Syzygies of secant varieties of curves of genus 2

The proof is essentially the same as that of Theorem 2.25 of [AN10]. However, the smoothness assumed there is not necessary because the ingredient is the Serre’s duality. ∎

By the construction of the secant varieties and Theorem 5.8 of [ENP20], we know that $H^{0}(\Sigma_{k},\mathcal{O}_{\Sigma_{k}}(1))=H^{0}(C,L)$, $H^{0}(\Sigma_{k},\omega_{\Sigma_{k}})=S^{k+1}H^{0}(C,K)$ and
${H^{0}(\Sigma_{k},\omega_{\Sigma_{k}}(1))=H^{0}(C,K+L)\otimes S^{k}H^{0}(C,K)}$, where $K$ is the canonical divisor on $C$ and the morphism $\phi$ becomes

$$H^{0}(C,L)\otimes S^{k+1}H^{0}(C,K)\to H^{0}(C,K+L)\otimes S^{k}H^{0}(C,K),$$

given by

$$x\otimes v_{1}\cdots v_{k+1}\mapsto\sum\limits_{i=1}^{k+1}v_{i}x\otimes v_{1}\cdots\widehat{v_{i}}\cdots v_{k+1}.$$

We still denote it by $\phi$ and assume $J=\mathrm{Im}(\phi)$. Write $H^{0}(C,K)=\mathrm{span}\{1,v\}$, where $1$ and $v$ are seen as rational functions on an affine open subset of $C$.

From now on I omit the symbol $C$ for short. As long as the multiplication morphism ${\psi_{0}:H^{0}(K)\otimes H^{0}(L)\to H^{0}(K+L)}$ is surjective, the elements in $H^{0}(K+L)\otimes S^{k}H^{0}(K)$ are spanned by elements of two types, $x\otimes v^{m}$ and $vx\otimes v^{m}$, where $0\leq m\leq k$ and $x\in H^{0}(L)$. Clearly the element $vx\otimes v^{k}$ is in $J$ since $\phi(x\otimes v^{k+1})=(k+1)vx\otimes v^{k}$ and $x\otimes 1$ is also in $J$ since $\phi(x\otimes 1)=(k+1)x\otimes 1$. We only need to show $x\otimes v^{m}\in J$ for $0\leq m\leq k$ and $x\in H^{0}(L)$. Once we have shown this, we would have $vx\otimes v^{m-1}\in J$ for all $x\in H^{0}(L),1\leq m\leq k$, because $\phi(x\otimes v^{m})$ is a combination of $x\otimes v^{m}$ and $vx\otimes v^{m-1}$ for all $1\leq m\leq k$. We prove it by induction on $m$.

For $x\otimes 1$, it was discussed above.

Assume that $x\otimes v^{m-1}\in J$ for all $x\in H^{0}(L)$ and $1\leq m\leq k$. We want to show $x\otimes v^{m}\in J$. If we further assume that $\psi_{1}:H^{0}(K)\otimes H^{0}(L-K)\to H^{0}(L)$ is surjective, we can write $x=vx_{1}+y_{1}$ for some $x_{1},y_{1}\in H^{0}(L-K)$. Note that $y_{1}\otimes v^{m}$ is in $J$ because $\phi(y_{1}\otimes v^{m})$ is a combination of $vy_{1}\otimes v^{m-1}$ and $y_{1}\otimes v^{m}$, and we have $vy_{1}\otimes v^{m-1}\in J$ by the induction hypothesis. Therefore we are left to show $vx_{1}\otimes v^{m}\in J$ for $x_{1}\in H^{0}(L-K)$.

