If $\mathbb{RP}^n$ can be immersed in $\mathbb{R}^{n+1}$, how do I see that $n$ must be of the form $2^r - 1$ or $2^r - 2$?

Question

If $\mathbb{RP}^n$ can be immersed in $\mathbb{R}^{n+1}$, how do I see that $n$ must be of the form $2^r - 1$ or $2^r - 2$? For starters, I know that if the $n$-dimensional $M$ can be immersed in $\mathbb{R}^{n+1}$, then each $w_i(M)$ is equal to the $i$-fold cup product $w_1(M)^i$.

Answer 1

Let $\alpha$ be the generator of the cohomology ring $H^{*}(\mathbb{RP}^n; \mathbb{Z}_2).$ Note that $w(\nu(\mathbb{RP}^n)) = 1$ or $1+\alpha,$ where $\nu(\mathbb{RP}^n)$ is the normal bundle (dimension $1$). Recall $w(\mathbb{RP}^n) = (1+\alpha)^{n+1}.$

In the first case, this implies $w(\mathbb{RP}^n) = 1,$ which means that $\binom{n+1}{i}$ is even for all $0<i<n+1,$ so $n=2^r - 1.$

In the second, $w(\mathbb{RP}^n) = (1+\alpha)^{-1} = 1+\alpha+\ldots + \alpha^n,$ which implies $n=2^r - 2.$

Answer 2

Suppose that you embed $RP^n$ in $R^{n+1}$ then the normal bundle $\tau$ is trivial or isomorphic to the canonical line bundle over $PR^n$ and $\tau+TPR^n$ is trivial. If $\tau$ is trivial the total stiefel whitney class of $\tau+TPR^n$ is $w(\tau)w(TPR^n)=w(TPR^n)=(1+a)^{n+1}=1$ where $a$ is the class of the canonical line bundle. This implies that $n+1=2^l$. if $\tau$ is isomorphic to the canonical line bundle, $w(\tau)=1+a$ thus $w(\tau+TPR^n)=(1+a)^{n+1}(1+a)=(1+a)^{n+2}=1$ this implies $n+2=2^l$.