Table of Contents

Introduction

Given a vector space $V_n$ of dimension $n$ over an algebraically closed field $K$ , the Grassmannian $Gr(r, V_n)$ is the collection of all $r$ -dimensional subspaces of $V_n$ with the quotient topology. It is immediate that $Gr(1, \mathbb{C}^n) = \mathbb{C}\mathbb{P}^{n-1}$ , hence we may think of it as a generalization of the usual projective spaces. In fact $Gr(r, V_n)$ is itself a projective variety (see the Plücker imbedding).

Since the Grassmannian has been such an important tool in algebraic geometry, unfortunately parts of this post will require some background knowledge of schemes and locally free sheaves over schemes.

Smooth structure

The simplest way to see that $Gr(r, V_n)$ is indeed a smooth manifold is to realize it as the quotient space by some group action. Let’s recall the following important theorem

Theorem 1 – Suppose G is a Lie group acting smoothly, freely and properly on a smooth manifold M. Then the orbit space M/G is a topological manifold of dimension dimM−dimG and has a unique smooth structure with the property that M→M/G is a smooth submersion.

An action is called free if only the identity element has fixed points, and is called proper if the map $(g, x) \mapsto (gx, x)$ is a proper map. For convenience, we note that if the group action is free and properly discontinuous the above quotient becomes a real covering map.

We may define $Gr(r, V_n)$ to be the quotient of all full rank matrices $M_{r\times n}$ modulo the left action of $GL(r)$ , where we think of the rows of $M \in M_{r\times n}$ as spanning the associated vector subspace, and the left action of $GL(r)$ as interchanging bases for $W$ .

Alternatively, pick a basis, or just an inner product for $V_n$ , assign to each $r$ -dimensional subspace $W_r\subset V_n$ an element in the unitary group $U_{W_r}\in U(n)$ , by taking any orthonormal basis for $W_r$ and extending to an orthonormal basis for $V_n$ . To make this correspondence bijective we need to mod out by any orthonormal base change of $W_r$ and $W_r^\perp$ . Thus as a set we have

$Gr(r, V_n) \simeq \frac{U(n)}{U(r)\times U(n-r)}$

where $U(r)\times U(n-r)$ denotes a block diagonal matrix.

The action of $U(r)\times U(n-r)$ on $U(n)$ , it is clearly smooth and free (check this!) and since $U(k)$ is compact and Hausdorff for any $k$ , the action is automatically proper, since any continuous map $f: X\to Y$ of compact Hausdorff spaces is proper.

The other way – The above construction seems slightly unsatisfactory as it gives us no idea of how the charts are actually constructed, without tracing the steps in the proof of Theorem 1. To remedy this shortcoming, fix a basis for $V$ and write every element in $W \in G(r, V_n)$ as matrices $M^W_{r\times n}$ where the rows represents the spanning vectors of $W$ . The Plucker imbedding gives us an imbedding

$Gr(r, V_n) \to \mathbb{P}(\bigwedge^rV_n)$

The local homogeneous coordinate charts inherited from $\mathbb{P}(\bigwedge^rV_n)$ gives the socalled Plücker coordinates (to be defined below) on $Gr(r, V_n)$ . Explicitly, it gives us a set of homogeneous coordinates on $Gr(r, V_n)$ given by

$M^W_{r\times n}\mapsto [|M_{I_1}|, ..., |M_{I_{n \choose r}}|]$

where $|M_{I_j}|$ ranges over all $r$ -minors of the matrix $M^W_{r\times n}$ . Restricting to the local chart $U_1$ , where $|M_{I_1}| \neq 0$ and letting $GL(r)$ act on $M^W_{k\times n}$ such that $M_{I_1} = I_{r\times r}$ we get a chart

$M^W_{r\times n} \mapsto \left( \frac{|M_{I_2}|}{|M_{I_1}|}, ..., \frac{|M_{I_{n\choose r}}|}{|M_{I_1}|}\right) = \left( |M_{I_2}|, …, |M_{I_{n\choose r}}| \right) .$

By fixing the first $r$ -minor of the $M^W_{r\times n}$ to be the identity matrix we may write every element in $U_1$ uniquely it as a matrix

$M^W_{r\times n} = \begin{bmatrix} 1 & 0 & ... & 0 & a_{11} &... & a_{1{n-r}} \\0 & 1 & ... &0 &a_{21} & ... & a_{2{n-r}} \\&& ... & &&&\\0 & 0 & … & 1 & a_{r1} &… & a_{r{n-r}} \end{bmatrix}$

This gives us a chart after the identification of $M_{r\times (n-r)} \simeq K^{r(n-r)}$ .

