translating coordinates, we may assume that $P$ contains the
$0\in\mathbb{R}^{3}$ and that $0$ lies in the interior of $\gamma_{0}$. Take a
large circle $\gamma_{R}$ of radius $R$ centred at $0$ in $P$ and, given
$c>0$, consider the family $\gamma_{\sqrt{R^{2}-2ct}}$ of circles depending on
$t$. For $c=1$ this curve is a solution to the usual curve shortening flow
(3.8) in the plane. In a fixed large annular region sufficiently far from the
origin we know that $\phi^{-1}$ is uniformly bounded from both below and
above. Thus, by choosing $c$ appropriately we can use these circles
$\gamma_{\sqrt{R^{2}-2ct}}$ as barriers in this region to ensure (again by the
maximum principle) that given an embedded curve satisfying (6.1) there exists
a compact set in $P$ so that the flow remains in this set. ∎

Lemmas 6.2 and 6.3 have the following important corollary.

######  Corollary 6.4.

Let $L_{0}$ be an embedded circle-invariant Lagrangian torus in an ALE or ALF
hyperkähler 4-manifold $X$ given by the Gibbons–Hawking ansatz as in Examples
2.4 or 2.6. Suppose further that the planar curve $\gamma_{0}=\pi(L_{0})$ lies
in a plane containing no singularities of $\phi$. Then, the solution
$\gamma_{t}$ to (6.1) starting at $\gamma_{0}$ shrinks to a point in finite
time, so the corresponding solution $L_{t}=\pi^{-1}(\gamma_{t})$ remains an
embedded torus until it collapses to a circle orbit in finite time.

Notice that there is no assumption on the initial Lagrangian torus other than
it is embedded and circle-invariant.

###### Proof.

Let $P$ be the plane containing $\gamma_{0}$ and hence the flow $\gamma_{t}$.
Since $P$ contains no singularities of $\phi$, $\phi^{-1}$ is uniformly
bounded below on $P$ by $a>0$. Moreover, by Lemma 6.3, the flow stays in a
bounded region in $P$ and hence $\phi^{-1}$ is bounded above by $b>0$ in this
region.

Therefore, the flow $\gamma$ is uniformly controlled by the curve shortening
flow (3.9) in the Euclidean plane $P$. Hence, by Grayson’s Theorem [Grayson],
the flow $\gamma$ will shrink to a point in finite time. ∎

In the setting of Corollary 6.4, the flow of planar curves starting at
$\gamma_{0}$ will become convex and then shrink to a round point in finite
time, i.e. it has a Type I singularity whose Type I blow-up is the shrinking
circle. Therefore, the Lagrangian mean curvature flow starting at $L_{0}$ has
a finite-time Type I singularity whose Type I blow-up is a shrinking cylinder
$S^{1}\times\mathbb{R}$. We now discuss flows where other singularities can
occur.

######  Example 6.4 (Clifford torus).

The Clifford torus in $\mathbb{R}^{4}$ is an embedded circle-invariant
Lagrangian torus which can be defined as $\pi^{-1}(\gamma_{0})$, where
$\gamma_{0}$ is a planar circle of radius $1$ centred at the origin in
$\mathbb{R}^{3}$. If we assume that

$\gamma_{t}=r(t)\gamma_{0}$

then

$\gamma_{t}^{\prime\prime}=-\frac{1}{r(t)}\gamma_{0}.$

Therefore, if we consider the Lagrangian mean curvature flow in Euclidean
$\mathbb{R}^{4}$ starting at the Clifford torus, we see by Example 2.2 and
(6.1) that the flow becomes the ODE:

$\dot{r}(t)=-\frac{2r(t)}{r(t)}=-2.$

Hence,

$r(t)=r(0)-2t.$

We deduce the well-known fact that the Clifford torus is a self-shrinker for
Lagrangian mean curvature flow, so it has a Type I singularity at the origin
in finite time and is its own Type I blow-up.

If instead we consider the Lagrangian mean curvature flow in Taub–NUT
$\mathbb{R}^{4}$ starting at the Clifford torus, we see by Example 2.5 and
(6.1) that the flow becomes:

$\dot{r}(t)=-\frac{1}{r(t)(m+\frac{1}{2r(t)})}=-\frac{2}{2mr(t)+1}.$

Hence, $r(t)$ is the positive solution to the quadratic equation

$mr(t)^{2}+r(t)=mr(0)^{2}+r(0)-2t.$

Notice that this reduces to the Euclidean case when $m=0$. One observes that
again the flow starting at the Clifford torus in Taub–NUT shrinks to the
origin in finite-time. Moreover, it is clear that the flow has a Type I
singularity whose Type I blow-up is the usual Clifford torus in
$\mathbb{R}^{4}$.

###  6.2. Long-time existence and stability (Proof of Theorem 6.1)

As we have seen, circle-invariant Lagrangian mean curvature flow is equivalent
to the flow (6.1) for curves $\gamma$ in a plane, which we may identify with
$\mathbb{R}^{2}$ (and we restrict $\phi$ to this plane). By Lemma 5.9, any
Lagrangian in Theorem 6.1 is of the form $L=\pi^{-1}(\gamma)$, where $\gamma$
is an embedded planar arc which has singularities $p_{1},p_{2}$ of $\phi$ as
endpoints and meet no other singularities $p_{i}$ for $i>2$ of $\phi$. In
proving Theorem 6.1, we may therefore restrict ourselves to the Dirichlet
problem (O) above for (6.1) for embedded arcs $\gamma$ in $\mathbb{R}^{2}$
which have endpoints at singularities $p_{1}$ and $p_{2}$ of $\phi$ in
$\mathbb{R}^{2}$, with initial curve $\gamma_{0}=\pi(L_{0})$.

We start with an elementary, but important observation.