Assume that we have reduced the problem to show $v^{p}x_{p}\otimes v^{m}\in J$ for all $x_{p}\in H^{0}(L-pK)$. If we further assume that $\psi_{p+1}:H^{0}(K)\otimes H^{0}(L-(p+1)K)\to H^{0}(L-pK)$ is surjective, then we can write $v^{p}x_{p}=v^{p}(vx_{p+1}+y_{p+1})$ for some $x_{p+1},y_{p+1}\in H^{0}(L-(p+1)K)$. Note that $v^{p}y_{p+1}\otimes v^{m}\in J$ since $\phi(v^{p}y_{p+1}\otimes v^{m})$ is a combination of $v^{p}y_{p+1}\otimes v^{m}$ and $v^{p+1}y_{p+1}\otimes v^{m-1}$ and the latter is in $J$ by the induction hypothesis. We are left to show $v^{p+1}x_{p+1}\otimes v^{m}\in J$. We do the step above for $k$ times. Then we are left to show $v^{k}x_{k}\otimes v^{m}\in J$ for $x_{k}\in H^{0}(L-kK)$.

To show $v^{k}x_{k}\otimes v^{m}\in J$, it suffices to show $v^{k-1}v_{k}\otimes v^{m+1}\in J$ since $\phi(v^{k-1}v_{k}\otimes v^{m+1})$ is a combination of $v^{k}x_{k}\otimes v^{m}$ and $v^{k-1}v_{k}\otimes v^{m+1}$. Similarly, to show $v^{k-1}v_{k}\otimes v^{m+1}\in J$, it suffices to show $v^{k-2}v_{k}\otimes v^{m+2}\in J$. Finally we are left to show $v^{m}x_{k}\otimes v^{k}\in J$. This is clear since $\phi(v^{m-1}x_{k}\otimes v^{k+1})=(k+1)v^{m}x_{k}\otimes v^{k}$.

For the deduction above, we need the surjectivity of the morphisms

$$\psi_{p+1}:H^{0}(K)\otimes H^{0}(L-(p+1)K)\to H^{0}(L-pK)$$

for $-1\leq p\leq k-1$. Note that the kernel of the morphism $H^{0}(K)\otimes\mathcal{O}_{C}\to K$ is a line bundle and hence it is $-K$. Consider the exact sequence

$$0\to L\otimes K^{-(p+2)}\to H^{0}(K)\otimes L\otimes K^{-(p+1)}\to L\otimes K^{-p}\to 0.$$

To verify the desired surjectivity, it suffices to check $H^{1}(L-(p+2)K)=0$ i.e. $H^{0}((p+3)K-L)=0$. Note that $\deg((p+3)K-L)\leq 2(k+2)-\deg(L)=2k+2-r<0$. Therefore its global section group vanishes. ∎

From the duality property we have $K_{r-2k-2,2k+2}(\Sigma_{k},\mathcal{O}_{\Sigma_{k}}(1))^{\vee}=K_{1,0}(\Sigma_{k},\omega_{\Sigma_{k}};\mathcal{O}_{\Sigma_{k}}(1))$, where the latter group is the homology of the sequence

$$\bigwedge\limits^{2}H^{0}(\Sigma_{k},\mathcal{O}_{\Sigma_{k}}(1))\otimes H^{0}(\Sigma_{k},\omega_{\Sigma_{k}}(-1))\to H^{0}(\Sigma_{k},\mathcal{O}_{\Sigma_{k}}(1))\otimes H^{0}(\Sigma_{k},\omega_{\Sigma_{k}})\to H^{0}(\Sigma_{k},\omega_{\Sigma_{k}}(1)).$$

We have $H^{0}(\omega_{\Sigma_{k}}(-1))=H^{2k+1}(\mathcal{O}_{\Sigma_{k}}(1))^{\vee}=0$ by Theorem 5.2 of [ENP20]. By Theorem 2.2, the map on the right is surjective, Therefore