Plücker imbedding

The Grassmannian is a projective variety, meaning it is given as the zero locus of a family of homogeneous polynomials of the same degree (called the Plücker relations). This is usually show by means of the Plücker embedding which I will state and prove here.

Let $V_n$ be an $n$ -dimensional vector space and let $Gr(r, V_n)$ be the Grassmannian of $r$ -dimensional subspaces. The Plücker imbedding is given by the map

$p: Gr(r, V_n) \to \mathbb{P}(\bigwedge^rV_n)$

sending a subspace $W = span\{e_1, ..., e_r\}$ to the (totally) decomposable tensor $[e_1\wedge ... \wedge e_r]$ .

The map is well defined since for any basis transformation $(e_1, ..., e_r) \mapsto (v_1, .., v_k)$ given by a linear map $T: W\to W$ , extend $T$ by the identity operator to the whole of $V$ , and note that

$det(T) e_1\wedge ... \wedge e_r = v_1 \wedge ...\wedge v_r.$

which are identified in $\mathbb{P}(\bigwedge^rV_n)$ . It is also easily seen to be injective, so as a set the Grassmannian are precisely the decomposable tensors of $\mathbb{P}(\bigwedge^rV_n)$ . To show it is closed in the Zariski topology I will follow the book of Harris (Algebraic Geometry – A First Course).

Some multilinear algebra prerequisites

For any $n$ -dimensional vector space $W$ over an algebraically closed field $K$ , fix a non-zero form $\sigma \in \bigwedge^nW$ . We can construct a bilinear map $\langle -, - \rangle : \bigwedge^r W \times \bigwedge^{n-r} W \to K$ given by

$\langle \alpha, \beta \rangle = \sigma (\alpha \wedge \beta).$

This pairing is non-degenerate, that is if $\langle \alpha, \beta \rangle = 0$ for all $\alpha$ then $\beta = 0$ . Hence we have an identification

$\bigwedge^r W \simeq \left(\bigwedge^{n-r} W\right)^\star \simeq \bigwedge^{n-r} W^\star$

given by the map $\alpha \mapsto \sigma(\alpha\wedge -)$ . Note how this is natural up to choice of form $\sigma$ , hence the identification

$\mathbb{P}(\bigwedge^rW) \simeq \mathbb{P}(\bigwedge^{n-r}W^\star)$

is natural.

In a basis $e_1, ..., e_n$ for $W$ with corresponding dual basis $e_1^\star, ..., e_n^\star$ and setting $\sigma(e_1\wedge ... \wedge e_n) = \lambda$ this identification sends

$e_I\mapsto \lambda\star(e^\star_I)$

where $\star(\cdot)$ is the Hodge star operator. For example, the identification sends

$e_1\wedge ... e_r \mapsto \lambda e^\star_{r+1}\wedge ... \wedge e_n^\star$

We will also need some basic facts about dual maps for the proof of the next theorem.

If $T: V\to W$ is a linear map with dual map $T^\star : W^\star \to V^\star$ (i.e. the map $\phi \mapsto \phi \circ T$ ), then if we fix a basis for $V$ and $W$ , the matrix representation of $T^\star$ with respect to the corresponding dual basis is the transposed of the matrix representation of $T$ with respect to said basis.

The rank of the map $T$ is the dimension of the image. Since the dimension of the columnspace and rowspace of matrix are equal, it follows that

$rank(T) = rank(T^\star)$

Lastly the following isomorphism is crucial

(1) $Im(T^\star) = (Ker(T))^\perp$

where $\perp$ denotes the annihilator subspace of $Ker(T)$ in $V^\star$ . The main result now is the following theorem

Theorem 2 – $p(Gr(r, V_n)) \subset \mathbb{P}(\bigwedge^rV_n)$ is closed in the Zariski topology.

