Proof. The fact that $\mathrm {Hameo}( M , \omega )$ is a normal subgroup of $\mathrm {Homeo}_0( M , \omega )$ is proven in [Reference Oh and Müller57]. The same statement for $\mathrm {FHomeo}( M , \omega )$ is proven in [Reference Cristofaro-Gardiner, Humilière and Seyfaddini18, Prop. 2.1], in the case where $ M $ is the disc; the same argument generalizes, in a straightforward way, to any $ M $ . This proves the first item.

The second item follows from the first and the inclusion $\mathrm {Hameo}( M ,\omega ) \subset \mathrm {FHomeo}( M ,\omega )$ .

The third item follows from a general argument, involving fragmentation techniques [Reference Epstein27, Reference Higman36, Reference Fathi28], which proves that any normal subgroup of $\mathrm {Homeo}_0( M , \omega )$ contains the commutator subgroup $[\mathrm {Homeo}_0( M , \omega ), \mathrm {Homeo}_0( M , \omega )]$ . A proof of this in the case where $ M=D^2 $ is presented in [Reference Cristofaro-Gardiner, Humilière and Seyfaddini18, Prop. 2.2]; the argument therein generalizes, in a straightforward way, to any $ M $ .

Proof of Theorem 1.1.

We now show that the spectral invariants of an equidistributed sequence of links satisfy the Calabi property.

We will suppose throughout the proof that $\int _{\Sigma } \omega =1$ . Denote by $L_1, \ldots , L_{k_m}$ the contractible components in $\underline {L}^m$ . These bound closed and pairwise disjoint discs $B_1, \ldots , B_{k_m}$ associated via (iii) above.

Now, fix $\varepsilon> 0$ . Then, since $\mathrm {diam}(\underline {L}^m)\to 0$ , for sufficiently large m we can find a smooth Hamiltonian $G_m$ such that

$$\begin{align*}G_m|_{B_i} = s_i(t), \quad \text{max} | H - G_m | \le \varepsilon ,\end{align*}$$

where each $s_i: [0,1] \to \mathbb {R}.$ We have that

$$\begin{align*}\left|\int_0^1 \int_{\Sigma} H \; \omega\, dt\, - c_{\underline{L}^m}(H) \right| \leqslant \end{align*}$$

$$\begin{align*}\left| \int_0^1 \int_{\Sigma} H - {G_m} \; \omega\, dt\, \right| + \left| \int_0^1 \int_{\Sigma} {G_m} \; \omega\, dt\, - c_{\underline{L}^m}({G_m}) \right| + \left| c_{\underline{L}^m}({G_m}) - c_{\underline{L}^m}(H) \right|,\end{align*}$$

and so we must bound the three terms from the previous line.

The term $\left | \int _0^1 \int _{\Sigma } H - {G_m} \; \omega \, dt\, \right |$ is bounded by $\varepsilon $ because $\max | H -{G_m} | \leqslant \varepsilon $ and $\mathrm {Area}(\Sigma ) = 1$ . Similarly, we have $\left | c_{\underline {L}^m}({G_m}) - c_{\underline {L}^m}(H) \right | \leqslant \varepsilon $ by the Hofer Lipschitz property from Theorem 1.13.

To bound the final term, use the Lagrangian control property of Theorem 1.13 to get

$$\begin{align*}c_{\underline{L}^m}({G_m}) = \frac{1}{k_m+\ell_m} \sum_{i=1}^{k_m}\int_0^1 s_i(t) \, dt + E_m ,\end{align*}$$

where $\ell _m$ is the number of noncontractible components of $\underline {L}^m$ and $E_m$ satisfies

$$\begin{align*}\frac{\ell_m}{k_m +\ell_m} (\min {H}-\varepsilon )\leqslant\frac{\ell_m}{k_m +\ell_m} \min {G_m} \leqslant E_m \leqslant \frac{\ell_m}{k_m +\ell_m} \max {G_m} \leqslant \frac{\ell_m}{k_m +\ell_m} (\max H +\varepsilon ). \end{align*}$$

In particular, since $\ell _m$ is bounded, $E_m$ converges to $0$ as $k_m$ goes to $\infty $ .

Now, noting that $ \int _0^1 s_i(t) dt = \frac 1{\alpha _m}\int _0^1\int _{B_i} {G_m} \, \omega \,dt$ , because $\mathrm {Area}(B_i)= \alpha _m$ , we can rewrite the above as

$$ \begin{align*} c_{\underline{L}^m}({G_m}) = \frac1{\alpha_m(k_m+\ell_m)}\sum_{i=1}^{k_m}\int_0^1 \int_{B_i} {G_m}\;\omega \, dt +E_m = \frac1{\alpha_m(k_m+\ell_m)}\int_0^1 \int_{\Sigma \setminus C_m} {G_m}\; \omega \; dt +E_m, \end{align*} $$

where $C_m$ denotes the complement $C_m := \Sigma \setminus \cup _{i=1}^{k_m} B_i$ . We claim that

(4) $$ \begin{align} \text{lim}_{m \to \infty} \frac{1}{\alpha_m(k_m + \ell_m)} = 1, \quad \text{lim}_{m \to \infty} \text{area}(C_m) = 0, \quad \text{lim}_{m \to \infty} k_m = \infty; \end{align} $$

from this, it follows by the third limit that $E_m$ converges to zero in view of the above and then from the first two limits that

$$\begin{align*}\left|c_{\underline{L}}({G_m})-\int_0^1 \int_{\Sigma} {G_m} \; \omega\, dt\right|\leqslant \varepsilon \end{align*}$$

for m large enough, as desired.

It remains to show equation (4).

