Proof. Let $X$ be the stabilisation of $D$ as above. By [Reference Hirschowitz and RamananHR98], it suffices to show that

$$\begin{eqnarray}K_{g-k+1,1}(X,\unicode[STIX]{x1D714}_{X})=0;\end{eqnarray}$$

see also [Reference AproduApr05, Proposition 7]. This is implied by the linear growth assumption on $C$ and the generality of the points $(x_{i},y_{i})$ ; see the proof of [Reference AproduApr05, Theorem 2].◻

Proof. The only reason why this lemma is not totally obvious is that $\mathfrak{Syz}$ was defined as the closure of smooth, marked curves with extra syzygies. However, the determinantal description from [Reference Farkas and KemenyFK, § 6] can be extended verbatim to the open locus $\overline{{\mathcal{M}}}_{2g-2k+3,2(2g-2k+3)}^{va}$ of marked stable curves $D^{\prime }$ such that the marking defines a very ample line bundle $N^{\prime }$ with $H^{1}(D^{\prime },N^{\prime })=0$ ; see also [Reference FarkasFar06, § 2] and [Reference FarkasFar09]. Indeed, the only thing which needs checking is that we continue to have $H^{1}(D^{\prime },\bigwedge ^{g-k}M_{N^{\prime }}\otimes {N^{\prime }}^{2})$ . This follows from the short exact sequence

$$\begin{eqnarray}0\rightarrow \bigwedge ^{g-k+1}M_{N^{\prime }}\otimes N^{\prime }\rightarrow \bigwedge ^{g-k+1}H^{0}(N^{\prime })\otimes N^{\prime }\rightarrow \bigwedge ^{g-k}M_{N^{\prime }}\otimes {N^{\prime }}^{2}\rightarrow 0\end{eqnarray}$$

and the assumption $H^{1}(D^{\prime },N^{\prime })=0$ .

Thus, we get a divisor $\mathfrak{Syz}^{va}\subseteq \overline{{\mathcal{M}}}_{2g-2k+3,2(2g-2k+3)}^{va}$ , which coincides with $\mathfrak{Syz}$ on ${\mathcal{M}}_{2g-2k+3,2(2g-2k+3)}^{va}$ . Now the fact that $L$ is very ample implies that $N$ is very ample; indeed, $H^{0}(C,L)\simeq H^{0}(D,N)$ , and $\unicode[STIX]{x1D719}_{N}:D\rightarrow \mathbb{P}^{g-2k+3}$ embeds $D$ as the union of the curve $C$ (embedded by $L$ ) together with $g-2k+3$ secant lines $R_{i}$ . So, $[D]\in \overline{{\mathcal{M}}}_{2g-2k+3,2(2g-2k+3)}^{va}$ . Riemann–Roch now gives $H^{1}(D,N^{\prime })=0$ . The point $[D]$ lies on precisely one boundary component of $\overline{{\mathcal{M}}}_{2g-2k+3,2(2g-2k+3)}$ , namely the component $\unicode[STIX]{x1D6FF}_{\text{irr}}$ whose general point is an integral curve with one node; see [Reference Arbarello and CornalbaAC99] for details of the boundary of $\overline{{\mathcal{M}}}_{g,n}$ .

Thus, it suffices to show that $\mathfrak{Syz}^{va}$ does not contain $\unicode[STIX]{x1D6FF}_{\text{irr}}$ . This follows easily from [Reference Farkas and KemenyFK, Theorem 1.8]. Indeed, it suffices to show that there exist integral, singular curves with nodal singularities in the linear system $|L|$ on the K3 surface $Z_{2g-2k+3}$ from [Reference Farkas and KemenyFK, § 3].Footnote ² For this, one can degenerate to the hyperelliptic K3 surface $\hat{Z}_{2g-2k+3}$ as in [Reference Farkas and KemenyFK, § 3], and take a general curve $A$ in the base-point-free linear system $|L-E|$ which meets a given elliptic curve $B\in |E|$ transversally. The nodal curve $A+B$ then deforms to an integral nodal curve in $|L|$ .◻

