# canonical bundle of relative Grassmannian

Let $X$ be a smooth projective variety and $\mathcal{E}$ a locally free sheaf of rank $r$ on it. Now we consider the relative Grassmannian $\mathrm{Grass}_X(l,\mathcal{E})$. What is the canonical bundle of the relative Grassmannian?

If you know the calculation of the canonical bundle of the usual Grassmannian (nice writeup here), you can adapt it directly to the relative situation. The result is that $K_G=\mathcal O_{\mathbf P}(-r)_{|G}$, where $\mathbf P=\mathbf P_X \left( \bigwedge^l \mathcal E \right)$ is the projective bundle into which the relative Grassmannian is embedded by the Plücker embedding.
Let $\pi:Grass(\mathcal{E})\rightarrow X$ be the projection, and $\mathbf{q}:\pi^*\mathcal{E}\twoheadrightarrow \mathcal{Q}$ the universal quotient rank $l$ quotient. One has an exact sequence
$0\rightarrow T_{G/X}\rightarrow T_{G}\rightarrow \pi^* T_{X}\rightarrow 0$.
Since the tangent space to the ordinary Grassmannian at a point $q:\mathbb{C}^n\twoheadrightarrow Q$ is $\text{Hom}(\ker q, Q)$, one finds that the relative tangent bundle $T_{G/X}$ is isomorphic to $\underline{Hom}(\ker\mathbf{q},\mathcal{Q})$. Now the exact sequence above allows us to express $\det T_G=K_G^{-1}$ as the tensor product of the determinants of the left and right guys in the sequence. Since $\det T_{X/G}=(\det\ker\mathbf{q})^{-l}\otimes(\det\mathcal{Q})^{r-l}$, we get an explicit formula for the canonical bundle on $G$ in terms of $K_X$ and the tautological bundles on the Grassmannian.
Note that in the case where $\det\mathcal{E}=\mathcal{O}$ one has an $\det\ker\mathbf{q}^{-1}=\det\mathcal{Q}$, so that one obtains $\det T_{G/X}=(\det\mathcal{Q})^r$. In other words, in terms of the Plucker embedding, $\det T_{G/X}=\mathcal{O}_G(r)$. In this special case, the relative canonical bundle is what was given in the previous answer.