Home » Mathematics » Algebraic combinatorics » Schubert Calculus I: Geometry of G(k,n)

# Schubert Calculus I: Geometry of G(k,n)

Goals for this post:

• coordinate systems
• line and vector bundles and divisors
• the functor of points perspective
• the homogeneous coordinate ring is a UFD

Coordinates

I’ll mostly think of the Grassmannian from the linear algebra perspective, that is,

$G(k,n) = \{ k\text{-dimensional vector subspaces of } \mathbb{C}^n \}$
$\ = \{ \text{ full-rank } k \times n \text{ matrices } \} / GL_k$

So, given a subspace $V \subset \mathbb{C}^n$, the basic approach is to represent $V$ by a $k \times n$ matrix $A$ of row vectors that span $V$.

For every maximal $k \times k$ minor of our matrix, corresponding to a choice of columns $I \in {[n] \choose k}$, we have the $I$-th Plücker coordinate $p_I$. These numbers are not well-defined, since we can multiply $A$ on the left by any $k \times k$ square matrix, which will rescale all the $p_I$ simultaneously by $\det(A)$. On the other hand, this simultaneous rescaling means that the $p_I$ do make sense as homogeneous coordinates, that is, the map

$Gr(k,n) \to \mathbb{P}^{{n \choose k} - 1},$

listing out the Plücker coordinates, is well-defined. (Note that some minor must be nonzero, since the matrix is full-rank.) In coordinate-free language, this is the map $G(k,n) \to \mathbb{P}(\bigwedge^k \mathbb{C}^n)$ that sends a subspace $V$ to $\bigwedge^k V \subset \bigwedge^k \mathbb{C}^n$.

This is not the only useful coordinate system on $G(k,n)$: there are also ${n \choose k}$ standard affine charts. In particular, consider the open set $U$ of matrices whose $I$-th Plücker coordinate is nonzero (fixing the choice of $I$). The columns $I$ therefore form an invertible $k \times k$ matrix in $A$. Let us multiply on the left by the inverse of this matrix, to get something of the form

$\begin{pmatrix} * & 1 & * & * & 0 & 0 & * \\ * & 0 & * & * & 1 & 0 & * \\ * & 0 & * & * & 0 & 1 & * \end{pmatrix}$

where the columns $I$ are now an identity matrix. (The above corresponds to $I = \{2,5,6\}$ in $G(3,7)$.)

The key observation is this: our new choice of representative is the unique representative having an identity matrix in the selected columns (since acting nontrivially by $GL_k$ on the left would change this identity matrix). In particular, this means that our set is an affine space: $U \cong \mathbb{A}^{k(n-k)}$. This gives a convenient affine chart for each choice of $I$.

Corollary. The Grassmannian $G(k,n)$ is locally isomorphic to affine space $\mathbb{A}^{k(n-k)}$. In particular, its dimension is $k(n-k)$, and it is irreducible and smooth.

Line Bundles, Vector Bundles and Divisors

Divisors on $G(k,n)$ are pretty easy to understand. We have the following:

Theorem. The Picard group of the Grassmannian is isomorphic to $\mathbb{Z}$, generated by the divisor class of a hyperplane section in the Plücker embedding, such as the vanishing of a Plücker coordinate.

Proof: We use the well-known right-exact sequence of Picard groups

$\mathbb{Z} \to \text{Pic}(X) \to \text{Pic}(U) \to 0$,

where $U \subset X$ is the complement of an integral Weil divisor $D$, and the first arrow maps $1 \mapsto [D]$. We take $D$ to be the vanishing locus of a Plücker coordinate (we will see next post that this is irreducible – in fact, birational to a smaller affine space) and $U = \mathbb{A}^{k(n-k)}$, so $\text{Pic}(U) = 0$. Finally, note that the first arrow is actually injective, since $[D]$ is an effective divisor on a projective variety, hence it can’t be a torsion element in the Picard group.

Wonderful! Now we understand how divisors and line bundles work: up to isomorphism, every line bundle $\mathcal{L}$ is isomorphic to a tensor power of the Plücker line bundle, which we often refer to as $\mathcal{O}(1)$.

