The dual curve, its class $d(d-1)$, and the counts of inflection points and bitangents

Let $X$ be a curve of degree $d$ in $\PP^2$. For each point $P \in X$, let $T_P(X)$ be the tangent line to $X$ at $P$ (I, Ex. 7.3). Considering $T_P(X)$ as a point of the dual projective plane $(\PP^2)^*$, the map $P \mapsto T_P(X)$ gives a morphism of $X$ to its dual curve $X^*$ in $(\PP^2)^*$ (I, Ex. 7.3).

Note that even though $X$ is nonsingular, $X^*$ in general will have singularities. We assume $\characteristic k=0$ below.

Fix a line $L \subseteq \PP^2$ which is not tangent to $X$. Define a morphism $\varphi: X \to L$ by $\varphi(P)=T_P(X) \intersect L$, for each point $P \in X$. Show that $\varphi$ is ramified at $P$ if and only if either (1) $P \in L$, or (2) $P$ is an inflection point of $X$, which means that the intersection multiplicity (I, Ex. 5.4) of $T_P(X)$ with $X$ at $P$ is $\geq 3$. Conclude that $X$ has only finitely many inflection points.
A line of $\PP^2$ is a multiple tangent of $X$ if it is tangent to $X$ at more than one point. It is a bitangent if it is tangent to $X$ at exactly two points. If $L$ is a multiple tangent of $X$, tangent to $X$ at the points $P_1, \ldots, P_r$, and if none of the $P_i$ is an inflection point, show that the corresponding point of the dual curve $X^*$ is an ordinary $r$-fold point, which means a point of multiplicity $r$ with distinct tangent directions (I, Ex. 5.3). Conclude that $X$ has only finitely many multiple tangents.
Let $O \in \PP^2$ be a point which is not on $X$, nor on any inflectional or multiple tangent of $X$. Let $L$ be a line not containing $O$. Let $\psi: X \to L$ be the morphism defined by projection from $O$. Show that $\psi$ is ramified at a point $P \in X$ if and only if the line $OP$ is tangent to $X$ at $P$, and in that case the ramification index is 2. Use Hurwitz’s theorem and (I, Ex. 7.2) to conclude that there are exactly $d(d-1)$ tangents of $X$ passing through $O$. Hence the degree of the dual curve (sometimes called the class of $X$) is $d(d-1)$.
Show that for all but a finite number of points of $X$, a point $O$ of $X$ lies on exactly $(d+1)(d-2)$ tangents of $X$, not counting the tangent at $O$.
Show that the degree of the morphism $\varphi$ of a. is $d(d-1)$. Conclude that if $d \geq 2$, then $X$ has $3d(d-2)$ inflection points, properly counted. (If $T_P(X)$ has intersection multiplicity $r$ with $X$ at $P$, then $P$ should be counted $r-2$ times as an inflection point. If $r=3$ we call it an ordinary inflection point.) Show that an ordinary inflection point of $X$ corresponds to an ordinary cusp of the dual curve $X^*$.
Now let $X$ be a plane curve of degree $d \geq 2$, and assume that the dual curve $X^*$ has only nodes and ordinary cusps as singularities (which should be true for sufficiently general $X$). Then show that $X$ has exactly $\frac{1}{2} d(d-2)(d-3)(d+3)$ bitangents. Hint: Show that $X$ is the normalization of $X^*$. Then calculate $p_a(X^*)$ two ways: once as a plane curve of degree $d(d-1)$, and once using (Ex. 1.8).
For example, a plane cubic curve has exactly 9 inflection points, all ordinary. The line joining any two of them intersects the curve in a third one.
A plane quartic curve has exactly 28 bitangents. (This holds even if the curve has a tangent with four-fold contact, in which case the dual curve $X^*$ has a tacnode.)

(Erratum to the degree-one setup and part (h))

The setup should assume $d\geq2$. For $d=1$ the Gauss map has a point as its image and the map $\varphi$ in part (a) is constant, so the ramification and degree assertions are not statements about a finite morphism of curves.

There is also a genuine inconsistency between parts (b) and (h). Part (b) defines a bitangent to be tangent at two distinct points. With that definition, part (h) is false for quartics having a hyperflex. The invariant statement is:

Every nonsingular plane quartic in characteristic $0$ has exactly $28$ generalized bitangent lines, where a line with section $4P$ is counted as a degenerate bitangent $2P+2P$. If $h$ is the number of hyperflex lines, then there are exactly $28-h$ bitangents in the strict sense of part (b).