######  Lemma 6.5.

If the curve $\gamma_{0}$ in $\mathbb{R}^{2}$ is almost calibrated (i.e. the
variation of the angle it makes with a fixed axis is less than $\pi$), then
the variation of the angle for the solution of (6.1) is non-increasing along
the flow, and $\gamma_{t}$ remains almost calibrated as long as the flow
exists.

###### Proof.

Recall that the Lagrangian angle $\beta$ of a zero Maslov circle-invariant
Lagrangian $L$ agrees with the angle the curve $\gamma=\pi(L)$ in
$\mathbb{R}^{2}$ makes with a fixed axis (see $\S$5.2.1). Since $\tau(L)=0$,
by translating $\beta$ by a constant if necessary, we can assume that the
almost calibrated condition is given by $\cos\beta>0$ (i.e.
$\mathop{\mathrm{Re}}\Omega|_{L}>0$). As $L_{0}$ is compact there exists some
$\varepsilon>0$ such that $\cos\beta_{0}\geq\varepsilon$.

Since the Lagrangian angle satisfies the heat equation along the flow (see
e.g. [ThomasYau, Lemma 2.3]), the maximum principle applied to the compact
manifold $L_{t}$ without boundary shows that the almost calibrated condition
is preserved as the minimum of $\cos\beta_{t}$ is non-decreasing. ∎

######  Remark 6.5.

To obtain the same result as Lemma 6.5 in [ThomasYau, Lemma 7.8], the authors
needed to use the stability assumption and a stronger initial bound on the
Lagrangian angle, together with a much more involved argument. However, one
can see that their proof can be greatly simplified, as here, in the case where
the Lagrangians are surfaces.

Lemma 6.5 gives two important consequences. The first is the following.

######  Lemma 6.6.

If the curve $\gamma_{0}$ in $\mathbb{R}^{2}$ is almost calibrated, then
$|\nabla^{\perp}_{\mathbb{R}^{3}}\log\phi|_{\mathbb{R}^{3}}^{2}$ remains
bounded as long as the curve does not reach any other singularities of $\phi$.

###### Proof.

Recall that the flow $\gamma$ stays in a compact region in $\mathbb{R}^{2}$ by
Lemma 6.3. If we take a compact subset $K$ of this region not containing any
singularities of $\phi$, then $\phi$ is uniformly bounded above and below away
from zero, and so $|\nabla_{\mathbb{R}^{3}}\log\phi|^{2}_{\mathbb{R}^{3}}$ is
bounded in $K$. Therefore, if we assume that $\gamma$ does not reach any
singularities $p_{i}$ of $\phi$ for $i>2$, we need only show that
$|\nabla^{\perp}_{\mathbb{R}^{3}}\log\phi|_{\mathbb{R}^{3}}^{2}$ remains
bounded at the endpoints $p_{1}$, $p_{2}$ of $\gamma$. Notice by (3.2) that
the dominant term in $\phi$ near each $p_{i}$ is rotationally symmetric around
$p_{i}$. Therefore for
$|\nabla^{\perp}_{\mathbb{R}^{3}}\log\phi|_{\mathbb{R}^{3}}^{2}$ to remain
bounded we must ensure that $\gamma$ does not wind around $p_{1}$ or $p_{2}$.

Since $\gamma_{0}$ is almost calibrated, we can choose two rays $\ell_{1}$ and
$\ell_{2}$ in $\mathbb{R}^{2}$ emanating from $p_{1}$ so that $\gamma_{0}$
does not meet $\ell_{1}$ and $\ell_{2}$ except at $p_{1}$, and the angle
between $\ell_{1}$ and $\ell_{2}$ is strictly less than $\pi$. As $\ell_{1}$
and $\ell_{2}$ are fixed by the flow (6.1) (they define minimal Lagrangians in
$X$), they act as barriers, and so $\gamma_{t}$ must not meet $\ell_{1}$ and
$\ell_{2}$ except at $p_{1}$ for all $t$. Therefore, $\gamma_{t}$ cannot wind
around $p_{1}$ and
$|\nabla^{\perp}_{\mathbb{R}^{3}}\log\phi|^{2}_{\mathbb{R}^{3}}$ must remain
bounded near $p_{1}$ along the flow. A similar argument works at $p_{2}$. ∎

######  Remark 6.6 (Flow stability of curves).

Let $L$ be an embedded, compact, almost calibrated, circle-invariant
Lagrangian in $(X,\omega,\Omega)$. Without loss of generality, suppose
$\pi(L)=\gamma$ is an embedded arc in the $\mu_{3}=0$ plane and
$\Omega=\omega_{1}+i\omega_{2}$. Suppose that the initial point of $\gamma$ is
$(0,0,0)$ and its final point is $(x,y,0)$. We find that

$\int_{L}\Omega=2\pi\int_{\gamma}d\mu_{1}+id\mu_{2}=2\pi(x+iy).$

Thus,

$e^{i\tau(L)}=\frac{x+iy}{\sqrt{x^{2}+y^{2}}}$

and hence $\tau(L)$ is the angle the straight line $\overline{\gamma}$ with
the same endpoints as $\gamma$ makes with the $\mu_{1}$-axis. Moreover,

$\int_{L}e^{-i\tau(L)}\Omega=2\pi\sqrt{x^{2}+y^{2}}=2\pi\mathrm{Length(\overline{\gamma})}.$

Note that, by Proposition 4.4, $\pi^{-1}(\overline{\gamma})$ is the area-
minimizer in $[L]\in H_{2}(M,\mathbb{Z})$ and
$2\pi\mathrm{Length(\overline{\gamma})}=\mathrm{Area}(\pi^{-1}(\overline{\gamma}))$.
We also see that $\tau(L)=0$ if and only if the endpoints of $\gamma$ lie on
the $\mu_{1}$-axis.