The Grassmannian has several important vector bundles as well. Most importantly, it has a “tautological short exact sequence” of bundles,

$0 \to \mathcal{S} \to \mathbb{C}^n \otimes \mathcal{O}_{G(k,n)} \to \mathcal{Q} \to 0$,

where $\mathbb{C}^n \otimes \mathcal{O}_{G(k,n)}$ is the trivial vector bundle whose fiber at every point is the vector space $\mathbb{C}^n$ where we started; $\mathcal{S}$ is the tautological subbundle, whose fiber at the point $V \in G(k,n)$ is the subspace $V$ itself; and lastly $\mathcal{Q}$ is the tautological quotient bundle, whose fiber at $V$ is the quotient vector space $\mathbb{C}^n / V$.

Note that $\mathcal{Q}$ is a globally-generated vector bundle of rank $n-k$, and $\mathcal{S}$ has rank $k$ and has no global sections at all.

Later on we’ll know the total Chern classes of these bundles, but for now it’s easy enough to compute their top Chern classes. Recall that for a vector bundle $\mathcal{E}$ of rank $r$, the top Chern class $c_1(\mathcal{E})$ is the locus where $r$ generic sections will fail to span, or equivalently, the wedge product of the sections is zero. Unwinding this definition a little, observe that the operation “take the determinant of columns $I$” is precisely the kind of wedge product necessary to compute $c_1(\mathcal{S}^*)$. In other words, the Plücker coordinates are (all the) global sections of $\det(\mathcal{S}^*)$! And from the identity $\det(\mathcal{S}^*) \cong \mathcal{O}(1)$ it follows that $c_1(\mathcal{S}) = -c_1(\mathcal{S}^*) = -[D]$, the negative of the Plücker divisor class.

And, by the tautological short exact sequence above, it also follows that $\det \mathcal{Q} = (\det \mathcal{S})^* \cong \mathcal{O}(1)$, that is, $c_1(\mathcal{Q}) = [D]$.

There’s one last relevant vector bundle: the tangent bundle $\mathcal{T}$, which has rank $k(n-k)$. The somewhat tricky important observation here is the following: every way to deform or “nudge” a point $V \in G(k,n)$ can be encoded as a linear map $\varphi: V \to \mathbb{C}^n$. If $V$ has basis $v_1, \ldots, v_n$, we consider the new space with basis $v_i + \varepsilon \varphi(v_i)$. (It’s possible to formalize this in terms of wedge products.) However, two different $\varphi$‘s will give the same tangent vector if their difference has image lying in $V$ itself, so really the tangent space is $\text{Hom}(V,\mathbb{C}^n / V)$. And so, the tangent bundle itself is

$\mathcal{T} \cong \mathcal{H}om(\mathcal{S},\mathcal{Q}) \cong \mathcal{S}^* \otimes \mathcal{Q},$

Now, we use the identity $\det(\mathcal{E} \otimes \mathcal{F}) = \det(\mathcal{E})^{\text{rk}(\mathcal{F})} \otimes \det(\mathcal{F})^{\text{rk}(\mathcal{E})}$ to conclude that the anticanonical divisor of the Grassmannian is

$\det(\mathcal{T}) \cong \mathcal{O}(1)^{\otimes(n-k)} \otimes \mathcal{O}(1)^{\otimes(k)} \cong \mathcal{O}(n).$

The Functor of Points Perspective

The “functor of points perspective” of scheme theory is the idea of understanding a scheme or variety by understanding maps to it, that is, given a scheme $X$, we wish to understand the functor $\text{Hom}(-,X)$ from schemes to sets. (A functor from schemes to sets that arises this way is called representable.)

For example, a map to $\mathbb{A}^1$ is the same as a globally-defined regular function: $\text{Hom}(-,\mathbb{A}^1) \cong H^0(\mathcal{O}_Y)$.

More relevant to our current discussion is the functor of points of projective space: a map $\varphi \in \text{Hom}(Y,\mathbb{P}^n)$ is the same as an $(n+1)$-tuple of global sections of a line bundle $\mathcal{L} \in \text{Pic}(Y)$ that globally generate $\mathcal{L}$. Given the global sections $s_i$, we can construct the map explicitly by building a projective space with hyperplane sections $X_i$ that restrict to $s_i$; conversely, given the map $\varphi$, we obtain the line bundle by pulling back $\mathcal{O}(1)$.