$\begin{array}[]{lcl}\dim K_{r-2k-2,2k+2}&=&\dim\ker(H^{0}(\mathcal{O}_{\Sigma_{k}}(1))\otimes H^{0}(\omega_{\Sigma_{k}})\to H^{0}(\omega_{\Sigma_{k}}(1)))\\ &=&\dim H^{0}(\mathcal{O}_{\Sigma_{k}}(1))\otimes H^{0}(\omega_{\Sigma_{k}})-\dim H^{0}(\omega_{\Sigma_{k}}(1))\\ &=&\dim H^{0}(L)\otimes S^{k+1}H^{0}(K)-\dim H^{0}(K+L)\otimes S^{k}H^{0}(K)\\ &=&(r+1)(k+2)-(r+3)(k+1)\\ &=&r-2k-1.\end{array}$
∎

We first check that when normalized such that $\dim K_{0,0}=1$, the Betti numbers of the pure diagrams satisfy the conditions in Theorem 3.5. In fact, we have

$$\frac{\dim K_{i,e_{i}-i}}{\dim K_{0,0}}=\frac{\prod\limits_{j\not=i,j\geq 0}\frac{1}{|e_{j}-e_{i}|}}{\prod\limits_{j\not=0}\frac{1}{|e_{j}|}}=|\prod\limits_{j\not=i,j\geq 1}\frac{e_{i}}{e_{j}-e_{i}}|.$$

Since the multiplicity is totally determined by the Betti numbers, and the Betti numbers of the pure diagrams divided by $\dim K_{0,0}$ satisfy the conditions in Theorem 3.5, the result of Theorem 3.5 is formally generalized. Then the multiplicity of $\displaystyle\frac{\beta(\mathbf{e})}{\dim K_{0,0}}$ is $\displaystyle\frac{\prod\limits_{i=1}^{n}e_{i}}{n!}$. As a result, the multiplicity of $\beta(\mathbf{e})$ is ${\dim K_{0,0}}\cdot\displaystyle\frac{\prod\limits_{i=1}^{n}e_{i}}{n!}=1$. ∎

The Hilbert series $HS_{V}(t)$ of $V$ is of the form $HS_{V}(t)=\sum\limits_{j}H_{V}(j)t^{j}$, where $H_{V}(t)$ is the Hilbert polynomial of $V$. Then $tHS_{V}(t)=\sum\limits_{j}H_{V}(j)t^{j+1}$. Taking the difference of these two equations, we get $(1-t)HS_{V}(t)\cong\sum\limits_{j}[H_{V}(j)-H_{V}(j-1)]t^{j}=\sum\limits_{j}H_{V\cap H_{1}}(j)t^{j}$. Here the symbol $\cong$ means that they are equal from a term of sufficiently large degree. Inductively, we get $(1-t)^{\dim V}HS_{V}(t)=\sum\limits_{j=0}^{N}c_{j}t^{j}+\sum\limits_{j=N+1}^{\infty}Dt^{j}$, where $D=\deg(V)$. Multiplied by $(1-t)$ on both sides, it is deduced that

$$HS_{V}(t)=\frac{(1-t)P(t)+Dt^{N+1}}{(1-t)^{\dim V+1}}=\frac{(1-t)P(t)+Dt^{N+1}}{(1-t)^{\dim S/I}}.$$

By the definition of the multiplicity, we see that it is equal to the degree. ∎

Consider $K_{r-2k-1,2k+2}$. For a pure diagram characterised by $\mathbf{e}$ contributing to $K_{r-2k-1,2k+2}$, we have $e_{r-2k-1}=r-2k-1+2k+2=r+1$. This implies that only those $\mathbf{i}$ with $i_{0}>0$ contribute to $K_{r-2k-1,2k+2}$. Applying Lemma 4.2 to $K_{r-2k-1,2k+2}$, we have

$$(k+2)/\deg(\Sigma_{k})=\sum\limits_{i_{0}>0}c_{\mathbf{i};d}\dim K_{r-2k-1,2k+2}(\pi_{k}(\mathbf{i};d)).$$

(1)

By the definition, $\dim K_{r-2k-1,2k+2}(\pi_{k}(\mathbf{i};d))=\displaystyle\frac{(r-2k-1)!}{(r+1)\prod\limits_{j\not=r-2k-1}|r+1-e_{j}|}$.
Since we have