For a concrete counterexample to the literal statement, take the Fermat quartic \[ X=V(x^4+y^4+z^4) \] over $\CC$. If $\zeta^4=-1$, then \[ P=[0:1:\zeta]\in X \] has tangent line \[ y+\zeta^3z=0. \] Restricting the quartic equation to this line gives $x^4=0$, so the tangent section is $4P$. Permuting the zero coordinate and choosing the four values of $\zeta$ gives $12$ hyperflexes. Step <1>14 below shows that this quartic therefore has $16$, not $28$, strict bitangents.

The final parenthetical in the source is also incorrect: the dual germ of a four-fold contact is unibranch of multiplicity $3$, with local leading parametrization $(t^3,t^4)$ and $\delta=3$. It is an $E_6$ cusp, not a tacnode. A tacnode has two branches. This local calculation is included in step <1>11.

Solution

Assume throughout that $d\geq2$. Let \[ \gamma:X\longrightarrow X^*\subset(\PP^2)^* \] denote the Gauss map $P\mapsto T_P(X)$.

1. If $P\in X$ and\[ r=I_P\bigl(X,T_P(X)\bigr), \]then, in suitable local coordinates, the Gauss map is\[ t\longmapsto \bigl(-h'(t),\,t h'(t)-h(t)\bigr), \qquad h(t)=a_rt^r+O(t^{r+1}),\quad a_r\ne0. \]

Choose affine coordinates with \[ P=(0,0), \qquad T_P(X)=V(y). \] Since $X$ is nonsingular at $P$, a local parameter $t=x$ writes the curve as \[ y=h(t), \] where $h(0)=h'(0)=0$. The order of $h$ is precisely the intersection multiplicity with the tangent line, so \[ h(t)=a_rt^r+O(t^{r+1}), \qquad r\geq2. \]

The tangent line at $(t,h(t))$ is \[ y-h(t)=h'(t)(x-t), \] or \[ -h'(t)x+y+t h'(t)-h(t)=0. \] In the dual affine chart where the coefficient of $y$ is $1$, this is exactly the displayed parametrization of $\gamma$.

2. The morphism $\varphi:X\to L$ of part (a) is ramified at $P$ if andonly if $P\in L$ or $P$ is an inflection point.

Keep the notation of step 1.

First suppose $P\notin L$. Apply a projective change of coordinates sending $L$ to the line at infinity. The tangent line in step 1 meets $L$ at \[ [1:h'(t):0]. \] Thus a local parameter on the target is $h'(t)$, and, because $\operatorname{char}k=0$, \[ e_P(\varphi) = \operatorname{ord}_t h'(t) =r-1. \] Hence $\varphi$ is ramified at such a point exactly when $r\geq3$, i.e. when $P$ is an inflection point.

Now suppose $P\in L$. Since $L$ is not tangent to $X$, choose the local coordinates so that \[ L=V(x), \qquad T_P(X)=V(y). \] The tangent line at $(t,h(t))$ meets $L$ at \[ (0,h(t)-t h'(t)). \] Since \[ h(t)-t h'(t) =(1-r)a_rt^r+O(t^{r+1}) \] and $1-r\ne0$ in characteristic $0$, \[ e_P(\varphi)=r\geq2. \] Thus every point of $X\cap L$ is ramified. These two cases prove the stated criterion.

3. The curve $X$ has only finitely many inflection points.

The map $\varphi$ is nonconstant. Indeed, if it were constant with value $Q\in L$, every tangent line to $X$ would pass through $Q$. Projection from $Q$ gives a nonconstant map from $X$ to $\PP^1$: it has degree $d$ if $Q\notin X$, and after removing the base point it has degree $d-1$ if $Q\in X$. Away from $Q$, its differential would vanish everywhere because the tangent line at every point passes through the centre of projection. That would make the projection inseparable, which is impossible in characteristic $0$.

Hence $\varphi$ is a nonconstant morphism of projective curves and therefore finite and separable. Its ramification locus is finite. By step 2 every inflection point outside the finite set $X\cap L$ lies in that ramification locus, so the set of inflection points is finite.

4. If a line $M$ is tangent to $X$ at distinct non-inflection points$P_1,\ldots,P_s$, then the corresponding point $[M]\in X^*$ is an ordinary$s$-fold point. Consequently $X$ has only finitely many multiple tangents.

