The 14th class of algebraic curves by Karl Christ
Last time we ended by introducing \(\operatorname{Sec}(X)\), the secant variety of \(X\), where \(X\subseteq \mathbb P^k\) is a smooth variety and \(\operatorname{dim}(X) = k\). We can obtain another similar space by considering the Gauss map
\[\begin{aligned} j:X&\to \mathbb{Gr}(k,\mathbb P^n) \\ p &\mapsto T_pX. \end{aligned}\]We recall the object \(\Sigma\subset \mathbb {Gr}(k,\mathbb P^n) \times \mathbb P^n\) from lecture 13 and the corresponding projections \(\pi_1:\mathbb {Gr}(k,\mathbb P^n)\times \mathbb P^n\to \mathbb {Gr}(k,\mathbb P^n)\) and \(\pi_2:\mathbb {Gr}(k,\mathbb P^n)\times \mathbb P^n \to \mathbbP^n\). Then we define the tangential variety \(\operatorname{Tan}(X) = \pi_2(\pi_1^{-1}(\overline{\operatorname{img} j}))\). It has dimension \(2k\). This makes geometric sense – you have a dimension \(k\) object for each point in \(X\). Note that \(\operatorname{Tan}(X)\subset \operatorname{Sec}(X)\) for curves (the tangent line at a point is the "secant line through that point twice").
This secant variety and tangential variety are enough to prove the following proposition.
Proof. We know we can always embed \(X\) in some \(\mathbb P^{g+1}\) by Riemann-Roch. The idea is to project down to \(\mathbb P^3\) in such a way that the composition with the embedding \(X\hookrightarrow \mathbb P^{g+1}\) remains an isomorphism.
Suppose you have an embedding \(X\hookrightarrow \mathbb P^r\) and consider then a projection \(\varphi_p:X\subset \mathbb P^r\to \mathbb P^{r-1}\) where \(p\in \mathbb P^r\) is a point which does not lie in \(X\)or in some chosen copy of \(\mathbb P^{r-1}\) in \(\mathbb P^r\). We obtain the projection by mapping \(x\in X\) to the point in \(\mathbb P^{r-1}\) intersecting the line between \(x\) and \(p\).
Then \(\varphi_p\) is generically finite, since we may assume that \(X\) is non-degenerate (i.e. that \(X\) is not embedded in \(\mathbb P^r\) as a subset of some smaller \(\mathbb P^d\)).
\(\varphi_p\) is \(1:1\) on its image if and only if \(p\) does not lie on any secant line of \(X\) and
\(\varphi_p\) is injective on tangent spaces if and only if \(p\) does not lie on a tangent line of \(X\).
We note that the map \(\varphi_p\) is given by linear series \(V\subset H^0(\mathbb P^r, \mathcal O_{\mathbb P^r}(1))\) that pass through \(p\). Then the above statements translate to
For any \(q_1,q_2 \in X\) with \(q_1\neq q_2\), there exists \(H\in V\) such that \(q_1\in H\) and \(q_2 \not\in H\).
For any \(q \in X\), there exists \(H\in V\) such that \(H\cap X \supseteq \{q\}\) but \(H.X \not\ni 2q\).
Thus \(\varphi_p\) is an isomorphism on its image if and only if \(p\not\in \operatorname{Sec}(X)\) and \(p\not\in \operatorname{Tan}(X)\). Since \(\operatorname{dim}(\operatorname{Sec}(X))\leq 3\) and \(\operatorname{dim}(\operatorname{Tan}(X)) \leq 2\), it follows that such a \(p\) always exists if \(r\geq 4\).
This proposition means you don't lose anything by simply studying curves in \(\mathbb P^3\). In general, you cannot realize \(X\) as a smnooth curve in \(\mathbb P^2\). However, the next best thing is true: Once can always map \(X\to \mathbb P^2\) birationally with nodal image.