	$\displaystyle\prod\limits_{j\not=r-2k-1}|r+1-e_{j}|\cdot r\cdots(r-k)\prod\limits_{j=0}^{k}(i_{j}+j)$	$\displaystyle=$	$\displaystyle r!$
	$\displaystyle\prod\limits_{j\not=r-2k-1}|r+1-e_{j}|$	$\displaystyle=$	$\displaystyle\frac{(r-k-1)!}{\prod\limits_{j=0}^{k}(i_{j}+j)},$

it is deduced that $\dim K_{r-2k-1,2k+2}(\pi_{k}(\mathbf{i};d))=\displaystyle\frac{\prod\limits_{j=0}^{k}(i_{j}+j)}{(r+1)\prod\limits_{j=r-2k}^{r-k-1}j}$.
It can be deduced from Proposition 5.10 of [ENP20] that, when $k\geq 1$, we have

	$\displaystyle\deg(\Sigma_{k})$	$\displaystyle=$	$\displaystyle\binom{r-k}{k+1}+2\binom{r-k-1}{k}+\binom{r-k-2}{k-1}$
		$\displaystyle=$	$\displaystyle\frac{(r^{2}+r-2k-2)\prod\limits_{j=r-2k}^{r-k-2}j}{(k+1)!}.$

Therefore the equation (1) becomes

$\displaystyle k+2$

$\displaystyle=$

$\displaystyle\sum\limits_{i_{0}>0}\frac{(r^{2}+r-2k-2)\prod\limits_{j=r-2k}^{r-k-2}j}{(k+1)!}\cdot c_{\mathbf{i};d}\cdot\frac{\prod\limits_{j=0}^{k}(i_{j}+j)}{(r+1)\prod\limits_{j=r-2k}^{r-k-1}j}$

$\displaystyle(k+2)!$

$\displaystyle=\sum\limits_{i_{0}>0}\frac{r^{2}+r-2k-2}{(r+1)(r-k-1)}c_{\mathbf{i};d}\prod\limits_{j=0}^{k}(i_{j}+j).$

(2)

Consider $K_{r-2k-2,2k+2}$. For the pure diagrams characterised by $\mathbf{e}$ contributing to $K_{r-2k-2,2k+2}$, we have $e_{r-2k-2}=r-2k-2+2k+2=r$, which implies that $e_{r-2k-1}=r+1\Rightarrow i_{0}=2$. So only $\pi_{k}((2,\cdots,2);d)$ contributes to $K_{r-2k-2,2k+2}$, meaning that

$$\frac{r-2k-1}{\deg(\Sigma_{k})}=c_{(2,\cdots,2);d}\dim K_{r-2k-2,2k+2}(\pi_{k}(2,\cdots,2);d).$$

By the definition, $\dim K_{r-2k-2,2k+2}(\pi_{k}(2,\cdots,2);d)=\displaystyle\frac{(r-2k-1)!}{r\prod\limits_{e_{p}\not=r}|r-e_{p}|}$.
Since we have

	$\displaystyle\prod\limits_{e_{p}\not=r}|r-e_{p}|\cdot(r-1)\cdots(r-k-1)\prod\limits_{j=0}^{k}|j+1|$	$\displaystyle=$	$\displaystyle(r-1)!$
	$\displaystyle\prod\limits_{e_{p}\not=r}|r-e_{p}|$	$\displaystyle=$	$\displaystyle\frac{(r-k-2)!}{(k+1)!}$

we know that $\dim K_{r-2k-2,2k+2}=\displaystyle\frac{(k+1)!}{r\prod\limits_{j=r-2k}^{r-k-2}j}.$
As a result,

