We first recall some basics about Lawlor necks [51][45], which are non-compact embedded exact special Lagrangians $L_{\phi,A}$ inside the standard Euclidean $\mathbb{C}^{n}$, asymptotic at infinity to the union of two planes

$$\Pi_{0}=\mathbb{R}^{n},\quad\Pi_{\phi}=(e^{i\phi_{1}},\ldots e^{i\phi_{n}})\mathbb{R}^{n},\quad 0<\phi_{i}<\pi,\quad\sum\phi_{i}=\pi.$$

Symplectic topologically, they can be viewed as a realisation of the Lagrangian handle $S^{n-1}\times\mathbb{R}$ that appears in the Lagrangian connected sum construction. This motivates the ansatz

$$L_{\phi,A}=\{(z_{1}(y)x_{1},...,z_{n}(y)x_{n}):y\in\mathbb{R},x_{k}\in\mathbb{R},x^{2}_{1}+\ldots+x_{n}^{2}=1\}.$$

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The special Lagrangian condition translates into an ODE system on the functions $z_{1}(y),\ldots z_{n}(y)$, which can be solved exactly as follows.

Let $n>2$ and $a_{1},...,a_{n}>0$, and define polynomials $p,P$ by

$$p(x)=(1+a_{1}x^{2})\ldots(1+a_{n}x^{2})-1,\quad P(x)=\frac{p(x)}{x^{2}}.$$

Define real numbers $\phi_{1},...,\phi_{n}$ and $A$ by

$$\phi_{k}=a_{k}\int_{-\infty}^{\infty}\frac{dx}{(1+a_{k}x^{2})\sqrt{P(x)}},\quad A=\int_{-\infty}^{\infty}\frac{dx}{2\sqrt{P(x)}}$$

Clearly $\phi_{k},A>0$, and elementary integration shows $\sum\phi_{i}=\pi$. This yields a 1-1 correspondence between $n$-tuples $(a_{1},\ldots,a_{n})$ with $a_{k}>0$, and $(n+1)$-tuples $(\phi_{1},\ldots,\phi_{n},A)$ with $\phi_{k}\in(0,\pi)$, $\sum\phi_{k}=\pi$ and $A>0$. Setting

$$z_{k}(y)=e^{i\psi_{k}(y)}\sqrt{a_{k}^{-1}+y^{2}},\quad\text{where }\psi_{k}(y)=a_{k}\int_{-\infty}^{y}\frac{dx}{(1+a_{k}x^{2})\sqrt{P(x)}},$$

yields the solution $(z_{1}(y),\ldots,z_{n}(y))$, hence the Lawlor necks $L_{\phi,A}$.

For fixed asymptotic planes $\Pi_{0},\Pi_{\phi}$, the Lawlor necks arise in a 1-parameter family, related to each other by the rescaling $\vec{z}\to\lambda\vec{z}$ in $\mathbb{C}^{n}$, and $A$ behaves like 2-dimensional area $A\to\lambda^{2}A$ under this scaling. One also observes that asymptotically near infinity, the Lawlor necks are graphs over $\Pi_{0}$ (resp. $\Pi_{\phi}$) of the differential $df$, where

$$|f|=O(|\vec{x}|^{2-n}),\quad|\partial^{k}f|=O(|\vec{x}|^{2-n-k}).$$

We say the Lawlor neck has asymptotic decay rate $2-n$. The upshot is that it approaches $\Pi_{0}\cup\Pi_{\phi}$ sufficiently fast.

Here is a sketch of their arguments:

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  Using the asymptotic assumption on the exact Lagrangian $L$, one can assign an analytic invariant $A(L)$ to $L$ as follows. Let $f_{L}:L\to\mathbb{R}$ be a primitive of the Liouville form $\lambda$, namely $df_{L}=\lambda|_{L}$, then $f_{L}$ converges to constants $c_{0},c_{\phi}$ at the two asymptotic ends along $\Pi_{0},\Pi_{\phi}$ respectively. Then one defines $A(L)=c_{\phi}-c_{0}$. If $L$ coincides with the Lawlor neck $L_{\phi,A}$, then $A(L)=A$.

• •

  Partially compactify $\mathbb{C}^{n}$ into a Liouville manifold identified as the plumbing $M$ of two cotangent bundles $T^{*}S^{n}$ with $T^{*}S^{n}$. Here the two copies of $S^{n}$ arise topologically as one-point compactifications of $\Pi_{0}$ and $\Pi_{\phi}$ by adding the points at infinity $\infty_{0}$ and $\infty_{\phi}$, and topologically $M$ is the union of $\mathbb{C}^{n}$ and the two cotangent fibres over $\infty_{0}$ and $\infty_{\phi}$ respectively. Under suitably fast decay condition at infinity, the unknown special Lagrangian $L$ can be compactified into an exact graded embedded Lagrangian $\bar{L}$ inside $M$. One would like to compare this to the Lagrangian $\bar{L}_{\phi,A}$ obtained by the compactification of the standard Lawlor necks $L_{\phi,A}$ inside $M$.

