Let $X$ be a smooth projective complex variety and $E$ a complex vector bundle of rank $n$ on $X$. Write $\mathbb G(k,E)$ for the fiber bundle whose fibers are the Grassmannians $\mathbb G(k,E_x)$ where $E_x$ is a fiber of $E$ at $x$. I managed to show that $\mathbb G(k,E)$ is a smooth complete variety ; to show properness it suffices to show that the morphism $\mathbb G(k,E) \to X$ is proper since $X$ is projective, and since properness is local on the base it suffices to check on a trivializing neighborhood $U$ for $E$, so that $$ \mathbb G(k,E|_U) \simeq U \times \mathbb G(k, \mathbb C^n) $$ is an isomorphism of $U$-schemes and $U \times \mathbb G(k, \mathbb C^n) \to U$ is proper since $\mathbb G(k, \mathbb C^n)$ is projective and this is a base extension.
How do I show that $\mathbb G(k,E)$ is projective? It seems to be assumed in a paper I am reading and I didn't believe it at first because I thought that the twists in the bundle $E$ could prevent the existence of an embedding (i.e. the existence of a very ample invertible sheaf). Any ideas?
Patrick,
May be without checking all the details, here is how it is done. I always think of rank $k$ quotients. Then, $G(k,E)$ and $G(k,E\otimes L)$ are naturally isomorphic for any line bundle $L$ on $X$. So, we may twist $E$ sufficiently and assume that $E$ is globally generated and so we have a surjection, $F=\mathcal{O}_X^N\to E$. From this, any rank $k$ quotient of $E$ is a rank $k$ quotient of $F$ and thus we get a closed embedding $G(k,E)\subset G(k,F)$. But, $G(k,F)=X\times G(k,N)$ and thus it is projective, if you knew that the usual Grassmannian (over a field) $G(k,N)$ is.