We deduce that, in the circle-invariant setting, the notion of flow stability
as in Definition 6.2 can be re-cast as follows. For any decomposition of
$\gamma$ as $\gamma_{1}\\#\gamma_{2}$, where $\gamma_{1},\gamma_{2}$ are
almost calibrated, we let $\overline{\gamma_{1}},\overline{\gamma_{2}}$ denote
the straight-lines with the same endpoints as $\gamma_{1}$ and $\gamma_{2}$
respectively. We assume that the endpoints of $\gamma$ are on the
$\mu_{1}$-axis and we let $\beta$, $\overline{\beta}_{1}$,
$\overline{\beta}_{2}$ denote the angles that $\gamma$,
$\overline{\gamma}_{1}$, $\overline{\gamma}_{2}$ make with the $\mu_{1}$-axis
respectively. Then $\gamma$, equivalently $L=\pi^{-1}(\gamma)$, is flow stable
if for all decompositions $\gamma=\gamma_{1}\\#\gamma_{2}$ we have

  * (a)

$[\overline{\beta_{1}},\overline{\beta}_{2}]\nsubseteq(\inf_{\gamma}\beta,\sup_{\gamma}\beta)$,
or

  * (b)

$\mathrm{Length}(\gamma)<\mathrm{Length}(\overline{\gamma_{1}})+\mathrm{Length}(\overline{\gamma_{2}})$

One can picture condition (a) in Figure 5.2. Condition (b) should be compared
with Remark 5.16.

For flow stable initial curves for the flow, we have a very important
consequence of the observations in Remark 6.6.

######  Lemma 6.7.

Let $p_{i}$ denote the singularities of $\phi$ as in Examples 2.4 and 2.6, and
recall that $p_{1},p_{2}$ are the fixed endpoints of the curve along flow.
There exists $\delta>0$ so that the flow (6.1) of curves $\gamma$ in
$\mathbb{R}^{2}$ starting at a flow stable curve remains outside of
$\cup_{i>2}\overline{B_{\delta}(p_{i})}$ for all time for which the flow is
defined. Hence, the flow (6.1) exists as long as $\phi^{-1}\kappa^{2}$ remains
bounded.

###### Proof.

Let $\gamma_{0}$ be the initial flow stable curve and let $\overline{\gamma}$
be the straight line between the endpoints $p_{1}$ and $p_{2}$ of
$\gamma_{0}$. By the reformulation of flow stability in Remark 6.6, we see
that the singularities $p_{i}$ for $i>2$ cannot lie in the interior of the
region bounded by $\gamma_{0}$ and $\overline{\gamma}$.

Suppose that $\gamma_{0}$ is flow stable in the sense of condition (a) in
Remark 6.6. Since the variation of the angle of $\gamma$ is non-increasing
along the flow (6.1) by Lemma 6.5, $\gamma$ satisfies (a) for all time for
which the flow exists. Then $p_{i}$ for $i>2$ can never lie in the interior of
the region bounded by $\gamma$ and $\overline{\gamma}$, and we deduce that
$\gamma$ must remain outside of a compact region containing the $p_{i}$ for
$i>2$.

Suppose instead that $\gamma_{0}$ is flow stable in the sense of condition (b)
in Remark 6.6. Lagrangian mean curvature flow is the gradient flow for the
area of the Lagrangian $\pi^{-1}(\gamma)$, so its area is decreasing along the
flow and hence, by (4.1), the length of $\gamma$ is decreasing along (6.1).
Therefore, $\gamma$ satisfies (b) for all time for which the flow exists, and
we again conclude as before that $\gamma$ must lie outside of
$\cup_{i>2}\overline{B_{\delta}(p_{i})}$ for some $\delta>0$.

Since we have established that the flow $\gamma$ never reaches any other
singularities of $\phi$, the final statement then follows from Lemmas 6.3 and
6.6. ∎

######  Remark 6.7.

Notice that if we bound the curvature $\kappa$ then we bound
$\phi^{-1}\kappa^{2}$ since $\phi^{-1}$ is bounded in the compact region in
which the flow is taking place. However, it is important to know that we
require $\phi^{-1}\kappa^{2}$ to blow up, not just $\kappa$ to blow up, for a
singularity to form. This is a key point where we differ from [ThomasYau], who
only assume $\kappa$ blows up at the singularity.

We now need to understand the possible blow-up behaviour of the flow. Suppose
we parametrise the flowing curves $\gamma=\gamma_{t}(s)$ by arclength so that
$s=0$ at $p_{1}$. Let $\kappa=\kappa_{t}(s)$ denote the curvature of $\gamma$.
Suppose the flow has a singularity at time $T$. Then there must exist a
sequence of spacetime points $(s_{i},t_{i})$ with $t_{i}\to T$ as $i\to\infty$
such that if $x_{i}=\gamma(s_{i})$ and
$\lambda_{i}=\phi(x_{i})^{-1/2}|\kappa_{t_{i}}(s_{i})|$ defined by

$\lambda_{i}=\max\\{\phi(\gamma_{t}(s))^{-1/2}|\kappa_{t}(s)|:s\geq 0,t\leq
t_{i}\\}$

satisfies $\lambda_{i}\to\infty$ as $i\to\infty$. By a diagonal argument or
taking a subsequence we can assume further that $(s_{i},t_{i})$ also
represents a maximum for $|\kappa_{t}(s)|$ for $t\leq t_{i}$.

We now perform the standard blow-up analysis and define curves
$\hat{\gamma}^{i}_{t}$ by

$\hat{\gamma}^{i}_{t}=\lambda_{i}(\gamma_{t_{i}+\lambda_{i}^{-2}t}-x_{i})$

for $t\in[-\lambda_{i}^{2}t_{i},\lambda_{i}^{2}(T-t_{i}))$. The flow cannot
have a Type I singularity, since any such singularity in our setting would be
modelled by a smooth self-shrinker, but by [Wang] any non-trivial smooth self-
shrinker would have angle variation more than $\pi$ (in fact, at least
$2\pi$), and so is excluded by Lemma 6.5. Thus, we have must have a Type II
singularity, i.e. $\lambda_{i}^{2}(T-t_{i})\to\infty$.

