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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 $RP^n$ (http://www.lehigh.edu/~dmd1/immtable). But the question I am asking is easier, hence there may be an answer and a proof I could follow.

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up vote 20 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 $H^{n}(A)\times H^{n}(B)=0$.

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this is actually essentially the same as the original argument of Thom, Theorem V.16 in "Espaces fibrés en sphères et carrés de Steenrod" numdam.org/item?id=ASENS_1952_3_69__109_0. 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
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