We claim the inequality

(5) $$ \begin{align} \frac1{k_m}\geqslant \alpha_m\geqslant \frac{1}{k_m +2N + 1}. \end{align} $$

Recall that N is the bound on $\ell _m$ which exists since the sequence $\underline {L}^m$ is equidistributed. The first inequality here is immediate. To see the second, consider the surface $C^{\prime }_m$ given by removing the noncontractible components of $\underline {L}^m$ from $C_m$ . Then, a coarse bound is that $C^{\prime }_m$ has at most $2N+1$ components, and so $C_m$ satisfies

$$\begin{align*}\mathrm{area}(C_m)\leqslant (2N+1)\alpha_m .\end{align*}$$

Using that $\mathrm {area}(C_m)+\sum \mathrm {area}(B_j)=1$ , we can now deduce equation (5).

To finish the proof of equation (4), since $\mathrm {diam}(\underline {L}^m)\to 0$ , we have $\alpha _m\to 0$ which, in combination with the inequality immediately above, gives the second limit in equation (4). It also gives, in combination with equation (5), the third limit. The first limit in equation (4) now follows from equation (5) since $\ell _m$ is bounded.

Proof of Proposition 3.3.

Define $\zeta : \mathrm {Ham}(\Sigma , \omega ) \rightarrow \mathbb {R}$ by

$$\begin{align*}\zeta(\varphi) = c_{\underline{L}}(\varphi) - c_{\underline{L'}}(\varphi).\end{align*}$$

We need to prove that $\zeta $ is uniformly continuous with respect to the $C^0$ distance.

Let $\varepsilon>0$ , and fix a closed disc $B\subset \Sigma \setminus (\underline {L} \cup \underline {L'})$ . ByFootnote ⁹ [Reference Cristofaro-Gardiner, Humilière and Seyfaddini19, Lemma 3.11], there exists a real number $\delta>0$ such that for any $g\in \mathrm {Ham}(\Sigma , \omega )$ satisfying $d_{C^0}(g,\mathrm {Id})<\delta $ , there exists $h\in \mathrm {Ham}(\Sigma , \omega )$ with support in B and

$$ \begin{align*} d_H(g,h)\leqslant \varepsilon. \end{align*} $$

Let $\phi _1, \phi _2\in \mathrm {Ham}(\Sigma , \omega )$ be such that $d_{C^0}(\phi _1,\phi _2)<\delta $ . We will prove that $|\zeta (\phi _1)-\zeta (\phi _2)|\leqslant 2\varepsilon $ , and this will conclude our proof.

Since $d_{C^0}(\phi _1\phi _2^{-1}, \mathrm {Id}) = d_{C^0}(\phi _1, \phi _2)\leqslant \delta $ , we may pick $h\in \mathrm {Ham}(\Sigma , \omega )$ supported in B and such that

(9) $$ \begin{align} d_{H}(\phi_1\phi_2^{-1},h)\leqslant \varepsilon. \end{align} $$

We now claim that

(10) $$ \begin{align} c_{\underline{L}}(h)=-c_{\underline{L}}(h^{-1})=c_{\underline{L'}}(h)=-c_{\underline{L'}}(h^{-1}). \end{align} $$

Indeed, this follows from the Lagrangian control property of Theorem 1.13, since we can find a mean normalized Hamiltonian H for h such that $H_t$ is constant in the complement of B, and so $h^{-1}$ has a mean normalized Hamiltonian equal to $-H$ in the complement of B.

Now, observe that

(11) $$ \begin{align} c_{\underline{L}}(\phi_1)&= c_{\underline{L}}(\phi_1\phi_2^{-1}\phi_2) \leqslant c_{\underline{L}}(\phi_1\phi_2^{-1} h^{-1}) + c_{\underline{L}}(h \phi_2) \nonumber \\[3pt]&\leqslant \varepsilon + c_{\underline{L}}(h \phi_2) \leqslant c_{\underline{L}}(h)+c_{\underline{L}}(\phi_2)+\varepsilon. \end{align} $$

Here, the first inequality holds by the triangle inequality property from above, the second holds by the Hofer Lipschitz property combined with equation (9), and the third holds by again applying the triangle inequality.

Similarly,

$$ \begin{align*} c_{\underline{L'}}(\phi_2)= c_{\underline{L'}}((\phi_1\phi_2^{-1})^{-1}\phi_1) \leqslant c_{\underline{L'}}(h^{-1}\phi_1)+\varepsilon \leqslant c_{\underline{L'}}(h^{-1})+c_{\underline{L'}}(\phi_1)+\varepsilon. \end{align*} $$

The above inequalities together with equation (10) give

$$ \begin{align*} \zeta(\phi_1) &= c_{\underline{L}}(\phi_1)-c_{\underline{L'}}(\phi_1)\\ &\leqslant c_{\underline{L}}(h)+c_{\underline{L}}(\phi_2)+\varepsilon +c_{\underline{L'}}(h^{-1})-c_{\underline{L'}}(\phi_2)+\varepsilon \\ &=\zeta(\phi_2) + 2\varepsilon. \end{align*} $$

Switching the roles of $\phi _1$ and $\phi _2$ , we obtain $|\zeta (\phi _1)-\zeta (\phi _2)| \leqslant 2\varepsilon $ , which shows that $\zeta $ is uniformly continuous.

Proof of Proposition 3.4.

