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

Hallo fellow mathematicians. I try to understand why the normal bundle of
$\mathbb{PR}^n$ is isomorphic (in the category of vector bundles) to the tautological line Bundle. More aptly, why $\nu_{\mathbb{RP}^{n+1}}^{\beta} \mathbb{RP}^n \cong L, \text{ where } L$ denotes the tautological line bundle over $\mathbb{RP}^n$ and $\beta \colon \mathbb{RP}^n \rightarrow \mathbb{RP}^{n+1}$ the inclusion induced by the inclusion $\mathbb{R}^{n+1} \subset \mathbb{R}^{n+2}$. I followed a few loose ends until now. One of them:

(1) In this script (pdf) I found a proof (p.8). I just don't understand, how that is supposed to be suffice as proof. Afaik under trivialisations I could identify every vector bundle of the same dimension, so what seals the proof here?

Another idea I came up with:

(2) On $\mathbb{PR}^{n+1}$ we have a riemannian metric, i.e. the tangent bundle allows a presentation as the whitney-sum: $\nu_{\mathbb{RP}^{n+1}}^{\beta} \mathbb{RP}^n \oplus T \mathbb{PR}^n \cong T\mathbb{RP}^{n+1}|_{\mathbb{PR}^n}$. According to Husemoller (Fibre Bundles, Theorem 2.7) we now (analogous to this) that two vector bundles - in order to be isomorphic - have to have cohomologous transition functions, i.e. they are related in a certain way to each other. Afaik we can split the transition functions of $T\mathbb{PR}^{n+1}$ into the direct sum (in the sense that $\phi_{ij}(p)=\begin{matrix} \phi_{ij}^1(p) & 0 \\ 0 & \phi_{ij}^2(p) \end{matrix}$ is the transition function of $T\mathbb{PR}^{n+1}$, where $\phi_{ij}^1$ is the trans. function of $T\mathbb{PR}^{n}$ and $\phi_{ij}^2$ the is trans. function of the normal bundle). As soon as I try to compute all this I encounter problems. So I took the transition functions of the "standard" atlas of $\mathbb{RP}^n$, i.e. with $U_i:=\{[x_1:\dots x_n]|x_i \neq 0 \}$ the transition function is $\phi_j \circ \phi_i^{-1} (y^1, \dots y^n)=(\frac{y^1}{y^j}, \dots \frac{y^{i-1}}{y^{j}}, \frac{1}{y^j}, \frac{y^{i+1}}{y^j}, \dots,\frac{y^n}{y^j})$. When I pass this to the Jacobian (which is the transition function of the tangent bundle, right?) things get ugly.

(3) Another idea I had (which is... regressing I suppose) is to use the orientiability of $\mathbb{RP}^{2n}$ and $\mathbb{RP}^{2n+1}$ and their stiefel whitney classes, which inplies in every case, that the normal bundle is not trivial. But now I would have to show that a line bundle is either trivial or isomorphic to the tautological bundle over $\mathbb{RP}^n$. Is that even true? I'm very greatful for any help you may be able to provide. If I made a mistake, you are more than welcome to correct me.

• What does "Afaik" mean? – Matt Samuel May 11 '15 at 22:15
• Afaik is an aconym and means "as far as I know". Sorry for the confusion! – Excalibur May 11 '15 at 22:33

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.

• Note that we use here $H^1(RP^n)=\{0,1\}$ which is what you tried to aim for in $(3)$ – Daniel Valenzuela May 11 '15 at 23:56
• Could you elaborate on your first sentence a little bit? I know that $\mathbb{RP}^\infty$ is the Eilenberg-Mclane Space $K(\mathbb{Z}_2, 1)$ and thusly we get the equationg above via the classyfing space. But what you stated is a little bit more general, isn't it? Also it would be very helpful if you could elaborate the second sentence as well ($L\rightarrow X$ is trivial iff orientable). Thank you for you answer so far! – Excalibur May 12 '15 at 0:26
• I edited my answer to address your further questions. Yes we get the equation how you say it, but it is a very natural one, i.e. factors through $\omega_1$. – Daniel Valenzuela May 12 '15 at 3:18