
## 26.22 Grassmannians

In this section we introduce the standard Grassmannian functors and we show that they are represented by schemes. Pick integers $k$, $n$ with $0 < k < n$. We will construct a functor

26.22.0.1
$$\label{constructions-equation-gkn} G(k, n) : \mathit{Sch}\longrightarrow \textit{Sets}$$

which will loosely speaking parametrize $k$-dimensional subspaces of $n$-space. However, for technical reasons it is more convenient to parametrize $(n - k)$-dimensional quotients and this is what we will do.

More precisely, $G(k, n)$ associates to a scheme $S$ the set $G(k, n)(S)$ of isomorphism classes of surjections

$q : \mathcal{O}_ S^{\oplus n} \longrightarrow \mathcal{Q}$

where $\mathcal{Q}$ is a finite locally free $\mathcal{O}_ S$-module of rank $n - k$. Note that this is indeed a set, for example by Modules, Lemma 17.9.8 or by the observation that the isomorphism class of the surjection $q$ is determined by the kernel of $q$ (and given a sheaf there is a set of subsheaves). Given a morphism of schemes $f : T \to S$ we let $G(k, n)(f) : G(k, n)(S) \to G(k, n)(T)$ which sends the isomorphism class of $q : \mathcal{O}_ S^{\oplus n} \longrightarrow \mathcal{Q}$ to the isomorphism class of $f^*q : \mathcal{O}_ T^{\oplus n} \longrightarrow f^*\mathcal{Q}$. This makes sense since (1) $f^*\mathcal{O}_ S = \mathcal{O}_ T$, (2) $f^*$ is additive, (3) $f^*$ preserves locally free modules (Modules, Lemma 17.14.3), and (4) $f^*$ is right exact (Modules, Lemma 17.3.3).

Proof. Set $F = G(k, n)$. To prove the lemma we will use the criterion of Schemes, Lemma 25.15.4. The reason $F$ satisfies the sheaf property for the Zariski topology is that we can glue sheaves, see Sheaves, Section 6.33 (some details omitted).

The family of subfunctors $F_ i$. Let $I$ be the set of subsets of $\{ 1, \ldots , n\}$ of cardinality $n - k$. Given a scheme $S$ and $j \in \{ 1, \ldots , n\}$ we denote $e_ j$ the global section

$e_ j = (0, \ldots , 0, 1, 0, \ldots , 0)\quad (1\text{ in }j\text{th spot})$

of $\mathcal{O}_ S^{\oplus n}$. Of course these sections freely generate $\mathcal{O}_ S^{\oplus n}$. Similarly, for $j \in \{ 1, \ldots , n - k\}$ we denote $f_ j$ the global section of $\mathcal{O}_ S^{\oplus n - k}$ which is zero in all summands except the $j$th where we put a $1$. For $i \in I$ we let

$s_ i : \mathcal{O}_ S^{\oplus n - k} \longrightarrow \mathcal{O}_ S^{\oplus n}$

which is the direct sum of the coprojections $\mathcal{O}_ S \to \mathcal{O}_ S^{\oplus n}$ corresponding to elements of $i$. More precisely, if $i = \{ i_1, \ldots , i_{n - k}\}$ with $i_1 < i_2 < \ldots < i_{n - k}$ then $s_ i$ maps $f_ j$ to $e_{i_ j}$ for $j \in \{ 1, \ldots , n - k\}$. With this notation we can set

$F_ i(S) = \{ q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q} \in F(S) \mid q \circ s_ i \text{ is surjective}\} \subset F(S)$

Given a morphism $f : T \to S$ of schemes the pullback $f^*s_ i$ is the corresponding map over $T$. Since $f^*$ is right exact (Modules, Lemma 17.3.3) we conclude that $F_ i$ is a subfunctor of $F$.

Representability of $F_ i$. To prove this we may assume (after renumbering) that $i = \{ 1, \ldots , n - k\}$. This means $s_ i$ is the inclusion of the first $n - k$ summands. Observe that if $q \circ s_ i$ is surjective, then $q \circ s_ i$ is an isomorphism as a surjective map between finite locally free modules of the same rank (Modules, Lemma 17.14.5). Thus if $q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q}$ is an element of $F_ i(S)$, then we can use $q \circ s_ i$ to identify $\mathcal{Q}$ with $\mathcal{O}_ S^{\oplus n - k}$. After doing so we obtain

$q : \mathcal{O}_ S^{\oplus n} \longrightarrow \mathcal{O}_ S^{\oplus n - k}$

mapping $e_ j$ to $f_ j$ (notation as above) for $j = 1, \ldots , n - k$. To determine $q$ completely we have to fix the images $q(e_{n - k + 1}), \ldots , q(e_ n)$ in $\Gamma (S, \mathcal{O}_ S^{\oplus n - k})$. It follows that $F_ i$ is isomorphic to the functor

$S \longmapsto \prod \nolimits _{j = n - k + 1, \ldots , n} \Gamma (S, \mathcal{O}_ S^{\oplus n - k})$

This functor is isomorphic to the $k(n - k)$-fold self product of the functor $S \mapsto \Gamma (S, \mathcal{O}_ S)$. By Schemes, Example 25.15.2 the latter is representable by $\mathbf{A}^1_\mathbf {Z}$. It follows $F_ i$ is representable by $\mathbf{A}^{k(n - k)}_\mathbf {Z}$ since fibred product over $\mathop{\mathrm{Spec}}(\mathbf{Z})$ is the product in the category of schemes.

