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27.22 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

27.22.0.1
\begin{equation} \label{constructions-equation-gkn} G(k, n) : \mathit{Sch}\longrightarrow \textit{Sets} \end{equation}

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 26.15.4. 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 jth 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 26.15.2 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 10.20.1) 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 (t)/\kappa (s). 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 27.22.2. 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 27.22.1 that these functors are indeed representable.

Lemma 27.22.3. Let n \geq 1. There is a canonical isomorphism \mathbf{G}(n, n + 1) = \mathbf{P}^ n_\mathbf {Z}.

Proof. According to Lemma 27.13.1 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


Comments (3)

Comment #3012 by Jize YU on

I suppose there might be a minor typo in the definition of . The element should be defined for , where appears in the -th position.

Comment #9792 by Lucas Henrique on

Where it reads "which is the direct sum of the coprojections corresponding to elements of ", shouldn't it be "corresponding to the elements "?


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