Let \(X\) be a curve of genus \(g\) embedded birationally in \(\PP^2\) as a curve of degree \(d\) with \(r\) nodes. Generalize the method of (Ex. 2.3) to show that \(X\) has \[ 6(g-1)+ 3d \] inflection points. A node does not count as an inflection point. Assume \(\characteristic k=0\).
Now let \(X\) be a curve of genus \(g\) embedded as a curve of degree \(d\) in \(\PP^n\), \(n \geq 3\), not contained in any \(\PP^{n-1}\). For each point \(P \in X\), there is a hyperplane \(H\) containing \(P\), such that \(P\) counts at least \(n\) times in the intersection \(H \intersect X\). This is called an osculating hyperplane at \(P\). It generalizes the notion of tangent line for curves in \(\PP^2\).
If \(P\) counts at least \(n+1\) times in \(H \intersect X\), we say \(H\) is a hyperosculating hyperplane, and that \(P\) is a hyperosculation point. Use Hurwitz’s theorem as above, and induction on \(n\), to show that \(X\) has \[ n(n+1)(g-1)+(n+1) d \] hyperosculation points.
If \(X\) is an elliptic curve, for any \(d \geq 3\), embed \(X\) as a curve of degree \(d\) in \(\PP^{d-1}\), and conclude that \(X\) has exactly \(d^2\) points of order \(d\) in its group law.
(Meaning of points of order d)
The \(d^2\) points in part (c) are the points annihilated by \(d\), i.e. the \(d\)-torsion subgroup \(X[d](k)\). The phrase “points of order \(d\)” in the source cannot mean points of exact group-theoretic order \(d\): for example, \(P_0\) is among these \(d^2\) points but has order \(1\).
(Characteristic scope)
The counting argument above uses the characteristic-zero hypothesis. More generally, if \(\characteristic k\) does not divide \(d\), the multiplication map \([d]\) is separable of degree \(d^2\), so \(X[d](k)\) again has \(d^2\) points. If \(\characteristic k=p\) divides \(d\), the finite group scheme \(X[d]\) still has degree \(d^2\), but its set of geometric points can have fewer than \(d^2\) elements. Thus the literal point-count conclusion requires characteristic zero, or at least \(\characteristic k\nmid d\).
Solution
We count inflection and hyperosculation points with their natural ramification weights. The characteristic-zero hypothesis is used in the Wronskian calculation below.
1. Let \(L\) be a line bundle of degree \(d\) on a smooth curve \(X\) of genus\(g\), and let\[ V\subseteq H^0(X,L), \qquad \dim V=n+1, \]be a base-point-free linear series. Its ramification divisor has degree\[ \boxed{ (n+1)d+n(n+1)(g-1). } \]
Fix a point \(P\in X\), a local parameter \(t\), and a local frame of \(L\). Write a basis of \(V\) locally as functions \[ f_0,\ldots,f_n. \] In characteristic zero form the Wronskian \[ W = \det\left( \frac{1}{i!}\frac{d^i f_j}{dt^i} \right)_{0\leq i,j\leq n}. \] Under a change of local frame of \(L\) the determinant transforms by the inverse \((n+1)\)st power of the transition function, and under a change of parameter it transforms by the inverse \[ 1+2+\cdots+n = \frac{n(n+1)}2 \] power of the canonical transition function. Hence the local Wronskians glue to a global section of \[ L^{\otimes(n+1)} \otimes \omega_X^{\otimes n(n+1)/2}. \] Its zero divisor therefore has degree \[ \begin{aligned} (n+1)\deg L &+ \frac{n(n+1)}2\deg\omega_X\\ &=(n+1)d +n(n+1)(g-1). \end{aligned} \]
To identify the zero order, choose a basis adapted to \(P\) with strictly increasing vanishing orders \[ 0\leq a_0(P)<a_1(P)<\cdots<a_n(P). \] The lowest term of the determinant is a nonzero Vandermonde multiple of \[ t^{\sum_i(a_i(P)-i)}, \] so \[ \operatorname{ord}_P(W) = \sum_{i=0}^n(a_i(P)-i). \] This is the usual ramification weight.
For \(n=1\) this is precisely the ramification divisor in Riemann–Hurwitz. Passing from order \(n-1\) jets to order \(n\) jets adds the line bundle \[ L\otimes\omega_X^{\otimes n}, \] whose degree is \(d+n(2g-2)\); thus the displayed formula is also exactly the induction on \(n\) requested in the exercise.
2. For a nondegenerate map to \(\PP^n\), a point is ahyperosculation point exactly when the Wronskian of step 1 vanishes there.