As in the previous proof, we start by letting $\varepsilon>0$ and fix a closed $B\subset \Sigma _0\setminus (L\cup L')$ . We then follow step by step the same argument until we arrive at inequality 11:

$$ \begin{align*} c_{\underline{L}}(\phi_1)\leqslant c_{\underline{L}}(h)+c_{\underline{L}}(\phi_2)+\varepsilon. \end{align*} $$

Since the Calabi homomorphism is 1-Lipschitz with respect to Hofer’s distance, inequality (9) yields

$$ \begin{align*} \mathrm{Cal}(\phi_1)\leqslant \mathrm{Cal}(\phi_2)+\mathrm{Cal}(h)+\varepsilon. \end{align*} $$

Now, by the shift property from Theorem 1.13, $c_{\underline {L}}(h)=-\mathrm {Cal}(h)$ , as can be seen by choosing a Hamiltonian for h that vanishes outside B and then mean normalizing. Thus, we obtain from the two previous inequalities:

$$\begin{align*}c_{\underline{L}}(\phi_1)+\mathrm{Cal}(\phi_1)\leqslant c_{\underline{L}}(\phi_2)+\mathrm{Cal}(\phi_2)+2\varepsilon .\end{align*}$$

We conclude by switching the roles of $\phi _1$ and $\phi _2$ as in the proof of Proposition 3.3.

Proof of Theorem 1.3.

We showed in Proposition 2.2 that $\mathrm {FHomeo}$ is a normal subgroup of $\overline {\mathrm {Ham}}$ . It remains to show that it is proper. To do this, we adapt the strategy from [Reference Cristofaro-Gardiner, Humilière and Seyfaddini18, Thm. 1.7], namely we construct an example of a Hamiltonian homeomorphism that does not have finite energy.

We first consider the case where $\Sigma $ is closed. Let $\underline {L}^m$ be an equidistributed sequence of Lagrangian links. Define $\zeta _m : \mathrm {Ham}(\Sigma , \omega ) \rightarrow \mathbb {R}$ by

$$ \begin{align*}\zeta_m(\varphi) = c_{\underline{L}^m}(\varphi) - c_{\underline{L}^1}(\varphi).\end{align*} $$

By Proposition 3.3, $\zeta _m$ admits a continuous extension to $\overline {\mathrm {Ham}}(\Sigma , \omega )$ .

We now claim that if $\phi \in \mathrm {FHomeo}$ , then $\zeta _m(\phi )$ remains bounded as m varies. To see this, let $\phi _i = \varphi ^1_{H_i}$ be a sequence of diffeomorphisms converging to $\phi $ such that the $H_i$ are mean normalized and have Hofer norm bounded by C. Then by the Hofer Lipschitz property from Theorem 1.13, applied with $H' = 0$ , we have that the $\zeta _m(\phi _i)$ are also bounded by C. Hence, by continuity, the $\zeta _m(\phi )$ are bounded as well.

Next, let $D \subset \Sigma \setminus \cup _{i=1}^{k_1}L_i^1$ be a smoothly embedded closed disc, which we identify with the disc of radius R in $\mathbb {R}^2$ centered at the origin with the area form $\omega = \frac {1}{2\pi } r dr \wedge d\theta $ . We now define an ‘infinite twist’ homeomorphism $\psi $ supported in D as follows. Let $(\theta ,r)$ denote polar coordinates. Let $f: (0,R] \to \mathbb {R}$ be a smooth function which vanishes near R, is decreasing, and satisfies

(12) $$ \begin{align} \int^1_0 r^3f(r)\, dr = \infty. \end{align} $$

We now define $\psi $ by $\psi (0) = 0$ and

(13) $$ \begin{align} \psi(r,\theta) = (r, \theta + 2 \pi f(r)) \end{align} $$

for $r> 0$ .Footnote ¹⁰ The heuristic behind the condition (12) is that it forces $\psi $ to have ‘infinite Calabi invariant’. Indeed, if f was defined on the closed interval $[0,R]$ , then $\psi $ would be a Hamiltonian diffeomorphism with Calabi invariant $ \int ^1_0 r^3f(r)\, dr$ .

We now claim that $\psi $ is a Hamiltonian homeomorphism with the property that $\zeta _m(\psi )$ diverges as m varies. By [Reference Cristofaro-Gardiner, Humilière and Seyfaddini18, Lem. 1.14],Footnote ¹¹ there are Hamiltonians $F_i$ , compactly supported in the interior of D, with the following properties:

1. 1. The sequence $\psi ^1_{F_i}$ converges in $C^0$ to $\psi $ .

2. 2. $F_i \le F_{i+1}.$

3. 3. $\text {lim}_{i \to \infty } \int ^1_0 \int _{\Sigma } F_i \omega = \infty .$

By the first property above, $\psi $ is a Hamiltonian homeomorphism. We now apply several properties from Theorem 1.13. By the shift property, $\zeta _m(\psi ^1_{F_i}) = c_{\underline {L}^m}(F_i) - c_{\underline {L}^1}(F_i)$ , and by the support control property from the same theorem, $c_{\underline {L}^1}(F_i) = 0.$ It then follows by continuity and the monotonicity property that

$$\begin{align*}\zeta_m(\psi) \ge c_{\underline{L}^m}(F_i),\end{align*}$$

hence by the Calabi property from Theorem 1.1,

$$\begin{align*}\text{lim}_{m \to \infty} \zeta_m(\psi) \ge \int^1_0 \int_{\Sigma} F_i \hspace{2 mm} \omega\end{align*}$$

for any i. Hence, by the third property above, the $\zeta _m(\psi )$ diverge.

In the case when $\Sigma $ is not closed, we reduce to the above by embedding $\Sigma $ into a closed surface $\Sigma '$ . Now, define an infinite twist exactly as above, except in addition the infinite twist is supported in $\Sigma $ : By the above, this map is not in $\mathrm {FHomeo}(\Sigma ',\omega ')$ , hence cannot be in $\mathrm {FHomeo}(\Sigma ,\omega ).$

Proof. Let $\phi \in \mathrm {Hameo}(\Sigma ,\omega )$ , and take an $H \in C^0([0,1]\times \Sigma )$ such that

$$ \begin{align*} \phi^1_{H_i} \xrightarrow{C^0} \phi, \text{ and } \| H- H_i \|_{(1, \infty)} \to 0,\end{align*} $$

where the $H_i$ are smooth Hamiltonians as in Definition 2.1. For future use, we record H in the notation by writing $\phi = \phi _H$ .

