Sections of associated bundles

Solution 1:

After many hours, I believe I have answered my own question: Claim:

There is a 1-1 correspondence \begin{align*} \{f:P\rightarrow V: f(pg)=\rho(g^{-1})f(p) \}\simeq \Gamma(P\times_G V). \end{align*} Proof:

Let $f:P\rightarrow V$ be as described. Define local sections $\tilde{s}_{\alpha}:U_{\alpha}\rightarrow P\times_G V$ by \begin{align*} \tilde{s}_{\alpha}(m)&=[(s_{\alpha}(m),f(s_{\alpha}(m))] \end{align*} Define $\sigma=\tilde{s}_{\alpha}$ on $U_{\alpha}$. Then $\sigma$ is a global section since the definition agrees on overlaps. Indeed, for $m\in U_{\alpha}\cap U_{\beta}$, \begin{align*} \tilde{s}_{\beta}(m)&=[(s_{\beta}(m),f(s_{\beta}(m))]\\ &=[(s_{\alpha}(m)g_{\alpha\beta}(m),f(s_{\alpha}(m)g_{\alpha\beta}(m))]\\ &=[(s_{\alpha}(m)g_{\alpha\beta}(m),\rho(g_{\beta\alpha}(m))f(s_{\alpha}(m))]\\ &=[((s_{\alpha}(m),f(s_{\alpha}(m))]\\ &=\tilde{s}_{\alpha}(m) \end{align*} So each $f$ corresponds to an element in $\Gamma(P\times_G V)$.

Conversely, suppose $\sigma\in \Gamma(P\times_G V)$. The section $\sigma$ allows us to construct a function $f_{\sigma}:P\rightarrow V$ in the following way: Let $p\in P$ be abritrary and write $m=\pi(p)$. We define the function $f_{\sigma}$ as \begin{align*} f_{\sigma}(p)&=v,\quad\text{such that } \sigma(m)=[(p,v)]. \end{align*} This is a well defined construction since if we choose another representative of $P\times_G V$ and write $\sigma(m)=[(q,w)]$ for some $q\in P$ and $w\in V$, we have that $q=pg$ for some $g\in G$ ($p$ and $q$ both lie in $\pi^{-1}(m)$), and then just set $v=\rho(g)w$.

This constructed $f_{\sigma}$ satisfies the desired property. Indeed, suppose $f(pg)=\tilde{v}$. That is, $\tilde{v}$ is the vector such that $\sigma(\pi(pg))=\sigma(m)=[(pg,\tilde{v})]$. Then we have \begin{align*} \sigma(m)=[(pg,\tilde{v})]&=[(p,\rho(g)\tilde{v})]. \end{align*} In other words, $f(p)=\rho(g)\tilde{v}=\rho(g)f(pg)$.

Would anyone be able to verify if that is correct?