Lemma: Take \(X\subseteq \mathbb P^3\), \(p \in \mathbb P^3\). Then \(\varphi_p|_X\) is birational with nodal image if and only if
\(p\) lies on finitely many tangent lines to \(X\)
\(p\not\in \operatorname{Tan}(X)\)
\(p\) does not lie on any multisecant (a secant line of \(X\) containing 3 or more points of \(X\))
\(p\) does not lie on a secant with coplanar tangent lines.
Note that
Above avoids the possibility that the image of \(X\) in \(\mathbb P^2\) has cusps
This ensures \(\varphi_p\) is defined on all of \(X\)
This condition ensures that we have at worst nodes in the image, we don't want any singularities with preimage consisting of 3 or more points
take two distinct points \(q_1,q_2 \in X\) with parallel (coplanar) tangent lines. Then if \(p\) lies on the secant line to these two points, the two tangent lines will be mapped to the same line in \(\mathbb P^2\). This condition ensures this doesn't happen.
This together with the proof of the previous proposition essentially prove the lemma.
Proof. We may assume \(X\subseteq \mathbb P^3\). We need to find \(p\in \mathbb P^3\) as in the lemma.
Take a rational map \(X\times X \to \mathbb {Gr}(1, \mathbb P^3)\) together with the projections of \(\Sigma\). Examining the image of a pair \((x, y)\in X\times X\) we see that there are only finitely many secant lines of \(X\) which pass through a general point \(p\in \mathbb P^3\).
We know that \(\operatorname{dim}(\operatorname{Tan}(X)) \leq 2\), thus a general point \(p\in \mathbb P^3\) does not live in \(\operatorname{Tan}(X)\).
For 3. and 4. we proceed in steps.
Step 1: If every secant line is a multisecant, then the tangent lines at any two points are coplanar. Suppose then that every secant line is a multisecant, consider \(p\in X\) and the projection \(\varphi_p:\mathbb P^3 \to \mathbb P^2\) (defined only away from \(p\), so this is a rational map). Then \(\varphi_p|_X\) has positive degree. If \(L_{pqr}\) is a multisecant line through \(p\), then \(d\varphi_p|X\) needs to identify the tangent spaces to \(q\) and \(r\). Thus \(T_qX\) and \(T_rX\) need to lie in \(d\varphi_p|_X(T_{\varphi_p(r)}X)\) which is a plane.
Step 2: If all tangent lines are coplanar, then there is a point \(r\in \mathbb P^3\) contained in every tangent line to \(X\). Let \(H\) be the plane spanned by \(T_pX\) and \(T_qX\). Call \(r = T_pX \cap T_qX\). Pick \(s \not\in H, s\in X\). Then \(T_sX \cap T_pX = \operatorname{pt}\) and \(T_sX \cap T_qX = \operatorname{pt}\) If \(T_sX \cap T_pX \neq T_sX\cap T_qX\) then \(T_sX\) would pass through 2 different points in \(H\). This would mean \(T_s'X\subseteq H\) hence \(s\in H\), which is a contradiction.
Step 3: This is not possible. Indeed, projecting from \(r\) would give a map \(X\to X'\) (some other curve) which is generically ramified, but the ramification locus of any map of smooth curves is a divisor of the upstairs curve, at least in characteristic 0 (you can have generically ramified maps in characteristic \(p\)!).
We conclude today with a remark.
Example:
If \(X\subseteq \mathbb P^3\) is non-degenerate, \(p\in \mathbb P^3\). Then \(\varphi_p(X) \subseteq \mathbb P^2\) is not smooth. This is because any smooth curve in \(\mathbb P^2\) needs to be embedded by a complete linear series. Suppose \(X\subseteq \mathbb P^3\) is a smooth curve. Then \(X\) is embedded by a complete linear series if and only if the induced map \(H^0(\mathbb P^2, \mathcal O_{\mathbb P^2}(1))\to H^0(X,\mathcal O_X(1))\) is surjective.
Given a non-degenerate smooth curve \(X\subseteq \mathbb P^3\), this implies \(g(X)\leq \frac{(d-1)(d-2)}{2}\) where \(d\) is the degree of \(X\subseteq \mathbb P^r\). This bounds the genus of all non-planar curves.