The inclusion $F_ i \subset F$ is representable by open immersions. Let $S$ be a scheme and let $q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q}$ be an element of $F(S)$. By Modules, Lemma 17.9.4. the set $U_ i = \{ s \in S \mid (q \circ s_ i)_ s\text{ surjective}\}$ is open in $S$. Since $\mathcal{O}_{S, s}$ is a local ring and $\mathcal{Q}_ s$ a finite $\mathcal{O}_{S, s}$-module by Nakayama's lemma (Algebra, Lemma 10.19.1) we have

$s \in U_ i \Leftrightarrow \left( \text{the map } \kappa (s)^{\oplus n - k} \to \mathcal{Q}_ s/\mathfrak m_ s\mathcal{Q}_ s \text{ induced by } (q \circ s_ i)_ s \text{ is surjective} \right)$

Let $f : T \to S$ be a morphism of schemes and let $t \in T$ be a point mapping to $s \in S$. We have $(f^*\mathcal{Q})_ t = \mathcal{Q}_ s \otimes _{\mathcal{O}_{S, s}} \mathcal{O}_{T, t}$ (Sheaves, Lemma 6.26.4) and so on. Thus the map

$\kappa (t)^{\oplus n - k} \to (f^*\mathcal{Q})_ t/\mathfrak m_ t(f^*\mathcal{Q})_ t$

induced by $(f^*q \circ f^*s_ i)_ t$ is the base change of the map $\kappa (s)^{\oplus n - k} \to \mathcal{Q}_ s/\mathfrak m_ s\mathcal{Q}_ s$ above by the field extension $\kappa (s) \subset \kappa (t)$. It follows that $s \in U_ i$ if and only if $t$ is in the corresponding open for $f^*q$. In particular $T \to S$ factors through $U_ i$ if and only if $f^*q \in F_ i(T)$ as desired.

The collection $F_ i$, $i \in I$ covers $F$. Let $q : \mathcal{O}_ S^{\oplus n} \to \mathcal{Q}$ be an element of $F(S)$. We have to show that for every point $s$ of $S$ there exists an $i \in I$ such that $s_ i$ is surjective in a neighbourhood of $s$. Thus we have to show that one of the compositions

$\kappa (s)^{\oplus n - k} \xrightarrow {s_ i} \kappa (s)^{\oplus n} \rightarrow \mathcal{Q}_ s/\mathfrak m_ s\mathcal{Q}_ s$

is surjective (see previous paragraph). As $\mathcal{Q}_ s/\mathfrak m_ s\mathcal{Q}_ s$ is a vector space of dimension $n - k$ this follows from the theory of vector spaces. $\square$

Definition 26.22.2. Let $0 < k < n$. The scheme $\mathbf{G}(k, n)$ representing the functor $G(k, n)$ is called Grassmannian over $\mathbf{Z}$. Its base change $\mathbf{G}(k, n)_ S$ to a scheme $S$ is called Grassmannian over $S$. If $R$ is a ring the base change to $\mathop{\mathrm{Spec}}(R)$ is denoted $\mathbf{G}(k, n)_ R$ and called Grassmannian over $R$.

The definition makes sense as we've shown in Lemma 26.22.1 that these functors are indeed representable.

Lemma 26.22.3. Let $n \geq 1$. There is a canonical isomorphism $\mathbf{G}(n, n + 1) = \mathbf{P}^ n_\mathbf {Z}$.

Proof. According to Lemma 26.13.1 the scheme $\mathbf{P}^ n_\mathbf {Z}$ represents the functor which assigns to a scheme $S$ the set of isomorphisms classes of pairs $(\mathcal{L}, (s_0, \ldots , s_ n))$ consisting of an invertible module $\mathcal{L}$ and an $(n + 1)$-tuple of global sections generating $\mathcal{L}$. Given such a pair we obtain a quotient

$\mathcal{O}_ S^{\oplus n + 1} \longrightarrow \mathcal{L},\quad (h_0, \ldots , h_ n) \longmapsto \sum h_ i s_ i.$

Conversely, given an element $q : \mathcal{O}_ S^{\oplus n + 1} \to \mathcal{Q}$ of $G(n, n + 1)(S)$ we obtain such a pair, namely $(\mathcal{Q}, (q(e_1), \ldots , q(e_{n + 1})))$. Here $e_ i$, $i = 1, \ldots , n + 1$ are the standard generating sections of the free module $\mathcal{O}_ S^{\oplus n + 1}$. We omit the verification that these constructions define mutually inverse transformations of functors. $\square$

Comment #3012 by Jize YU on

I suppose there might be a minor typo in the definition of $f_j$. The element $f_j$ should be defined for $j\in\{1,\cdots,n-k\}$, $f_j\colon=(0,\cdots,1\cdots,0)$ where $1$ appears in the $j$-th position.

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