• •

  By analyzing the intersection pattern with the two cotangent fibres at infinity, and using the classification results of Abouzaid and Smith [1], one shows that inside $D^{b}Fuk(M)$, the Lagrangian object $\bar{L}$ is isomorphic to one of the two Lagrangian connected sums of the two $S^{n}$ with suitable gradings, and in fact the assumption on Floer degrees $\sum\phi_{i}=\pi$ singles out $\bar{L}\simeq\bar{L}_{\phi,A}\in D^{b}Fuk(M)$, the opposite surgery corresponding to $\sum\phi_{i}=(n-1)\pi$. This step needs $n\geq 3$. For contradiction, we assume $\bar{L}$ does not coincide with $\bar{L}_{\phi,A}$ for any choice of parameter $A>0$.

• •

  By a modification of the Thomas-Yau argument, one shows that after a small Hamiltation perturbation $\bar{L}^{\prime\prime}$ of $\bar{L}_{\phi,A}$, we can ensure $\bar{L}^{\prime\prime}$ is transverse to $\bar{L}$, there is no degree $0,n$ intersection points in $\bar{L}^{\prime\prime}\cap\bar{L}$ inside $\mathbb{C}^{n}$, and there is precisely one intersection point $p$ and $q$ in $\bar{L}^{\prime\prime}\cap\bar{L}$ on each of the two cotangent fibres at infinity respectively. Morever, the $D^{b}Fuk(M)$ class and the analytic invariants of $\bar{L}^{\prime\prime}$ agree with that of $\bar{L}_{\phi,A}$.

  Remark 2.8.

  The subtlety at infinity prevents one from removing degree $0,n$ intersections outside the $\mathbb{C}^{n}$ region, so one does not reach an immediate contradiction as in the Thomas-Yau argument. This technical failure is necessary, because the Lawlor necks with fixed asymptotic planes are not unique, but do arise in a 1-parameter family. It is in overcoming this technical problem that holomorphic curves appear in [40].

• •

  Now suppose the Lawlor neck is chosen with the parameter $A=A(L)$, which presumes $A(L)>0$.

  Lemma 2.7.

  [40, Thm 2.15] Assume $J$ is a generic almost complex structure on $M$ compatible with the Liouville structure. There exists a $J$-holomorphic strip $\Sigma$ with boundary on $\bar{L}$ and $\bar{L}^{\prime\prime}$ and two corners at $p$ and $q$ respectively.

  Proof.

  Consider the Floer cup product with mod 2 coefficients

  $$HF^{0}(\bar{L}^{\prime\prime},\bar{L})\otimes HF^{0}(\bar{L},\bar{L}^{\prime\prime})\to HF^{0}(\bar{L},\bar{L}),$$

  which can be identified as the cup product

  $$H^{0}(\bar{L},\bar{L})\otimes H^{0}(\bar{L},\bar{L})\to H^{0}(\bar{L},\bar{L}),\quad 1_{\bar{L}}\cup 1_{\bar{L}}=1_{\bar{L}},$$

  and thus must be nontrivial. However, at chain level this Floer product comes from the $A_{\infty}$ operation

  $$m_{2}:CF^{0}(\bar{L}^{\prime\prime},\bar{L})\otimes CF^{0}(\bar{L},\bar{L}^{\prime\prime})\to CF^{0}(\bar{L},\bar{L}),$$

  which must be nontrivial. The counting interpretation implies there are intersection points $p^{\prime}\in CF^{0}(\bar{L},\bar{L}^{\prime\prime})$ and $q^{\prime}\in CF^{0}(\bar{L}^{\prime\prime},\bar{L})\simeq CF^{n}(\bar{L},\bar{L}^{\prime\prime})^{\vee}$ and some holomorphic strip in between. Since degree $0,n$ intersection points cannot occur inside $\mathbb{C}^{n}$, they can only occur at infinity, so we must have $\{p,q\}=\{p^{\prime},q^{\prime}\}$. ∎

  Now the area of the $J$-holomorphic curve can be computed cohomologically. Using the choice of parameter $A$,

  $$\int_{\Sigma}\omega=\int_{\partial\Sigma}\lambda=\int_{p^{\prime}\to q^{\prime}}df_{\bar{L}}+\int_{q^{\prime}\to p^{\prime}}df_{\bar{L}^{\prime\prime}}=\pm(A(L)-A(L^{\prime\prime}))=\pm(A-A)=0.$$

  This contradicts the positivity of area of the holomorphic curve, which proves $L$ must coincide with $L_{\phi,A}$.