We now define

(14) $$ \begin{align} \mathrm{Cal}(\phi) := \int_{0}^1 \int_{\Sigma} H \, \omega \, dt. \end{align} $$

We claim this is well defined. To show this, it suffices to show that, if $\phi = \text {Id}$ , then

(15) $$ \begin{align} \int_{0}^1 \int_{\Sigma} H \, \omega \, dt = 0 \end{align} $$

since $\text {Cal}$ is a homomorphism on $\mathrm {Ham}(\Sigma ,\omega ).$ In other words, we will show that if $\phi ^1_{H_i} \xrightarrow {C^0} \mathrm {Id}$ and $\Vert H-H_i \Vert _{(1, \infty )} \to 0$ , then equation (15) holds.

As in Proposition 3.4, embed $\Sigma $ into a closed surface $\Sigma '$ , choose a sequence of equidistributed Lagrangian links $\underline {L}^m$ in $\Sigma '$ and consider $\xi _m: \mathrm {Ham}(\Sigma , \omega ) \rightarrow \mathbb {R}$ by

$$\begin{align*}\xi_m(\varphi) = c_{\underline{L}^m}(\varphi) + \mathrm{Cal}(\varphi).\end{align*}$$

By Proposition 3.4, $\xi _m$ extends continuously to $\overline {\mathrm {Ham}}(\Sigma , \omega )$ . This in particular implies that

(16) $$ \begin{align} \lim_{j \to \infty}\xi_m(\phi^1_{H_j}) = 0. \end{align} $$

For any fixed $m, i$ , we can write

$$ \begin{align*} \left\vert \int_{0}^1 \int_{\Sigma} H \, \omega \, dt \right\vert &\leqslant \left\vert \int_0^{1} \int_{\Sigma} H \, \omega \, dt \; - \mathrm{Cal}(\phi^1_{H_i}) \right\vert \\[5pt] & \quad + \left\vert \mathrm{Cal}(\phi^1_{H_i}) - \xi_m(\phi^1_{H_i}) \right\vert + \left\vert \xi_m(\phi^1_{H_i}) \right\vert. \end{align*} $$

The right-hand side of the above inequality is a sum of three terms. We know that

$$ \begin{align*} \left\vert \int_{0}^1 \int_{\Sigma} H \, \omega - \mathrm{Cal}(\phi^1_{H_i}) \, dt \right\vert \le || H - H_i ||_{(1,\infty)}, \end{align*} $$

since $H_i $ are smooth and compactly supported Hamiltonians and so $\mathrm {Cal}(\phi ^1_{H_i}) = \int _{0}^1 \int _{\Sigma } H_i \, \omega \, dt.$ We claim that the third term has the same bound. Indeed, by the Hofer Lipschitz property from Theorem 1.13, we have $| \xi _m(\phi ^1_{H_j}) - \xi _m(\phi ^1_{H_i}) | \le || H_j - H_i ||_{(1,\infty )}$ for all $i, j$ , and then fixing i and taking the limit as $j \to \infty $ gives

$$ \begin{align*} |\xi_m (\phi^1_{H_i}) | \le || H - H_i ||_{(1, \infty)} \end{align*} $$

by equation (16). Hence, whatever m, the first and third terms of the above inequality can be made arbitrarily small by choosing i sufficiently large. As for the second term, for fixed i, this can be made arbitrarily small by choosing m sufficiently large by the Calabi property proved in Theorem 1.1.

Hence, $\text {Cal}$ is well defined. It remains to show that it is a homomorphism. The fact that $\mathrm {Cal}$ is a homomorphism if well defined was in fact previously shown in [Reference Oh54], so we will be brief. Let $\psi _1$ and $\psi _2$ be elements of $\mathrm {Hameo}(\Sigma ,\omega )$ , and choose corresponding H, G. By reparametrizing, we can assume that H and G vanish near $0$ and $1$ , and we can then form the concatenation

$$\begin{align*}K(t,x) = \begin{cases} 2 H(2t,x), &{\quad\text{if }} t\in [0, \frac{1}{2}]\\ 2 G(2t-1,x),&{\quad\text{if }} t\in [\frac{1}{2}, 1] \end{cases}. \end{align*}$$

One now checks that $\phi _K = \phi _G \circ \phi _H$ , and it now follows immediately from this formula for K and equation (14) that $\mathrm {Cal}(\phi _G \circ \phi _H) = \mathrm {Cal}(\phi _H) + \mathrm {Cal}(\phi _G).$ The proof that $\mathrm {Cal}((\phi _H)^{-1}) = - \mathrm {Cal}(\phi _H)$ is similar.

Proof of Corollary 4.6 assuming Lemma 4.5.

The submanifold

$$ \begin{align*} \widetilde{L}_{a_1,\dots,a_k,\varepsilon }:= \left\{|x_i-a_i|\left|\prod_{j \neq i} (a_j-a_i)\right|=\varepsilon \text{ for all }i\right\} \cap \tilde{G} \end{align*} $$

is a product of circles, and $\pi (\widetilde {L}_{a_1,\dots ,a_k,\varepsilon })$ is precisely $\operatorname {Sym}(\underline {L}_{a_1,\dots ,a_k,\varepsilon })$ . Since the isotopy $\Phi ^t$ constructed in Lemma 4.5 is supported in $\tilde {G}$ and $\pi |_{\tilde {G}}: \tilde {G} \to G$ is a diffeomorphism, we can descend $\Phi ^t$ to a family of diffeomorphisms $\Phi ^t_G$ supported in G such that $\Phi ^1_G(\operatorname {Sym}(\underline {L}_{a_1,\dots ,a_k,\varepsilon }))=L_{\varepsilon }$ and $\Phi ^t_G(D_i)=D_i$ for all i.