Slightly more is going on here: in fact we have a whole short exact sequence of bundles to pull back:

$0 \to \mathcal{O}(-1) \to \mathbb{C}^{n+1} \to \mathcal{Q} \to 0$,

consisting of the tautological line bundle, trivial vector bundle, and tautological quotient bundle of projective space. Note that $\mathcal{L}$ is pulled back from the dual of the tautological bundle.

Maps to the Grassmannian directly generalize this setup. A map from a scheme or variety $Y$ to the Grassmannian $G(k,n)$ is given by the choice of a rank-$k$ vector bundle $\mathcal{V}$, together with a choice of $n$ sections $s_1, \ldots, s_n$ that globally generate it. (Equivalently, a locally-free rank-$k$ quotient sheaf of a trivial rank-$n$ sheaf.) Explicitly, at each point $y \in Y$, the sections generate a $k$-dimensional vector space (since $\mathcal{V}$ has rank $k$), which is a quotient of our space $\mathbb{C}^n$ of global sections. The dual is a $k$-dimensional subspace of the dual $\mathbb{C}^n$ (this is also the reason for the dualizing step in the last paragraph).

In terms of vector bundles, this means that $\mathcal{V}$ is obtained by pulling back $\mathcal{S}^*$, the dual of the tautological sub-bundle on $G(k,n)$.

Note: sometimes the functor of points of $G(k,n)$ is instead described as the set of rank-$k$ locally free subsheaves of a rank-$n$ trivial sheaf, whose cokernel is also locally-free.

This is a nice ‘physical’ description: maps to the Grassmannian are given by assigning to each point an actual $k$-dimensional subspace of a fixed $n$-dimensional vector space. That said, I like my above description a bit better for its explicitness and similarity to the case of $\mathbb{P}^n$. (Also, the weird-sounding ‘quotient is also locally free’ bit doesn’t come up when you phrase things the other way: if you start with a locally free quotient $\mathcal{O}_X^{\oplus n} \to \mathcal{V} \to 0$, the kernel is automatically locally free, by Nakayama’s lemma.)

The Homogeneous Coordinate Ring is a UFD

I will probably not have time for this in the mini-course, but if I did, I would prove the following:

Theorem. The homogeneous coordinate ring of the Grassmannian in its Plücker embedding is a unique factorization domain.

Proof sketch: This is in three steps. First, we show that the coordinate ring is the same as the ring of invariants $k[a_{ij} : 1 \leq i \leq k, 1 \leq j \leq n]^{SL_k}$ (thinking of $a_{ij}$ as the entries of our $k \times n$ matrix.) It’s clear that the Plücker coordinates themselves are in this ring; to show that there’s nothing else, we use a combinatorial proof, basically using a monomial ordering on the $a_{ij}$.

Second, we use the representation theory of $SL_k$: namely, all $SL_k$-representations are restrictions of representations of $GL_k$, and all the 1-dimensional representations of the latter are powers of the determinant. In particular, they are trivial on $SL_k$.

Third, we prove the theorem. Fix $f$ in the ring of invariants, and factor $f = f_1 \cdots f_k$ in the “upstairs” ring. We will show that this factorization already consists of $SL_k$-invariant factors. Observe first that $SL_k$ must act by permuting the factors (and possibly introducing scalars). In particular, the (closed) subgroup of $SL_k$ that fixes the factors is a finite-index subgroup. But $SL_k$ is connected, so this subgroup must be all of $SL_k$. So all the factors are fixed. But then the only thing that can change is a scalar, and the map $SL_k \to \mathbb{C}^*$ given by $g \mapsto \frac{f^g}{f}$ is a homomorphism. By step 2, it must be trivial, so the scalar is 1, and so the factorization descends to the invariant ring.

That completes the (very cool, in my opinion) proof, and incidentally provides a second proof that the Picard group is isomorphic to $\mathbb{Z}$, since this is true any time the homogeneous coordinate ring is a UFD.