The curves $\hat{\gamma}^{i}$ satisfy the equation

(6.2) $\frac{\partial}{\partial
t}\hat{\gamma}^{i}=\left(m+\sum_{j=1}^{k}\frac{1}{|\lambda_{i}^{-1}\hat{\gamma}^{i}+x_{i}-p_{j}|}\right)^{-1}(\hat{\gamma}^{i})^{\prime\prime},$

using the notation of Examples 2.4 and 2.6.

Since the flowing curves are compact and remain so, we have that the points
$x_{i}=\gamma_{t_{i}}(s_{i})$ admit a convergent subsequence converging to a
point $x_{\infty}$, say. There are two cases: when the blow-up happens in the
interior of the curve (i.e. $x_{\infty}\notin\\{p_{1},p_{2}\\}$) or at $p_{1}$
or $p_{2}$. We begin with the first.

######  Lemma 6.8.

In the setting above, if there is a singularity at time $T$, then $x_{\infty}$
must be $p_{1}$ or $p_{2}$.

###### Proof.

Suppose that $x_{\infty}$ is not $p_{1}$ or $p_{2}$. Then, also using Lemma
6.7, there exists some $\delta>0$ so that for $i$ sufficiently large, we have
that $B_{1/\sqrt{\lambda_{i}}}(x_{i})$ is disjoint from
$\cup_{j=1}^{k}\overline{B_{\delta}(p_{j})}$. Restricting to the balls
$B_{1/\sqrt{\lambda_{i}}}(x_{i})$, our curves $\hat{\gamma}^{i}_{t}$ are
defined on the ball of radius $B_{\sqrt{\lambda_{i}}}(0)$. Therefore,
$|\lambda_{i}^{-1}\hat{\gamma}^{i}_{t}|\to 0$ on this sequence of balls, and
so (passing to a subsequence as necessary), from (6.2) we will obtain an
eternal solution $\hat{\gamma}^{\infty}_{t}$ to

$\frac{\partial}{\partial
t}\hat{\gamma}^{\infty}_{t}=\left(m+\sum_{j=1}^{k}\frac{1}{|x_{\infty}-p_{j}|}\right)^{-1}(\hat{\gamma}^{\infty})^{\prime\prime}=c(\hat{\gamma}^{\infty}_{t})^{\prime\prime}$

on $\mathbb{R}^{2}$, where $c>0$ is constant. Since $\hat{\gamma}^{\infty}$
satisfies the curve shortening flow on $\mathbb{R}^{2}$ up to a constant
reparametrization of time, the blow-up analysis for the curve shortening flow
is valid (see e.g. [GageHamilton]) and shows that a singularity can only
develop if the variation of the angle of $\hat{\gamma}^{\infty}_{t}$ is at
least $\pi$. This possibility is excluded by Lemma 6.5 and so the result
follows. ∎

Now we know, by Lemma 6.8, that any singularity of the flow must occur,
without loss of generality, at $p_{1}$. By changing coordinates we can assume
$p_{1}=0$. Therefore, our points $x_{i}$ must satisfy $|x_{i}|\to 0$ as
$i\to\infty$, and thus $\phi(x_{i})^{-1}\approx|x_{i}|$ for $i$ large.
However, we also have $\phi(x_{i})^{-1}\kappa_{t_{i}}(s_{i})^{2}\to\infty$.
There are therefore two possibilities: there is a subsequence so that
$\phi(x_{i})^{-1}|\kappa_{t_{i}}(s_{i})|\to\infty$ (i.e. the points move
“slowly” towards the singularity) or $\phi(x_{i})^{-1}|\kappa_{t_{i}}(s_{i})|$
is bounded (so the points $x_{i}$ move “quickly” towards the singularity). We
now exclude each case in turn.

######  Lemma 6.9.

In the setting above, there does not exist a subsequence so that
$\phi(x_{i})^{-1}|\kappa_{t_{i}}(s_{i})|\to\infty$.

###### Proof.

Suppose not. Then consider the balls $B(x_{i},\frac{1}{2}|x_{i}|)$ for $i$
sufficiently large so that they do not contain $p_{1}=0$ (which is clear by
construction) or $p_{2}$. Translating $x_{i}$ to the origin and rescaling by
$|\kappa_{t_{i}}(s_{i})|$ (not by $\lambda_{i}$), we obtain balls
$B(0,\frac{1}{2}|x_{i}\kappa_{t_{i}}(s_{i})|)$ which will cover
$\mathbb{R}^{2}$ as $i\to\infty$ since $|x_{i}|\approx\phi(x_{i})^{-1}$.
Moreover, using $|\kappa_{t_{i}}(s_{i})|$ in our blow-up sequence instead of
$\phi(x_{i})^{-1/2}|\kappa_{t_{i}}(s_{i})|$ we obtain a sequence of curves
$\hat{\gamma}^{i}$ with curvature $1$ at the origin for all $i$. Taking the
limit as $i\to\infty$ then gives a smooth curve $\hat{\gamma}^{\infty}$ in
$\mathbb{R}^{2}$ with curvature $1$ at $0$. However, the corresponding blow-up
sequence $\hat{L}^{i}$ of Lagrangian mean curvature flows (using
$|\kappa_{t_{i}}(s_{i})|$) will have a limit $\hat{L}^{\infty}$ which is a
plane, since we are blowing up at a faster rate (since $\phi(x_{i})^{-1/2}\to
0$ as $i\to\infty$) than the norm of the second fundamental form, recalling
the proof of Lemma 4.6. This contradicts the fact that the projection of
$\hat{L}^{\infty}$ to $\mathbb{R}^{2}$, which is $\hat{\gamma}^{\infty}$, is
not a straight line. ∎

######  Lemma 6.10.

In the notation above, we cannot have that
$\phi(x_{i})^{-1}|\kappa_{t_{i}}(s_{i})|$ is bounded.