Proof

We will show the image $p(Gr(r, V_n))$ is the zero locus of a family of homogeneous polynomials of degree 2 (called ‘quadrics’). For each $\omega \in \bigwedge^r V_n$ define the two maps

(1) $\begin{align*} \phi_\omega&: V\to \bigwedge^{r+1}V_n \qquad v \mapsto \omega \wedge v \\ \psi_{\omega^\star}&: V^\star \to \bigwedge^{n-r+1} V^\star_n \qquad v^\star \mapsto \omega^\star \wedge v^\star \end{align*}$

We make the following observation

$\omega$ is totally decomposable if and only if the rank of $\phi_\omega$ (or equivalently $\psi_{\omega^\star}$ ) are as small as possible, which here means

$rank(\phi_\omega) = n-r$

and

$rank(\psi_{\omega^\star}) \leq r.$

To see this note that $\omega$ is totally decomposable, say $\omega = e_1\wedge ...\wedge e_r$ for some basis, then $Ker(\phi_\omega)= Span\{e_1, ..., e_r\}$ . This implies $Rank(\phi_\omega) = dim(ker(\phi_\omega)^\perp) = n-r$ . Now assume $rank(\phi_\omega) = n-r$ , let $v_1, .., v_n \subset V_n$ be a basis such that $v_1, .., v_r$ is a basis for $Ker(\phi_\omega)$ . Let $\omega = \sum_I \lambda_I v_I$ be the usual multivector basis expansion of $\omega$ , Since $0= v_i \wedge \omega = \sum_I\lambda_I v_i \wedge v_I$ it follows that $\lambda_I=0$ for each multiindex $I$ not containing $v_i$ , hence $\omega = \lambda_{(1,2,...,r)}v_1\wedge ...\wedge v_r$ . The proof for $\psi_{\omega^\star}$ is analogous.

Now if we dualize these maps and writing $\langle a, b\rangle = a(b)$ for the correspondence between $V$ and $V^\star$ we claim that $\omega$ is totally decomposable if and only if

$\langle \phi_\omega^\star (b), \psi_{\omega^\star}^\star (a) \rangle = 0 \qquad \forall ~ a\in V^\star, ~ b\in V$

which happens if and only if $\phi_\omega\psi_{\omega^\star}^\star= 0$ by non-degeneracy. To prove this claim, assume $\omega = e_1\wedge … \wedge e_r$ is totally decomposable, and using equation (1) we get

$\begin{align*}Im(\psi_{\omega^\star}^\star) & = \left( Ker(\psi_{\omega^\star}) \right)^\perp = \left( span\{ e_{r+1}^\star, .., e_n^\star\} \right)^\perp \\ & = span\{e_1, ..., e_r\} = Ker(\phi_\omega)\end{align*}$

Conversely assume $\phi_\omega\psi_{\omega^\star}^\star =0$ and (for contradiction) that $\omega$ is not decomposable. Then by previous observations $rank(\psi_{\omega^\star}^\star) = rank(\psi_{\omega^\star}) > r$ while $dim(Ker(\phi_\omega)) = n - rank(\phi_\omega) < r$ , which which implies that the image of $\psi_{\omega^\star}^\star$ would not fit inside the kernel of $\phi_\omega$ . A contradiction has arisen!

It follows from this that the Grassmannian can be written as the zero locus of the family

$\sum_k\phi^{ir}_\omega \psi^{jr}_{\omega^\star}$

where $\phi^{mn}$ are the entries of the matrix representations, each of which is a linear function of $\omega$ , hence they are all quadratic polynomials in $\omega$

. Now a final check to see it is actually a projective variety –

Theorem 3 – $p(Gr(r, V_n)) \subset \mathbb{P}(\bigwedge^rV_n)$ is irreducible

Proof

The map $K^{rn} \simeq M_{r\times n} \to K^{n\choose r}$ sending a matrix to all it’s r-minors is a parametrization of the affine cone over $p(Gr(r, V_n)) \subset \mathbb{P}(\bigwedge^r V_n)$ , which is the space

$\{ (x_1, ..., x_{n\choose r}) ~|~ [x_1 : … : x_{n\choose r}] \in p(Gr(r, V_n)) \} \cup \{0 \}.$

Since $K^{rn}$ is irreducible, we must have that the image is irreducible (or else the preimages would be a decomposition of $K^{rn}$ into closed disjoint sets). Lastly a projective variety is irreducible if and only if its affine cone is irreducible. This concludes the proof.

