Erratum to: Modular properties of nodal curves on $K3$ surfaces

1 Erratum to: Math. Z. (2012) 270:871–887 DOI 10.1007/s00209-010-0830-2

In the original publication [3], the Theorem 3.1 and Theorem 4.2 are erroneous.

• (i) First, the proof of Theorem 3.1 is incorrect: The fault is at the Step 2 of the proof. In the meantime, the result has been proved in [4] with better bounds.

• (ii) Second, I correct Theorem 4.2. At p. 884, the last row of the diagram A.1 should be tensored by ${\fancyscript{O}}_E(-2E)$. This error affects the subsequent computations from Lemma A.2 onward, which are used in the proof of the Theorem.

The corrected version is provided below.

2 The modified Wahl map

Recall that $\hat{C}\in |{\fancyscript{L}}={\fancyscript{A}}^d|$ is a nodal curve with nodes $\fancyscript{N}:=\{\hat{x}_1,\ldots ,\hat{x}_\delta \}$ on the polarized $K3$ surface $(S,{\fancyscript{A}})$, such that ${\fancyscript{A}}\in \mathop {\mathrm{Pic}}\nolimits (S)$ is not divisible, ${\fancyscript{A}}^2=2(n-1)$. (Note that the article [3] deals only with $K3$ surfaces with cyclic Picard group). Let $\sigma :\tilde{S}\rightarrow S$ be the blow-up of $S$ at $\fancyscript{N}$, and denote by $E^{a}$, ${a}=1,\ldots ,\delta$, the exceptional divisors, and $E:=E^1+\cdots +E^\delta$. The normalization $C$ of $\hat{C}$ fits into

(1)

$\tilde{u}$ is an embedding, and $K_C=\sigma ^*{\fancyscript{L}}(-E)\otimes {\fancyscript{O}}_C$. The curve $C$ carries the divisor

$$\begin{aligned} \varDelta :=x_{1,1}+x_{1,2}+\cdots +x_{\delta ,1}+x_{\delta ,2}, \end{aligned}$$

where $\{x_{{a},1},x_{{a},2}\}=E^{a}\cap C$ is the pre-image of $\hat{x}_a\in \hat{C}$ by $\nu$.

In general, if $V^{\prime }$ is a subscheme of some variety $V$, ${\fancyscript{I}}_V(V^{\prime })$ or ${\fancyscript{I}}(V^{\prime })$ stands for its sheaf of ideals, and $DV^{\prime }\subset V\times V$ denotes the diagonally embedded $V^{\prime }$.

Let $(X,\varDelta _X)$ be an arbitrary smooth, irreducible curve together with $\delta$ pairwise disjoint pairs of points $\varDelta _X=\{\{x_{1,1},x_{1,2}\},\ldots ,\{x_{\delta ,1},x_{\delta ,2}\}\}\subset X.$ The exact sequence $0\rightarrow {\fancyscript{I}}(DX)^2\rightarrow {\fancyscript{I}}(DX)\rightarrow K_X\rightarrow 0$ yields the Wahl map

$$\begin{aligned} w_X:H^0\left( X\times X,{\fancyscript{I}}(DX)\otimes K_{X\times X}\right) {\rightarrow } H^0\left( X,K_X^3\right) . \end{aligned}$$

The vector space $H^0({\fancyscript{I}}(DX)\otimes K_{X\times X})$ splits into

$$\begin{aligned} H^0({\fancyscript{I}}(DX)\otimes K_{X\times X})\cap \mathop {\mathrm{Sym}}\nolimits ^2H^0(K_X) \oplus \mathop {\bigwedge }\limits ^{2}H^0(K_X), \end{aligned}$$

and $w_X$ vanishes on the first direct summand, as it is skew-symmetric. Denote

$$\begin{aligned} {P}_{\varDelta _X}:={\mathop {\mathop {\bigcup }\limits ^{\delta }}\limits _{a=1}} \{x_{a,1},x_{a,2}\}\times \{x_{a,1},x_{a,2}\}\subset X\times X, \end{aligned}$$

(2)

and let $w_{X,\varDelta _X}$ be the restriction of $w_X$ to $H^0\big ({\fancyscript{I}}({P}_{\varDelta _X})\cdot {\fancyscript{I}}(DX)\otimes K_{X\times X}\big ) \cap \mathop {\bigwedge }\limits ^{2}H^0(K_X)$. (Thus, $w_{X,\varDelta _X}$ is a punctual modification of the usual Wahl map.) With this notation, we replace [3], Theorem4.2] by the following.

Theorem 1

(i) Let $(S,{\fancyscript{A}})$, ${\fancyscript{A}}^2\geqslant 6$, be as above. Consider a nodal curve $\hat{C}\,{\in }\,|d{\fancyscript{A}}|$ with

$$\begin{aligned} \delta \leqslant \min \left\{ \frac{d^2{\fancyscript{A}}^2}{3(d+4)}, \delta _\mathrm{max}(n,d)\right\} \end{aligned}$$

(3)

nodes and let $(C,\varDelta )$ be as above ($\delta _\mathrm{max}(n,d)$ is defined in [3], p. 872]; the minimum is the first expression, except a finite number of cases). Then, the homomorphism $w_{C,\varDelta }$ is not surjective.