At a non-inflection point, step 1 has $r=2$, so \[ h(t)=a_2t^2+O(t^3), \qquad a_2\ne0. \] Writing the dual coordinates in step 1 as $(u,v)$ gives \[ u=-2a_2t+O(t^2), \qquad v=a_2t^2+O(t^3). \] Thus the image of a neighbourhood of $P_i$ is a nonsingular branch of $X^*$. Its tangent line is $v=0$, which intrinsically is the line \[ P_i^*\subset(\PP^2)^* \] parametrizing all primal lines through $P_i$. Since the $P_i$ are distinct, the dual lines $P_i^*$ are distinct. Therefore the $s$ branches through $[M]$ are smooth and have pairwise distinct tangent directions, which is exactly an ordinary $s$-fold point in the sense of Exercise I.5.3.

By step 3 there are only finitely many inflectional tangent lines. Every other multiple tangent gives a singular point of the projective curve $X^*$ by the preceding paragraph. A projective curve has only finitely many singular points, hence there are only finitely many multiple tangents.

5. Let $O\notin X$ lie on no inflectional or multiple tangent. Projectionfrom $O$ gives a degree-$d$ morphism\[ \psi:X\longrightarrow\PP^1 \]whose ramification points are exactly the points $P$ for which $OP=T_P(X)$,and each has ramification index $2$.

A fibre of the projection is the intersection of $X$ with a line through $O$. At a point $P$, choose a local coordinate on the pencil of lines through $O$ whose value at $P$ corresponds to the line $OP$. The difference from that target value is, up to a unit, a local equation for $OP$ restricted to $X$. Hence \[ e_P(\psi)=I_P(X,OP). \] It follows that $e_P(\psi)>1$ exactly when $OP=T_P(X)$. Since $O$ lies on no inflectional tangent, the contact order is then exactly $2$. The hypothesis that $O$ lies on no multiple tangent ensures that different ramification points give different tangent lines through $O$.

Because $O\notin X$, a general line through $O$ meets $X$ in $d$ points, counted with multiplicity, so $\deg\psi=d$.

6. Exactly $d(d-1)$ tangent lines to $X$ pass through such a point $O$,and\[ \deg X^*=d(d-1). \]

5. Its ramificationdivisor has degree\[ \deg R_\psi =2g(X)-2+2d =d^2-d =d(d-1). \]Every ramification point is simple and corresponds to exactly one tangentthrough $O$, so this is the desired number of tangents.

In the dual plane, the lines through $O$ form a line $O^*$. Its intersection with $X^*$ consists of the tangent lines to $X$ passing through $O$. At each of the $d(d-1)$ points just found, $X^*$ is smooth by step 4, and its tangent line is $P^*$ for the corresponding tangency point $P$. Since $O\ne P$, the lines $O^*$ and $P^*$ are distinct, so the intersection is transverse. Therefore \[ O^*\cdot X^*=d(d-1), \] which is the degree of $X^*$.

7. For all but finitely many $O\in X$, exactly\[ (d+1)(d-2) \]tangent lines to $X$, other than $T_O(X)$, pass through $O$.

By steps <1>3–<1>4 there are only finitely many inflectional and multiple tangent lines. Exclude from $X$ all inflection points and all points lying on one of those finitely many exceptional lines. This removes only finitely many points.

Fix a remaining point $O$. The pencil of lines through $O$, after removing the fixed point $O$ from each line section, defines a morphism \[ \psi_O:X\longrightarrow\PP^1 \] of degree $d-1$. It is unramified at $O$: in coordinates \[ O=(0,0),\qquad T_O(X)=V(y),\qquad y=a_2x^2+O(x^3), \] a local coordinate on the pencil is the slope \[ \frac yx=a_2x+O(x^2), \] which has order $1$.

At a point $P\ne O$, the same argument as in step 5 shows that $\psi_O$ is ramified exactly when $OP=T_P(X)$. By the choice of $O$, such a tangent is neither inflectional nor multiple, so every ramification index is $2$ and distinct ramification points give distinct tangent lines. Riemann–Hurwitz therefore gives \[ \begin{aligned} \#\{\text{tangents through }O\text{ other than }T_O(X)\} &=2g(X)-2+2(d-1)\\ &=d^2-d-2\\ &=(d+1)(d-2). \end{aligned} \]

8. The morphism $\varphi:X\to L$ of part (a) has degree\[ d(d-1). \]

Choose a general point $Q\in L$. We may arrange that $Q\notin X$ and that $Q$ lies on no inflectional or multiple tangent. By step 6 there are exactly $d(d-1)$ tangent lines to $X$ through $Q$.