	$\displaystyle c_{(2,\cdots,2);d}$	$\displaystyle=$	$\displaystyle\frac{r-2k-1}{\deg(\Sigma_{k})\dim K_{r-2k-2,2k+2}}$
		$\displaystyle=$	$\displaystyle(r-2k-1)\cdot\frac{(k+1)!}{(r^{2}+r-2k-2)\prod\limits_{j=r-2k}^{r-k-2}j}\cdot\frac{r\prod\limits_{j=r-2k}^{r-k-2}j}{(k+1)!}$
		$\displaystyle=$	$\displaystyle\frac{r^{2}-2kr-r}{r^{2}+r-2k-2}.$

We now plug the value of $c_{(2,\cdots,2);d}$ in the equation (2):

	$\displaystyle(k+2)!$	$\displaystyle=$	$\displaystyle\sum\limits_{i_{0}>0}\frac{r^{2}+r-2k-2}{(r+1)(r-k-1)}c_{\mathbf{i};d}\prod\limits_{j=0}^{k}(i_{j}+j)$
	$\displaystyle(k+2)!$	$\displaystyle=$	$\displaystyle\frac{r^{2}+r-2k-2}{(r+1)(r-k-1)}\frac{r^{2}-2kr-r}{r^{2}+r-2k-2}(k+2)!+$
			$\displaystyle\sum\limits_{i_{0}=1}\frac{r^{2}+r-2k-2}{(r+1)(r-k-1)}c_{\mathbf{i};d}\prod\limits_{j=0}^{k}(i_{j}+j)$

Moving the term with $(k+2)!$ to the left hand side and canceling the factor $r^{2}+r-2k-2$, we get

$\displaystyle(k+2)![1-\frac{r^{2}-2kr-r}{(r+1)(r-k-1)}]=\sum\limits_{i_{0}=1}\frac{r^{2}+r-2k-2}{(r+1)(r-k-1)}c_{\mathbf{i};d}\prod\limits_{j=0}^{k}(i_{j}+j)$

(3)

For the right hand side, note that for each $\mathbf{i}$, we have $i_{0}=1$ and $i_{j}\leq 2$ for $1\leq j\leq k$. Therefore

$$\prod\limits_{j=0}^{k}(i_{j}+j)\leq(1+0)(2+1)\cdots(2+k)=\frac{(k+2)!}{2}.$$

Applying this inequality to (3), we know that

	$\displaystyle(k+2)![1-\frac{r^{2}-2kr-r}{(r+1)(r-k-1)}]$	$\displaystyle\leq$	$\displaystyle\sum\limits_{i_{0}=1}\frac{r^{2}+r-2k-2}{(r+1)(r-k-1)}c_{\mathbf{i};d}\frac{(k+2)!}{2}$
	$\displaystyle(k+2)!(rk+r-k-1)$	$\displaystyle\leq$	$\displaystyle\sum\limits_{i_{0}=1}(r^{2}+r-2k-2)c_{\mathbf{i};d}\frac{(k+2)!}{2}$
		$\displaystyle=$	$\displaystyle\frac{(k+2)!}{2}(r^{2}+r-2k-2)\sum\limits_{i_{0}=1}c_{\mathbf{i};d}$
		$\displaystyle\leq$	$\displaystyle\frac{(k+2)!}{2}(r^{2}+r-2k-2)(1-c_{(2,\cdots,2);d})$
		$\displaystyle=$	$\displaystyle\frac{(k+2)!}{2}(2kr+2r-2k-2).$

The inequalities are actually equalities! This implies, firstly, only the coefficient of $(1,2,\cdots,2)$ is non-zero among $\mathbf{i}$ with $i_{0}=1$, and secondly, $\sum\limits_{i_{0}=1}c_{\mathbf{i};d}=c_{(1,2,\cdots,2);d}$ is exactly $1-c_{(2,\cdots,2);d}$. In other words, the Betti diagram of $\Sigma_{k}$ is decomposed into the combination of 2 pure diagrams, which are represented by $(2,\cdots,2)$ and $(1,2,\cdots,2)$.