Grassmannians - Part 1

In this posting, we follow Vakil's notes to start discussing about Grassmannians. Let us fix the base scheme $S$ once and for all. Given an $S$-scheme $Y$ and integers $0 \leq r \leq n,$ we denote by $\mathrm{Gr}_{S}(r, n)(Y)$ the set of isomorphism classes exact sequences of the form $$\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0,$$ where $\mathscr{Q}$ is a locally free sheaf (or more precisely $\mathscr{O}_{X}$-module) of rank $r$ on $Y.$

Unimportant remark. The latter $\mathscr{Q}$ is supposed to be "Q", for "quotient".

Given any epimorphism $\phi : \mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ of $S$-schemes where $\mathscr{Q}$ is a locally free sheaf of rank $r$ on $Y,$ its kernel $\ker(\phi)$ is locally free of rank $n - r.$ (See Vakil 13.5.B.(a) for details.) Thus, we see that $\mathrm{Gr}_{S}(r, n)(Y)$ can be described as the set of isomorphism classes of short exact sequences of the form $$0 \rightarrow \mathscr{S} \rightarrow \mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0,$$ where $\mathscr{S}$ has rank $n - r$ and $\mathscr{Q}$ has rank $r.$

Remark. Given an exact sequence $$0 \rightarrow \mathscr{S} \rightarrow \mathscr{O}_{Y}^{\oplus n}$$ with a locally free sheaf $\mathscr{S}$ of rank $r,$ its cokernel is not necessarily locally free. An example (which we learn from Vakil 13.4.1) is as follows: take $$Y = \mathbb{A}^{1}_{k} = \mathrm{Spec}(k[t]),$$ where $k$ is a field. If we take $n = 1$ and $\mathscr{S}$ the locally free sheaf of rank $r = 1$ associated to the $k[t]$-submodule $k[t]t$ of $k[t]$ generated by $t.$ Then the quotient $\mathscr{O}_{Y}/\mathscr{S}$ is the coherent sheaf associated to $k$ by giving $t$-action as multiplication by $0$, which is not locally free.

We want to make $\mathrm{Gr}_{S}(r, n)$ into a functor $\textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$ so let's describe where a morphism $\pi : X \rightarrow Y$ of $S$-schemes is mapped to. Given an exact sequence $$\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{Q} \rightarrow 0$$ of $\mathscr{O}_{Y}$-modules, since $\pi^{*}\mathscr{O}_{Y} = \mathscr{O}_{X}$ and $\pi^{*}$ is right-exact, we have an exact sequence $$\mathscr{O}_{X}^{\oplus n} = \pi^{*}(\mathscr{O}_{Y}^{\oplus n}) \rightarrow \pi^{*}\mathscr{Q} \rightarrow 0$$ of $\mathscr{O}_{X}$-modules. Since $\mathscr{Q}$ is a locally free $\mathscr{O}_{Y}$-module of rank $r,$ it follows that $\pi^{*}\mathscr{Q}$ is a locally free $\mathscr{O}_{X}$-module of rank $r$ (e.g., Vakil 16.3.7.(3)). If there is another map $\rho : Y \rightarrow Z$ of $S$-schemes, applying $\rho^{*}$ to the last sequence gives $$\mathscr{O}_{Z} = (\pi \circ \rho)^{*}(\mathscr{O}_{Y}^{\oplus n}) = \rho^{*}\pi^{*}(\mathscr{O}_{Y}^{\oplus n}) \twoheadrightarrow \rho^{*}\pi^{*}\mathscr{Q} = (\pi \circ \rho)^{*}\mathscr{Q},$$ which is the same as applying $(\pi \circ \rho)^{*}$ to $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{Q}$ we began with. It is not hard to see that our discussion is compatible with isomorphisms of sequences, so we have constructed a functor $$\mathrm{Gr}_{S}(r, n) : \textbf{Sch}_{S}^{\mathrm{op}} \rightarrow \textbf{Set},$$ which we call the Grassmanian of $r$-subspaces in $n$-spaces over $S.$