Proof of Lemma 4.5.

For simplicity of notation, we will give the proof in the case in which $a_i \in \mathbb {R}$ for each i.

Let $u_i+\sqrt {-1} v_i=x_i-a_i$ . Then $\widetilde {D}_i \cap \widetilde {G}=\{(u,v)\in \widetilde {G}: u_i=v_i=0\}$ . The system of equations

(17) $$ \begin{align} (x_i-a_i)\left((1-t)\prod_{j \neq i} (a_j-a_i)+t\prod_{j \neq i} (x_j-a_i)\right)\ =\ \alpha_i +\sqrt{-1} \beta_i \qquad 1\leqslant i\leqslant k \end{align} $$

for $t \in [0,1]$ and $\alpha _i, \beta _i \in \mathbb {R}$ becomes, in the $u_i, v_i$ coordinates,

(18) $$ \begin{align} (u_i+\sqrt{-1} v_i)\left((1-t)\prod_{j \neq i} (a_j-a_i)+t\prod_{j \neq i} (u_j+\sqrt{-1} v_j+a_j-a_i)\right)=\alpha_i +\sqrt{-1} \beta_i. \end{align} $$

Taking real and imaginary parts, we obtain

$$ \begin{align*} u_i \prod_{j \neq i} (a_j-a_i) + t H_{u_i}(u,v)=\alpha_i,\\ v_i \prod_{j \neq i} (a_j-a_i) + t H_{v_i}(u,v)= \beta_i, \end{align*} $$

where $H_{u_i}(u,v)$ and $H_{v_i}(u,v)$ are polynomials in $u_j,v_j$ in which each term has degree at least two.

Let $\rho : \mathbb {R}_{\ge 0} \to \mathbb {R}_{\ge 0}$ be a cutoff function such that $\rho (s)=1$ for $s<\frac {\varepsilon ^2}{2}$ , $\rho (s)=0$ for $s>\varepsilon ^2$ and $|\rho '(s)|<\frac {C}{\varepsilon ^2}$ for some constant C independent of $\varepsilon $ and for all s. We denote $\sum _i u_i^2+v_i^2$ by $|(u,v)|^2$ . Let

$$ \begin{align*} F_{u_i}^{t}(u,v,\alpha,\beta):=u_i \prod_{j \neq i} (a_j-a_i) + t\rho(|(u,v)|^2) H_{u_i}(u,v)-\alpha_i,\\ F_{v_i}^{t}(u,v,\alpha,\beta):=v_i \prod_{j \neq i} (a_j-a_i) + t\rho(|(u,v)|^2) H_{v_i}(u,v)-\beta_i. \end{align*} $$

The $2k \times 2k$ square matrix

$$ \begin{align*} D_{u,v}F^t:= \begin{pmatrix} \frac{\partial F_{u_1}^{t}}{\partial u_1} & \frac{\partial F_{u_1}^{t}}{\partial v_1} & \dots & \frac{\partial F_{u_1}^{t}}{\partial u_k} &\frac{\partial F_{u_1}^{t}}{\partial v_k} \\ \frac{\partial F_{v_1}^{t}}{\partial u_1} & \frac{\partial F_{v_1}^{t}}{\partial v_1} & \dots & \frac{\partial F_{v_1}^{t}}{\partial u_k} & \frac{\partial F_{v_1}^{t}}{\partial v_k} \\ \vdots & \vdots & & \vdots & \vdots \\ \frac{\partial F_{u_k}^{t}}{\partial u_1} & \frac{\partial F_{u_k}^{t}}{\partial v_1} & \dots & \frac{\partial F_{u_k}^{t}}{\partial u_k} & \frac{\partial F_{u_k}^{t}}{\partial v_k} \\ \frac{\partial F_{v_k}^{t}}{\partial u_1} & \frac{\partial F_{v_k}^{t}}{\partial v_1} & \dots & \frac{\partial F_{v_k}^{t}}{\partial u_k} & \frac{\partial F_{v_k}^{t}}{\partial v_k} \end{pmatrix} \end{align*} $$

can be written as a sum $A+t\rho B_1 + t\rho ' B_2$ , where A is the diagonal matrix with entries $\prod _{j \neq i} (a_j-a_i)$ at both the $(2i-1,2i-1)^{th}$ and $(2i,2i)^{th}$ positions, for each $i=1,\dots ,k$ , and where the entries of $B_{\ell }$ are polynomials, with each nonzero term having degree at least $1$ when $\ell =1$ and degree at least $3$ when $\ell =2$ .

Since the support of $\rho $ is $[0,\varepsilon ^2]$ , when $\varepsilon>0$ is small we have

$$ \begin{align*} \|D_{u,v}F^t-A\|=t\,O(\varepsilon ) \end{align*} $$

for all $t \in [0,1]$ and for all points $(u,v)$ . By the implicit function theorem, there exists a unique $g^t(\alpha ,\beta )$ such that

(19) $$ \begin{align} F_{u_i}^{t}(g^t(\alpha,\beta),\alpha,\beta)=0 \end{align} $$

