# Grassmannians: Varieties swept out by linear spaces (Eisenbud & Harris: 3264 and All That)

I have a couple of questions on statements from Harris' and Eisenbuds's lecture "3264 and All That" at page 145, Section 4.2.3: Varieties swept out by linear spaces.

The content of 4.2.3 answers the Keynote Question (b) from page 131:

Let $$C \subset \mathbb{G}(1,3) \subset \mathbb{P}^5$$ (the last inclusion is Plucker emb) be a twisted cubic curve twisted cubic curve contained in the Grassmannian $$\mathbb{G}(1,3) \subset \mathbb{P}^5$$ of lines in $$\mathbb{P}^5$$, and let

$$S= \bigcup_{ [\Lambda] \in C} \Lambda \subset\mathbb{P}^3$$

be the surface swept out by the lines corresponding to points of $$C$$. What is the degree of $$S$$? (Answer on page 145.)

Recall the degree of a $$r$$ dimensional subvariety $$V \subset \mathbb{P}^n$$ is the degree (=number of points) of of the intersection between $$V$$ and a general $$n-r$$ dimensional hyperplane $$H \subset \mathbb{P}^n$$ that intersects with $$V$$ transversally: for such $$H$$ we have $$deg(X) = \# (V \cap H)$$.

what says section 4.2.3:

Let $$C \subset \mathbb{G}(1,3)$$ be an irreducible curve, and consider the variety $$X \subset \mathbb{P}^n$$ swept out by the linear spaces corresponding to points of $$C$$; that is,

$$X= \bigcup_{ [\Lambda] \in C} \Lambda \subset\mathbb{P}^3$$

(See Figure 4.1). We would like to relate the geometry of $$X$$ to that of $$C$$; in particular, Keynote Question (b) asks us to find the degree of $$X$$ when $$C \subset \mathbb{G}(1,3) \subset \mathbb{P}^5$$ is a twisted cubic curve.

To begin with, observe that $$X$$ is indeed a closed subvariety of $$\mathbb{P}^n$$: If

$$\Phi=\{(\Lambda,p) \in \mathbb{G}(k,n) \times \mathbb{P}^n \vert p \in \lambda \}$$

is the universal $$k$$-plane $$\mathbb{G}(k,n)$$, as described in Section 3.2.3, and $$\alpha: \Phi \to \mathbb{G}(k,n)$$ and $$\beta: \Phi \to \mathbb{P}^n$$ are the projections, then we can write

$$X= \beta(\alpha^{-1}(C)).$$

Now, suppose that a general point $$x \in X$$ lies on a unique $$k$$-plane $$\Lambda \in C$$ - that is, the restricted map $$\beta:\alpha^{-1}(C) \to X \subset \mathbb{P}^n$$ is birational (???), so that in particular $$\dim(X)=k+1$$. The degree of $$X$$ is the number of points of intersection of $$X$$ with a general $$(n-k-1)$$-plane $$\Gamma \subset \mathbb{P}^n$$; since each of these points is a general point of $$X$$, and so lies on a unique $$k$$-plane $$\Lambda$$, the number is the number of $$k$$-planes $$\Lambda$$ that meet $$\Gamma$$. In other words, we have

$$deg(X)= \#(X \cap \Gamma) \\ = \#(C \cap \Sigma_1(\Gamma)) \\ = \deg([C] \cdot \sigma_1 \text{ (by Kleiman's thm)} \\ =\deg(C),$$

the notations for $$\Sigma_1(\Gamma)$$ and $$\sigma_1$$ are explained in chapter 3: Intro to Grassmannians and lines (page 85).

where by the degree of $$C$$ we mean the degree under the Plucker embedding of $$\mathbb{G}(k,n)$$.

These ideas allow us to answer Keynote Question (b): The surface $$X \subset \mathbb{P}^n$$ swept out by the lines corresponding to a twisted cubic $$C \subset \mathbb{G}(k,n)$$, times the degree of the map $$\beta$$ defined above, is equal to $$3$$. Thus the surface X itself has degree $$3$$ or $$1$$. (???)

In the latter case, the curve $$C$$ would be contained in a Schubert cycle $$\Sigma_{1,1}$$ and as we have seen in the description on page 138, this Schubert cycle is contained in the $$2$$-plane in $$\mathbb{P}^5$$ defined by the vanishing of three Plucker coordinates. Since a twisted cubic is not contained in a $$2$$-plane, this shows that the surface $$X$$ has degree $$3$$. More of the geometry [...]

Questions:

Q_1: On "suppose that a general point $$x \in X$$ lies on a unique $$k$$-plane $$\Lambda \in C$$ —that is, the map $$\beta$$ (see above) is birational": I not understand this argument.

The "general" condition says that there exist an open dense subset $$U \in X$$ such that the restriction of $$\beta$$ to $$\beta^{-1}(U)$$ is bijective. Why this implies also birationality?

Recall, birational means for a map $$f: A \to B$$ between varieties $$A,B$$ that there exist open dense $$V \subset B$$ such that the restriction of $$f$$ to $$f^{-1}(V)$$ is an isomorphism between $$f^{-1}(V)$$ and $$V$$ in category of varieties resp. schemes. Here we observe that the restriction $$\beta \vert _{\beta^{-1}(U)}$$ is only a bijection on sets. Why this imply the birationality?

Q_2: I not understand the sentences "surface $$X \subset \mathbb{P}^n$$ swept out by the lines corresponding to a twisted cubic $$C \subset \mathbb{G}(k,n)$$, times the degree of the map $$\beta$$ defined above, is equal to $$3$$" and "Thus the surface $$X$$ itself has degree $$3$$ or $$1$$".

Literally it says $$X$$ ( probably the author means the degree of $$X$$?) times degree of $$\beta$$ (since $$\beta$$ birat., it's degree is well defined) is equal $$3$$, i.e. $$\deg(\beta) \cdot \deg(X) =3$$. Why does it hold?

In addition, haven't we above already showed that $$deg(X)=deg(C)=3$$ as $$C$$ twisted cubic? That is it seems to me that at this point we are already done, or not?

Why we need an additional argument to exclude that $$deg(X)=1$$ from page 138? Isn't it redundant as we already know $$deg(X)=deg(C) (=3)$$? Do I miss here some point?