Reduction to the triple-contact points

Proof. We first check the last assertion directly in the exchange of blow-ups. Near a tangency of intersection length two, choose analytic coordinates $(z,u,v)$ on the target threefold in which the two curves are $$C=(u=v=0),\qquad B=(u=0,\ v=z^2).$$ Such coordinates exist because the minimum order of the two normal coordinates of $B$ along $C$ is exactly two. Blow up $C$, using $u=v\xi$. The strict transform of $B$ has ideal $(\xi,v-z^2)$. On the relevant chart of its blow-up put $v-z^2=\xi w$. The exceptional surface is $F=(\xi=0)$, with coordinates $(z,w)$, and its ruling has base coordinate $z$. The ideal of $B$ on the original target pulls back to $$(u,v-z^2) =\xi\,(z^2+\xi w,w).$$ Thus the remaining indeterminacy on $F$ is the ideal $(w,z^2)$. Its tangent direction is $w=0$, which is transverse to the ruling fibre $z=0$. This also identifies the two successive point blow-ups that resolve the double marked scheme. Reduced points arising from different intersections of $B$ and $C$ lie on different ruling fibres.

Let $\widehat F$ be the common smooth surface resolution, and write $$\rho:\widehat F\longrightarrow F,\qquad \eta:\widehat F\longrightarrow F_B$$ for its maps to the original exceptional surface and to the exceptional surface of the blow-up of $B$ on the target. Every exceptional class used below is a total transform class. In particular these classes have square $-1$ and are pairwise orthogonal, also when two centres are infinitely near.

For the first extraction, write $c$ for the negative section of $F_B\cong\mathbb F_1$ and $f$ for its ruling fibre. Let $e_1,e_2$ be the total exceptional classes of $\eta$. Pullback to $\widehat F$ of the restriction of the nef divisor $M_2=H_Q+2A-J$ is $$P_2=2c+3f-e_1-e_2, \qquad \rho^*s=c+f-e_1-e_2, \qquad \rho^*f=f.$$ For two reduced centres the curves contracted by $\rho$ are $f-e_1$ and $f-e_2$. They are its total exceptional classes $\varepsilon_1,\varepsilon_2$. For a double centre, $\eta$ first blows up its support and then its specified tangent direction. The map $\rho$ contracts first $L=f-e_1$ and then the image of $K=e_1-e_2$. Its total exceptional classes, in the reverse order of blow-up, are therefore $$\varepsilon_1=K+L=f-e_2,\qquad \varepsilon_2=L=f-e_1.$$ In either case the restriction has the same expression $$P_2=\rho^*(2s+3f)-\varepsilon_1-\varepsilon_2.$$ The intermediate surface in the double case has an $A_1$ point, but $M_2$ restricts to an invertible sheaf on it. Its pullback to $\widehat F$ is the displayed class; no fractional exceptional coefficient is introduced. Since $M_2$ is nef, so is $P_2$.

For the second extraction use $F_B\cong\mathbb P^1\times\mathbb P^1$ and the classes $f,g$ already used in the trisecant calculation. With $e_1,e_2,e_3$ the total exceptional classes of $\eta$, the nef divisor $D_2=A+h$ restricts on the common resolution as $$P_2=2f+2g-e_1-e_2-e_3, \qquad \rho^*s=g+f-e_1-e_2-e_3, \qquad \rho^*f=f.$$ If all three centres are reduced, the total exceptional classes of $\rho$ are $\varepsilon_i=f-e_i$. For type $2a+b$, label $e_1,e_2$ by the two successive blow-ups over the double centre and $e_3$ by the simple centre. The total exceptional classes of $\rho$ are $$\varepsilon_a=f-e_2,\qquad \varepsilon_{a,1}=f-e_1,\qquad \varepsilon_b=f-e_3.$$ The first two are obtained from the same contraction of $L$ followed by $K$ as above. Hence, in both configurations, $$P_2=\rho^*(2s+3f)-\sum_i\varepsilon_i.$$ It is nef by the nefness of $D_2$. The displayed formulas for $\rho^*s$ can also be recovered without a position assumption: its intersection with a ruling fibre is one, it is orthogonal to every curve contracted by $\rho$, and its square is $-1$. Orthogonality forces coefficient $-1$ for every total class $e_i$, and the square then forces the displayed coefficient of $f$. Thus these calculations use only the order-of-blow-up exchange and the nef divisors; they do not assume any position of the marked centres relative to $s$.