Proof. From the inclusion $L-C_{2(g-2k+3-j)}-K_{C}\subseteq C_{k-1+j}-C_{g-k-j}$ , we have that, for every effective divisor $D$ of degree $2(g-2k+3-j)$ , there exists an effective divisor $E$ of degree $k-1+j$ such that

$$\begin{eqnarray}[L-(D+E)]\in K_{C}-C_{g-k-j}.\end{eqnarray}$$

As $h^{0}(C,K_{C})=g$ , this implies that $L-(D+E)$ has at least $k+j$ sections. This is equivalent to

$$\begin{eqnarray}[D+E]\in V_{2g-3k-j+5}^{2g-3k-j+4}(L).\end{eqnarray}$$

Let $C^{(i)}$ denote the $i$ th symmetric product of $C$ . There are only finitely many possible $D^{\prime }\in C^{(2(g-2k+3-j))}$ such that we have the equality of divisors

$$\begin{eqnarray}[D+E]=[D^{\prime }+E^{\prime }]\in C^{(2g-3k-j+5)}\end{eqnarray}$$

for some effective divisor $E^{\prime }$ of degree $k-1+j$ . Hence, the dimension of $V_{2g-3k-j+5}^{2g-3k-j+4}(L)$ is at least $2(g-2k+3-j)$ .◻

Proof. Firstly note that, if $0\leqslant j<g-2k+3$ , then all secant loci $V_{2g-3k-j+5}^{2g-3k-j+4}(L)$ are nonempty by [Reference Arbarello, Cornalba, Griffiths and HarrisACGH85, p. 356]. For $j=g-2k+3$ , the secant locus $V_{g-k+2}^{g-k+1}(L)=\emptyset$ , by the assumption that $L$ is $(g-k+1)$ -very ample. Suppose that there exists $0\leqslant j<g-2k+3$ with $\dim V_{2g-3k-j+5}^{2g-3k-j+4}(L)>2g-2j-4k+5$ . By [Reference Aprodu and SernesiAS15, Remark 4.2], $\dim V_{g-k+3}^{g-k+2}(L)\geqslant 2$ .

Consider the Abel–Jacobi map $\unicode[STIX]{x1D70B}:V_{g-k+3}^{g-k+2}(L)\rightarrow \text{Pic}^{g-k+3}(C)$ . We claim that $\unicode[STIX]{x1D70B}$ is finite. Indeed, otherwise we would have a one-dimensional family of $[D_{t}]\in V_{g-k+3}^{g-k+2}(L)$ with ${\mathcal{O}}_{C}(D_{t})$ constant. Then the line bundle $K_{C}-L+D_{t}$ is independent of $t$ , and furthermore it is effective, since $[D_{t}]\in V_{g-k+3}^{g-k+2}(L)$ . Let $Z\in |K_{C}-L+D_{t}|$ and $s\in \text{Supp}(Z)$ ; the assumption that $k\geqslant 3$ ensures that $\deg (K_{C}-L+D_{t})\geqslant 1$ . There exists some $t^{\prime }$ such that $s\in \text{Supp}(D_{t^{\prime }})$ ; let $D^{\prime }:=D_{t^{\prime }}-s$ . Then $Z-s\in |K_{C}-L+D^{\prime }|$ , so $K_{C}-L+D^{\prime }$ is effective, and $D^{\prime }\in V_{g-k+2}^{g-k+1}(L)$ , contradicting that $V_{g-k+2}^{g-k+1}(L)=\emptyset$ . Thus, we have that $\unicode[STIX]{x1D70B}$ is finite.

We will now apply [Reference Fulton, Harris and LazarsfeldFHL84] to see that $V_{g-k+2}^{g-k+1}(L)\neq \emptyset$ (cf. [Reference Aprodu and SernesiAS15, Remark 4.4] and the proof of [Reference Farkas and KemenyFK, Theorem 1.5]). This contradiction will finish the proof. Indeed, for any point $p\in C$ , we can find an irreducible, closed curve $S\subseteq V_{g-k+3}^{g-k+2}(L)$ such that, for all $s\in S$ , the corresponding divisor $[D_{s}]\in V_{g-k+3}^{g-k+2}(L)$ passes through $p$ , so that $D_{s}^{\prime }:=D_{s}-p$ is an effective divisor. Now consider the Abel–Jacobi map

$$\begin{eqnarray}\displaystyle p:V_{g-k+3}^{g-k+2}(L) & \rightarrow & \displaystyle \text{Pic}^{k-2}(C)\nonumber\\ \displaystyle [D] & \mapsto & \displaystyle K_{C}-L+D.\nonumber\end{eqnarray}$$