The $K$ -scheme associated with this variety $Gr(r, V_n)$ is hence

$Proj(K[x_1, ..., x_{n \choose r}])/ \langle f_1, .., f_r \rangle$

where $f_i$ are the “Plücker relations” (the homogeneous degree 2 polynomials) constructed above.

Homogeneous coordinates

The Plücker imbedding gives us a chance to write the Grassmannian in the homogeneous coordinates of $\mathbb{P}^n$ , often called the Plücker coordinates. First we need to fix a basis $e_1, ..., e_n \in V_n$ and assume $W = span\{ v_1,..., v_r\}$ where $v_i = \sum_{j=1}^n \lambda^i_j e_j$ are linearly independent and denote by $e_{I_1},..., e_{I_{n \choose r}}$ the usual basis for $\bigwedge^rV_n$ giving them any ordering you like (like the lexicographical). Now the imbedding of $W$ into $\mathbb{P}(\bigwedge^rV_n)$ is

(2) $\begin{align*}[W] &\mapsto [v_1\wedge ... \wedge v_r] \\ &= [ \sum_{j=1}^n \lambda^1_j e_j\wedge ... \wedge \sum_{j=1}^n \lambda^r_j e_j] \\ &= [ \sum_{i= 1}^{{n \choose r}} |M_{I_i}| e_{I_i}]\\&=:[ |M_{I_1}| : ... :|M_{I_{n\choose r }}|]\end{align*}$

where $M$ is the matrix $(\lambda_j^i)$ (the matrix whose $i$ ‘th column is $v_i$ ), $|M_{I_i}|$ is the determinant of the $r\times r$ matrix determined by the multiindex $I_i$ . Note that not all these $r\times r$ minors can be zero, since the matrix $\lambda^i_j$ has rank $r$ .

Functorial view

Be warned, this section is pretty sketchy as I don’t really think I will ever find a use for this construction in my work, but it’s good to at least know this material exists.

Recall that the lemma of Yoneda let’s us associate to any object in a category $C\in \mathcal{C}$ , a functor $Hom(-, C)$ which uniquely determines the element $C$ . Explicitly the lemma states that the above map induces a full and faithful imbedding

$\mathcal{C}\to Fun(\mathcal{C}^{op} \to (Sets)).$

Throughout this section let $X = Gr(r, V_n)$ be our Grassmannian which you can think of as a projective variety, a scheme over the scalar field $K$ or just a smooth manfiold over $K = \mathbb{R}$ if you have no need for the added generality. The associated functor of the Grassmannian $H(-, X)$ has a particularly nice equivalent definition, which will be presented shortly.

To prove this equivalence, we need to define the universal bundle over $X = Gr(r, V_n)$ , (or equivalently the universal exact sequence $0 \mapsto R \mapsto X\times V_n \mapsto Q \mapsto 0$ ), which is the map

$\phi_u: X\times V_n \to Q$

where $Q$ is the quotient of the trivial bundle $X\times V_n$ by the tautological bundle $R = \{(x, v) \in X\times V_n ~|~ v \in X\}$ , that is, each fiber of Q at a point $x\in X$ is just $V/x$ .

The following important theorem which holds (mutatis mutandi) over any base scheme, but we restrict our attention to $K$ -schemes where $K$ is the (algebraically closed) scalar field of $V_n$ , will be the essential step in the proof of the equivalence.

Theorem 4 – We have a bijection between the following sets

$\psi : Y \to Gr(r, V_n)$ The set of all morphisms of K-schemes
$(L_r, s_1, …, s_n)$ locally free rank k sheaves on $Y$ with r-nonvanish. sections
$(L_r, s_1, …, s_n)$ rank r vector bundles on $Y$ with r-nonvanish. sections
$\phi: Y\times V_n \twoheadrightarrow L_r$ , where $L_r$ is a rank $r$ vector bundle on $Y$
$\phi: O^{\oplus n}_Y\twoheadrightarrow L_r$ , where $L_r$ is a rank r locally free sheaf on $Y$

