Algebraic Curves Lecture 13:

The 13th lecture of algebraic curves by Karl Christ

Grassmanians Continued

We have a correspondence between maps to $\mathbb P^n$ and line bundles, so an embedding $\varphi:C\hookrightarrow \mathbb P^3$ gives a line bundle $L = \varphi_L^*\mathcal O_{\mathbb P^3}(1) = \mathcal O_{\mathbb P^3}(1)|_C = K_C$ and each line bundle $L$ gives an embedding $\varphi:C\hookrightarrow \mathbb P^3$ . CHECK THIS AFTER CLASS

Grassmanians Continued

An element $\lambda\in \bigwedge^n V$ is decomposable if $\lambda = v_1\wedge...\wedge v_k$ . We have a map $G(K,V)\to \mathbb P\left(\bigwedge^k V\right)$ called the Plücker embedding:

\begin{aligned} \mathbb{Gr}(k,V) &\to \mathbb P(\bigwedge^k V) \\ W &\mapsto v_1\wedge ... \wedge v_k \end{aligned}

Plücker Relations

Fix a basis for $V$ . The coordinates of a $k$ -dimensional subspace $W\subseteq V$ are given by the $K\times K$ minors of the $n\times k$ matrix with columns basis vectors of $W$ .

For any two sequences $i_1 < ... < i_{k-1}$ and $j_1 < ... < j_{k+1}$ with $i\leq i_s, j_\ell \leq n$ , it holds that

\begin{aligned} \sum^{k+1}_{\ell = 1}(-1)^{\ell} W_{i_1...i_{k-1}j_\ell} W_{j_1...\hat{j_\ell}...j_{k+1}} = 0 \end{aligned}

where $W_{i_1...i_{k-1}j_\ell}$ is the coordinate corresponding to the $k\times k$ minor with rows in $i_1...i_{k-1}j_\ell$ under the Plücker embedding. Such relations are called the Plücker relations.

Example: Consider the Plücker embedding

\varphi: G(2,4) \to \bigwedge^2 V

and choose

W\in G(2,4)

. Choose a basis

v_1,v_2

W

where

\begin{aligned} v_1 &= a_1e_1 + a_2e_2 + a_3e_3 + a_4e_4 \\ v_2 &= b_1e_1 + b_2e_2 + b_3e_3 + b_4e_4. \end{aligned}

These are the image of the matrix

\begin{aligned} M = \begin{pmatrix} a_1 & b_1 \\ a_2 & b_2 \\ a_3 & b_3 \\ a_4 & b_4 \end{pmatrix}. \end{aligned}

Taking

v_1\wedge v_2

gives me

\begin{aligned} v_1 \wedge v_2 &= (a_1e_1 + a_2e_2 + a_3e_3 + a_4e_4)\wedge (b_1e_1 + b_2e_2 + b_3e_3 + b_4e_4) \\ &= b_1a_2\cdot (e_2 \wedge e_1) + b_2\cdot a_1 \cdot (e_1 \wedge e_2) ... \\ &= (b_1 \cdot a_2 - b_2 \cdot a_1)(e_1\wedge e_2) + ... \end{aligned}

Now take sequences

i_1 = 1

(only option) and

j_1 = 1, j_2 = 2, j_3 = 3

(

k = 2

and

n=4

). Then the Plücker relation is given

\begin{aligned} - 0 + W_{12} \cdot W_{13} - W_{13}\cdot W_{12} = 0, \end{aligned}

which is always true. Taking

i_1 = 1

j_1 = 2, j_3 = 3, j_4 = 4

and we get

\begin{aligned} - W_{12}\cdot W_{34} + W_{13}\cdot W_{24} - W_{14}\cdot W_{23} = 0, \end{aligned}

which is a nonempty statement. This is the only Plücker relation in this case. This realizes

G(2,4)

as a quadric hypersurface in

\mathbb P^5

Claim: $\lambda \in \bigwedge^2 V$ is decomposable if and only if $\lambda \wedge \lambda = 0$ .