This image $p(S)$ is a closed curve, each point of which parametrises an effective line bundle. By [Reference Fulton, Harris and LazarsfeldFHL84], there exists an $s\in S$ with $K_{C}-L+D_{s}-p=K_{C}-L+D_{s}^{\prime }$ effective. But this is the same as saying that $D_{s}^{\prime }\in V_{g-k+2}^{g-k+1}(L)$ .◻

Proof. Assume that $K_{g-k,2}(C,L)\neq 0.$ Then $K_{g-k+2,0}(C,\unicode[STIX]{x1D714}_{C};L)\neq 0$ , by Koszul duality [Reference GreenGre84]. Likewise, $K_{g-k,2}(D,N)=0$ if and only if $K_{g-k+2,0}(D,\unicode[STIX]{x1D714}_{D};N)$ . Note that $H^{0}(D,N)\simeq H^{0}(C,L)$ , and the proof of Koszul duality using kernel bundles goes through unchanged in our case, even though $D$ is nodal; see [Reference Aprodu and NagelAN10, Theorem 2.24]. Restriction induces natural inclusions

$$\begin{eqnarray}\displaystyle H^{0}(D,N) & {\hookrightarrow} & \displaystyle H^{0}(C,L),\nonumber\\ \displaystyle H^{0}(D,\unicode[STIX]{x1D714}_{D}) & {\hookrightarrow} & \displaystyle H^{0}\biggl(C,\unicode[STIX]{x1D714}_{C}\biggl(\mathop{\sum }_{i=1}^{g-2k+3}x_{i}+y_{i}\biggr)\biggr),\nonumber\\ \displaystyle H^{0}(D,N\otimes \unicode[STIX]{x1D714}_{D}) & {\hookrightarrow} & \displaystyle H^{0}\biggl(C,L\otimes \unicode[STIX]{x1D714}_{C}\biggl(\mathop{\sum }_{i=1}^{g-2k+3}x_{i}+y_{i}\biggr)\biggr).\nonumber\end{eqnarray}$$

We thus get the following commutative diagram, where both vertical arrows are injective:

We have an isomorphism $H^{0}(D,N)\simeq H^{0}(C,L)$ , and $H^{1}(C,L)=0$ , so Riemann–Roch implies that $H^{1}(D,N)=0$ .

The image of the restriction map $H^{0}(D,\unicode[STIX]{x1D714}_{D}){\hookrightarrow}H^{0}(C,\unicode[STIX]{x1D714}_{C}(\sum _{i}x_{i}+y_{i}))$ includes $H^{0}(C,\unicode[STIX]{x1D714}_{C})\subseteq H^{0}(C,\unicode[STIX]{x1D714}_{C}(\sum _{i}x_{i}+y_{i}))$ . We have a natural commutative diagram, where the vertical arrows are injective:

Thus, if $K_{g-k+2,0}(C,\unicode[STIX]{x1D714}_{C};L)\neq 0$ , then there exists a nonzero element of $\text{Ker}(\tilde{d}_{g-k+2,0})$ which lies in the image of

$$\begin{eqnarray}\bigwedge ^{g-k+2}H^{0}(D,N)\otimes H^{0}(D,\unicode[STIX]{x1D714}_{D})\rightarrow \bigwedge ^{g-k+2}H^{0}(C,L)\otimes H^{0}\biggl(C,\unicode[STIX]{x1D714}_{C}\biggl(\mathop{\sum }_{i}x_{i}+y_{i}\biggr)\biggr)\end{eqnarray}$$

and thus $d_{g-k+2,0}$ is noninjective, so $K_{g-k+2,0}(D,\unicode[STIX]{x1D714}_{D};N)\neq 0$ .◻

Proof of Theorem 1.4.

