Why is this linear vector bundle not trivial?

Question

I need to prove that any smooth section of the following vector bundle vanishes at some point: $$E_{1,n+1}(\mathbb{R})=\left\{([x],v):[x]\in \mathbb{RP}^n,v=\lambda x\in \mathbb{R}^{n+1}\right\} $$ \begin{align*}\pi:E_{1,n+1}(\mathbb{R})&\to \mathbb{RP}^n\\ ([x],v)&\mapsto [x] \end{align*} I am not given a differential structure on $E_{1,n+1}(\mathbb{R})$, but I guess if $(\mathcal{U}_i,\varphi_i)$ are the usual charts for the projective space one could adjoin those with the chart sending $v=(v_1,\dots,v_{n+1})$ to $v_i$, i.e. $$\hat{\varphi}_i([x],v)= \left(\left(\frac{x_j}{x_i}\right)_{j\neq i},v_i\right)\in \mathbb{R}^{n+1},\qquad ([x],v)\in \pi^{-1}(\mathcal{U}_i) $$ In this case the local trivializations may be given by $$\phi_i([x],v)=([x],v_i),\qquad ([x],v)\in \pi^{-1}(\mathcal{U}_i) $$ whose local charts expression

$\sigma:\mathbb{RP}^n\to E_{1,n+1}(\mathbb{R})$ is a section if and only if it of the form $$ [x]\mapsto ([x],v([x]))$$ for some map $v:\mathbb{RP}^n\to \mathbb{R}^{n+1}$ such that $v([x])=\lambda x$. If $\sigma$ vanishes nowhere, then $\lambda \neq 0$, which means that $[v([x])]=[x]$. Moreover, $\sigma$ is smooth if and only if the map $$[x]\mapsto v([x]),\qquad [x]\in \mathbb{RP}^n $$ is smooth. I really cannot see how to proceed from here. In fact, I would say that if we choose $v$ so that it maps each $[x]\in \mathbb{RP}^n$ to its unique representative having norm $1$ with $x_1\geq 0$, then this is a smooth section that vanishes nowhere, since in local coordinates $$v(\varphi_i^{-1}(x))=\begin{cases}\frac{(x_1,\dots, x_{i-1},1,x_{i+1},\dots, x_{n})}{\sqrt{1+|x|^2}}& x_1\geq 0 \\ -\frac{(x_1,\dots, x_{i-1},1,x_{i+1},\dots, x_{n})}{\sqrt{1+|x|^2}} & x_1<0\end{cases}, \qquad x\in \varphi_i(\mathcal{U}_i) $$ which is smooth.

Answer 1

To show that every section for $E_n$ vanishes, we can use the intermediate value theorem. In particular, consider a section $s: \mathbb RP^n \to E_n$.

but we can similarly see that precomposing with the quotient $S^n \to \mathbb RP^n$ yields a map $g:S^n \to E_n$ given by

$$g:x \mapsto ( x,\lambda(x)x),$$ where $t(x) \in \mathbb R$ is a choice of scalar.

Since this map factors through $\mathbb RP^n$, we must have that $g(-x)=g(x)$, or in other words $ (-x,\lambda(-x)(-x))=(x,\lambda(x)x)$ implying that $$\lambda(-x)=-\lambda(x).$$ Hence, we have a map $\lambda:S^n \to \mathbb R$ that is odd, and so vanishes by the intermediate value theorem.

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By the way, there are algebraic topological tools available for this question about real vector bundles. If a real vector bundle possesses a nonvanishing section, then the top stiefel-whitney class must vanish. This can be found in Milnor's book "Characteristic Classes."

Answer 2

As you noticed, if the section is not zero we can assume it has norm $1$, else we could take $\frac{\sigma}{\|\sigma\|}$. The choice of the vector is not unique though, because for a given line $\ell$ there are two unitary vectors $v,-v$ on $\ell$.

It is now clear that constructing the section boils down to construct a map $\sigma : \Bbb RP^n \to S^n$ with $\pi \circ \sigma = \text{id}_{\Bbb RP^n}$ where $\pi : S^n \to \Bbb RP^n$ is the canonical projection. Such map does not exist, else the covering $S^n \to \Bbb RP^n$ would be trivial, in particular $S^n$ would be disconnected.