The fibre $\varphi^{-1}(Q)$ consists exactly of their tangency points, since \[ \varphi(P)=Q \quad\Longleftrightarrow\quad Q\in T_P(X). \] The choice of $Q$ makes those points distinct and unramified in this fibre. Thus a general fibre has $d(d-1)$ points and \[ \deg\varphi=d(d-1). \]

9. The inflection points of $X$, counted with weight\[ I_P(X,T_P(X))-2, \]have total weight\[ 3d(d-2). \]

Choose the line $L$ in part (a) generally enough that it contains no inflection point. Since it is not tangent to $X$, it meets $X$ transversely in exactly $d$ points. At each $P\in X\cap L$, step 2 gives \[ e_P(\varphi)=2, \] so these points contribute $d$ to the ramification divisor.

If $P\notin L$ is an inflection point with \[ r=I_P(X,T_P(X))\geq3, \] step 2 gives \[ e_P(\varphi)=r-1, \] so its ramification contribution is $r-2$. Step 2 shows that there is no other ramification.

By step 8 and Riemann–Hurwitz, \[ \begin{aligned} \deg R_\varphi &=2g(X)-2+2d(d-1)\\ &=d^2-3d+2d^2-2d\\ &=3d^2-5d. \end{aligned} \] Subtracting the $d$ contributions from $X\cap L$ leaves \[ 3d^2-6d=3d(d-2), \] which is exactly the asserted weighted inflection count.

10. An ordinary inflection point of $X$ maps to an ordinary cusp of$X^*$.

At an ordinary inflection point, step 1 has \[ h(t)=a_3t^3+O(t^4), \qquad a_3\ne0. \] Hence the local parametrization of the dual branch is \[ u=-3a_3t^2+O(t^3), \qquad v=2a_3t^3+O(t^4). \] Thus the branch has multiplicity $2$ and its tangent line meets it with multiplicity $3$. Equivalently, after formal changes of local coordinates it has parametrization $(t^2,t^3)$, the ordinary cusp.

11. The Gauss map $\gamma:X\to X^*$ is the normalization of $X^*$.If $P$ is a hyperflex, then $\gamma(P)$ is a unibranch multiplicity-$3$singularity with $\delta=3$, not a tacnode.

The Gauss map is nonconstant by the argument of step 3, hence finite onto its image. By step 4 there are only finitely many multiple tangents, so a general point of $X^*$ is the tangent line at exactly one point of $X$. Therefore $\gamma$ is generically one-to-one and hence birational. Since $X$ is nonsingular and thus normal, the finite birational map \[ \gamma:X\longrightarrow X^* \] is the normalization morphism.

Now suppose \[ I_P(X,T_P(X))=4. \] Step 1 gives \[ h(t)=a_4t^4+O(t^5), \qquad a_4\ne0, \] and hence \[ u=-4a_4t^3+O(t^4), \qquad v=3a_4t^4+O(t^5). \] This is one branch, of multiplicity $3$. Its value semigroup has initial values $3$ and $4$. Hence it contains the numerical semigroup $\langle3,4\rangle$, which contains every integer at least $6$. Conversely, no element of the completed local ring can have valuation $1$ or $2$, and an element of valuation $5$ cannot occur: below valuation $6$, the only nonconstant monomials in the two local coordinates have the distinct valuations $3$ and $4$, so there are no equal leading terms whose cancellation could create valuation $5$. Thus the only gaps are \[ 1,2,5, \] and the normalization quotient has length \[ \delta=3. \] This is the $E_6$ cusp (characteristic parametrization $(t^3,t^4)$). In particular it cannot be a tacnode, because a tacnode has two branches.

12. If $X^*$ has only nodes and ordinary cusps, then the number $b$ ofbitangents in the strict sense of part (b) is\[ b=\frac12d(d-2)(d-3)(d+3). \]

Under the stated hypothesis every inflection point is ordinary. Indeed, if its tangent contact had order $r\geq4$, step 1 would give a dual branch of multiplicity $r-1\geq3$, which is neither a node nor an ordinary cusp. Therefore step 9 shows that the number of cusps of $X^*$ is \[ \kappa=3d(d-2). \]

Under the stated singularity hypothesis, a multiple tangent cannot contain an inflection point: otherwise the corresponding point of $X^*$ would have at least one singular branch together with another branch, hence would be neither a node nor an ordinary cusp. It also cannot be tangent at three or more distinct points, because step 4 would give an ordinary multiple point with at least three branches. Thus every multiple tangent is tangent at exactly two non-inflection points, i.e. is a strict bitangent, and step 4 shows that it gives an ordinary node of $X^*$. Conversely, by step 11 the two branches of a node lift under the normalization to two distinct points of $X$ having the same tangent line. Hence the number of nodes is exactly $b$.