(ii) For generic a generic curve $X$ of genus $g\geqslant 12$ with generic markings $\varDelta _X$, such that $\delta \leqslant \frac{g-1}{2}$, the homomorphism $w_{X,\varDelta _X}$ is surjective.

This is a nonsurjectivity property for the pair $(C,\varDelta )$, rather than for $C$ itself. I conclude the note (see Sect. 4) with some evidence toward the nonsurjectivity of the Wahl map $w_C$ itself, and comment on related work in [4].

3 Relationship between the Wahl maps of $C$ and ${\tilde{S}}$

Lemma 2

• (i) The following diagram has exact rows and columns:

  

  (4)

  In other words, we have ${\fancyscript{I}}:= {\fancyscript{I}}_{C\times C}(D\varDelta )\cdot {\fancyscript{I}}_{C\times C}(DC)=w_C^{-1}(K_C(-\varDelta ))$. Also, the involution $\tau _C$ which interchanges the factors of $C\times C$ leaves ${\fancyscript{I}}$ invariant.

• (ii) $H^0\bigl ( C\times C, {\fancyscript{I}}\otimes K_{C\times C}) \bigr ) = w_C^{-1}\bigl (\;H^0(C,K_C^3(-\varDelta ))\;\bigr )$.

• (iii) For $\varLambda := \bigl \{ s-\tau _C^*(s)\mid s\in H^0\bigl ( {\fancyscript{I}}\otimes K_{C\times C}) \bigr ) \bigr \}$ holds

  $$\begin{aligned} \varLambda \mathop {=}\limits ^{(\star )} H^0\bigl ({\fancyscript{I}}\otimes K_{C\times C}\bigr ) \;\cap \; {\mathop {\bigwedge }\limits ^{2}}\,H^0(K_C) \mathop {\subset }\limits ^{(\star \star )} H^0\bigl ({\fancyscript{I}}_{C\times C}(DC)\otimes K_{C\times C}\bigr ). \end{aligned}$$

  (iv) $w^{\prime }_{C}(\varLambda ) = w^{\prime }_{C}\bigl (\; H^0\bigl ({\fancyscript{I}}\otimes K_{C\times C}\bigr )\bigr ).$

Proof

• (i) The middle column is exact because ${\fancyscript{I}}_{C\times C}(DC)$ is locally free. We check the exactness of the first row around each point $(o,o)\in D\varDelta$. Let $u$ be a local (analytic) coordinate on $C$ such that $o=0$, and $u_1,u_2$ be the corresponding coordinates on $C\times C$. Then, the first row becomes $0\rightarrow \langle u_2-u_1\rangle ^2\rightarrow \langle u_2-u_1\rangle \cdot \langle u_1,u_2\rangle \rightarrow \langle u\rangle \cdot \mathrm{d}u\rightarrow 0,$ with $\mathrm{d}u:=(u_2-u_1)\mathrm{mod} (u_2-u_1)^2$, which is exact. The second statement is obvious.

• (ii) We tensor (4) by $K_{C\times C}$, and take the sections in the last two columns. An elementary diagram chasing yields the claim.

• (iii) Let us prove $(\star )$. The vector space $\varLambda$ is contained in ${\mathop {\bigwedge }\limits ^{2}}\,H^0(K_C)$ by the very definition, and also in $H^0\bigl ({\fancyscript{I}}\otimes K_{C\times C}\bigr )$ because the sheaf ${\fancyscript{I}}(C,\varDelta )$ is $\tau _C$-invariant. For the inclusion in the opposite direction, take $s$ in the intersection. As $s\in {\mathop {\bigwedge }\limits ^{2}}\;H^0(K_C)$, it follows $\tau _C^*(s)=-s$, so $s=1/2\cdot (s-\tau _C^*(s))\in \varLambda$. The inclusion $(\star \star )$ is obvious.

• (iv) Indeed, the Wahl map is anti-commutative: $w_C({\sum }_{i}s_i\otimes t_i)=-w_C({\sum }_{i}t_i\otimes s_i)$.

$\square$

Lemma 3

Let $\varXi :=\{(x_{{a},1},x_{{a},2}),(x_{{a},2},x_{{a},1})\mid {a}=1,\ldots ,\delta \}\subset C\times C$, and consider the sheaf of ideals ${\fancyscript{I}}(C,\varDelta ):={\fancyscript{I}}(\varXi )\cdot {\fancyscript{I}} \mathop {=}\limits ^{(2)} {\fancyscript{I}}({P}_\varDelta )\cdot {\fancyscript{I}}(DC) \subset {\fancyscript{I}}.$ Furthermore, denote

$$\begin{aligned} \varLambda (\varDelta ):=H^0({\fancyscript{I}}(C,\varDelta )\otimes K_{C\times C}) \,\cap \, \mathop {\bigwedge }\limits ^{2}\,H^0(K_C). \end{aligned}$$

