Let $G = \mathrm{GL}_n$, let $B$ be a Borel subgroup of $G$.
The set $\mathcal{B}$ of $G$-conjugates of $B$ can be given an algebraic variety structure and is known as the variety of Borel subgroups or the complete flag variety. It is isomorphic to $G/N_G(B) = G/B$.
A classic result is the following:
Theorem: $\bigcup_{g\in G} g B g^{-1} = G$.
Can this result be understood by the existence of a fibre bundle $$ \pi: G \to B $$ with fibres isomorphic to $\mathcal{B}$? If so, what is this map and where can I find a reference?
Edit: It was pointed out to me that it might be easier to instead consider the quotient map $$ \pi: G \to \mathcal{B} \cong G/B $$ and show that it is a vector bundle with fibres isomorphic to $B$. This would also be a possible way of understanding the theorem. But still, I'm more interested in finding references to this.
I still haven't found any references but I have shown that the answer is yes. I'm going to assume $G = \mathrm{GL}_2(\mathbb{C})$ for ease of notation; let $B = \{b\in G\ : \ b_{12}=0\}$.
To begin, we have to understand the following definition:
Def. The Chevalley quotient of an affine algebraic group $G$ by a closed subgroup $B$ is a quasi-projective homogeneous $G$-space $X$ with a basepoint $x_0$ such that the fibers of $G\to X$ are the cosets of $H$.
We give $G/B$ a structure of Chevalley quotient by taking $(X,x_0) = (\mathbb{P}^1(\mathbb{C}), [0:1])$ via the map $g \mapsto [g_{12} : g_{22}]$. This is a projective homogeneous $G$-space and one can check that $gB \mapsto [g_{12} : g_{22}]$ is well-defined; so it is a Chevalley quotient.
The definition of vector bundle that I'll be using is the following:
Def. A vector bundle of rank $r$ over an algebraic variety $X$ is a variety $E$ together with a morphism $\pi:E\to X$ and an open cover $\{U_i\}_{i\in I}$ of $X$ such that
I'm now ready to show the claim.
Lemma. The quotient map $\pi: G \to X$ is a vector bundle with fibers isomorphic to $B$.
Proof. Define $U_1 = \{[x_1:x_2]\in X\ : \ x_1\ne0\}$ and $U_2 = \{[x_1:x_2]\in X\ : \ x_2\ne0\}$. Then $\{U_1, U_2\}$ is an open cover of $X$. We have $\pi^{-1}(U_2) = \{g \in G \ : \ g_{22}\ne0\}$. Write this set as $$ \left\{ \begin{pmatrix} g_{11} & g_{12}\\ g_{21} & g_{22} \end{pmatrix} : g_{22}\ne0\right\} \cong \left\{ \begin{pmatrix} g_{11} & 0\\ g_{21} & g_{22} \end{pmatrix} : g_{22}\ne0\right\}\times \{g_{12} \in \mathbb{C}\}. $$ The first factor is simply $B$; the second factor is isomorphic to $U_2$. Similarly for $U_1$.
The transition function between $U_1$ and $U_2$ is given by $[x_1:x_2] \mapsto [x_2 : x_1]$, which is a regular function in their intersection. $\blacksquare$