(Sketchy) Proof

(4)=(5) and (2)=(3) follow from the equivalence of category of locally free sheaves and vector bundles over a common base scheme (like $K$ ), which sends the trivial bundle to the free sheaf (of the same rank). Let us try to show (1)=(3) which is a generalization of Theorem 16.4.1 of [2]. Given a bundle $L_r$ on $Y$ and global sections $s_1,.., s_n$ , construct a morphism $\phi: Y \to Gr(r, V_n)$ by the map

$y \mapsto s_1(y), ..., s_n(y) \subset V_n.$

This gives us the desired morphism. The inverse correspondence is determined by pullback of the universal bundle. The fact that (3)=(4) is left as an exercise :p

The phrase “ $r$ -non vanishing sections” is misleading, but you can only be so precise in the margin. It means that at each point $p \in Y$ the sections of the $L_r$ -sheaf or vector bundle on $Y$ , $(s_1,..., s_n)$ generate $L_r$ near $p$ . In the case of a vector bundle this means they are local trivializations of the bundle, while for sheaves this means they generate the free sheaf of rank $r$ $L_k|_U = O^{\oplus r}|_{U}$ , where $U$ is a neighborhood of $p$ .

With Theorem 4 at our disposal, it is not that difficult to show that the Grassmannian functor is naturally equivalent to the following functor

(3) $\begin{align*} F_X(Y) & = \{ \phi: O^{\oplus n}Y \twoheadrightarrow L_r \} / iso \\ &= \{ \phi: Y\times V_n \twoheadrightarrow L_r \}/iso \\ & = \{ L{n-r} \subset O_Y\otimes_K V_n ~| ~ O_Y\otimes_K V_n/ L_{n-r} \text{ is loc free of rk r} \}/iso \\ & =\{ 0 \to L_{n-r} \to O^{\oplus n}_Y \to L_r \to 0 | \text{ exact }\}/iso \end{align*}$

where $L_k$ are either locally free sheaves of rank k or vector bundles of rank $k$ over $Y$ , depending on the working category, $O_Y$ is the structure sheaf of $Y$ treated as an $O_Y$ -module and $Y\times V_n$ is the trivial vector bundle over the scheme $Y$ . These identifications again rely on the equivalence of categories of locally free sheaves and vector bundles over a fixed base scheme, and the fact that any surjection $\phi: O_Y^{\oplus n}\twoheadrightarrow L_r$ has a locally finite kernel of rank $n -r$ .

If this makes no sense, it is strongly recommended that one sticks to the category of varieties with ordinary vector bundles. The reason I mentioned it is that in this identification we send trivial bundles of rank $n$ to free sheaves of rank $n$ . Hence one often identifies $O^{\oplus n}_X \simeq X\times V_n$ in the literature. We also used $O_X\otimes V_n \simeq O_X^{\oplus n}$ . Now finally –

Theorem 5 The functor $F_X$ is naturally equivalent to the functor $Hom(-, X)$ .

(Very sketchy) Proof

Let $f: Y'\to Y$ be a morphism of (schemes|varieties), then

$\begin{matrix}Hom(Y, X) & \xrightarrow{ g\circ f}& Hom(Y', X)\\ \downarrow{g^*\phi_u} & & \downarrow{g^*\phi_u} \\\{ \phi_Y: Y\times V_n \twoheadrightarrow L_r^Y\} &\xrightarrow[\phi \mapsto f^\star\phi_Y]{} &{ \phi_{Y'}: {Y'}\times V_n \twoheadrightarrow L_r^{Y'}} \end{matrix}$

The commutativity of this diagram follows by the fact that $(g\circ f)^\star = f^\star g^\star$ . The maps $g\mapsto g^\star \phi_u$ (the universal bundle defined above) is actually invertible. Let $\phi: Y\times V_n \twoheadrightarrow L_r^Y$ , now the map $g_{L_r^Y}: Y \to X,$ given by $y \mapsto L_r^Y(y)$ is an inverse to the map $g \mapsto g^\star \phi_u$ .

To be continued….

Subsequent post will cover the tangent space of $Gr(r,V_n)$ , some subvarietes, and introduce the Schubert cells.

[1] – Harris. Algebraic geometry: A first course.
[2] – R. Vakil, Fundations of algebraic geometry : The rising sea (November 18, 2017 edition)

The Grassmannian (1)