(5)

Then, the following statements hold:

• (i) ${\fancyscript{I}}(C,\varDelta )$ is $\tau _C$-invariant, so $w^{\prime }_C\bigl (\,\varLambda (\varDelta )\,\bigr ) =w^{\prime }_C\bigl (\;H^0({\fancyscript{I}}(C,\varDelta )\otimes K_{C\times C})\;\bigr ).$

• (ii) ${\fancyscript{I}}(C,\varDelta )+{\fancyscript{I}}_{C\times C}(DC)^2={\fancyscript{I}}$, and ${\fancyscript{I}}(C,\varDelta )\cap {\fancyscript{I}}(DC)^2= {\fancyscript{I}}(\varXi )\cdot {\fancyscript{I}}(DC)^2$. Therefore, the various sheaves introduced so far fit into the commutative diagram

  

  (6)

Proof

• (i) The proof is identical to Lemma 2(iv).

• (ii) The inclusion $\subset$ is clear. For the reverse, notice that ${\fancyscript{O}}_{C\times C}={\fancyscript{I}}(\varXi )+{\fancyscript{I}}$, so ${\fancyscript{I}}\subset {\fancyscript{I}}(C,\varDelta )+{\fancyscript{I}}^2 \subset {\fancyscript{I}}(C,\varDelta )+{\fancyscript{I}}(DC)^2$. The second claim is analogous.

$\square$

Now, we compare the Wahl maps of $C$ and ${\tilde{S}}$. Let $\rho :\!{\fancyscript{I}}_{\tilde{S}\times \tilde{S}}(D\tilde{S})\rightarrow {\fancyscript{I}}_{C\times C}(DC)$ be the restriction homomorphism, and ${\fancyscript{M}}:=\sigma ^*{\fancyscript{L}}(-E)$. The diagram below relates various objects involved in the definition of $w_C$ and $w_{\tilde{S}}$:

(7)

The rightmost column corresponds to the first-order expansions of the sections along $E$ and at $\varDelta$. By using $0{\rightarrow }{\fancyscript{O}}_E(1){\rightarrow }{\fancyscript{O}}_{2E}{\rightarrow }{\fancyscript{O}}_E{\rightarrow }0$, we deduce that it fits into:

(8)

Along each $E^{a}\subset \tilde{S}$, we consider local coordinates $u,v$ as follows: $v$ is the coordinate along $E^{a}$, and $u$ is a coordinate in the normal direction to $E^{a}$ (so $E^{a}$ is given by $\{u=0\}$). Moreover, we assume that $C$ is given by $\{v=0\}$ around the intersection points $\{x_{{a},1},x_{{a},2}\}=E^{a}\cap C$. Then, any element $\tilde{s}\in H^0({\fancyscript{M}})$ can be expanded as

$$\begin{aligned} \tilde{s}=\tilde{s}^{a}_0(v)+u\tilde{s}^{a}_1(v)+O\left( u^2\right) , \end{aligned}$$

(9)

and its image in $H^0({\fancyscript{M}}\otimes {\fancyscript{O}}_{2E^{a}})$ is $\tilde{s}^{a}_0(v)+u\tilde{s}^{a}_1(v)$. Finally, observe that the values of $w_{\tilde{S}}$ are sections of $\varOmega ^1_{\tilde{S}}\otimes {\fancyscript{M}}^2$, and the restriction of this latter to $E$ fits into

$$\begin{aligned} 0\rightarrow \underbrace{{\fancyscript{O}}_E(3)} _{\mathrm{normal~component}} \rightarrow \varOmega ^1_{\tilde{S}}\otimes {\fancyscript{M}}^2|_{E}\rightarrow \underbrace{\varOmega ^1_E\otimes {\fancyscript{M}}^2_E={\fancyscript{O}}_E} _{\mathrm{tangential~component}} \rightarrow 0. \end{aligned}$$

Lemma 4

Let the notation be as in (9). We consider $\tilde{e}= {\sum }_{i}\,\tilde{s}_i\wedge \tilde{t}_i\in \mathop {\bigwedge }\limits ^{2}\,H^0({\fancyscript{M}})$, and let $e:=\rho (\tilde{e})={\sum }_{i}\, s_i\wedge t_i$. Then, the following statements hold:

• (i) $w_{\tilde{S}}(\tilde{e})\in H^0(\varOmega ^1_{\tilde{S}}(-E)\otimes {\fancyscript{M}}^2)$ if and only if:

  $$\begin{aligned} \left\{ \begin{array}{ccl} (\star ) &{}\quad &{} \mathop {\sum }\limits _{i}\, \bigl ( s_{i}(x_{{a},1}) t_{i}(x_{{a},2}) - t_{i}(x_{{a},1}) s_{i}(x_{{a},2}) \bigr ) =0,\quad \text {and}\\ (\star \star ) &{}\quad &{} \mathop {\sum }\limits _{i}\, \bigl ( \tilde{s}_{i,0}^{a}\tilde{t}_{i,1}^{a}-\tilde{t}_{i,0}^{a}\tilde{s}_{i,1}^{a}\bigr ) =0,\quad \forall \,a=1,\ldots ,\delta . \end{array} \right. \end{aligned}$$

