I'm interested in proving the fact of the title and I was following the reasoning in page 8 here (at the end).

There is a step which are totally unclear to me, namely the identification of the normal bundle associated to the usual embedding (denote such bundle with $\nu$) with the dual of the tautological bundle

In symbols we have the following chain of isomorphisms $$ \nu \overset{?}{\cong} \hom(\gamma_n^1,\epsilon_n^1) =: (\gamma_n^1)^*\cong \gamma_n^1$$

where the question mark indicates where my first doubt lies. The construction given in the notes is totally unclear to me. They start with a line $l \in \mathbb{R}P^{n+1}$ and then define the map $\lambda_1$. Problem if we choose another line $l'$ such that $\pi(l)=\pi(l')$ (surely possible), we end up with a possibly different $\lambda_1$, and therefore this association may be not well-defined. How do we get rid of this problem?


Here's a proof of the result, but not along the lines of the one you're following in the book.

For notation, write $H^\ast(\mathbb{R}P^n;\mathbb{Z}/2\mathbb{Z})\cong \mathbb{Z}/2\mathbb{Z}[a]/a^{n+1}$ and $H^\ast(\mathbb{R}P^{n+1};\mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}[b]/b^{n+2}$

The inclusion map $i:\mathbb{R}P^n\rightarrow \mathbb{R}P^{n+1}$ is non-trivial on $\pi_1$, hence nontrivial on $H_1$, hence non-trivial on $H^1$. In particular, $i^\ast(b) = a$.

Also, recall that the tangent bundle of $\mathbb{R}P^n$ has total Stiefel-Whitney class $(1+a)^{n+1}$.

In addition, line bundles over a connected manifold $M$ are classified by their first Stiefel-Whitney class. So, it is enough to show that the the first Stiefel-Whitney class of the normal bundle $\nu$ of $\mathbb{R}P^n\subseteq \mathbb{R}P^{n+1}$ is non-trivial.

To do that, start with $T\mathbb{R}P^{n+1}$ and pull it back to $\mathbb{R}P^n$. Then we see that $i^\ast(T\mathbb{R}P^{n+1})\cong T\mathbb{R}P^n \oplus \nu$. Since pullbacks commute with Stiefel-Whitney classes, and using the Whitney sum formula, we have \begin{align*} w_1\left(i^\ast T\mathbb{R}P^{n+1}\right) &= w_1\left( T\mathbb{R}P^n\oplus \nu\right)\\ i^\ast(w_1(T\mathbb{R}P^{n+1})) &= w_1(T\mathbb{R}P^n) + w_1(\nu)\\ (n+2)i^\ast(b) &= (n+1)a + w_1(\nu) \\ (n+2)a &= (n+1)a + w_1(\nu)\\ a&=w_1(\nu).\end{align*}

  • $\begingroup$ thanks for the answer! I'll wait some time to accept it, since I was interested in a more geometric approach, if no other (more geometric) answers come, I'll accept it! $\endgroup$ – Luigi M Apr 10 '16 at 8:20
  • $\begingroup$ Take your time - I'm ok if you never accept this answer. (I'd also like to see a more geometric answer.) Incidentally, here's another approach math.stackexchange.com/questions/1277823/…. $\endgroup$ – Jason DeVito Apr 10 '16 at 21:09

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