###### Proof.

If $\phi(x_{i})^{-1}|\kappa_{t_{i}}(s_{i})|$ is bounded then for $i$
sufficiently large, there is some constant $A$ so that

(6.3) $|x_{i}|<\max\left\\{A|\kappa_{t_{i}}(s_{i})|^{-1},1\right\\},$

since $|x_{i}|\approx\phi(x_{i})^{-1}$. We may then consider the balls
$B(x_{i},A\phi(x_{i})|\kappa_{t_{i}}(s_{i})|^{-1})$ which will contain the
origin since, by (6.3), $|x_{i}|<1$ whereas
$A\phi(x_{i})|\kappa_{t_{i}}(s_{i})|^{-1}>1$ for $i$ sufficiently large.
Rescaling by $\lambda_{i}=\phi(x_{i})^{-1/2}|\kappa_{t_{i}}(s_{i})|$ and
translating by $x_{i}$, we obtain balls $B(0,A\phi(x_{i})^{1/2})$ which will
cover $\mathbb{R}^{2}$ as $i\to\infty$. Moreover, the image $y_{i}$ of $0$
under this transformation will lie at distance
$\phi(x_{i})^{-1/2}|x_{i}||\kappa_{t_{i}}(s_{i})|$ from $0$, which tends to
$0$ as $i\to\infty$ by (6.3).

By construction, above the origin the norm of the second fundamental form of
the blow-up sequence $\hat{L}^{i}=\pi^{-1}(\hat{\gamma}^{i})$ of Lagrangian
mean curvature flows is $1$. Sending $i\to\infty$ we obtain, as a limit, a
non-trivial, smooth, embedded, almost calibrated, ancient solution
$\hat{L}^{\infty}$ to Lagrangian mean curvature flow in $\mathbb{R}^{4}$,
which projects to a curve $\hat{\gamma}^{\infty}$ through $0$. Moreover, since
$\pi^{-1}(y_{i})$ is a point in $\hat{L}^{i}$ for all $i$ and $|y_{i}|\to 0$
as $i\to\infty$, we have that $\pi^{-1}(0)\in\hat{L}^{\infty}$ is a point.
Hence, $\hat{L}^{\infty}$ cannot topologically be a cylinder
$S^{1}\times\mathbb{R}$. We also know that $\hat{L}^{\infty}$ must be an exact
Lagrangian by [LambertLotaySchulze].

We now observe that, since $\hat{\gamma}^{\infty}$ is almost calibrated, it
must be graphical over a straight line, and its blow-down must be at most two
(multiplicity one) rays emanating from the origin. In particular, the blow-
down cannot be a multiplicity two ray by the almost calibrated condition. This
means that the blow-down of $\hat{L}^{\infty}$ is either a single plane or a
pair of planes intersecting transversely at a point. The exact, almost
calibrated, ancient solutions to Lagrangian mean curvature flow in
$\mathbb{R}^{4}$ whose blow-downs are planes were classified in
[LambertLotaySchulze]. The only such ancient solutions whose blow-down is a
single plane is the plane itself, which is ruled out because
$\hat{L}^{\infty}$ is non-trivial. If instead the blow-down is a pair of
transverse planes, the only possibilities are the Lawlor neck (a special
Lagrangian described, for example, in [LambertLotaySchulze]) or the pair of
planes themselves. The pair of planes is again excluded because
$\hat{L}^{\infty}$ is non-trivial, and the Lawlor neck is excluded because it
is topologically $S^{1}\times\mathbb{R}$.

We have therefore reached our desired contradiction, and the result follows. ∎

Combining our lemmas gives our long-time existence and convergence result.

######  Corollary 6.11.

The flow (6.1) starting at a flow stable curve with endpoints $p_{1}$ and
$p_{2}$ exists for all time and converges smoothly to the straight line
connecting $p_{1}$ and $p_{2}$.

###### Proof.

The long-time existence of the flow (6.1) of curves $\gamma$ follows directly
from Lemmas 6.7–6.10.

It is then a standard argument, as one may see in [ThomasYau], using the fact
that the Lagrangian angle satisfies the heat equation and that circle-
invariance is preserved along Lagrangian mean curvature flow, to deduce that
the Lagrangian mean curvature flow $L=\pi^{-1}(\gamma)$ converges smoothly to
a circle-invariant special Lagrangian $L_{\infty}=\pi^{-1}(\gamma_{\infty})$.
Theorem 5.8 then gives that $\gamma_{\infty}$ is the straight line between
$p_{1}$ and $p_{2}$. ∎

Corollary 6.11 completes the proof of Theorem 6.1.

##  Appendix A Curvature of special Lagrangian spheres

In this appendix, we prove the following sufficient condition for positive
curvature of the induced metric of a special Lagrangian sphere in a
hyperkähler 4-manifold $X$ given by the Gibbons–Hawking ansatz, with harmonic
function $\phi$ as in Examples 2.4 or 2.6 with singularities
$p_{1},\ldots,p_{k}$.

######  Proposition A.1.

Let $\gamma\subseteq\mathbb{R}^{3}$ be a straight line between $p_{1}$ and
$p_{2}$, let $2d=\text{\emph{Length}}(\gamma)$, and let $q$ by the midpoint of
$\gamma$. Suppose that the Euclidean distance from $p_{i}$ to $q$ is strictly
greater than $sd$ for $s\geq\max\\{4,\sqrt{(k-2)/2}\\}$ and all $i>2$. Then
$\pi^{-1}(\gamma)$ is a smooth minimal $2$-sphere and its induced metric is
positively curved.