Assume that

$$\begin{eqnarray}L-K_{C}\notin C_{g-k+2}-C_{k-3}.\end{eqnarray}$$

We need to show that $K_{g-k,2}(C,L)=0$ . From Lemma 2.5, it suffices to prove that $K_{g-k,2}(D,N)=0$ . From Lemma 2.2 and Proposition 2.1, it suffices to show that the marked curve $[D]\in \overline{{\mathcal{M}}}_{2g-2k+3,2(2g-2k+3)}$ lies outside $\overline{\mathfrak{Sec}}$ . For this, it is sufficient to show that

$$\begin{eqnarray}H^{0}\biggl(D,\bigwedge ^{g-k}M_{K_{D}}(2K_{D}-N)\biggr)=0,\end{eqnarray}$$

by [Reference Farkas, Mustaţă and PopaFMP03, Proposition 3.6]. Here $M_{K_{D}}$ is the kernel bundle, defined by the exact sequence

$$\begin{eqnarray}0\rightarrow M_{K_{D}}\rightarrow H^{0}(D,K_{D})\otimes {\mathcal{O}}_{D}\rightarrow K_{D}\rightarrow 0.\end{eqnarray}$$

Equivalently, if $\unicode[STIX]{x1D719}_{K_{D}}:D\rightarrow \mathbb{P}^{2g-2k+2}$ is the canonical morphism, then $M_{K_{D}}\simeq \unicode[STIX]{x1D719}_{K_{D}}^{\ast }\unicode[STIX]{x1D6FA}_{\mathbb{P}^{2g-2k+2}}(1)$ . Note that $\unicode[STIX]{x1D719}_{K_{D}}$ is not an embedding; indeed, each component $R_{i}$ is contracted to a point.

We define subcurves of $D$ as such: for $1\leqslant k<g-2k+3$ , let

$$\begin{eqnarray}D_{k}:=C\cup R_{k+1}\cup \cdots R_{g-2k+3}\end{eqnarray}$$

and set $D_{g-2k+3}=C$ . Define $N_{i}:=N_{|_{D_{i}}}$ . The Mayer–Vietoris sequence gives

$$\begin{eqnarray}\displaystyle 0 & \rightarrow & \displaystyle \bigwedge ^{g-k}M_{K_{D}}\otimes (2K_{D}-N)\rightarrow \bigwedge ^{g-k}M_{K_{D_{1}}(x_{1}+y_{1})}(2K_{D_{1}}-N_{1}+2x_{1}+2y_{1})\oplus {\mathcal{O}}_{R_{1}}(-1)^{\binom{2g-2k+2}{g-k}}\nonumber\\ \displaystyle & \rightarrow & \displaystyle \bigwedge ^{g-k}M_{K_{D}}\otimes (2K_{D}-N)_{|_{x_{1},y_{1}}}\rightarrow 0,\nonumber\end{eqnarray}$$

using that $M_{{K_{D}}_{|_{D_{1}}}}\simeq M_{K_{D_{1}}(x_{1}+y_{1})}$ (note that restriction induces an isomorphism $H^{0}(D,K_{D})\simeq H^{0}(D_{1},K_{D_{1}}(x_{1}+y_{1}))$ ).

So, it suffices to show that the evaluation map

$$\begin{eqnarray}\displaystyle & & \displaystyle H^{0}\biggl(D_{1},\bigwedge ^{g-k}M_{K_{D_{1}}(x_{1}+y_{1})}(2K_{D_{1}}-N_{1}+2x_{1}+2y_{1})\biggr)\nonumber\\ \displaystyle & & \displaystyle \quad \rightarrow H^{0}\biggl(D,\bigwedge ^{g-k}M_{K_{D_{1}}(x_{1}+y_{1})}(2K_{D_{1}}-N_{1}+2x_{1}+2y_{1})_{|_{x_{1},y_{1}}}\biggr)\nonumber\end{eqnarray}$$

is injective or

$$\begin{eqnarray}H^{0}\biggl(D_{1},\bigwedge ^{g-k}M_{K_{D_{1}}(x_{1}+y_{1})}(2K_{D_{1}}-N_{1}+x_{1}+y_{1})\biggr)=0.\end{eqnarray}$$

