Algebraic curves lecture 26 and 27

Algebraic curves lecture 26 and 27 by Karl Christ

Construction of Brill-Noether varieties
Determinantal Varieties
1. Examples
Tangent spaces of the generic determinantal variety.
Start of Lecture 27
1. Why do we care about determinental varieties?
2. Examples

Construction of Brill-Noether varieties

Question: For what triples $(g,r,d)$ are there maps $f:X\to \mathbb P^r$ such that $f$ has $\deg d$ and $X$ has genus $g$ ?

Idea: Look at the locus of line bundles

\begin{aligned} W^r_d(X) = \{L\mid r(L) \geq r\} \subseteq \operatorname{Pic}^d(X) \end{aligned}

whose rank is at least $r$ .

We'll need a few ingredients.

Determinantal varieties
Some facts about the Picard scheme $\operatorname{Pic}^d(X)$

Determinantal Varieties

Let $M_K$ denote the locus of matrices of rank $\leq k$ , and note that this is a subset of $M(n,m)$ . This is called the generic determinantal variety. We can realize $M_K$ as a variety in the following way. Define this auxiliary object

\begin{aligned} \tilde{M}_k = \{(A, W) \in M(n,m) \times Gr(n-k, n) \mid A\cdot W = 0\} \end{aligned}

consisting of all tuples $(A,W)$ where $A$ is a $n\times m$ matrix and $W$ is a $(n-k)$ -dimensional subset of $k^n$ where $W\subset \operatorname{ker} A$ . We then get a map $\varphi:\tilde{M}_k \to M_k$ , and it is generically $1$ to $1$ : if $A\in M_k$ has rank $k$ then $\varphi^{-1}(A) = (A, \operatorname{ker} A)$ .

Let $\pi_2$ denote the projection onto the second component:

\begin{aligned} \pi_2:M(n,m)\times Gr(n-k,M) \to Gr (n-k,n). \end{aligned}

For $H\in Gr(n-k, n)$ , the fiber of $\pi_2|_{\tilde{M}_k}$ consists of all the $A\in M(n,m)$ which vanish on $H$ . This is isomorphic to $\mathbb A^{m - k}$ (to see this, row reduce $A$ ).

It is true that $\tilde{M}_k$ is irreducible, and hence $M_k$ is irreducible since $\varphi$ is surjective. We (meaning those of us in class) don't know why $\tilde{M}_k$ is irreducible, it would follow from the properness of $\pi_2$ except that $\pi_2$ isn't proper, so for now we simply accept that $\tilde{M}_k$ is irreducible.

Lemma:

Gr(e, \ell)

has dimension

e\cdot (\ell - e)

Proof. Fix $H \in Gr(\ell - e, \ell)$ and let $V$ be the $\ell$ -dimensional vector space whose $\ell-e$ dimensional subspaces $Gr(\ell - e, \ell)$ parameterizes. Let $\lambda\in (\bigwedge^e V)^*$ and $U\subseteq Gr(e, \ell)$ be the set given by $\lambda \neq 0$ . Then $U$ is the subset of $e$ -dimensional subspaces transversal to $H$ . Any element in $U$ can be identified with the graph of a linear function $V/H \to H$ . Elaborating: each choice of $H$ gives you a split short exact sequence $H\to V\to V/H$ , which in turn gives you a map $V/H \to H$ . Conversely, if you have a surjective map $V \to B$ with $B$ isomorphic to $V/H$ , you get a subspace $H$ in $V$ so that $H\oplus B \cong V$ .

This means

\begin{aligned} U \cong \operatorname{Hom}(V/H, H) \cong \mathbb C^{(\ell - e)\cdot e}. \end{aligned}

Note that

(\ell - e)\cdot e

is the product of the dimension of

H

with the dimension of an element in

U

. We have a Plucker embedding

Gr(e, \ell) \hookrightarrow (\bigwedge^e V)^*

, and so these

U

's actually give us the affine charts of

Gr(e, \ell)

\square

Cor: -

\operatorname{dim} \tilde{M}_k = \operatorname{dim} (M_k) = m\cdot k + k\cdot (n - k) = k\cdot (n + m - k)

\operatorname{codim}(M_k) = n\cdot m - (n + m - k)k = (n - k)(m - k)

Theorem: (Second fundamental theorem of invariant theory.)

M_k

is cut out scheme-theoretically by the

(k+1)\times (k+1)

-minors.

