Tautological line bundle over $\mathbb{RP}^n$ isomorphic to normal bundle? Also "splitting" of transition functions

Solution 1:

Yes it is true, that $\omega_1:Vect_1X \to H^1(X;Z/2)$ is an isomorphism for general (paracompact) $X$. Hence you have to show that the normal bundle is not trivial. To do that note that $L\to X$ is trivial iff orientable iff $L_0$ disconnected (if $X$ is connected).

Now we have a very natural identification of the total space of $L$, namely $RP^{n+1}-*$ where we choose a point in the interior of the attached $n+1$-disk. So we have $L_0=(RP^{n+1}-*)-RP^n=D^{n+1}-*$, hence connected.

Intuitively speaking, we use that the attaching map $S^n \to RP^n$ is the two fold covering, which gives you the easy situation, that you have 2 ways to go out from the zero section, which both end in the same component $L_0$. Maybe imagine the normal bundle as $RP^n$ and gluing $S^n\times [0,1)$ along it.

Edit concerning the first sentence of this answer.

So consider we have the following: $$ \begin{array}{ccc} [X,RP^\infty] && \\ \downarrow & \searrow&\\ Vect_1X &\xrightarrow{\omega_1}&H^1(X;Z/2) \end{array} $$

Observe that the diagonal map ${[f] \mapsto f^*\alpha}$ is the map which makes the Eilenberg Maclane isomorphism. But this triangle is so natural that it easily commutes and since the vertical map is an isomorphism, we get an isomorphism on the horizontal map. Those are all set-theoretic isomorphisms, i.e. bijections. Now note that $Vect_1X$ becomes a group under $\otimes$ and that the Stiefel Whitney class behaves nicely. So we get a group isomorphism $Vect_1X \cong H^1(X;Z/2)$. Hence there is up to isomorphism only one non trivial line bundle, where we already know one, namely the tautological one. That is why it suffices to show that our line bundle is of non-trivial isomorphism type.

Now note that for a line bundle being of trivial type is equivalent to being orientable. A continous choice of orientation, gives you by renormalizing a nowhere vanishing section, i.e. a trivialization. For the other direction note that a trivialization gives you a continous choice of a basis.