## Tag `089R`

## 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 \begin{equation} \tag{26.22.0.1} 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).

Lemma 26.22.1. Let $0 < k < n$. The functor $G(k, n)$ of (26.22.0.1) is representable by a scheme.

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 calledGrassmannian over $S$. If $R$ is a ring the base change to $\mathop{\mathrm{Spec}}(R)$ is denoted $\mathbf{G}(k, n)_R$ and calledGrassmannian 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. $\square$

The code snippet corresponding to this tag is a part of the file `constructions.tex` and is located in lines 4805–5012 (see updates for more information).

```
\section{Grassmannians}
\label{section-grassmannian}
\noindent
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
\begin{equation}
\label{equation-gkn}
G(k, n) : \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.
\medskip\noindent
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 \ref{modules-lemma-set-isomorphism-classes-finite-type-modules}
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 \ref{modules-lemma-pullback-locally-free}),
and (4) $f^*$ is right exact
(Modules, Lemma \ref{modules-lemma-exactness-pushforward-pullback}).
\begin{lemma}
\label{lemma-gkn-representable}
Let $0 < k < n$.
The functor $G(k, n)$ of (\ref{equation-gkn}) is representable by a scheme.
\end{lemma}
\begin{proof}
Set $F = G(k, n)$. To prove the lemma we will use the criterion of
Schemes, Lemma \ref{schemes-lemma-glue-functors}.
The reason $F$ satisfies the sheaf property for the
Zariski topology is that we can glue sheaves, see Sheaves,
Section \ref{sheaves-section-glueing-sheaves} (some details omitted).
\medskip\noindent
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 \ref{modules-lemma-exactness-pushforward-pullback})
we conclude that $F_i$ is a subfunctor of $F$.
\medskip\noindent
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 \ref{modules-lemma-map-finite-locally-free}).
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 \ref{schemes-example-global-sections}
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 $\Spec(\mathbf{Z})$ is the product in the category of schemes.
\medskip\noindent
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 \ref{modules-lemma-finite-type-surjective-on-stalk}.
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 \ref{algebra-lemma-NAK}) 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 \ref{sheaves-lemma-stalk-pullback-modules})
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.
\medskip\noindent
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.
\end{proof}
\begin{definition}
\label{definition-grassmannian}
Let $0 < k < n$. The scheme $\mathbf{G}(k, n)$ representing the functor
$G(k, n)$ is called {\it Grassmannian over $\mathbf{Z}$}.
Its base change $\mathbf{G}(k, n)_S$ to a scheme $S$ is called
{\it Grassmannian over $S$}. If $R$ is a ring the base change
to $\Spec(R)$ is denoted $\mathbf{G}(k, n)_R$ and called
{\it Grassmannian over $R$}.
\end{definition}
\noindent
The definition makes sense as we've shown in
Lemma \ref{lemma-gkn-representable}
that these functors are indeed representable.
\begin{lemma}
\label{lemma-projective-space-grassmannian}
Let $n \geq 1$. There is a canonical isomorphism
$\mathbf{G}(n, n + 1) = \mathbf{P}^n_\mathbf{Z}$.
\end{lemma}
\begin{proof}
According to Lemma \ref{lemma-projective-space} 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.
\end{proof}
\input{chapters}
```

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