Special case: rank $1$. Consider the case $r = 1.$ Locally free sheaves of rank $1$ are line bundles by definition. Given $1 \leq i \leq n,$ denote by $$e_{i} = (0, \dots, 0, 1, 0, \dots, 0) \in \mathscr{O}_{Y}(Y)^{\oplus n},$$ where $1$ only appears in the $i$-th spot. For any open subset $U \subset Y,$ this element restricts to an element $\mathscr{O}_{Y}(U)^{\oplus n}$ with the same description. Fix a line bundle $\mathscr{L}.$ Then a map $\mathscr{O}_{Y}^{\oplus n} \rightarrow \mathscr{L}$ of sheaves is an epimorphism if and only if its localization $\mathscr{O}_{Y, y}^{\oplus n} \rightarrow \mathscr{L}_{y}$ is surjective for all $y \in Y.$ If we denote by $s_{i} \in \mathscr{L}(Y)$ the image of $e_{i}$ under the map $\mathscr{O}_{Y}(Y)^{\oplus n} \rightarrow \mathscr{L}(Y),$ the global section of the given sheaf map, then the last statement is equivalent to saying that one of $s_{1}, \dots, s_{n}$ must have a nonzero image in $\mathscr{L}_{y}$ for each $y \in Y$ (because $\mathscr{L}_{y} \simeq \mathscr{O}_{Y,y},$ so any nonzero element would generate the whole module), or in other words, the global sections $s_{1}, \dots, s_{n}$ do not share a common zero in $Y.$

This gives rise to an $S$-scheme map $[s_{1} : \cdots : s_{n}] : Y \rightarrow \mathbb{P}^{n-1}_{S},$ which can be painlessly obtained by first working over $\mathrm{Spec}(\mathbb{Z})$ and then base change over $S.$ Given another line bundle $\mathscr{L}'$ on $Y,$ an isomorphism of two maps $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{L}$ and $\mathscr{O}_{Y}^{\oplus n} \twoheadrightarrow \mathscr{L}'$ is an isomorphism $\mathscr{L} \rightarrow \mathscr{L}'$ of $\mathscr{O}_{Y}$-modules such that $s_{i} \mapsto s'_{i}$ for $1 \leq i \leq n,$ where $s'_{i} \in \mathscr{L}'(Y)$ is the image of $e_{i} \in \mathscr{O}_{Y}(Y)^{\oplus n}.$ This implies that for each $y \in Y,$ there is an affine open $U \ni y$ in $Y$ such that $\mathscr{L}|_{U} \simeq \mathscr{O}_{Y}|_{U}$ and for $1 \leq i \leq n,$ the image of $s'_{i}$ is the product of the image of $s_{i}$ and an invertible element in $\mathscr{O}_{Y}(U).$ This implies that $[s_{1} : \cdots : s_{n}] = [s'_{1} : \cdots : s'_{n}].$

On the other hand, any $S$-scheme map $\pi : Y \rightarrow \mathbb{P}^{n-1}_{S}$ can be obtained as $\pi = [\pi^{*}x_{0} : \cdots : \pi^{*}x_{n-1}],$ where $x_{0}, \dots, x_{n-1}$ are given by the global sections of $\mathscr{O}_{\mathbb{P}^{n-1}_{S}}(1).$ Explicitly, if $\mathrm{Spec}(R) \subset S$ is an affine open subset, then $$x_{0}, \dots, x_{n-1} \in R[x_{0}, \dots, x_{n-1}].$$ Let's review the meaning of the pullback of a section of a line bundle:

What do we mean by pullback of a global section of a line bundle? Given a global section $s \in \mathscr{L}(Y)$ of a line bundle $\mathscr{L}$ on $Y,$ we can consider the map $\mathscr{O}_{Y}(Y) \rightarrow \mathscr{L}(Y)$ given by $1 \mapsto s.$ This extends to a unique $\mathscr{O}_{Y}$-module map $\mathscr{O}_{Y} \rightarrow \mathscr{L}$ per each $s.$ This way, the section $s$ can be viewed as a map $\mathscr{O}_{Y} \rightarrow \mathscr{L}.$ Now, if $\pi : X \rightarrow Y$ is a map of $S$-schemes, then we may consider the pullback $\pi^{*}s : \mathscr{O}_{X} = \pi^{*}\mathscr{O}_{Y} \rightarrow \pi^{*}\mathscr{L}.$ The image of $1 \in \mathscr{O}_{X}(X)$ under the map $$\mathscr{O}_{X}(X) = (\pi^{*}\mathscr{O}_{Y})(X) \rightarrow (\pi^{*}\mathscr{L})(X)$$ given by $\pi^{*}s$ is also denoted as $\pi^{*}s.$