Let $I$ be the set of centres of $\rho$ that lie on the successive strict transform of $s$. Since $s$ is smooth, that strict transform has class $$\widehat s=\rho^*s-\sum_{i\in I}\varepsilon_i.$$ Using total exceptional classes gives $$0\le P_2\cdot\widehat s =(2s+3f)\cdot s-|I|=1-|I|.$$ Thus $|I|\le1$. Two reduced points on $s$ would give $|I|\ge2$. If a double scheme supported on $s$ were tangent to $s$, both its support and its infinitely near centre would lie on the successive strict transform, again giving $|I|\ge2$. Finally, a double support on $s$ and a further simple point on $s$ would give the same contradiction. This proves all the assertions. ◻

Proof of Proposition 61.2. Let $V$ be a prime divisor over $X$ with $\beta_X(V)\leq0$. By [40, Main Theorem], its centre is a closed point $p\in D$. Blow up the fibre $\ell$ of $E\to C$ through $p$. If the associated length-two scheme is reduced, the first or second estimate of Lemma 63.2 applies. The required position of the marked points is supplied by Lemma 65.1. If the scheme is nonreduced and supported off the negative section, apply the third estimate, using the absence of a ruling component in $R$. In these cases Abban–Zhuang adjunction gives $$\frac{A_X(V)}{S_X(V)}\geq \min\left\{\frac{2}{19/12},\delta(F;W)\right\}>1,$$ contradicting the choice of $V$. Thus the scheme is nonreduced and its support lies on the negative section. By Lemma 63.5, $p\in\mathop{\mathrm{Sing}}(D)$.

Now blow up $\ell'_p$. Since this line is tangent to $E$ at $p$ and $E\ell'_p=3$, either $E|_{\ell'_p}=2p+p_1$ with $p_1\ne p$, or it is $3p$. In the first case the double marked point on the second exceptional surface lies on its negative section: the other ruling $\ell$ is tangent to $E'$ at $p$. Lemma 65.1 shows that the double direction is transverse to the negative section and to the ruling, and that the simple point lies off the negative section. Thus all the hypotheses of the double-plus-simple surface calculation hold. Lemmas 64.5 and 64.7 give, by the strict form of adjunction, $$\beta_X(V)\geq \left(2-\frac{13}{8}\right)\mathop{\mathrm{ord}}_V(F)=\frac38\mathop{\mathrm{ord}}_V(F)>0.$$ Here $\mathop{\mathrm{ord}}_V(F)>0$ because the centre on $X$ is the point $p$ on the blown-up line. This is again a contradiction. Hence $p\in\Sigma_X$. If $\Sigma_X$ is empty, all prime divisors have positive beta invariant, so the valuative criterion gives K-stability. Smoothness of $D$ implies $\Sigma_X=\varnothing$ and gives the stated special case. ◻

Proof. Put $\mathcal M=\mathcal L\otimes\mathcal A^{-1}$, of degree two. The base-point-free pencil trick gives a surjection $H^0(\mathcal A)\otimes H^0(\mathcal M)\to H^0(\mathcal L)$, whose kernel has dimension one. Evaluation at the projective point $P$ gives a matrix $H^0(\mathcal A)\to H^0(\mathcal M)^*$ of size $2\times3$. If its rank were one, there would be a nonzero $m\in H^0(\mathcal M)$ such that $P$ annihilates $mH^0(\mathcal A)=H^0(\mathcal L(-Z_2))$, where $Z_2=(m=0)$ has length two. This says $P\in\langle Z_2\rangle$, contrary to the choice of $P$. Rank zero is excluded by surjectivity of multiplication. Thus the matrix has rank two. Its one-dimensional kernel determines a unique section $a$ of $\mathcal A$. The condition that $P$ annihilates $aH^0(\mathcal M)=H^0(\mathcal L(-(a=0)))$ is exactly $P\in\langle(a=0)\rangle$. In families over $\operatorname{Pic}^3(\mathcal C)$, the constant-rank kernel is a line bundle, so this construction is a morphism, and its inverse is the Abel map. Projection turns each such plane into its trisecant line. No line has intersection length four with the projected curve, since a length-four divisor of an elliptic normal quintic spans a projective three-space, whereas the inverse image of a line is a plane. ◻