Examples

Suppose $m = n$ , $k = m-1$ . Then $M(n,m) \ M_{m-1} = GL(m)$ and $M_{m-1}$ is given by the vanishing of the determinant.
Suppose $k = 1$ and that $A$ has rank at most $1\iff A = u^t\cdot v$ . Then $M_1(n, m)$ is the image of the Segre map $\mathbb A^n \times \mathbb A^m \to \mathbb A^{n-m}$ . A quadric surface in $\mathbb P^3$ is the projectivization of $M_1(2,2)$ . (This is what we keep encountering, a quadric surface embeds in $\mathbb P^1\times \mathbb P^1\subset \mathbb P^3$ embedded in $\mathbb P^3$ via the Segre embedding).
The following is a determinantal variety, and it is true that $X_{a_1,a_2} = M_1(2, r + 1) \cap H_i$ where $H_i$ is a hyperplane given by $x_{1,2} = x_{2,1}, x_{1,3} = x_{2,2},...$

\begin{aligned} X_{a_1,a_2} = \begin{pmatrix} x_0 & x_1 & x_2 & \dots & x_{a_1-1} & y_0 & y_1 & \dots & y_{a_2 - 1} \\ x_1 & x_2 & x_3 & \dots & x_{a_1} & y_0 & y_1 & \dots & y_{a_2} \end{pmatrix}. \end{aligned}

Tangent spaces of the generic determinantal variety.

For a variety $X$ over $k$ and $p\in X$ a point, we define the tangent space of $X$ at $p$ by

\begin{aligned} T_pX = \operatorname{Hom}((\operatorname{Spec} k[\epsilon]/(\epsilon^2), (\epsilon)), (X, p)). \end{aligned}

Example: Consider $T_H Gr(k, V)$ . Given a map

\begin{aligned} \varphi:\operatorname{Spec} k[\epsilon]/(\epsilon^2) \to Gr (k,V) \end{aligned}

and a basis $\omega_1,...,\omega_k$ a basis of $H$ , we obtain a relative basis $\omega_i + \epsilon\cdot v_i$ . Conversely, a map $\omega \mapsto v_i$ gives a linear map $H\to V/H$ . This should be enough to convince oneself that

\begin{aligned} T_H Gr(k,V) = \operatorname{Hom}(H, V/H). \end{aligned}

Start of Lecture 27

I arrived late and so these notes ought to be cross checked with Karl's.

Last time we saw that $T_WGr(\ell,V) \cong \operatorname{Hom}(W, V/W)$ .

\begin{aligned} T_{(W,W)}\widetilde{M}_k = \{(\varphi,B) ~\mid~ B\in M, \varphi\in \operatorname{Hom}(W,V/W) \text{ s.t. } B|_W = A\circ \varphi\}. \end{aligned}

Proposition:

\widetilde{M}_k

is smooth.

Take $\pi:\widetilde{M}_k\to M_k$ to be the projection map. Then $\pi_*T_{(A,W)}\widetilde{M}_k \subseteq T_AM_k$ .

The forward direction is obvious, if $B|_W = -A\circ \varphi$ then $B\cdot W \subset \operatorname{img}(A)$ . Likewise, if $B\cdot W \subset \operatorname{img}(A)$ , we can define a $\varphi$ by setting for any $B(w) = A(v)$ $\varphi(w) = -v$ . This defines a map $\varphi:W\to V/W$ .

\pi_* T_{(A,W)}\widetilde{M}_k = \{B\in M ~\mid~ B\cdot W\subseteq \operatorname{img}(A)\}.

\begin{aligned} \operatorname{dim}(\pi_*T_{(A,W)}\widetilde{M}) &= \operatorname{dim}(\operatorname{Hom}(W,\operatorname{img}(A))) + k\cdot m \\ &= (n-k)k + k\cdot m = (n+m - k)k \\ &= \operatorname{dim}(M_k) = \operatorname{dim}(\widetilde{M}_k) \end{aligned}

Proposition:

M_k\setminus M_{k-1}

is smooth.

M_k

is singular along

M_{k-1}

with Zariski tangent space equal to

M(n,m)

Why do we care about determinental varieties?

If $\varphi:F\to E$ is a map of vector bundles of ranks $n, m$ on $X$ . There exists a $U\subset X$ which simultaneously trivializes these bundles; $\varphi|_U:\mathcal O^{\oplus n}_X \to \mathcal O^{\oplus m}_X$ . Then $\varphi$ gives a matrix whose entries are regular functions on $U$ , i.e. functions $\psi:U\to M$ , $U_k(\varphi):=\psi^{-1}M_k$ . The set $U_k(\varphi)$ does not depend on the choice of $U$ . This means there exists some $X_k(\varphi)\subset X$ that restricts to $U_k(\varphi)$ on $U$ . This is called the $k$ th determinantal locus of $\varphi$ .

Proposition:

X_k(\varphi)

is either empty or of codimension at most

(m-k)(n-k)

Examples

If $F\cong \mathcal O_X$ then $\varphi:F\to E$ corresponds to a global section of $E$ . $X_0(\varphi)$ is the vanishing locus of such a section. Hence the expected codimenion is $\operatorname{codim}(X_0(\varphi)) = (m - k)(n - k) = (1 - 0)(\operatorname{rank}(E) - 0) = \operatorname{rank}(E)$ .

If $E = \mathcal O_{\mathbb P^2} \oplus \mathcal O_{\mathbb P^2}(1)$ , then for a section $(\sigma, \tau)\in E$ . If $\sigma \neq 0$ then $V(s) = \emptyset$ and if $\sigma = 0$ then $V(s)$ is a line which has codimension $1$ in $\mathbb P^2$ .