Let \(\varphi: \PP_k^1 \rightarrow \PP_k^3\) be the 3-uple embedding (I, Ex. 2.12). Let \(\mathcal{I}\) be the sheaf of ideals of the twisted cubic curve \(C\) which is the image of \(\varphi\). Then \(\mathcal{I} / \mathcal{I}^2\) is a locally free sheaf of rank 2 on \(C\), so \(\varphi^*\left(\mathcal{I} / \mathcal{I}^2\right)\) is a locally free sheaf of rank 2 on \(\PP^1\). By (2.14), therefore, \[ \varphi^*\left(\mathcal{I} / \mathcal{I}^2\right) \cong \mathcal{O}(l) \oplus \mathcal{O}(m) .\] for some \(l, m \in \ZZ\). Determine \(l\) and \(m\).
Repeat part (a) for the embedding \(\varphi: \PP^1 \rightarrow \PP^3\) given by \(x_0=t^4\), \(x_1=t^3 u\), \(x_2=t u^3\), \(x_3=u^4\), whose image is a nonsingular rational quartic curve.
Answer: If \(\operatorname{char} k \neq 2\), then \(l=m=-7\); if \(\operatorname{char} k=2\), then \(l, m=-6,-8\).
Solution
For \(n\ge2\), consider the parametrized rational curve \[ \varphi_n:\PP^1\longrightarrow\PP^3, \qquad [t:u]\longmapsto [t^{n+1}:t^nu:tu^n:u^{n+1}]. \] Thus \(n=2\) is the twisted cubic of part (a), and \(n=3\) is the quartic of part (b).
1. The image \(C_n\) lies on the smooth quadric\[ Q=V(x_0x_3-x_1x_2)\subseteq\PP^3, \]and under\[ Q\cong\PP^1\times\PP^1 \]it is the graph of the power map\[ [t:u]\longmapsto[t^n:u^n]. \]
The parametrization satisfies \[ x_0x_3-x_1x_2 = t^{n+1}u^{n+1}-t^nu\,tu^n =0. \] The quadric is smooth because the four partial derivatives of its equation are \[ x_3,\qquad -x_2,\qquad -x_1,\qquad x_0, \] which have no common projective zero.
Using the Segre coordinates \[ [a:b]\times[c:d] \longmapsto [ac:ad:bc:bd], \] the curve is obtained by \[ [a:b]=[t^n:u^n], \qquad [c:d]=[t:u]. \] The second projection therefore identifies \(C_n\) with \(\PP^1\), even when the first power map is inseparable.
2. Let\[ E_n=\varphi_n^*(\mathcal I_{C_n}/\mathcal I_{C_n}^2). \]There is an exact sequence\[ \boxed{ 0\longrightarrow\OO_{\PP^1}(-2n-2) \longrightarrow E_n \longrightarrow\OO_{\PP^1}(-2n) \longrightarrow0.} \]
For the regular immersions \[ C_n\subset Q\subset\PP^3, \] the conormal sequence is \[ 0 \longrightarrow \mathcal I_Q/\mathcal I_Q^2|_{C_n} \longrightarrow \mathcal I_{C_n}/\mathcal I_{C_n}^2 \longrightarrow \mathcal I_{C_n/Q}/\mathcal I_{C_n/Q}^2 \longrightarrow0. \]
The quadric is a hypersurface of degree \(2\), so \[ \mathcal I_Q/\mathcal I_Q^2|_{C_n} \cong \OO_{C_n}(-2). \] Since \[ \varphi_n^*\OO_{\PP^3}(1)=\OO_{\PP^1}(n+1), \] its pullback is \[ \OO_{\PP^1}(-2n-2). \]
As a divisor on \(Q\cong\PP^1\times\PP^1\), the graph \(C_n\) has type \((1,n)\), hence \[ C_n^2=2n. \] Therefore \[ \mathcal I_{C_n/Q}/\mathcal I_{C_n/Q}^2 \cong \OO_{C_n}(-C_n) \cong \OO_{\PP^1}(-2n). \] Pulling the conormal sequence back along the isomorphism \(\varphi_n:\PP^1\xrightarrow{\sim}C_n\) gives the displayed sequence.
3. The extension class of step 2 is\[ \boxed{(n-1)\eta} \]where\[ 0\ne\eta\in H^1(\PP^1,\OO_{\PP^1}(-2)) \]is a generator.