Up to a rigid motion of $\mathbb{R}^{3}$ we may assume that $\gamma$ is a
straight line connecting $p_{\pm}=(0,0,\pm a)$ and we re-label the remaining
singularities of $\phi$ by $q_{1},\ldots,q_{k-2}$. Then, using $r_{l}$ to
denote the Euclidean distance to the point $q_{l}$, at the points of $\gamma$,
where $\mu_{1}=0=\mu_{2}$ and $\mu_{3}\in(-a,a)$, we find that

(A.1)
$\phi=m+\sum_{l=1}^{k-2}\frac{1}{2r_{l}}+\frac{1}{2(\mu_{3}+a)}-\frac{1}{2(\mu_{3}-a)}=\tilde{\phi}+\frac{a}{a^{2}-\mu_{3}^{2}}.$

######  Lemma A.2.

On $\gamma$, the function $\tilde{\phi}=m+\sum_{l=1}^{k-2}\frac{1}{2r_{l}}$
satisfies

$|\partial_{\mu_{3}}\tilde{\phi}|\leq\sum_{l=1}^{k-2}\frac{1}{2r_{l}^{2}},\quad\text{and}\quad\partial^{2}_{\mu_{3}}\tilde{\phi}=\sum_{l=1}^{k-2}\frac{1}{r_{l}^{3}}>0.$

###### Proof.

Using $\partial_{\mu_{3}}r_{l}=\tfrac{\mu_{3}-q_{l}^{3}}{r_{l}}$, where
$q_{l}^{3}$ denotes the $\mu_{3}$-component of $q_{l}$, we find from (A.1)
that

$\partial_{\mu_{3}}\tilde{\phi}=-\sum_{l=1}^{k-2}\frac{\partial_{\mu_{3}}r_{l}}{2r_{l}^{2}}=-\sum_{l=1}^{k-2}\frac{\mu_{3}-q_{l}^{3}}{2r_{l}^{3}}.$

Given that
$r_{l}=\sqrt{(q_{l}^{1})^{1}+(q_{l}^{2})^{2}+(\mu_{3}-q_{l}^{3})^{2}}\geq|\mu_{3}-q^{3}_{l}|$
on $\gamma$, we deduce that

$|\partial_{\mu_{3}}\tilde{\phi}|\leq\sum_{l=1}^{k-2}\frac{|\mu_{3}-q_{l}^{3}|}{2r_{l}^{3}}\leq\sum_{l=1}^{k-2}\frac{1}{2r_{l}^{2}}$

as claimed. As for the second derivatives of $\tilde{\phi}$, we find

$\partial^{2}_{\mu_{3}}\tilde{\phi}=-\sum_{l=1}^{k-2}\left(\frac{1}{2r_{l}^{3}}-3\frac{\mu_{3}-q^{3}_{l}}{2r_{l}^{4}}\partial_{\mu_{3}}r_{l}\right)=-\sum_{l=1}^{k-2}\frac{1}{2r_{l}^{3}}+\frac{3}{2}\sum_{l=1}^{k-2}\frac{(\mu_{3}-q^{3}_{l})^{2}}{r_{l}^{5}}=\sum_{l=1}^{k-2}\frac{1}{r_{l}^{3}},$

which gives the result. ∎

An orthonormal coframing of $\pi^{-1}(\gamma)$ for the induced metric is given
by

$f^{0}=\phi^{-1/2}\eta,\ \ f^{1}=\phi^{1/2}d\mu_{3}.$

The Cartan structure equations yield that the connection $1$-form $\alpha$ is
determined by

$df^{0}=-\alpha\wedge f^{1},\ \ df^{1}=\alpha\wedge f^{0},$

and the curvature $2$-form by

$d\alpha=Kf^{0}\wedge f^{1},$

where $K$ is the Gauss curvature of $\pi^{-1}(\gamma)$. We compute

$df^{0}=-\tfrac{1}{2}\phi^{-3/2}(\partial_{\mu_{3}}\phi)d\mu_{3}\wedge\eta=\tfrac{1}{2}\phi^{-2}(\partial_{\mu_{3}}\phi)\eta\wedge
f^{1}\quad\text{and}\quad df^{1}=0.$

We deduce that

$\alpha=-\frac{1}{2\phi^{2}}\frac{\partial\phi}{\partial\mu_{3}}\eta.$

Thus

$d\alpha=-\frac{\partial}{\partial\mu_{3}}\left(\frac{1}{2\phi^{2}}\frac{\partial\phi}{\partial\mu_{3}}\right)d\mu_{3}\wedge\eta=-\frac{\partial^{2}}{\partial\mu_{3}^{2}}\left(\frac{1}{2\phi}\right)f^{0}\wedge
f^{1},$

and so the Gaussian curvature of $\pi^{-1}(\gamma)$ is given by

(A.2)
$K=-\frac{\partial^{2}}{\partial\mu_{3}^{2}}\left(\frac{1}{2\phi}\right).$

Using our formula (A.1) for $\phi$ we find that

$\phi^{-1}=\frac{a^{2}-\mu_{3}^{2}}{a+\tilde{\phi}(a^{2}-\mu_{3}^{2})}.$

From this we compute

$\displaystyle\frac{\partial}{\partial\mu_{3}}\left(\frac{1}{\phi}\right)$
$\displaystyle=\frac{-2\mu_{3}(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))+2\tilde{\phi}\mu_{3}(a^{2}-\mu_{3}^{2})-(\partial_{\mu_{3}}\phi)(a^{2}-\mu_{3}^{2})^{2}}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{2}}$
$\displaystyle=\frac{-2a\mu_{3}-(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2}}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{2}}$