• (ii) $\varLambda (E):= w_{\tilde{S}}^{-1}\bigl (H^0(\varOmega ^1_{\tilde{S}}(-E)\otimes {\fancyscript{M}}^2)\bigr ) \,\cap \,\mathop {\bigwedge }\limits ^{2}H^0({\fancyscript{M}})$ has the property

  $$\begin{aligned} w_{\tilde{S}}(\varLambda (E)) = w_{\tilde{S}}\bigl (\; w_{\tilde{S}}^{-1}\bigl (H^0(\varOmega ^1_{\tilde{S}}(-E) \otimes {\fancyscript{M}}^2)\bigr )\bigr ). \end{aligned}$$

• (iii) $\rho \big (\varLambda (E)\big )\subset \varLambda (\varDelta )$, where the right hand side is defined by (5).

Proof

• (i) The element $w_{\tilde{S}}(\tilde{e})$ vanishes along $E$ if and only if both its tangential and normal components along each $E^{a}\subset E$ vanish. A short computation shows that the normal component is $(\star \star )$. The tangential component is ${\sum }_{i}\left( \tilde{s}^{a}_{i,0}\left( \tilde{t}^{a}_{i,0}\right) ^{\prime }-\tilde{t}^{a}_{i,0}\left( \tilde{s}^{a}_{i,0}\right) ^{\prime } \right)$. But $\tilde{s}^{a}_{i,0},\tilde{t}^{a}_{i,0}\in H^0({\fancyscript{O}}_{E^{a}}(1)),$ that is they are linear polynomials in $v$, so

  $$\begin{aligned} \tilde{s}^{a}_{i,0}\left( \tilde{t}^{a}_{i,0}\right) ^{\prime }-\tilde{t}^{a}_{i,0}\left( \tilde{s}^{a}_{i,0}\right) ^{\prime } = \tilde{s}^{a}_{i,0}\left( x_{{a},1}\right) \tilde{t}^{a}_{i,0}\left( x_{{a},2}\right) -\tilde{t}^{a}_{i,0}\left( x_{{a},1}\right) \tilde{s}^{a}_{i,0}\left( x_{{a},2}\right) , \end{aligned}$$

  up to a constant factor. Also, we have $\tilde{s}^{a}_{i,0}(x_{{a},j})=\tilde{s}^{a}(x_{{a},j}) =s(x_{{a},j})$, and $(\star )$ follows.

• (ii) The vector space $w_{\tilde{S}}^{-1}\bigl (H^0(\varOmega ^1_{\tilde{S}}(-E) \otimes {\fancyscript{M}}^2)\bigr )$ is invariant under the involution $\tau _{\tilde{S}}$ of $\tilde{S}\times \tilde{S}$ which switches the two factors. As $w_{\tilde{S}}$ is anti-commutative, the claim follows as in Lemma 2.

• (iii) Take $\tilde{e}\in \varLambda (E)$ and $e:=\rho (\tilde{e})$. Then, $e(x_{{a},1},x_{{a},2})=-e(x_{{a},2},x_{{a},1})\mathop {=}\limits ^{(\star )}0$, and also $w_C(e)(x_{{a},j})$ equals the expression $(\star \star )$ at $x_{{a},j}$ (so it vanishes), for $j=1,2$. $\square$

Now. we consider the commutative diagram:

(10)

It is the substitute in the case of nodal curves for [3], diagram (4.2)].

Lemma 5

Assume $\mathop {\mathrm{Pic}}\nolimits (S)=\mathbb {Z}{\fancyscript{A}}$. Then, $\rho _\varDelta :\varLambda (E)\rightarrow \varLambda (\varDelta )$ is surjective.

Proof

The restriction $H^0({\tilde{S}},{\fancyscript{M}})\rightarrow H^0(C,K_C)$ is surjective (see [3], LemmaA.1]), and the kernel of $\mathop {\bigwedge }\limits ^{2}\,H^0(\tilde{S},{\fancyscript{M}})\rightarrow \mathop {\bigwedge }\limits ^{2}\,H^0(C,K_C)$ consists of elements of the form $\tilde{t}\wedge (\tilde{s}_C\tilde{s}_E)$, where $\tilde{t}\in H^0({\fancyscript{M}})$ and $\tilde{s}_C,\tilde{s}_E$ are the canonical sections of ${\fancyscript{O}}_{\tilde{S}}(C)$ and ${\fancyscript{O}}_{\tilde{S}}(E)$, respectively. (See the middle column of (7).)