(20) $$ \begin{align} F_{v_i}^{t}(g^t(\alpha,\beta),\alpha,\beta)=0 \end{align} $$

for all $i=1,\dots ,k$ . Define a smooth isotopy starting at the identity by

$$\begin{align*}\Phi^t(u,v):=g^t \circ (g^0)^{-1}(u,v). \end{align*}$$

We can control the $C^1$ -norm of the isotopy as follows. We have $D\Phi ^t=D(g^t) \circ D((g^0)^{-1})$ . Since $g^t$ solves the equations (19) and (20) for each i, by differentiating with respect to $\alpha $ and $\beta $ , we have

$$ \begin{align*} Dg^t = -(D_{u,v}F^t)^{-1} \begin{pmatrix} \frac{\partial F_{u_1}^{t}}{\partial \alpha_1} & \frac{\partial F_{u_1}^{t}}{\partial \beta_1} & \dots & \frac{\partial F_{u_1}^{t}}{\partial \beta_k} \\ \frac{\partial F_{v_1}^{t}}{\partial \alpha_1} & \frac{\partial F_{v_1}^{t}}{\partial \beta_1} & \dots & \frac{\partial F_{v_1}^{t}}{\partial \beta_k} \\ \vdots & & & \vdots \\ \frac{\partial F_{v_k}^{t}}{\partial \alpha_1} & \frac{\partial F_{v_k}^{t}}{\partial \beta_1} & \dots & \frac{\partial F_{v_k}^{t}}{\partial \beta_k} \\ \end{pmatrix}=(D_{u,v}F^t)^{-1}. \end{align*} $$

As a result, we have

$$ \begin{align*} \|Dg^t-A^{-1}\|=\frac{t\,O(\varepsilon )}{\|A\|^2}. \end{align*} $$

Moreover, when $t=0$ , we have exactly $Dg^0=A^{-1}$ . Therefore, we have

$$ \begin{align*} \|D\Phi^t-\mathrm{Id}\|=\frac{t\, O(\varepsilon )}{\|A\|}, \end{align*} $$

so the $C^1$ -norm of $\Phi ^t$ is smaller than the prescribed $\kappa $ whenever $\varepsilon $ is sufficiently small.

We now check the remaining conditions. Equation (18) implies that

$$\begin{align*}(g^t)^{-1}(\{u_i=v_i=0\}) \subset \{\alpha_i=\beta_i=0\}, \end{align*}$$

so

$$\begin{align*}\Phi^t(\widetilde{D}_i \cap \widetilde{G})=\widetilde{D}_i \cap \widetilde{G} \end{align*}$$

for every $t, i$ . It is also clear from the construction, cf. equation (17), that there exists $0<\varepsilon ' \ll \varepsilon $ such that

$$ \begin{align*} \Phi^1\left( \left\{|x_i-a_i|\left|\prod_{j \neq i} (a_j-a_i)\right|=\varepsilon ' \text{ for all }i\right\}\right)=\widetilde{L}_{\varepsilon '}. \end{align*} $$

Therefore, replacing $\varepsilon $ by $\varepsilon '$ , the final claim of the statement holds.

Proof. First, we consider the special case that $\underline {L}=\underline {L}_{a_1,\dots ,a_k,\varepsilon }$ for small $\varepsilon $ . Since $\operatorname {Sym}(\underline {L})$ is smoothly isotopic to $L_{\varepsilon }$ , there is an isomorphism of relative homology groups

(24) $$ \begin{align} H_2(X,\operatorname{Sym}(\underline{L})) \cong H_2(X, L_{\varepsilon }). \end{align} $$

Since $\Phi ^t_G$ is $C^1$ -small and being totally real is an open condition, we can take the isotopy to be through totally real tori, in which case the isomorphism (24) preserves the Maslov class [Reference Robbin and Salamon65]. Furthermore, the isotopy $\Phi ^t_G$ is supported away from the anticanonical divisor $\cup _{i=1}^{k+1} D_i$ , so the isomorphism (24) does not change the intersection number with $D_i$ . Since $L_{\varepsilon }$ is a Clifford torus, it is known that $H_2(X,L_{\varepsilon })$ admits a basis $\{[u_i]\}_{i=1}^{k+1}$ such that $[u_i] \cdot D_j=\delta _{ij}$ . Moreover, it is also known that $\mu (u_i)=2$ for all i [Reference Cho15]. Hence, the same is true for $\operatorname {Sym}(\underline {L})$ .

For general $\underline {L}$ in $\mathbb {P}^1$ such that $B_i$ are discs for $i=1,\dots ,k=s-1$ , we can find a smooth family of $(\underline {L}_t)_{t \in [0,1]}$ in $\mathbb {P}^1$ such that $\underline {L}_0=\underline {L}$ and $\underline {L}_1=\underline {L}_{a_1,\dots ,a_k,\varepsilon }$ for some small $\varepsilon>0$ . Moreover, we can assume that $\underline {L}_t$ is disjoint from $\{a_1,\dots ,a_k\}$ for all t. Therefore, we get a smooth family of Lagrangian tori $\operatorname {Sym}(\underline {L}_t)$ that is disjoint from $D_i$ for all i and all t. The result follows.

Proof. Let $\widehat {S}'=\sqcup _{j=1}^k S_j$ , where $S_j=S$ is a unit disc for each j. For $i=1,\dots ,k$ , let $v^{\prime }_i:\widehat {S}' \to S^2$ be a map such that $v^{\prime }_i|_{S_j}$ is a constant map to a point in $L_j$ if $j \neq i$ and $v^{\prime }_i|_{S_i}$ is a biholomorphism to $B_i$ . Let $\pi _{\widehat {S}'}:\widehat {S}' \to S$ be the trivial covering map. Since $[v_i'] \cdot a_j =\delta _{ij}$ for $j=1,\dots ,k+1$ (see Remark 4.4), the map $u^{\prime }_i:S \to X$ obtained from the tautological correspondence from $(v^{\prime }_i,\pi _{\widehat {S}'})$ satisfies $[u^{\prime }_i] \cdot D_j=\delta _{ij}$ for $j=1,\dots ,k+1$ . Therefore, $u_i'$ represents the class $[u_i]$ . Since $\pi _{\widehat {S}'}$ is a trivial covering and $\{v_i'|_{S_j}\}_{j=1}^k$ have pairwise disjoint images, the image of $u^{\prime }_i$ is disjoint from $\Delta $ so we have $[u^{\prime }_i]\cdot \Delta =0$ for $i=1,\dots ,k$ .