We have a short exact sequence

$$\begin{eqnarray}0\rightarrow \bigwedge ^{g-k}M_{K_{D_{1}}}\rightarrow \bigwedge ^{g-k}M_{K_{D_{1}}(x_{1}+y_{1})}\rightarrow \bigwedge ^{g-k-1}M_{K_{D_{1}}}(-x_{1}-y_{1})\rightarrow 0;\end{eqnarray}$$

see for instance [Reference BeauvilleBea03, Remark on p. 345]. So, it is enough to show the following two vanishings:

$$\begin{eqnarray}\displaystyle H^{0}\biggl(D_{1},\bigwedge ^{g-k}M_{K_{D_{1}}}(2K_{D_{1}}-N_{1}+x_{1}+y_{1})\biggr) & = & \displaystyle 0,\nonumber\\ \displaystyle H^{0}\biggl(D_{1},\bigwedge ^{g-k-1}M_{K_{D_{1}}}(2K_{D_{1}}-N_{1})\biggr) & = & \displaystyle 0.\nonumber\end{eqnarray}$$

For any semistable curve $Y$ and vector bundle $E$ on $Y$ , we define $\unicode[STIX]{x1D6E9}_{E}$ as the set of line bundles $M$ with

$$\begin{eqnarray}H^{0}(Y,E\otimes M)\neq 0.\end{eqnarray}$$

We define $C^{\text{sm},i}$ as the intersection of $C$ with the smooth locus of $D_{i}$ . As $x_{1},y_{1}$ are general, it is enough to satisfy the following two conditions:

(3) $$\begin{eqnarray}\displaystyle \displaystyle 2K_{D_{1}}-N_{1}+C_{2}^{\text{sm},1} & \nsubseteq & \displaystyle \unicode[STIX]{x1D6E9}_{\bigwedge ^{g-k}M_{K_{D_{1}}}},\end{eqnarray}$$

(4) $$\begin{eqnarray}\displaystyle 2K_{D_{1}}-N_{1} & \nsubseteq & \displaystyle \unicode[STIX]{x1D6E9}_{\bigwedge ^{g-k-1}M_{K_{D_{1}}}},\end{eqnarray}$$

where the notation $C_{d}^{\text{sm},i}$ refers to the $d$ th symmetric product of $C^{\text{sm},i}$ .

Using the Mayer–Vietoris sequence

$$\begin{eqnarray}0\rightarrow {\mathcal{O}}_{D_{i-1}}\rightarrow {\mathcal{O}}_{D_{i}}\oplus {\mathcal{O}}_{R_{i}}\rightarrow {\mathcal{O}}_{x_{i},y_{i}}\rightarrow 0\end{eqnarray}$$

together with

$$\begin{eqnarray}0\rightarrow \bigwedge ^{p}M_{K_{D_{i}}}\rightarrow \bigwedge ^{p}M_{K_{D_{i}}(x_{i}+y_{i})}\rightarrow \bigwedge ^{p-1}M_{K_{D_{i}}}(-x_{i}-y_{i})\rightarrow 0,\end{eqnarray}$$

and the generality of $x_{1},y_{1},\ldots ,x_{2(2g-2k+3)},y_{2(2g-2k+3)}$ , we see that, in order to verify the conditions

(5) $$\begin{eqnarray}\displaystyle 2K_{D_{i}}-N_{i}+C_{2(i-j)}^{\text{sm},i}\nsubseteq \unicode[STIX]{x1D6E9}_{\bigwedge ^{g-k-j}M_{K_{D_{i}}}},\quad 0\leqslant j\leqslant i, & & \displaystyle\end{eqnarray}$$

it is enough to verify that

(6) $$\begin{eqnarray}\displaystyle 2K_{D_{i+1}}-N_{i+1}+C_{2(i+1-j)}^{\text{sm},i+1}\nsubseteq \unicode[STIX]{x1D6E9}_{\bigwedge ^{g-k-j}M_{K_{D_{i+1}}}},\quad 0\leqslant j\leqslant i+1. & & \displaystyle\end{eqnarray}$$

Hence, it is enough to verify that

(7) $$\begin{eqnarray}\displaystyle 2K_{C}-L+C_{2(g-2k+3-j)}\nsubseteq \unicode[STIX]{x1D6E9}_{\bigwedge ^{g-k-j}M_{K_{C}}},\quad 0\leqslant j\leqslant g-2k+3 & & \displaystyle\end{eqnarray}$$

or

$$\begin{eqnarray}\displaystyle L-C_{2(g-2k+3-j)}-K_{C}\nsubseteq C_{k+j-1}-C_{g-k-j},\quad 0\leqslant j\leqslant g-2k+3, & & \displaystyle \nonumber\end{eqnarray}$$

by Serre duality and [Reference Farkas, Mustaţă and PopaFMP03, Proposition 3.6]. This follows from Lemmas 2.3 and 2.4. ◻

Proof of Theorem 1.5.

