Checking that a linear vector bundle on $\mathbb{RP}^n$ is trivial Consider $\mathbb{RP}^n$ with standard charts $(\mathcal{U}_i,\varphi_i)_{i=1}^{n}$ and the linear vector bundle $E_d$ constructed from the transition maps
$$f_{jk}([x])= \left(\frac{x_k}{x_j}\right)^d \in GL(1,\mathbb{R}),\qquad [x]\in \mathcal{U}_j\cap \mathcal{U}_k$$
In other words, we set 
\begin{align*}E_d&:=\left[\bigsqcup_{j=1}^{n}\left(\left\{j\right\}\times \mathcal{U}_j\times \mathbb{R}\right)\right]/\sim\\ 
(j,p,g_j)\sim (k,p,g_k)&\iff g_j=f_{jk}(p)g_k,\; p\in \mathcal{U}_j\cap 
\mathcal{U}_k\\ 
&\pi([(j,p,g_j)]_{\sim})=p
\end{align*}
I need to prove that if $d$ is even, then $E_d$ is trivial, by proving that there exists a smooth global section vanishing nowhere. I do not understand why this is true only if $d$ is even. Intuitively, it is because $\left(\frac{x_k}{x_j}\right)^d$ is always positive only if $d$ is even, so it has to do with connectedness. 

But still, if I define the section  $$\sigma([x])=\left[
 (i([x]),[x],1)\right]_{\sim}=\left\{\left(j,[x],\left(\frac{x_j}{x_{i([x])}}\right)^d\right):[x]\in
 \mathcal{U}_j\cap \mathcal{U}_{i([x])}\right\}\in \pi^{-1}([x])\subset
 E_d$$ Where $i:\mathbb{RP}^n\to \left\{1,\dots,n\right\}$ is any
   function such that $[x]\in \mathcal{U}_{i([x])}$ for all $[x]$.
   Clearly $\sigma$ vanishes nowhere, even if $d$ is odd, is everywhere
  defined, and is smooth because the function $[x]\mapsto 1$ is smooth.

EDIT: As noted, $\sigma$ is not smooth globally but only in the level sets of the function $i:[x]\mapsto i([x])$. But this seems to be the case regardless of whether $d$ is odd or even, so I have no idea how to construct a smooth non-vanishing section when $d$ is even.
 A: Once you choose a function $i$, you can only conclude that $\sigma$ is smooth on open sets on which $i$ is constant. It is not continuous globally: 
Fix $[x_0]$, let $j \neq i([x_0])$ and pick $[x_1]\in \mathcal{U}_{i([x_0])}\cap \mathcal{U}_j$ with $f_{i([x_1])j}([x])=-1$. Then $\sigma([x_1])=[i([x_1]),[x_1],1]=[j,[x_1],-1]$. Next let $[x_2]\in\mathcal{U}_j$ with $i([x_2])=j$, then $\sigma([x_2])=[j,[x_2],1]$. Connect $[x_1]$ and $[x_2]$ with a curve $c\subset \mathcal{U}_j$. Then $\sigma(c(t))=0$ for some $t$ by the intermediate value theorem, which is a contradiction.

Edit: Here is how you can define a global, nowhere vanishing section, if $d$ is even: View  $\mathbb{R}P^n$ as quotient of $S^{n}$. Then if $d$ is even,  the functions
$$
a_k:S^{n}\rightarrow \mathbb{R}, x=(x_0,\dots,x_n)\mapsto x_k^d
$$
descend to functions $\bar a_k\colon \mathbb{R}P^n\rightarrow \mathbb{R}$ and we can define for $0\le k \le n$:
$$
\sigma_k([x])=[k,[x],\bar a_k([x])]\quad \text{for } [x]\in \mathcal{U}_k=\{x_k\neq 0\}.
$$
Clearly each $\sigma_k$ is smooth and nonvanishing. I claim that all sections glue together to a global section $\sigma$. Take $j,k$ and $[x]\in \mathcal{U}_j\cap \mathcal{U}_k$. Then
$$
\sigma_j([x])=[j,[x],\bar a_j([x])]=[k,[x],f_{jk}([x])\bar a_j([x])], \quad \text{ and } \quad \sigma_k([x])=[k,[x],\bar a_k([x])].
$$
But $f_{jk}([x])\bar a_j([x])=(x_k/x_j)^d\cdot x_j^d = x_k^d = \bar a_k([x])$, hence $\sigma_j([x])  = \sigma_k([x])$ and we are done.