• •

  Finally one needs to a priori justify $A(L)>0$. This relies on a slightly more complicated holomorphic polygon counting argument, and the main upshot is that one can produce a nontrivial holomorphic triangle from a distinguished triangle in $D^{b}Fuk(M)$, with the three edges on $\bar{L}$, $\Pi_{0}\cup\{\infty_{0}\}$ and $\Pi_{\phi}\cup\{\infty_{\phi}\}$. Then one shows $A(L)$ has the interpretation as its area, so must be positive.

In [40] the perturbations involved in the partial compactification and the genericity of the almost complex structure makes the holomorphic curves rather difficult to visualize.¹⁴¹⁴ 14 A typical feature of Floer theory, is that completely realistic examples about holomorphic curves are also non-explicit. We now present a heuristic way to see holomorphic triangles with the edges on $\bar{L}_{\phi,A}$, $\Pi_{0}\cup\{\infty_{0}\}$ and $\Pi_{\phi}\cup\{\infty_{\phi}\}$, by restricting attention to $\mathbb{C}^{n}$ with the standard complex structure, and we imagine the two vertices $\infty_{0},\infty_{\phi}$ as the intersection points at infinity.

We choose any $\vec{x}=(x_{1},\ldots x_{n})\in S^{n-1}$. Assume first that $x_{k}\neq 0$. Then coordinatewise, we have a real curve in $\mathbb{C}$ swept out by $z_{k}(y)x_{k}$ as $y$ varies from $-\infty$ to $+\infty$, and two straight rays emanating from the origin defined by $\text{sign}(x_{k})\mathbb{R}_{+}$ and $\text{sign}(x_{k})e^{i\phi_{k}}\mathbb{R}_{+}$. Inside $\mathbb{C}$, these three real curves enclose a noncompact holomorphic triangle, with one vertex at the origin, and two idealized intersection points at the infinity of $\text{sign}(x_{k})\mathbb{R}_{+}$ and $\text{sign}(x_{k})e^{i\phi_{k}}\mathbb{R}_{+}$. In the product space $\mathbb{C}^{n}$, this gives rise to a holomorphic triangle $\Sigma_{\vec{x}}$ with boundary on $L_{\phi,A},\Pi_{0},\Pi_{\phi}$, and corners at $0,\infty_{0},\infty_{\phi}$. Now in case some $x_{k}=0$, there is still a holomorphic triangle in the product space that makes sense; the $k$-th projection of this triangle is simply the origin. What happens when $x_{k}\to 0$, is simply that the $k$-th projection $\pi_{k}(\Sigma_{\vec{x}})$ becomes very thinly concentrated near the two rays $\text{sign}(x_{k})\mathbb{R}_{+}$ and $\text{sign}(x_{k})e^{i\phi_{k}}\mathbb{R}_{+}$, and its area shrinks to zero. Morever for any given $\epsilon$, the subset $\Sigma_{\vec{x}}\cap\pi_{k}^{-1}(|z|>\epsilon)$ disappears into the infinity of $\mathbb{C}^{n}$ as $x_{k}\to 0$. Thus when we restrict to any compact subset of $\mathbb{C}^{n}$, the holomorphic discs behave continuously as $x_{k}\to 0$.

Using any of the holomorphic triangles $\Sigma_{\vec{x}}$ parametrised by $S^{n-1}$, we can calculate its area cohomologically by

$$\int_{\Sigma}\omega=\int_{\partial\Sigma}\lambda=\int_{\partial\Sigma\cap L_{\phi,A}}df_{L_{\phi,A}}=A,$$

which is the intuitive explanation of why $A$ must be positive, an important ingredient of [40].

We want to draw attention also to a different aspect not explicit in [40]: that these holomorphic triangles naturally arise in an $(n-1)$-dimensional moduli, rather than as isolated triangles. Consequently, the universal family of such holomorphic triangles is naturally $(n+1)$-dimensional. A generic point on $\Pi_{0}\cup\Pi_{\phi}\cup L_{\phi,A}$ is swept out precisely once by some $\partial\Sigma_{\vec{x}}$. When the orientations are taken into account, then the total space of this universal family gives rise to an $(n+1)$-dimensional integration current, which provides a bordism current between the integration cycles of $L_{\phi,A}$ and $\Pi_{0}\cup\Pi_{\phi}$. Producing bordism currents via universal families of holomorphic curves will be essential to our proposals concerning the Thomas-Yau conjecture.