On the other hand, if $\widehat {S}"=\mathbb {P}^1$ , the Riemann–Hurwitz formula shows that a simple k-fold branched covering $\pi _{\widehat {S}"}: \widehat {S}" \to \mathbb {P}^1$ has $2(k-1)$ branch points because by definition, every critical point of a simple branched covering is of multiplicity $2$ and there is at most one critical point over each branch point. Let $v":\widehat {S}" \to \mathbb {P}^1$ be a biholomorphism, and $u" :\mathbb {P}^1 \to X$ be the $J_X$ -holomorphic map tautologically corresponding to $(v", \pi _{\widehat {S}"})$ . We know that $u"$ represents the class $\sum _{i=1}^{k+1} [u_i]$ because $[v"] \cdot a_j=[u"] \cdot D_j=1$ for every j. Since $v"$ is a biholomorphism, $u"(z) \in \Delta $ if and only if z is a branch point of $\pi _{\widehat {S}"}$ . The assumption that $\pi _{\widehat {S}"}$ is a simple branched covering guarantees that the intersection multiplicity between u and $\Delta $ at every branch point of $\pi _{\widehat {S}"}$ is $1$ (this fact can be checked by a local calculation). Therefore, we know that $[u"] \cdot \Delta =2(k-1)$ because $\pi _{\widehat {S}"}$ is a simple branched covering with $2(k-1)$ branch points. As a result, $[u_{k+1}] \cdot \Delta =([u"]-\sum _{i=1}^k [u_i']) \cdot \Delta =2(k-1)$ .

Proof. Let A denote the image of $\pi _2(X, \operatorname {Sym}(\underline {L})) \to H_2(X,\operatorname {Sym}(\underline {L}))$ , and $\zeta :A\to \mathbb {Z}^s$ the linear map given by

$$\begin{align*}a\mapsto(a\cdot D_1,\dots, a\cdot D_k).\end{align*}$$

Equation (26) implies that $(\zeta (u_1),\dots , \zeta (u_s))$ is the standard basis of $\mathbb {Z}^s$ . In particular, $\zeta $ is surjective. To prove that the $[u_i]$ freely generate A, there remains to prove that $\zeta $ is injective hence an isomorphism.

We have the following commutative diagram with exact rows.

In this diagram, the rows correspond to the relative long exact sequences of the pair $(X,\operatorname {Sym}(\underline {L}))$ respectively in homotopy and homology, and the vertical arrows are given by the Hurewicz map. The top-left entry is $0$ since $\operatorname {Sym}(\underline {L})$ is a torus hence has vanishing $\pi _2$ .

Let $I:=\operatorname {im}(\pi _2(X) \to \pi _2(X,\operatorname {Sym}(\underline {L})))$ and $K:=\ker (\pi _1(\operatorname {Sym}(\underline {L})) \to \pi _1(X))$ , so we have a short exact sequence

$$ \begin{align*} 0 \to I \to \pi_2(X,\operatorname{Sym}(\underline{L})) \to K \to 0. \end{align*} $$

The image of $\pi _2(X) \to H_2(X)$ is isomorphic to $\mathbb {Z}$ (see [Reference Bertram and Thaddeus4, Theorem 9.2]). Therefore, the rank of the image of $I \to H_2(X,\operatorname {Sym}(\underline {L}))$ is at most $1$ .

On the other hand, we have isomorphisms $\pi _1(X)=H_1(X)=H_1(\Sigma )$ (see, e.g., [Reference Ozsváth and Szabó58, Lemma 2.6]), $\pi _1(\operatorname {Sym}(\underline {L}))=H_1(\operatorname {Sym}(\underline {L}))= H_1(\underline {L})$ . Moreover, the map $\pi _1(\operatorname {Sym}(\underline {L})) \to \pi _1(X)$ can be identified with the map $H_1(\underline {L}) \to H_1(\Sigma )$ via the commutative diagramFootnote ¹⁵

Now, the map $H_1(\underline {L}) \to H_1(\Sigma )$ also sits inside the relative long exact sequence for the pair $(\Sigma ,\underline {L})$

$$ \begin{align*} H_2(\Sigma) \xrightarrow{f_1} H_2(\Sigma, \underline{L}) \xrightarrow{f_2} H_1(\underline{L}) \xrightarrow{f_3} H_1(\Sigma), \end{align*} $$

where $H_2(\Sigma )=\mathbb {Z}$ , $H_2(\Sigma , \underline {L})=\mathbb {Z}^s$ and $f_1$ is injective. Therefore, we have $K=\ker (f_3)=\operatorname {im}(f_2)=\operatorname {coker}(f_1)=\mathbb {Z}^{s-1}$ .

Since K is free, we have $\pi _2(X,\operatorname {Sym}(\underline {L})) \simeq I \oplus K$ . Therefore, the image of $\pi _2(X,\operatorname {Sym}(\underline {L})) \to H_2(X,\operatorname {Sym}(\underline {L}))$ (which we called A) is isomorphic to the image of a linear map $\mathbb {Z}^s \to H_2(X,\operatorname {Sym}(\underline {L}))$ . Composing with $\zeta $ , we obtain a linear map

$$\begin{align*}\mathbb{Z}^s\to A \stackrel{\zeta}{\to}\mathbb{Z}^s\end{align*}$$

which is surjective (as a composition of two surjective maps), hence an isomorphism. As a consequence, the first map $\mathbb {Z}^s \to A$ is injective. This implies that the first map is an isomorphism, and hence $\zeta $ is an isomorphism as well.