Proof. Choose an Abel coordinate $z$ at $c$ and a local analytic lift $v(z)=v_0+zv_1+z^2v_2+\cdots$ of the embedding. Write $P=\rho_0v_0+\rho_1v_1+\rho_2v_2$; here $\rho_2\ne0$, because the tangent line is contained in the secant variety. Near the divisor $3c$, write the three points as the roots of $$f(z,t)=z^3-tz^2+a(t)z-b(t).$$ The parameter $t$ is an Abel coordinate on $\operatorname{Pic}^3(\mathcal C)$, by the preceding lemma. Reducing $v(z)$ modulo this cubic gives, to first order in $(t,a,b)$, the three spanning vectors $$v_0+bv_3,\qquad v_1-av_3+bv_4,\qquad v_2+tv_3-av_4+bv_5.$$ Modulo $\langle v_0,v_1,v_2\rangle$, constancy of $P$ therefore implies $$(\rho_0b'-\rho_1a'+\rho_2)[v_3] +(\rho_1b'-\rho_2a')[v_4]+\rho_2b'[v_5]=0.$$ If $\mathcal L\not\cong\mathcal O(5c)$, the vanishing sequence is $0,1,2,3,4$, so $[v_3],[v_4]$ are independent. Setting $b'=0$ would force $a'=0$ and then $\rho_2=0$, a contradiction. Thus $b'\ne0$, which makes the incidence equation $f=0$ smooth at $(z,t)=(0,0)$.

If $\mathcal L\cong\mathcal O(5c)$, the vanishing sequence is $0,1,2,3,5$. Write $[v_4]=\alpha[v_3]$; then $[v_5]$ is independent of $[v_3]$. The same equation forces $b'=0$ and $(\rho_1+\alpha\rho_2)a'=\rho_2$. In particular $a'\ne0$. The quadratic part of $f$ is $a'tz-b''t^2/2$, a product of two distinct linear forms. Hence the incidence curve has an ordinary node.

To verify compatibility explicitly, take $c$ as the origin of a Weierstrass model and use the $|5c|$ embedding with coordinates $[1:x:y:x^2:xy]$. The osculating plane at $c$ is $X_0=X_1=0$. Its intersection with the secant variety is, set-theoretically, the union of the lines $X_2=0$ and $X_3=0$. The first is the tangent line at $c$. For any other secant, vanishing of $X_0$ and $X_1$ forces its two endpoints to have the same $x$-coordinate; they are opposite points of the elliptic curve, and the corresponding intersection point has $X_3=0$. Conversely these opposite-point secants sweep the second line, including their tangent limits at two-torsion points. Thus points $P=[0:0:\rho_2:\rho_1:\rho_0]$ with $\rho_1\rho_2\ne0$ lie outside the secant variety and give exactly the exceptional configuration. ◻

Proof. The complete linear series has five sections, and the given embedding uses a four-dimensional subspace; its annihilator is $P$. Smoothness of the given embedding implies $P\notin\operatorname{Sec}(\mathcal C)$, including its tangent lines. Each fibre $\ell'$ of $q|_{E'}$ maps isomorphically under $\pi$ onto a line $\Lambda\subset\mathbb P^3$, because $H\cdot\ell'=1$. The equality $\mathcal I_C\mathcal O_X=\mathcal O_X(-E)$ identifies, as schemes, $E|_{\ell'}$ with $\Lambda\cap C$. These intersections have length three. Conversely, the strict transform of any trisecant line has $B$-degree $3-3=0$, so is contracted by $q$ and is one of these fibres. A trisecant line cannot have intersection length four, by the last assertion of Lemma 65.2.

The finite flat family $D\to C'$ therefore defines a morphism $C'\to\operatorname{Hilb}^3(C)=\operatorname{Sym}^3(C)$, whose image is exactly the family of trisecant divisors. Distinct fibres give distinct lines: off $C$, the map $\pi$ is an isomorphism, so a line has only one strict transform. Consequently their divisors are distinct as well, since a length-three divisor spans its trisecant line. By Lemma 65.2, composition with the Abel map is a bijective morphism $C'\to\operatorname{Pic}^3(C)$. It is an isomorphism, since both curves are smooth and the characteristic is zero. By the defining property of the Hilbert scheme, $D$ is the pullback of the universal length-three divisor; the preceding scheme-theoretic fibre identification is precisely this Hilbert morphism. This proves the assertion, including nonreduced fibres and singular points of $D$. ◻

Proof. Combine Lemmas 65.4 and 65.3 with the definition of $\Sigma_X$. There are exactly twenty-five points $c$ with $\mathcal O(5c)\cong\mathcal L$, because multiplication by five on an elliptic curve has degree twenty-five. For each such $c$, the divisor $3c$ determines at most one incidence point. ◻