom in the above-linked paper uses the Alexander-Spanier cohomology, which is why he can apply Mayer-Vietoris. For singular cohomology the triad in question might be not excisive. $\endgroup$ – Igor Belegradek Jan 29 '18 at 2:29

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