Consider $e={\sum }_{i}\;(s_i\otimes t_i-t_i\otimes s_i) \in \varLambda (\varDelta )$, and let $\tilde{e}={\sum }_{i}\;(\tilde{s}_i\otimes \tilde{t}_i-\tilde{t}_i\otimes \tilde{s}_i) \in \mathop {\bigwedge }\limits ^{2}\,H^0({\fancyscript{M}})$ be such that $\rho (\tilde{e})=e$. The proof of 4(i) shows that, for all ${a}$, the tangential component of $w_{\tilde{S}}(\tilde{e})|_{E^{a}}$ equals $e(x_{{a},1},x_{{a},2})=0$, so $w_{\tilde{S}}(\tilde{e})|_{E}$ is a section of $\varOmega ^1_{E/{\tilde{S}}}\otimes {\fancyscript{M}}_E^2\cong {\fancyscript{O}}_E(3)$. Since $w_{\tilde{S}}(\tilde{e})|_{E}$ vanishes at the points of $\varDelta$, it is actually determined up to an element in $H^0({\fancyscript{O}}_E(1))$. We claim that this latter can be canceled by adding to $\tilde{e}$ a suitable element of the form $\tilde{t}\wedge (\tilde{s}_C\tilde{s}_E)$. A short computation yields

$$\begin{aligned} w_{\tilde{S}}\left( \,\tilde{t}\wedge \left( \tilde{s}_C\tilde{s}_E\right) \right) |_E = \tilde{t}_E\cdot \left( \tilde{s}_C|_E\right) \cdot \left( \mathrm{d}\tilde{s}_E\right) |_E\in {\fancyscript{O}}_E(3), \end{aligned}$$

where $\tilde{t}_E\in H^0({\fancyscript{O}}_E(1))$, $\tilde{s}_C|_E\in H^0({\fancyscript{O}}_E(2))$ vanishes at $\varDelta =E\cap C$, and $(\mathrm{d}\tilde{s}_E)|_E\in H^0({\fancyscript{O}}_E)$ (it is a section of $\varOmega ^1_{\tilde{S}}|_E$ with vanishing tangential component). Thus, these two latter factors are actually (nonzero) scalars.

The previous discussion shows that $\tilde{e}+\tilde{t}\wedge (\tilde{s}_C\tilde{s}_E)\in \varLambda (E)$ as soon as $\tilde{t}\in H^0({\fancyscript{M}})$ satisfies $\tilde{t}_E=-w_{\tilde{S}}(\tilde{e})|_E\in H^0({\fancyscript{M}}_E)$. According to Corollary 8, such an element $\tilde{t}$ exists because the restriction $H^0({\fancyscript{M}})\rightarrow H^0({\fancyscript{M}}_E)$ is surjective. $\square$

Proof of Theorem 1

(i) Case $\mathop {\mathrm{Pic}}\nolimits (S)=\mathbb {Z}{\fancyscript{A}}$. If $w_{C,\varDelta }$ is surjective; then, the homomorphism $b$ in the diagram (10) is surjective too. Now, we follow the same pattern as in [3], p. 884,top]: $b$ is the restriction homomorphism at the level of sections of

$$\begin{aligned} 0\rightarrow K_C\rightarrow \varOmega ^1_{\tilde{S}}\bigr |_C\otimes K_C^2(-\varDelta ) \rightarrow K_C^3(-\varDelta )\rightarrow 0, \end{aligned}$$

and its surjectivity implies that this sequence splits. This contradicts [3], Lemma4.1].

General case. It is a deformation argument. We consider

$$\begin{aligned} \begin{array}{l} {\fancyscript{K}}_n:= \left\{ \left( S,{\fancyscript{A}}\right) \mid {\fancyscript{A}}\in \mathop {\mathrm{Pic}}\nolimits (S)\text { is ample, not divisible},{\fancyscript{A}}^2=2(n-1)\right\} ,\\ {\fancyscript{V}}^d_{n,\delta }:=\left\{ \left( (S,{\fancyscript{A}}),\hat{C} \right) \mid \left( S,{\fancyscript{A}}\right) \in {\fancyscript{K}}_n,\; \hat{C}\in |d{\fancyscript{A}}|\text { nodal curve with }\delta \text { nodes} \right\} . \end{array} \end{aligned}$$

Then, the natural projection $\kappa :{\fancyscript{V}}^d_{n,\delta }\rightarrow {\fancyscript{K}}_n$ is submersive onto an open subset of ${\fancyscript{K}}_n$. (See [3], Theorem 1.1(iii)] and the reference therein.)

Hence, for any $((S,{\fancyscript{A}}),\hat{C})\in {\fancyscript{V}}^d_{n,\delta }$ there is a smooth deformation $((S_t,{\fancyscript{A}}_t),\hat{C}_t)$ parameterized by an open subset $T\subset {\fancyscript{K}}_n$. The points $t\in T$ such that $\mathop {\mathrm{Pic}}\nolimits (S_t)=\mathbb {Z}{\fancyscript{A}}_t$ are dense; for these $w_{C_t,\varDelta _t}$ are nonsurjective. Since the nonsurjectivity condition is closed, we deduce that $w_{C,\varDelta }$ is nonsurjective too.