Use the affine chart \(x_0\ne0\) with coordinates \[ a=\frac{x_1}{x_0}, \qquad b=\frac{x_2}{x_0}, \qquad c=\frac{x_3}{x_0}. \] Put \[ q=c-ab, \qquad h=b-a^n. \] Then \(q,h\) generate the ideal of \(C_n\) on this chart. Along the parametrization, with \[ s=\frac ut, \] one has \[ a=s, \qquad b=s^n, \qquad c=s^{n+1}. \]
On the chart \(x_3\ne0\), put \[ \alpha=\frac{x_1}{x_3}, \qquad \beta=\frac{x_2}{x_3}, \qquad \gamma=\frac{x_0}{x_3}, \] and take generators \[ q'=\gamma-\alpha\beta, \qquad h'=\alpha-\beta^n. \] On the overlap, \[ \alpha=\frac a c, \qquad \beta=\frac b c, \qquad \gamma=\frac1c. \] Thus, modulo the square of the ideal \((q,h)\), \[ q'=\frac{q}{c^2} \equiv s^{-2n-2}q. \]
For the second generator, \[ h' = \frac{a c^{n-1}-b^n}{c^n}. \] Substitute \[ b=a^n+h, \qquad c=ab+q=a^{n+1}+ah+q \] and retain only terms linear in \(q,h\). The numerator becomes \[ -a^{n^2-n}h + (n-1)a^{n^2-n-1}q \pmod{(q,h)^2}. \] Since \(c^n\equiv a^{n^2+n}\) to zeroth order, one gets \[ h' \equiv -s^{-2n}h + (n-1)s^{-2n-1}q \pmod{(q,h)^2}. \]
2. After factoring them out, the off-diagonal term is represented by\[ (n-1)s^{-1}. \]On the standard two-chart cover of \(\PP^1\), \(s^{-1}\) represents a generatorof\[ H^1(\PP^1,\OO(-2)). \]Hence the extension class is \((n-1)\eta\).
4. If the extension class in step 3 is zero, then\[ E_n \cong \OO(-2n)\oplus\OO(-2n-2). \]If it is nonzero, then\[ \boxed{E_n\cong\OO(-2n-1)\oplus\OO(-2n-1).} \]
The zero-class assertion is exactly the definition of a split extension.
Assume the class is nonzero and twist the exact sequence of step 2 by \(\OO(2n)\): \[ 0 \longrightarrow \OO(-2) \longrightarrow E_n(2n) \longrightarrow \OO \longrightarrow0. \] Its connecting map \[ H^0(\OO)\longrightarrow H^1(\OO(-2)) \] is the nonzero extension class. Both spaces are one-dimensional, so this map is an isomorphism. Consequently \[ H^0(E_n(2n))=0. \]
By Birkhoff–Grothendieck, write \[ E_n\cong\OO(l)\oplus\OO(m), \qquad l\ge m. \] Step 2 gives \[ l+m=-4n-2. \] The vanishing of \(H^0(E_n(2n))\) implies \[ l+2n<0, \] so \[ l\le-2n-1. \] But \(l\) is at least the average \((l+m)/2=-2n-1\). Hence \[ l=m=-2n-1. \]
5. For the twisted cubic,\[ \boxed{l=m=-5.} \]
Here \(n=2\). By step 3 the extension class is \[ (2-1)\eta=\eta\ne0 \] over every field. Step 4 therefore gives \[ E_2\cong\OO(-5)\oplus\OO(-5). \] This proves part (a).
6. For the rational quartic,\[ \boxed{ E_3\cong \begin{cases} \OO(-7)\oplus\OO(-7),&\operatorname{char}k\ne2,\\ \OO(-6)\oplus\OO(-8),&\operatorname{char}k=2. \end{cases}} \]
Here \(n=3\). Step 3 gives extension class \[ 2\eta. \] If \(\operatorname{char}k\ne2\), this is nonzero, and step 4 gives \[ E_3\cong\OO(-7)\oplus\OO(-7). \] If \(\operatorname{char}k=2\), the class vanishes, so the sequence splits: \[ E_3 \cong \OO(-6)\oplus\OO(-8). \] This is exactly the answer stated in part (b).
7. Q.E.D.
Step 5 proves part (a), and step 6 proves part (b).