Going back to the discussion, open sets of the form $\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R))$ cover $Y,$ where $U_{i} = D_{\mathbb{P}^{n-1}}(x_{i})$ and $\mathrm{Spec}(R) \subset S$ is affine open. The restriction $$\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R)) \rightarrow U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R) = \mathrm{Spec}(R[x_{0}/x_{i}, \dots, x_{n-1}/x_{i}])$$ of $\pi$ corresponds to the ring map $$R[x_{0}/x_{i}, \dots, x_{n-1}/x_{i}] \rightarrow \Gamma(\pi^{-1}(U_{i} \times_{\mathrm{Spec}(\mathbb{Z})} \mathrm{Spec}(R)), \mathscr{O}_{Y})$$ such that $x_{j}/x_{i} \mapsto \pi^{*}x_{j}/\pi^{*}x_{i}.$ Some more details can be found in (the proof of) Vakil 16.4.1. Hence, we have established an explicit bijection $$\mathrm{Gr}_{S}(1, n)(Y) \leftrightarrow \mathrm{Hom}_{S}(Y, \mathbb{P}^{n-1}_{S}),$$ which we can briefly write as $(s_{1}, \dots, s_{n}) \mapsto [s_{1} : \cdots : s_{n}].$ Given any map $\rho : X \rightarrow Y,$ the induced map $\mathrm{Gr}_{S}(1, n)(Y) \rightarrow \mathrm{Gr}_{S}(1, n)(X)$ can be described as $(s_{1}, \dots, s_{n}) \mapsto (\rho^{*}s_{1}, \dots, \rho^{*}s_{n}),$ while $\mathrm{Hom}_{S}(Y, \mathbb{P}^{n-1}_{S}) \rightarrow \mathrm{Hom}_{S}(X, \mathbb{P}^{n-1}_{S})$ is given by $\phi \mapsto \rho \circ \phi.$ We can write $$\phi = [s_{1} : \cdots : s_{n}] = [\phi^{*}x_{1} : \cdots : \phi^{*}x_{n}]$$ and $$\begin{align*}\rho \circ [s_{1} : \cdots : s_{n}] &= \phi \circ \rho \\ &= [(\phi \circ \rho)^{*}x_{1} : \cdots : (\phi \circ \rho)^{*}x_{n}] \\ &= [\rho^{*}\phi^{*}x_{1} : \cdots : \rho^{*}\phi^{*}x_{n}] \\ &= [\rho^{*}s_{n} : \cdots : \rho^{*}s_{n}].\end{align*}.$$ This means that our bijection is now checked to be a natural trasformation so that it is an isomorphism of functors: $$\mathrm{Gr}_{S}(1, n)(-) \simeq \mathrm{Hom}_{S}(-, \mathbb{P}^{n-1}_{S}).$$ This, by definition, means that the functor $\mathrm{Gr}_{S}(1, n)(-)$ is representable by the projective space $\mathbb{P}^{n-1}_{S}$ over $S.$

Example. Take $S = \mathrm{Spec}(k),$ where $k$ is a field. Then we have $$\mathrm{Gr}(1, n)(k) \simeq \mathbb{P}^{n-1}(k),$$ given by $$(a_{1}, \dots, a_{n}) \mapsto [a_{1} : \cdots : a_{n}],$$ where $a_{i} \in k,$ at least one of which is nonzero. We understand the right-hand side quite well, so let's consider the left-hand side. An element is given by (the isomorphism class of) a map $\mathscr{O}_{\mathrm{Spec}(k)}^{\oplus n} \twoheadrightarrow \mathscr{L}.$ This is nothing more than a surjective $k$-linear map $k^{n} \twoheadrightarrow k,$ so the image of each $e_{i}$ is $a_{i} \in k$ we are considering.

Next time. We describe what happens to $\mathrm{Gr}_{S}(r, n),$ when $r$ may be larger than $1.$