(ii) Now let $(X,\varDelta _X)$ be a generic marked curve of genus at least 12. By [1], the Wahl map $w_X\,{:}\,\mathop {\bigwedge }\limits ^{2}\,H^0(K_X)\rightarrow H^0(K_X^3)$ is surjective; thus, $\widetilde{w}\,^{\prime }_X\,{:=}\,H^0(w^{\prime }_X\otimes K_{X\times X})$ in (6) is surjective as well (see lemma 2(ii)). As $\delta \leqslant \frac{g-1}{2}$, the evaluation homomorphism $H^0(K_X)\rightarrow K_{X}\otimes {\fancyscript{O}}_{\varDelta _X}$ is surjective for generic markings, so the same holds for

$$\begin{aligned} H^0(K_X)^{\otimes 2}\rightarrow \mathop {\mathop {\bigoplus }\limits ^{\delta }}\limits _{a=1}\left( K_{X,x_{a,1}}\oplus K_{X,x_{a,2}}\right) ^{\otimes 2}. \end{aligned}$$

The restriction to the anti-symmetric part (on both sides) yields the surjectivity of

$$\begin{aligned} \mathrm{ev}_\varXi :\mathop {\bigwedge }\limits ^{2}H^0(K_X)\rightarrow \mathop {\mathop {\bigoplus }\limits ^{\delta }}\limits _{a=1}K_{X,x_{a,1}}\otimes K_{X,x_{a,2}} =\mathop {\mathop {\bigoplus }\limits ^{\delta }}\limits _{a=1} K_{X\times X, \left( x_{a,1},\,x_{a,2}\right) }. \end{aligned}$$

(For $s\in \mathop {\bigwedge }\limits ^{2}H^0(K_X)$, $\mathrm{ev}_\varXi (s)$ takes opposite values at $(x_{a,1},x_{a,2})$ and $(x_{a,2},x_{a,1})$.)

The diagram (6) yields

(11)

A straightforward diagram chasing shows that $w_{X,\varDelta _X}$ is surjective if

$$\begin{aligned} \mathrm{ev}_\varXi ^{\prime \prime }: \underbrace{H^0\bigl (\,{\fancyscript{I}}(DX)^2\cdot K_{X\times X}\,\bigr ) \,\cap \, {\mathop {\bigwedge }\limits ^{2}}\,H^0(K_X)}_{:=G} \rightarrow \underbrace{ {\mathop {\mathop {\bigoplus }\limits ^{\delta }}\limits _{a=1}} K_{X\times X,(x_{a,1},\,x_{a,2})}}_{:=H_{\varXi }} \end{aligned}$$

is so, or equivalently when the induced $h_\varXi :\mathop {\bigwedge }\limits ^{\delta }\,G\rightarrow \mathop {\bigwedge }\limits ^{\delta }\,H_\varXi$ is nonzero. This is indeed the case for generic markings.

Claim ${\bigcap }_{\varDelta _X}\mathrm{Ker}(h_\varXi )=0$. ($h_\varXi$ depends on $\varDelta _X$.) Indeed, since $\dim G\geqslant \delta$, we have

$$\begin{aligned} 0\ne \mathop {\bigwedge }\limits ^{\delta }\,G \subset \mathop {\bigwedge }\limits ^{\delta }\bigl (\mathop {\bigwedge }\limits ^{2}H^0(K_X)\bigr ) \subset H^0(K_{X\times X})^{\otimes \delta } = H^0\big ((X^2)^{\delta },K_{X\times X}\boxtimes \cdots \boxtimes K_{X\times X}\big ). \end{aligned}$$

The wedge is a direct summand of the tensor product (appropriate skew-symmetric sums), and $h_\varXi$ is induced by the evaluation map

$$\begin{aligned} \mathrm{ev}^{\delta }: H^0\big ((X^2)^\delta ,K_{X\times X}\boxtimes \cdots \boxtimes K_{X\times X}\big ) \otimes {\fancyscript{O}} \rightarrow K_{X\times X}\boxtimes \cdots \boxtimes K_{X\times X} \end{aligned}$$

at $\big ((x_{1,1},x_{1,2}),\ldots ,(x_{\delta ,1},x_{\delta ,2})\big )\in (X^2)^\delta$. If $e\in \mathop {\bigwedge }\limits ^{\delta }G$ belongs to the intersection above, then $e\in H^0\big (\mathrm{Ker}(\mathrm{ev}^{\delta })\big )=\{0\}$. Hence, for any $e_1,\ldots ,e_\delta \in G$ with $e_1\wedge \cdots \wedge e_\delta \ne 0$, there are markings $\varDelta _X$ such that $\mathrm{ev}_\varXi ^{\prime \prime }(e_1),\ldots ,\mathrm{ev}_\varXi ^{\prime \prime }(e_\delta )$ are linearly independent in $H_\varXi$ (thus, they span it). $\square$

4 Multiple point Seshadri constants of $K3$ surfaces with cyclic Picard group

This section is independent of the rest. Here we determine a lower bound for the multiple point Seshadri constants of ${\fancyscript{A}}$, which is necessary for proving Lemma 5.