Put \[ m=\deg X^*=d(d-1). \] The arithmetic genus of the plane curve $X^*$ is \[ p_a(X^*)=\frac{(m-1)(m-2)}2. \] Since $X$ is its normalization, Exercise IV.1.8 gives \[ g(X)=p_a(X^*)-b-\kappa, \] because a node and an ordinary cusp each have $\delta$-invariant $1$. Substituting \[ g(X)=\frac{(d-1)(d-2)}2, \qquad m=d(d-1), \qquad \kappa=3d(d-2) \] gives \[ \begin{aligned} 2b &=(m-1)(m-2)-(d-1)(d-2)-6d(d-2)\\ &=d^4-2d^3-9d^2+18d\\ &=d(d-2)(d-3)(d+3). \end{aligned} \] Dividing by $2$ proves part (f).

13. A plane cubic has exactly nine inflection points, all ordinary, andthe line through any two of them meets the cubic in a third inflection point.

For $d=3$, step 9 gives total inflection weight \[ 3\cdot3\cdot(3-2)=9. \] A tangent line to a nonsingular cubic cannot have contact order greater than $3$ by Bézout, so every inflection has contact order exactly $3$ and weight $1$. Hence there are exactly nine of them, all ordinary.

Let $P$ and $Q$ be distinct inflection points, and let a line through them meet $X$ in the divisor \[ P+Q+R. \] The third point is distinct from $P,Q$: if the line were tangent at either, then because that point is a flex its intersection multiplicity there would already be $3$, leaving no room for the other point.

Let $H$ denote a line section. The flex tangents give \[ 3P\sim H, \qquad 3Q\sim H, \] while the line through $P,Q$ gives \[ P+Q+R\sim H. \] Therefore \[ 3R \sim3H-3P-3Q \sim H. \] For a plane cubic, $g=1$ and $\deg H=3$, so Riemann–Roch gives $\ell(H)=3$. The three-dimensional space of ambient linear forms restricts injectively to $H^0(X,\OO_X(H))$, hence it is the complete linear system $|H|$. Thus the effective divisor $3R\in|H|$ is cut out by a line, and that line has contact order $3$ at $R$. Hence $R$ is also an inflection point.

14. For an arbitrary nonsingular plane quartic, if $h$ is the number ofhyperflex lines and $b$ the number of strict bitangents, then\[ b=28-h. \]Consequently there are always exactly $28$ generalized bitangent lines whenhyperflexes are included.

For a quartic, every tangent contact has order at most $4$. Let $f$ be the number of ordinary flexes and $h$ the number of hyperflexes. Step 9 gives the weighted relation \[ f+2h=24. \]

11. A singularpoint of $X^*$ with at least two normalization preimages is a line tangentto $X$ at at least two distinct points; Bézout forces the contact divisor tobe $2P+2Q$, so it is a strict bitangent. A singular point with onenormalization preimage comes from a point where the local Gaussparametrization of step 1 is singular, hence from contact order $3$ or$4$. Therefore the singularities of $X^*$ are exactly:

the $b$ nodes coming from strict bitangents, each with $\delta=1$;
the $f$ ordinary cusps coming from ordinary flexes, each with $\delta=1$;
the $h$ hyperflex cusps of step <1>11, each with $\delta=3$.

The genus formula of Exercise IV.1.8 now gives \[ 3 =55-(b+f+3h), \] so \[ b+f+3h=52. \] Using $f=24-2h$ yields \[ b=52-(24-2h)-3h=28-h. \] Thus $b+h=28$, proving the corrected form of part (h).

For the Fermat quartic in the erratum, the $12$ displayed hyperflexes already contribute total inflection weight $24$, so step 9 shows there are no others. Hence $h=12$ and \[ b=28-12=16, \] which explicitly disproves the literal strict-bitangent reading of part (h).

15. Q.E.D.

Steps <1>1–<1>3 prove part (a), step 4 proves part (b), steps <1>5–<1>6 prove part (c), step 7 proves part (d), steps <1>8–<1>10 prove part (e), step 12 proves part (f), step 13 proves part (g), and step 14 proves the corrected form of part (h). Step 11 supplies the local correction to the source’s four-fold-contact parenthetical.

Source Collections

Hartshorne, Algebraic Geometry, IV.2.3