MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

So first for n even, $RP^n$ is not orientable, hence can not be embedded in $\mathbb{R}^{n+1}$.

For odd n, $RP^{n}$ is orientable, hence the normal bundle is trivial. Now using stiefel-Whitney classes, one can prove when $n$ not of the form $2^k - 1$, it can not be embedded in $\mathbb{R}^{n+1}$. I would appreciate if someone can give an more elementary proof in this case.

Then for the left cases. $RP^3$ can not be embedded proving by homology thoery (Alexander sphere duality, lefschetz duality and a long exact sequence).

For the other cases, I do not know how to prove. I realized Don Davis has a table for the immersion and embedding of $RP^n$ ( But the question I am asking is easier, hence there may be an answer and a proof I could follow.

share|cite|improve this question
up vote 22 down vote accepted

I'll use cohomology with coefficients $\mathbb{Z}/2$ everywhere.

Suppose that the space $P=\mathbb{R}P^{n-1}$ embeds in $S^{n}$ (where $n>2$). Recall that $$ H^*(P)=(\mathbb{Z}/2)[x]/x^{n} = (\mathbb{Z}/2)\{1,x,\dotsc,x^{n-1}\} $$ By examining the top end of the long exact sequence of the pair $(S^{n},P)$ we find that $H^{n}(S^{n},P)$ has rank two. Lefschetz duality says that this group is isomorphic to $H_0(S^{n}\setminus P)$, so we see that $S^{n}\setminus P$ has two connected components. (I don't need any orientation conditions here as I am working mod 2.) Let $A$ and $B$ be the closures of these components, so $A\cap B=P$ and $A\cup B=S^{2n}$. Lefschetz duality also gives $H^{n}(A)\times H^{n}(B)=H^{n}(S^{n}\setminus P)=H_0(S^{n},P)=0$.

We now have a Mayer-Vietoris sequence relating the cohomology groups of $A$, $B$, $P$ and $S^{n}$. As $H^1(S^{n})=H^2(S^{n})=0$ this gives an isomorphism $H^1(A)\times H^1(B)\to H^1(P)=\{0,x\}$. After exchanging $A$ and $B$ if necessary, we can assume that $H^1(B)=0$ and that there is an element $a\in H^1(A)$ that maps to $x$ in $H^1(P)$. It follows that $a^{n-1}$ maps to $x^{n-1}$, which generates $H^{n-1}(P)$, so the Mayer-Vietoris connecting map $H^{n-1}(P)\to H^{n}(S^{n})=\mathbb{Z}/2$ must be zero. This contradicts exactness at the next stage, because $H^{n}(A)\times H^{n}(B)=0$.

share|cite|improve this answer
this is actually essentially the same as the original argument of Thom, Theorem V.16 in "Espaces fibrés en sphères et carrés de Steenrod" He observed that if $P^{n-1}$ embeds into $S^n$ then from Mayer-Vietoris $H^\star(P)$ is a direct sum as a ring of images of $H^\star(A)$ and $H^\star (B)$. Since $H^*(\mathbb {RP}^{n-1},\mathbb Z_2)$ doesn't split like that unless one of the summands is zero the result follows. – Vitali Kapovitch May 10 '12 at 22:23
Thanks, Neil, Very nice proof. I forgot to using the Ring structure of $H^{\ast}(RP^n)$ – Xiaolei Wu May 11 '12 at 0:06
Good work Neil (exactly how I would prove it). Comment: this shows $\Bbb RP^{3}$ doesn't embed in $\Bbb R^4$. A nice elementary exercise for first year graduate students is to show that it does embed in $\Bbb R^5$ (one can do it explicitly). – John Klein May 11 '12 at 2:43

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.