Definition 6

(See [2], Section 6] for the original definition) The multiple point Seshadri constant of ${\fancyscript{A}}$ corresponding to $\hat{x}_1,\ldots ,\hat{x}_\delta \in S$ is defined as

$$\begin{aligned} \varepsilon =\varepsilon _{S,\delta }({\fancyscript{A}}):=\inf _{Z} \frac{Z\cdot {\fancyscript{A}}}{{{\sum _{a=1}^{\delta }}\mathrm{mult}_{\hat{x}_a}(Z)}} = \sup \bigl \{ c\in \mathbb {R}\mid \sigma ^*{\fancyscript{A}}-cE\text { is ample on }{\tilde{S}}\bigr \}. \end{aligned}$$

(12)

The infimum is taken over all integral curves $Z\subset S$ which contain at least one of the points $\hat{x}_a$ above. Throughout this section, we assume that $Z\in |z{\fancyscript{A}}|$, with $z\geqslant 1$.

As the self-intersection number of any ample line bundle is positive, the upper bound $\varepsilon \leqslant \frac{\sqrt{{\fancyscript{A}}^2}}{\sqrt{\delta }}$ is automatic. We are interested in finding a lower bound.

Theorem 7

Assume that $\mathop {\mathrm{Pic}}\nolimits (S)=\mathbb {Z}{\fancyscript{A}}$, ${\fancyscript{A}}^2=2(n-1)\geqslant 4$, and $\delta \geqslant 1$. Then, the Seshadri constant (12) satisfies $\varepsilon \geqslant \frac{2{\fancyscript{A}}^2}{\delta +\sqrt{\delta ^2+4\delta (2+{\fancyscript{A}}^2)}},$ for any points $\hat{x}_1,\ldots ,\hat{x}_\delta \in S$.

Our proof is inspired from [5], which treats the case $\delta =1$.

Proof

We may assume that the points are numbered such that

$$\begin{aligned} \mathrm{mult}_{\hat{x}_a}(Z)\geqslant 2,\text { for }a=1,\ldots ,\alpha , \quad \mathrm{mult}_{\hat{x}_a}(Z)=1,\text { for }a=\alpha +1,\ldots ,\beta , \quad (\beta \leqslant \delta ). \end{aligned}$$

We denote $p:=\mathop {\mathop {\sum }\limits ^{\alpha }}\limits _{a=1}\mathrm{mult}_{\hat{x}_a}(Z)\geqslant 2\alpha$ and $m:=\mathop {\mathop {\sum }\limits ^{\delta }}\limits _{a=1}\mathrm{mult}_{\hat{x}_a}(Z)\leqslant p+\delta -\alpha$.

If $\alpha =0$, then $\frac{z\cdot {\fancyscript{A}}^2}{m}\geqslant \frac{{\fancyscript{A}}^2}{\delta }$ satisfies the inequality, so we may assume $\alpha \geqslant 1$. A point of multiplicity $m$ lowers the arithmetic genus of $Z$ by at least $\bigl (\begin{array}{c}{m}\\ {2}\end{array}\bigr )$; hence,

$$\begin{aligned} p_a(Z)=\frac{z^2{\fancyscript{A}}^2}{2}+1\geqslant \frac{1}{2}\mathop {\mathop {\sum }\limits ^{\alpha }}\limits _{a=1} \bigl (\mathrm{mult}_{\hat{x}_a}(Z)^2-\mathrm{mult}_{\hat{x}_a}(Z)\bigr ) \mathop {\mathop {\geqslant }\limits ^{\mathrm{Jensen}}}\limits _{\mathrm{inequality}} \frac{1}{2}\bigl (\frac{p^2}{\alpha }-p\bigr ), \end{aligned}$$

so $p\leqslant \frac{\alpha +\sqrt{\alpha ^2+4\alpha (2+z^2{\fancyscript{A}}^2)}}{2}$. We deduce the following inequalities:

$$\begin{aligned} \frac{z{\fancyscript{A}}^2}{m}\geqslant & {} \frac{z{\fancyscript{A}}^2}{p-\alpha +\delta }\geqslant \underbrace{ \frac{z{\fancyscript{A}}^2}{\delta +\frac{\sqrt{\alpha ^2+4\alpha \left( 2+z^2{\fancyscript{A}}^2\right) }-\alpha }{2}} }_{\mathrm{decreasing~in}\,\alpha } \geqslant \underbrace{ \frac{z{\fancyscript{A}}^2}{\delta +\frac{\sqrt{\delta ^2+4\delta \left( 2+z^2{\fancyscript{A}}^2 \right) }-\delta }{2}} }_{\mathrm{increasing~in}\,z} \\\geqslant & {} \frac{2{\fancyscript{A}}^2}{\delta +\sqrt{\delta ^2+4\delta \left( 2+{\fancyscript{A}}^2\right) }}. \end{aligned}$$

$\square$

Corollary 8

$H^0({\fancyscript{M}})\rightarrow H^0({\fancyscript{M}}_E)$ is surjective, for ${\fancyscript{A}}^2\geqslant 6$ and $\delta \leqslant \frac{d^2{\fancyscript{A}}^2}{3(d+4)}$.

