The Stacks Project


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 $j \in \{1, \ldots, k\}$ we denote $f_j$ the global section of $\mathcal{O}_S^{\oplus k}$ 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. $\square$

    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 $j \in \{1, \ldots, k\}$
    we denote $f_j$ the global section of $\mathcal{O}_S^{\oplus k}$
    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}

    Comments (1)

    Comment #3012 by Jize YU on December 4, 2017 a 5:50 pm UTC

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

    Add a comment on tag 089R

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?