This shows that $\{[u_i]\}_{i=1}^s$ freely generates the image of $\pi _2(X,\operatorname {Sym}(\underline {L})) \to H_2(X,\operatorname {Sym}(\underline {L}))$ . Moreover, we know that the image of $I \to H_2(X,\operatorname {Sym}(\underline {L}))$ is isomorphic to $\mathbb {Z}$ . To conclude the proof, it suffices to find a continuous map $u:\mathbb {P}^1 \to X$ representing the class $\sum _{i=1}^s[u_i]$ .

We can construct u using tautological correspondence. Let $\widehat {S}=\Sigma $ and $v:\widehat {S} \to \Sigma $ be the identity map. Let $\pi _{\widehat {S}}:\widehat {S} \to \mathbb {P}^1$ be a topological k-fold simple branched covering. The map $u= v \circ \pi _{\widehat {S}}^{-1}$ satisfies $[u] \cdot D_i=1$ for all $i=1,\dots ,s$ , so we have $[u]=\sum _{i=1}^s[u_i]$ .

Proof. It suffices to prove that $\mu (u_i)=2$ for all i. Since $\mathcal {J}_{\Delta }$ is connected, it suffices to consider $J_X$ .

Let $(v_i,\pi _{\widehat {S}_i})$ tautologically correspond to $u_i$ . Recall that $\widehat {S}_i=B_i \sqcup _{j=1}^{k-\tau _i} S_j$ , $v_i|_{B_i}$ is the identity map and the $v_i|_{S_j}$ are constant maps. It follows that $u_i$ factors through the following holomorphic embedding (i.e., $\mathrm {Im}(u_i)$ lies inside the image of the following map)

(29) $$ \begin{align} \operatorname{Sym}^{\tau_i}(B_i) \times \prod_{L_j \nsubseteq \partial B_i} D^*L_j &\to \operatorname{Sym}^{k}(\Sigma)=X \end{align} $$

(30) $$ \begin{align} ([x_1,\dots,x_{\tau_i}], p_1,\dots,p_{k-\tau_i})&\mapsto [x_1,\dots,x_{\tau_i},p_1,\dots,p_{k-\tau_i}], \end{align} $$

where $D^*L_j$ is a neighborhood of $L_j \subset \Sigma $ such that $\{B_i \} \cup \{ D^*L_j\}_{L_j \nsubseteq \partial B_i}$ are pairwise disjoint. With respect to the product decomposition of the left-hand side of equation (29), we can write $u_i=(\bar {u}_i, c_1,\dots ,c_{k-\tau _i})$ where $\bar {u}_i:(S,\partial S) \to (\operatorname {Sym}^{\tau _i}(\Sigma ), \operatorname {Sym}(\partial B_i))$ and $c_i$ are constant maps. It follows that $u_i^*(TX, T\operatorname {Sym}(\underline {L}))$ has $k-\tau _i$ trivial factors, which contribute $0$ to the Maslov index. Therefore, it suffices to prove that $\mu (\bar {u}_i)=2$ . Notice that $\bar {u}_i$ tautologically corresponds to the pair $(v_i|_{B_i}, \pi _{\widehat {S}_i}|_{B_i})$ . Since $B_i$ is a planar domain, we may choose an embedding $B_i \hookrightarrow \mathbb {P}^1$ to obtain a map (of the same Maslov index) $\bar {u}_i:S \to (\operatorname {Sym}^{\tau _i}(B_i), \operatorname {Sym}(\partial B_i)) \subset (\operatorname {Sym}^{\tau _i}(\mathbb {P}^1), \operatorname {Sym}(\partial B_i))$ . By Corollary 4.10, the Maslov index of $\bar {u}_i$ is $2$ .

Proof. We use the notation of the proof of Lemma 4.17. Since $v_i|_{S_j}$ are constant maps, we have $[u_i]\cdot \Delta =[\bar {u}_i]\cdot \bar {\Delta }$ , where $\bar {\Delta }$ is the diagonal in $\operatorname {Sym}^{\tau _i}(B_i)$ . By regarding $\bar {u}_i$ as a map from S to $\operatorname {Sym}^{\tau _i}(B_i) \subset \operatorname {Sym}^{\tau _i}(\mathbb {P}^1)$ , we can apply Lemma 4.11 and equation (25) to conclude the result.

Proof. Suppose $J \in \mathcal {J}$ . By Lemma 4.12, $[u]$ is a multiple of $\sum _{i=1}^s [u_i]$ . By positivity of intersection with $D_i$ , u is a positive multiple of $\sum _{i=1}^s [u_i]$ . Hence, the result follows from Lemma 4.17 because $B_i$ being all planar implies that $s \ge 2$ .

Now, suppose $J \in \mathcal {J}_{\Delta }$ . If the image of u is contained in $\Delta $ , then u is actually $J_X$ -holomorphic and we reduce to the previous case. If the image of u is not contained in $\Delta $ , we have positivity of intersection between u and $\Delta $ , so $[u]$ is still a positive multiple of $\sum _{i=1}^s [u_i]$ . The result follows from Lemma 4.18.

Proof. It is easy to check that $\omega _X(u_i)=A_i$ (see Definition 4.14 and equation (27)). Therefore, equation (31) is a direct consequence of applying Lemmas 4.12, 4.17 and 4.18 to all the $[u_i]$ . The last sentence follows from the positivity of $\omega _X(u)$ and nonnegativity of $\eta [u] \cdot \Delta $ for a nonconstant J-holomorphic disc u such that $J \in \mathcal {J}_{\Delta }$ .

Proof. For $J \in \mathcal {J}$ , we have positivity of intersection between $D_i$ and a J-holomorphic disc u with boundary on $\operatorname {Sym}(\underline {L})$ . Therefore, Lemma 4.17 guarantees that u has nonpositive Maslov index if and only if $[u]=0$ . In this case, u is a constant map.