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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?

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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.

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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.

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