Solutions to Exercises 3 and 6

Math 8205
Spring 1998

Solutions to Exercises 3 and 6

Last updated Thursday, June 18, 1998.

Solution to Exercise 3a
Solution to Exercise 3b

Solution to Exercise 6a
Solution to Exercise 6b

Solution to Exercise 3a By Exercise 6.4 in the text, every line in Q = G(1,3)

P⁵ is of the form L = $\Sigma$ _p,H,
where p is a point in P³ and H is a plane in P³. Therefore, we obtain a map of sets $\varphi$ : W -> F₁(Q)

G(1,5) by sending (p,H) --> $\Sigma$ _p,H.
So, the main point is to show that it is a regular map, i.e. that the Plücker coordinates of L $\in$ G(1,5) are given, at least locally,
as polynomials in the coordinates of p and H. Now, in W

P³(P³)* we have an equality of points: (p,H) = ([x₀,..., x₃],[y₀,..., y₃]). and the inclusion p $\in$ H is expressed by the following equation: x₀y₀ + ... + x₃y₃ = 0. Since regularity is a local property, we may assume that p is in the open subset x₀ $\neq$ 0. Then [y₀,..., y₃] $\neq$ [1,0,0,0],
so we may assume without loss of generality that H $\in$ (P³)* lies in the open subset y₃ $\neq$ 0.
To find the Plücker coordinates of L $\in$ G(1,5), it is enough to find the coordinates of two points of L,
or equivalently, the Plücker coordinates of two lines, say $\Lambda$ ₁ and $\Lambda$ ₂, which lie in H and pass through p.
To specify $\Lambda$ ₁ and $\Lambda$ ₂, it suffice sto find two points of H such that p doesn't lie on the line joining them.
One possibility is to choose the points [0, y₃, 0,-y₁] and [0, 0, y₃, -y₂] respectively.
This gives us the Plücker coordinates of $\Lambda$ ₁ and $\Lambda$ ₂ as the 2 by 2 minors of the matrices :

and

respectively. These minors obviously are polynomials in the homogeneous coordinates of p and H.
To finish, we use these minors as the entries of a 2 by 6 matrix whose 2 by 2 minors will be the Plücker coordinates of L.
Those, again, will certainly be polynomials in the homogeneous coordinates of p and H.

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Solution to Exercise 3b
It's fairly clear that our map W -> F₁(Q) is bijective. Therefore we only need to find dim(W).
We consider the map $\pi$ : W -> (P³)* induced by the projection of P³(P³)* to the second fiber.
Clearly, $\pi$ is surjective. Since the inverse image of H $\in$ (P³)* is (isomorphic to) the plane H P³,
every fiber of $\pi$ is of dimension = 2. We conclude that dim(W) = 3 + 2 = 5.

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Solution to Exercise 6a The tangential variety TX = Tan(X) has dimension $\leq$ 2r, where r = dim(X).
Also recall that the differential (d $\pi$ )_p : T_p(X) -> T_(p)(P^{2r
- 1}) is not injective if and only if L $\cap$ t_X,p is nonempty,
where t_X,p is the embedded tangent space of X at p.

Case 1: dim(TX) < 2r. In this case, L $\cap$ TX is empty for a general line L P^2r+1,

Case 2: dim(TX) = 2r. In this case, L $\cap$ TX is a finite set of points for a general line L P^2r+1.

provided that

L $\cap$ TX

To check this last point, consider the closed subset Y X $\times$ P^2r+1, consisting of all pairs (p,z) such that z $\in$ t_X,p.

By considering the projection to the first factor, we see that dim(Y) = 2r.
TX is the image of Y under the projection to the second factor. Let $\varphi$ : Y -> TX be induced by this projection.

$\in$

$\varphi$

^-1

$\in$

t_X,p

$\varphi$

^-1

₀

L $\cap$ V

₀

^2r+1

Choosing L so that L $\cap$ TX is finite and L $\cap$ V₀ is empty, we conclude that there are at most finitely many points p such that (d $\pi$ )_p is not injective.

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Solution to Exercise 6b The secant variety Sec(X) has dimension $\leq$ 2r + 1, where r = dim(X).

Case 1: dim(Sec X) < 2r. In this case, L $\cap$ Sec(X) is empty for a general line L P^2r+1, and $\pi$ = $\pi$ _L is injective.
Case 2: dim(Sec X) = 2r. In this case, L $\cap$ Sec(X) is a finite set of points for a general line L P^2r+1.

L $\cap$

Case 3: dim(Sec X) = 2r + 1. In this case Sec(X) = P^2r+1, so that L Sec(X).

Let W₀

(X $\times$ X - $\Delta$ ) $\times$ P^2r+1 be the set of triples (p,q,z) such that z $\in$ <p,q> (= the line joining p and q),
and let W be the closure of W₀ in X $\times$ X $\times$ P^2r+1. By considering the projection to X $\times$ X - $\Delta$ , we see that dim(W) = dim(W₀ ) = 2r + 1.
Now, let $\psi$ : Y -> Sec(X) be induced by projection of the product to P^2r+1. Then a general fiber of $\psi$ is finite.
Let W₁ W be the set of points where the local fiber dimension is $\geq$ 1. Clearly, dim(W₁) < dim(W) = 2r + 1.
We consider the restriction of $\psi$ to W₁. The key point that dim( $\psi$ (W₁)) < dim(W₁).
(Just note that a general fiber of the restriction has dimension $\geq$ 1.) It follows that dim( $\psi$ (W₁)) $\leq$ 2r - 1.
Therefore, if L is a general line in P^2r+1, the intersection L $\cap$ $\psi$ (W₁) is empty, and the double point set of $\pi$ _L is of dimension = 1.

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Math 8205 Spring 1998

Solutions to Exercises 3 and 6

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Math 8205
Spring 1998