Is the associated G/B fibration to a G-torsor projective?

Question

Let $X$ be a smooth projective variety over $\mathbb{C}$, $G$ a reductive group, and $P \to X$ a $G$-torsor. Let $B \subset G$ be a Borel subgroup. Is the associated $G/B$ fibre bundle $$ Y=G/B \times_{G} P $$ projective?

Answer 1

A similar, but shorter answer. Choose a $G$-equivariant projective embedding $G/B \to \mathbb{P}(V)$. It gives a closed embedding $$ G/B \times_G P \hookrightarrow \mathbb{P}(V) \times_G P $$ over $X$ whose target is obviously projective, hence so is the source.

Edit. Let me explain how the projectivity of $\mathbb{P}(V)$ can be deduced.

First, assume that $V$ is a representation of $G$. Then $\mathbb{P}(V)$ is the projectivization over $X$ of the vector bundle $V \times_G P$, and a projectivization of a vector bundle over a projective variety is projective.

Now, assume that $V$ is only a projective representation of $G$. Then $\mathrm{Sym}^d(V)$ is a representation of $G$ for some positive $d$, hence the relative Veronese embedding realizes the target as a closed subset of the projectivization of a vector bundle associated with $\mathrm{Sym}^d(V)$, hence it is projective.

Edit 2. Let me explain why a symmetric power of a projective representation is a linear representation.

An action of $G$ on $\mathbb{P}(V)$ is given by a homomorphism $$ G \to \mathrm{Aut}(\mathbb{P}(V)) = \mathrm{PGL}(V). $$ Consider the central extension $$ 1 \to \mu_N \to \mathrm{SL}(V) \to \mathrm{PGL}(V) \to 1 $$ and its pullback $$ 1 \to \mu_N \to \tilde{G} \to G \to 1, $$ where $N = \dim(V)$. Then the action of $G$ on $\mathbb{P}(V)$ lifts to an action of $\tilde{G}$ on $V$ such that the subgroup $\mu_N$ acts by a character. Clearly, the action of $\mu_N$ on $\mathrm{Sym}^N(V)$ is trivial, hence the action of $\tilde{G}$ on $\mathbb{P}(\mathrm{Sym}^N(V))$ factors through $G$. This shows that $\mathrm{Sym}^N(V)$ is a linear representation of $G$.