and

$\displaystyle\frac{\partial^{2}}{\partial\mu_{3}^{2}}\left(\frac{1}{\phi}\right)$
$\displaystyle=\frac{(-2a-\partial_{\mu_{3}}((\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2}))(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{2}}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{4}}$
$\displaystyle\ \ \ \
+\frac{(4a\mu_{3}+2(\partial_{\mu_{3}}\phi)(a^{2}-\mu_{3}^{2})^{2})(-2\tilde{\phi}\mu_{3}+(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2}))(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{4}}$
$\displaystyle=\frac{(-2a-\partial_{\mu_{3}}((\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2}))(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}$
$\displaystyle\ \ \ \
+\frac{(4a\mu_{3}+2(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2})(-2\tilde{\phi}\mu_{3}+(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2}))}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}$
$\displaystyle=\frac{-2a^{2}-2a\tilde{\phi}(a^{2}-\mu_{3}^{2})-a\partial_{\mu_{3}}((\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2})-\tilde{\phi}(a^{2}-\mu_{3}^{2})\partial_{\mu_{3}}((\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2})}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}$
$\displaystyle\ \ \ \
+\frac{-8a\tilde{\phi}\mu_{3}^{2}+4a\mu_{3}(a^{2}-\mu_{3}^{2})\partial_{\mu_{3}}\tilde{\phi}-4\mu_{3}\tilde{\phi}(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2}+2(\partial_{\mu_{3}}\tilde{\phi})^{2}(a^{2}-\mu_{3}^{2})^{3}}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}$
$\displaystyle=-\frac{2a^{2}+2a\tilde{\phi}(a^{2}-\mu_{3}^{2})+a(\partial_{\mu_{3}}^{2}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2}+\tilde{\phi}(\partial_{\mu_{3}}^{2}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{3}+8a\tilde{\phi}\mu_{3}^{2}}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}$
$\displaystyle\ \ \ \
+\frac{2(\partial_{\mu_{3}}\tilde{\phi})^{2}(a^{2}-\mu_{3}^{2})^{3}+8a\mu_{3}(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}.$

Hence,

$\frac{\partial^{2}}{\partial\mu_{3}^{2}}\left(\frac{1}{\phi}\right)=M+N$

where

(A.3)
$N=-\frac{2a^{2}+2a\tilde{\phi}(a^{2}-\mu_{3}^{2})+8a\tilde{\phi}\mu_{3}^{2}+a(\partial_{\mu_{3}}^{2}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{2}+\tilde{\phi}(\partial_{\mu_{3}}^{2}\tilde{\phi})(a^{2}-\mu_{3}^{2})^{3}}{(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}}<0$

by Lemma A.2 and the fact that $\mu_{3}\in(-a,a)$, and the mixed sign term $M$
is determined by $(a+\tilde{\phi}(a^{2}-\mu_{3}^{2}))^{3}M=m_{1}+m_{2}$ where

(A.4) $m_{1}=2(\partial_{\mu_{3}}\tilde{\phi})^{2}(a^{2}-\mu_{3}^{2})^{3},\ \
\ m_{2}=8a\mu_{3}(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2}).$

To prove Proposition A.1 we need to show that $N+M<0$ by (A.2).

Note that the assumptions in Proposition A.1 imply that $r_{l}>sa$. For
$m_{1}$, we see that

$|\partial_{\mu_{3}}\tilde{\phi}|\leq\sum_{l=1}^{k-2}\frac{1}{2r_{l}^{2}}\leq\max_{l\in\\{1,\ldots,k-2\\}}\frac{1}{r_{l}^{2}}\frac{k-2}{2}\leq\frac{k-2}{2s^{2}a^{2}}$

by Lemma A.2, and hence, by (A.4),

$0\leq m_{1}\leq 2a^{2}\left(\frac{k-2}{2s^{2}}\right)^{2}.$

Since we are assuming that $s\geq\sqrt{(k-2)/2}$, the term given by $m_{1}$ in
$M$ can be absorbed in the first term in $N$ in (A.3) so that the sum remains
negative.

To analyze $m_{2}$ in (A.4), we observe by Lemma A.2 that

$|\mu_{3}(\partial_{\mu_{3}}\tilde{\phi})|\leq\sum_{l=1}^{k-2}\frac{|\mu_{3}|}{2r_{l}^{2}}\leq\max_{l\in\\{1,\ldots,k-2\\}}\frac{|\mu_{3}|}{r_{l}}\tilde{\phi}\leq\frac{|\mu_{3}|\tilde{\phi}}{sa}\leq\frac{\tilde{\phi}}{s},$

where we have used $r_{l}>sa$ and $|\mu_{3}|\leq a$. Then, inserting this
inequality in $m_{2}$ in (A.4) gives:

$|m_{2}|\leq
8a|\mu_{3}|(\partial_{\mu_{3}}\tilde{\phi})(a^{2}-\mu_{3}^{2})\leq
8s^{-1}a\tilde{\phi}(a^{2}-\mu_{3}^{2}).$

As $s\geq 4$ by assumption, we see that the term in $M$ given by $m_{2}$ can
be absorbed into the second term appearing in $N$ in (A.3) so that the sum
remains negative.

We conclude that $\partial^{2}_{\mu_{3}}\phi^{-1}=M+N<0$ and so, by (A.2) ,
$K>0$ and Proposition A.1 is proved.

##  Appendix B Hessian of circle-invariant functions

Let $X$ be a hyperkähler 4-manifold given by the Gibbons–Hawking ansatz and
let $f:X\rightarrow\mathbb{R}$ be a smooth ${\rm U}(1)$-invariant function.
The Hessian of $f$ at a point $p$ is given by $\mathrm{Hess}(f)=\nabla df$, so
if $\\{e_{\mu}\\}_{\mu=0}^{3}$ denotes an orthonormal frame on $X$ as in Lemma
2.2, we have

(B.1)
$\mathrm{Hess}(f)(e_{\mu},e_{\nu})=\nabla_{e_{\mu}}\nabla_{e_{\nu}}f-(\nabla_{e_{\mu}}e_{\nu})\cdot
f$

and $e_{0}\cdot f=0$ by the ${\rm U}(1)$-invariance of $f$. From the formulae
for $\nabla_{e_{\mu}}e_{\nu}$ given in Lemma 2.2, we find (recalling the
permutation symbol $\varepsilon_{ijk}$ and the fact that we are using
summation convention):