The hyperplanes in \(\PP^n\) correspond to the sections in \(V\). At \(P\), the largest possible order of contact of a hyperplane section is \[ a_n(P). \] Because the \(a_i(P)\) are distinct nonnegative integers, \[ a_i(P)\geq i. \] Thus an osculating hyperplane has contact at least \(n\), and hyperosculation occurs exactly when \[ a_n(P)\geq n+1. \] This happens if and only if some \(a_i(P)>i\), which by step 1 is equivalent to \[ \operatorname{ord}_P(W)>0. \] Its natural multiplicity as a hyperosculation point is the ramification weight \[ \sum_i(a_i(P)-i). \]
3. In part (a), the total inflection weight is\[ \boxed{ 6(g-1)+3d. } \]
Let \[ \nu:X\longrightarrow C\subseteq\PP^2 \] be the normalization of the nodal plane model, and put \[ L=\nu^*\OO_C(1). \] Then \[ \deg L=d, \] and the three coordinate linear forms give a base-point-free \(g^2_d\) on \(X\). Apply step 1 with \(n=2\): \[ \deg R_V = 3d+6(g-1). \]
At a point whose branch in the plane has tangent-contact order \(m\geq2\), the vanishing sequence of this \(g^2_d\) is \[ 0,1,m. \] Its ramification weight is therefore \[ m-2, \] which is exactly the usual inflection multiplicity. In particular an ordinary node contributes no inflection merely by being a node: each branch has the generic initial vanishing \(0,1,2\) unless it has additional tangent contact, and only such excess contact contributes.
Thus the ramification divisor is precisely the properly counted inflection divisor, and its degree is \[ 6(g-1)+3d. \] Using \[ g=\frac{(d-1)(d-2)}2-r, \] the same number is \[ 3d(d-2)-6r. \]
4. In part (b), the total hyperosculation weight is\[ \boxed{ n(n+1)(g-1)+(n+1)d. } \]
Take \[ L=\OO_X(1), \] so \(\deg L=d\), and let \(V\) be the \((n+1)\)-dimensional space obtained by restricting hyperplanes of \(\PP^n\) to \(X\). Nondegeneracy means this series has projective dimension \(n\).
By step 2 its ramification divisor is exactly the hyperosculation divisor, with the proper weights. Step 1 gives its degree as \[ (n+1)d+n(n+1)(g-1), \] which is the required formula.
5. For the elliptic curve in part (c), the complete linear system\[ \abs{dP_0} \]embeds \(X\) in \(\PP^{d-1}\) as a curve of degree \(d\).
Put \[ L=\OO_X(dP_0). \] Since \(g(X)=1\) and \(\deg L=d>0\), Riemann–Roch gives \[ h^0(X,L)=d. \] For any \(P,Q\in X\), allowing \(P=Q\), Riemann–Roch also gives \[ h^0(X,L(-P))=d-1, \qquad h^0(X,L(-P-Q))=d-2, \] because both line bundles have positive degree. Thus sections of \(L\) separate distinct points, and taking \(Q=P\) shows that they also separate tangent directions. Hence \(L\) is very ample, so the complete linear system gives a closed immersion \[ X\hookrightarrow\PP H^0(X,L)^* = \PP^{d-1}. \] The pullback of a hyperplane is \(L\), hence the embedded curve has degree \[ \deg L=d. \]
6. A point \(P\in X\) is a hyperosculation point for the embedding of step5 if and only if\[ \boxed{d(P-P_0)=0} \]in the elliptic-curve group law.
Here \[ n=d-1. \] A hyperosculating hyperplane has contact at least \[ n+1=d \] at \(P\). Since a hyperplane section has total degree \(d\), this means its intersection divisor is exactly \[ dP. \] Such a hyperplane exists if and only if \[ dP\sim dP_0. \] Under \[ X\cong\Pic^0(X), \qquad Q\longmapsto[Q-P_0], \] this is equivalent to \[ d(P-P_0)=0. \]
7. Every point satisfying \(d(P-P_0)=0\) has hyperosculation weight exactly\(1\).
For \[ 0\leq m\leq d-1, \] the line bundle \[ L(-mP) \] has positive degree \(d-m\), so Riemann–Roch on the elliptic curve gives \[ h^0(X,L(-mP))=d-m. \]
If \(d(P-P_0)=0\), then \[ L(-dP)\cong\OO_X, \] so \[ h^0(X,L(-dP))=1, \qquad h^0(X,L(-(d+1)P))=0. \] Therefore the vanishing sequence at \(P\) is \[ 0,1,\ldots,d-2,d. \] Its ramification weight is \[ d-(d-1)=1. \]
If \(d(P-P_0)\ne0\), then the degree-zero line bundle \(L(-dP)\) is nontrivial, so it has no nonzero global section. The vanishing sequence is then \[ 0,1,\ldots,d-1, \] and the weight is \(0\).
8. In characteristic zero, the subgroup of points killed by \(d\) hasexactly\[ \boxed{d^2} \]elements.
Apply step 4 with \[ g=1, \qquad n=d-1, \qquad \deg X=d. \] The total hyperosculation weight is \[ n(n+1)(g-1)+(n+1)d = 0+d^2 = d^2. \] By steps <1>6–<1>7 the hyperosculation divisor is reduced and its support is exactly \[ X[d](k) = \{P\in X(k):d(P-P_0)=0\}. \] Hence this set has exactly \(d^2\) points.
9. Q.E.D.
Step 3 proves part (a), step 4 proves part (b), and steps <1>5–<1>8 prove part (c).