Proof. One direction is immediate. For the other direction, suppose

\lambda

is indecomposable and write it

\lambda = e_1\wedge e_2 + e_3\wedge e_4

for some choice of coordinates on

V

. Then

\begin{aligned} (\lambda \wedge \lambda) &= 2 (e_1\wedge e_2\wedge e_3 \wedge e_4) \neq 0. \end{aligned}

Note that

\lambda\wedge\lambda

lives in

\bigwedge^4 V

\square

Let's return to the above example of $G(2,4)$ and the Plücker relation

\begin{aligned} - W_{12}\cdot W_{34} + W_{13}\cdot W_{24} - W_{14}\cdot W_{23} = 0. \end{aligned}

We now have a characterization of decomposable elements, so for

\begin{aligned} \lambda &= \lambda_{12}(e_1\wedge e_2) + \lambda_{13}(e_1\wedge e_3) + \lambda_{14}(e_1\wedge e_4) \\ &\hspace{0.5cm} + \lambda_{23}(e_2\wedge e_3) + \lambda_{24}(e_2\wedge e_4) + \lambda_{34}(e_3\wedge e_4), \end{aligned}

we get

\begin{aligned} \lambda \wedge \lambda &= (\lambda_{12} \cdot \lambda_{34} - \lambda_{13}\cdot \lambda_{24}+\lambda_{14}\cdot \lambda_{23})(e_1\wedge e_2 \wedge e_3 \wedge e_4). \end{aligned}

The coefficient $\lambda_{12} \cdot \lambda_{34} - \lambda_{13}\cdot \lambda_{24}+\lambda_{14}\cdot \lambda_{23}$ gives the Plücker relation. Set

\begin{aligned} \Sigma = \left\{(x,H) ~\middle|~ x\in H\right\} \subseteq \mathbb P^n\times G(k,\mathbb P^n) \end{aligned}

and consider the projections $\pi_1:\Sigma \to \mathbb P^n$ and $\pi_2:\Sigma\to G(k,\mathbb P^n)$ . We call $\Sigma$ an incidence variety. It is a subvariety since the condition $x \in H$ means that $x \wedge b_1 \wedge ... \wedge b_k = 0$ where $b_1,...,b_k$ is a basis for $H$ , and hence

\begin{aligned} \Sigma = \left\{(x,h) ~\middle|~ x\wedge b_1 \wedge ... \wedge b_k = 0\right\} \end{aligned}

meaning that $\Sigma$ is a variety since $x \wedge b_1 \wedge ... \wedge b_k$ is a polynomial equation.

For any subvariety $Y\subseteq G(k,\mathbb P^n)$ , I get a subvariety $\pi_1(\pi_2^{-1}(Y))$ of $\mathbb P^n$ . If I take two subvarieties $X,Y\subseteq \mathbb P^n$ and consider the rational map

\begin{aligned} \varphi: X\times Y&\to G(1,\mathbb P^n) \\ (x,y)&\mapsto x\wedge y \end{aligned}

defined outside the set $x = y$ in $X\times Y$ , then we can define two new varieties

\begin{aligned} J(X,Y) &= \pi_1(\pi_2^{-1}(\overline{\operatorname{img} \varphi})) \hspace{1cm}&\text{"join of $X$ and $Y$"}\\ \textrm{Sec}(X) &= J(X,X)\hspace{1cm}&\text{"secant variety"}. \end{aligned}

These have dimensions satisfying

\begin{aligned} \operatorname{dim}(J(X,Y)) &\leq \operatorname{dim} (X) + \operatorname{dim}(Y) + 1\\ \operatorname{dim}(\textrm{Sec}(X)) &\leq 2\operatorname{dim} X + 1. \end{aligned}

Example: Take $X$ to be a curve in $\mathbb P^n$ with $n\geq 3$ .

$\operatorname{dim}(\textrm{Sec}(X)) = 1$ if and only if $X$ is a line.
$\operatorname{dim}(\textrm{Sec}(X)) = 2$ if and only if $X$ is contained in a plane but not a plane.
$\operatorname{dim}(\textrm{Sec}(X)) = 3$ if and only if $X$ is not contained in a plane.