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 [...]


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?


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