(B.2) $\displaystyle(\nabla_{e_{0}}e_{0})\cdot f$
$\displaystyle=\frac{1}{2\phi^{2}}\frac{\partial\phi}{\partial\mu_{i}}\frac{\partial
f}{\partial\mu_{i}},$ (B.3) $\displaystyle(\nabla_{e_{i}}e_{0})\cdot f$
$\displaystyle=\frac{1}{2\phi^{2}}\varepsilon_{ijk}\frac{\partial\phi}{\partial\mu_{j}}\frac{\partial
f}{\partial\mu_{k}}=(\nabla_{e_{0}}e_{i})\cdot f,$ (B.4)
$\displaystyle(\nabla_{e_{i}}e_{j})\cdot f$
$\displaystyle=\frac{1}{2\phi^{2}}\left(\frac{\partial
f}{\partial\mu_{i}}\frac{\partial\phi}{\partial\mu_{j}}-\delta_{ij}\frac{\partial\phi}{\partial\mu_{k}}\frac{\partial
f}{\partial\mu_{k}}\right).$

We may also compute

(B.5)
$\displaystyle\nabla_{e_{i}}\nabla_{e_{j}}f=\nabla_{e_{i}}\frac{1}{\phi^{1/2}}\frac{\partial
f}{\partial\mu_{j}}=\frac{1}{\phi}\frac{\partial^{2}f}{\partial\mu_{i}\partial\mu_{j}}-\frac{1}{2\phi^{2}}\frac{\partial\phi}{\partial\mu_{i}}\frac{\partial
f}{\partial\mu_{j}}.$

Moreover, since both $f$ and $\phi$ are ${\rm U}(1)$-invariant, we have that

(B.6) $\nabla_{e_{0}}\nabla_{e_{\mu}}f=0=\nabla_{e_{\mu}}\nabla_{e_{0}}f.$

Inserting (B.2)–(B.6) in (B.1) yields:

(B.7) $\displaystyle\mathrm{Hess}(f)_{00}$
$\displaystyle=-\frac{1}{2\phi^{2}}\frac{\partial\phi}{\partial\mu_{i}}\frac{\partial
f}{\partial\mu_{i}},$ (B.8) $\displaystyle\mathrm{Hess}(f)_{i0}$
$\displaystyle=-\frac{1}{2\phi^{2}}\varepsilon_{ijk}\frac{\partial\phi}{\partial\mu_{j}}\frac{\partial
f}{\partial\mu_{k}}=\mathrm{Hess}(f)_{0i},$ (B.9)
$\displaystyle\mathrm{Hess}(f)_{ij}$
$\displaystyle=\frac{1}{\phi}\frac{\partial^{2}f}{\partial\mu_{i}\partial\mu_{j}}+\frac{\delta_{ij}}{2\phi^{2}}\frac{\partial\phi}{\partial\mu_{k}}\frac{\partial
f}{\partial\mu_{k}}-\frac{1}{2\phi^{2}}\left(\frac{\partial
f}{\partial\mu_{i}}\frac{\partial\phi}{\partial\mu_{j}}+\frac{\partial\phi}{\partial\mu_{i}}\frac{\partial
f}{\partial\mu_{j}}\right).$

In particular, if we let $f=r^{2}=\mu_{1}^{2}+\mu_{2}^{2}+\mu_{3}^{2}$ then
(B.7)–(B.9) give:

(B.10) $\displaystyle\mathrm{Hess}(r^{2})_{00}$
$\displaystyle=-\frac{\mu_{i}}{\phi^{2}}\frac{\partial\phi}{\partial\mu_{i}},$
(B.11) $\displaystyle\mathrm{Hess}(r^{2})_{i0}$
$\displaystyle=\varepsilon_{ijk}\frac{\mu_{j}}{\phi^{2}}\frac{\partial\phi}{\partial\mu_{k}}=\mathrm{Hess}(r^{2})_{0i},$
(B.12) $\displaystyle\mathrm{Hess}(r^{2})_{ij}$
$\displaystyle=\delta_{ij}\left(\frac{2}{\phi}+\frac{\mu_{k}}{\phi^{2}}\frac{\partial\phi}{\partial\mu_{k}}\right)-\frac{1}{\phi^{2}}\left(\mu_{i}\frac{\partial\phi}{\partial\mu_{j}}+\mu_{j}\frac{\partial\phi}{\partial\mu_{i}}\right).$

As an application of this computation we give the following result.

######  Example B.1 (No compact minimal submanifolds in Euclidean or Taub–NUT
$\mathbb{R}^{4}$).

Let $\phi$ be as in (2.6) or (2.9), which gives Euclidean or Taub–NUT
$\mathbb{R}^{4}$. Then

$\frac{\partial\phi}{\partial\mu_{i}}=-\frac{\mu_{i}}{2r^{3}}.$

Inserting this in the equations (B.10)–(B.12) for the Hessian of $r^{2}$
gives:

$\displaystyle\mathrm{Hess}(r^{2})_{00}$ $\displaystyle=\frac{1}{2r\phi^{2}},$
$\displaystyle\mathrm{Hess}(r^{2})_{i0}$
$\displaystyle=-\varepsilon_{ijk}\frac{\mu_{j}\mu_{k}}{2r^{3}\phi^{2}}=0=\mathrm{Hess}(r^{2})_{0i},$
$\displaystyle\mathrm{Hess}(r^{2})_{ij}$
$\displaystyle=\delta_{ij}\left(\frac{2}{\phi}-\frac{1}{2r\phi^{2}}\right)+\frac{\mu_{i}\mu_{j}}{r^{3}\phi^{2}}=\frac{\delta_{ij}}{2r\phi^{2}}(4r\phi-1)+\frac{\mu_{i}\mu_{j}}{r^{3}\phi^{2}}$

One may now explicitly compute that $\mathrm{Hess}(r^{2})$ is a positive
definite matrix for $r>0$, and so $r^{2}$ is a strictly convex function for
$r>0$. Hence, there are no compact minimal submanifolds (including geodesics)
in Euclidean or Taub–NUT $\mathbb{R}^{4}$.

## References