In the case $p>g-k$ , then each line bundle $L\in \text{Pic}^{2g+p-k+3}(C)$ fails to be $(p+1)$ -very ample [Reference Farkas and KemenyFK]. Thus, by the known direction of the secant conjecture, $K_{p,2}(C,L)\neq 0$ . So, assume that $p\leqslant g-k$ , $H^{1}(C,2K_{C}-L)=0$ , $H^{1}(C,L)=0$ and that $V_{g-p-3}^{g-p-4}(2K_{C}-L)$ has the expected dimension $g-k-p-1$ . Note that this implies that $L$ is $(p+1)$ -very ample. Indeed, otherwise

$$\begin{eqnarray}L-K_{C}\in C_{p+2}-C_{k-3},\end{eqnarray}$$

which implies that

$$\begin{eqnarray}2K_{C}-L-C_{g-k-p}\subseteq K_{C}+C_{k-3}-C_{g-k+2},\end{eqnarray}$$

which gives $\dim V_{g-p-3}^{g-p-4}(2K_{C}-L)\geqslant g-k-p,$ using the assumption that $H^{1}(C,2K_{C}-L)=0$ . In fact, this last inclusion is equivalent to $\dim V_{g-p-3}^{g-p-4}(2K_{C}-L)\geqslant g-k-p$ ; use that a one-dimensional family of divisors must pass through any given point. In particular, the previous discussion shows that $L$ is base-point free. For a general, effective divisor $D$ of degree $g-k-p$ , the argument above gives $L(D)-K_{C}\notin C_{g-k+2}-C_{k-3}$ . By Theorem 1.4, we have

$$\begin{eqnarray}K_{g-k,2}(C,L(D))=0.\end{eqnarray}$$

By [Reference Farkas and KemenyFK, Proposition 2.1], this implies that $K_{p,2}(C,L)=0$ .◻

Proof of Corollary 1.6.

As we are assuming that $k\geqslant 3$ , the inequality $p\leqslant g-k$ implies that $\deg (L-K_{C})\leqslant g-1$ , so we have $H^{1}(C,2K_{C}-L)=0$ for a general $L\in \text{Pic}^{2g+p-k+3}(C)$ . To show that the condition ‘ $V_{g-p-3}^{g-p-4}(2K_{C}-L)$ has the expected dimension $g-k-p-1$ ’ holds, for $C$ a general $k$ -gonal curve and $L$ general with $H^{1}(C,L)=H^{1}(C,2K_{C}-L)=0$ , we need to show that

$$\begin{eqnarray}2K_{C}-L-C_{g-k-p}\nsubseteq K_{C}+C_{k-3}-C_{g-k+2};\end{eqnarray}$$

see [Reference Farkas and KemenyFK]. This is equivalent to showing that

$$\begin{eqnarray}L-K_{C}+C_{g-k-p}\nsubseteq C_{g-k+2}-C_{k-3}.\end{eqnarray}$$

For this, we may specialise $C$ to a hyperelliptic curve, as the $k$ -gonality stratum in ${\mathcal{M}}_{g}$ contains the locus of hyperelliptic curves. In this case, the condition

$$\begin{eqnarray}L-K_{C}+C_{g-k-p}\subseteq C_{g-k+2}-C_{k-3}\end{eqnarray}$$

implies that

$$\begin{eqnarray}L-K_{C}\in C_{p+2}-C_{k-3},\end{eqnarray}$$

by [Reference Farkas and KemenyFK, Proposition 2.7]. Under the assumption that $p\leqslant g-k$ , if $L$ is a general line bundle of degree $2g+p-k+3$ , then $L-K_{C}$ does not lie in $C_{p+2}-C_{k-3}$ . This completes the proof. ◻