Proof

Indeed, it is enough to check that $H^1({\tilde{S}},{\fancyscript{M}}(-E))= H^1({\tilde{S}},K_{\tilde{S}}\otimes {\fancyscript{M}}(-2E))$ vanishes. By the Kodaira vanishing theorem, this happens as soon as ${\fancyscript{M}}(-2E)=\sigma ^*{\fancyscript{A}}^d(-3E)$ is ample. The previous theorem implies that, in order to achieve this, is enough to impose $\frac{3}{d}\leqslant \frac{2{\fancyscript{A}}^2}{\delta +\sqrt{\delta ^2+4\delta (2+{\fancyscript{A}}^2)}}$, which yields $\delta \leqslant \frac{d^2({\fancyscript{A}}^2)^2}{3(d{\fancyscript{A}}^2+3{\fancyscript{A}}^2+6)}$. $\square$

5 Concluding remarks

(I) Evidence for the nonsurjectivity of $\varvec{w_C}$ Theorem 1 is a nonsurjectivity property for the Wahl map of the pointed curve $(C,\varDelta )$, rather than that of the curve $C$ itself.

Claim. In order to prove the nonsurjectivity of the Wahl map $w_C$, is enough to have the surjectivity of the evaluation homomorphism

$$\begin{aligned} { H^0({\fancyscript{I}}(DC)^2\otimes K_{C\times C}) \rightarrow \mathop {\mathop {\bigoplus }\limits ^{\delta }}\limits _{a=1} K_{C\times C,(x_{a,1},x_{a,2})} \oplus K_{C\times C,(x_{a,2},x_{a,1})}. } \end{aligned}$$

(13)

(For $\delta$ in the range (3), corollary 8 implies that $K_C={\fancyscript{M}}_C$ separates $\varDelta$, consequently $\mathop {\bigwedge }\limits ^{2}H^0(K_C)\rightarrow {\bigoplus }_{a=1}^{\delta } K_{C\times C,(x_{a,1},x_{a,2})}$ is surjective. The surjectivity of (13) yields that of $\mathrm{ev}_\varXi ^{\prime \prime }$ in (11), which is relevant for us.)

For the claim, observe that one has the following implications (see (4), (11)):

$$\begin{aligned} w_C\text { surjective }\;\Rightarrow \; w_C^{\prime }\text { surjective } \;\mathop {\mathop {\Rightarrow }\limits ^{(13)}}\limits _\mathrm{surj.}\; w_{C,\varDelta }\text { surjective, a contradiction.} \end{aligned}$$

(14)

The surjectivity of (13) is clearly a positivity property for ${\fancyscript{I}}(DC)^2\otimes K_{C\times C}$. We use again the Seshadri constants to argue why this is likely to hold. The $\varXi$-pointed Seshadri constants of the self-product of a very general curve $X$ at very general points $\varXi$ (as in Lemma 3) satisfy (see [6], p. 65 below Theorem 1.6, and Lemma 2.6]):

$$\begin{aligned} \varepsilon _{X\times X\!,\,\varXi }({\fancyscript{I}}(DX)^2\otimes K_{X\times X})\geqslant & {} 2(g-2)\varepsilon _{\mathbb {P}^2\!,\,g+\delta } ({\fancyscript{O}}_{\mathbb {P}^2}(1))\nonumber \\> & {} \frac{2(g-2)}{\sqrt{g+\delta }}\sqrt{1-\frac{1}{8(g+\delta )}}, \end{aligned}$$

(15)

$$\begin{aligned} \varepsilon _{X\times X\!,\,\varXi }({\fancyscript{I}}(DX)^4\otimes K_{X\times X})\geqslant & {} 4\cdot \frac{g-3}{2}\cdot \varepsilon _{\mathbb {P}^2\!,\,g+\delta } ({\fancyscript{O}}_{\mathbb {P}^2}(1))\nonumber \\> & {} \frac{2(g-3)}{\sqrt{g+\delta }}\sqrt{1-\frac{1}{8(g+\delta )}} =:\varphi (g,\delta ). \end{aligned}$$

(16)

The equation (16) implies (see [2], Proposition6.8]) that ${\big ({\fancyscript{I}}(DX)^2\otimes K_{X\times X}\big )}^2$ generates the jets of order $\lfloor \varphi (g,\delta )\rfloor -2$ at $\varXi \subset X\times X$. (We only need the generation of jets of order zero for ${\fancyscript{I}}(DX)^2\otimes K_{X\times X}$; also, note that $\varphi (g,\delta )$ grows linearly with $\sqrt{g}$ as long as $\delta$ is small compared with $g$ (see (3)).) This discussion suggests that ${\fancyscript{I}}(DX)^2\otimes K_{X\times X}$ is ‘strongly positive/generated.’ However, the passage to (13) above requires even more control.

(II) Related work In [4], the author extensively studies the properties of nodal curves on $K3$ surfaces. Among several other results, he proves the nonsurjectivity of a marked Wahl map (different from the one introduced in here) for nodal curves on $K3$ surfaces.