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A simple triod is the union of three arcs that share one endpoint and are otherwise disjoint. The Moore triod theorem states that every pairwise disjoint family of simple triods in the plane is countable. One proof chooses data from a countable basis of open disks around each branch point; the Jordan curve theorem shows that two disjoint triods cannot determine the same finite basis data, giving an injection into a countable set of finite tuples.

For bond percolation, let $C(0)$ be the percolation cluster of the origin. Its percolation probability isWe prove $\theta (1/2)=0$ by alternating arms at the self-dual parameter, using the uniqueness of the infinite percolation cluster proved in Question 4. Suppose instead that $\theta (1/2)>0$.

At $p=1/2$, declaring a dual edge open exactly when its crossing primal edge is closed produces independent dual bond percolation with the same parameter on ${\mathbb{Z}}^2+(1/2,1/2)$. It therefore also has an infinite percolation cluster almost surely, and at most one. The two preliminary arguments below imply that, for a sufficiently large square, the probability of two primal infinite exterior arms starting through its top and bottom sides exceeds $3/4$. Applying the same arguments to the dual lattice gives probability greater than $3/4$ for dual infinite exterior arms starting through the left and right sides. Thus all four arms occur together with positive probability, by the union bound.

Here are precise boundary conventions for the alternating exterior-arm separation lemma. Use the primal graph vertex box $[-N,N{]}^2$, the dual graph vertex box $[-N+1/2,N-1/2{]}^2$, and the common continuous square with sides at $\pm (N+1/4)$. The primal arms first cross its top and bottom; the dual arms first cross its left and right. Subsequent graph vertices of every arm lie outside this continuous square. Large $N$ gives the required probabilities for both boxes, by the same rotational symmetry and boundary-side estimate. Primal open paths and dual open paths cannot cross, since the two states of a crossing edge are complementary.

The two primal arms cannot meet in the exterior. If they did, a finite simple primal crosscut in the exterior would join the top and bottom entry points. Together with one of the two arcs of the square boundary it would enclose a bounded region containing one of the dual entry points. That dual exterior arm could not reach infinity without crossing the primal crosscut or the forbidden square interior. This is the alternating exterior-arm separation lemma. Thus the two primal arms are disjoint.

Now modify only the primal edges with both endpoints in $[-N,N{]}^2$, making them open. Their grid supplies a simple interior path joining the two primal arm roots. The unchanged exterior arms and this connector form a proper doubly infinite simple primal path. Such a path separates the plane into two sides, by the Jordan curve theorem applied to its one-point compactification in the sphere. The left and right dual arms lie on opposite sides. Preserve their tails from their first exterior dual graph vertices; none of those tails crosses a modified edge. Their original boundary attachment edges may be destroyed, but the infinite exterior tails themselves remain. They now belong to two distinct infinite dual clusters, since any dual connection between them would have to cross the doubly infinite primal path.

This modification has positive probability by the finite-energy property of Bernoulli percolation: condition on the exterior configuration and open the finitely many interior edges. The exterior projections of the four-arm event have positive probability. The resulting positive-probability event with two infinite dual clusters contradicts uniqueness of the infinite percolation cluster on the dual lattice. Therefore

Only the selected paths are drawn. The black connector lies in the finitely modified box; the two blue infinite dual tails remain outside and on opposite sides.

The proof uses self-duality, symmetry, positive association, finite-energy property of Bernoulli percolation and uniqueness; it does not assume the Harris-Kesten theorem as a black box. The boundary language means an infinite ray whose first graph vertex is on the side and all later graph vertices are outside the graph vertex box. It cannot mean literal intersection of two disjoint sets, a closed box boundary and its strict complement.

A parametrization is $X(\theta ,z)=(\cosh z\cos \theta ,\cosh z\sin \theta ,z)$. It gives a homeomorphism $S^1\times \mathbb{R}\to S$, with inverse obtained from the height and angular position. Its first fundamental form isFor a surface of revolution with radius $r(z)$, the Gaussian curvature is $-r^{\prime \prime }/[r(1+r^{\prime 2}{)}^2]$. Thus $\boxed{K=-{\mathrm{sech}}^{4}z<0}$. The waist $z=0$ is a closed geodesic: ${\Gamma }_{\theta \theta }^z=-\tanh z$ vanishes there, and $\theta$ can be parametrized at constant speed.

For the uniqueness of a closed geodesic on a negatively curved cylinder, use Gauss-Bonnet theorem with piecewise geodesic boundary:A geodesic disk with smooth boundary would require curvature integral $2\pi$. A disk bounded by a geodesic monogon would require $\pi +\alpha >0$, and a geodesic bigon would require $\alpha +\beta >0$. All are impossible when $K<0$.

Here is the topological reduction needed to apply this observation without assuming simplicity. Lift a closed geodesic to the universal covering plane of the cylinder. A self-intersection of its lift supplies an innermost monogon by the Jordan curve theorem, so the lift is embedded. A contractible closed geodesic would similarly supply a geodesic disk or monogon, so its winding number is nonzero. Its lift is consequently a proper line, invariant under a nonzero power of the deck translation. Two distinct translates of that line cannot intersect: periodicity would repeat an intersection, and innermost arcs between intersections would bound a bigon. The disjoint translates are ordered transversely, with the deck translation preserving that order. A finite cyclic permutation of this ordered family is impossible, so the lift is invariant under the primitive deck translation too. Thus the original geodesic has a simple image and is at most a repeated traversal of it.

Likewise two distinct simple essential closed geodesics cannot intersect: their periodic lifts would produce a bigon. If disjoint, the Jordan curve theorem on the cylinder says they bound a compact annulus. Its two boundaries have zero geodesic curvature and $\chi =0$, so Gauss-Bonnet gives $\int KdA=0$, again impossible. Therefore there is at most one closed geodesic image. Reparametrizations, reversed orientation and multiple traversals are understood to describe the same geometric geodesic.

On $S_{0,3}$, a simple proper arc joining two distinct specified punctures is unique up to isotopy relative to its ends. Indeed, a small regular neighborhood of the arc and its two ends has one boundary curve separating those two punctures from the third; the Jordan curve theorem gives the unique such separation, and the disc it bounds gives the isotopy between any two choices.

Choose the three joining arcs, one for each puncture pair, with disjoint interiors. They form an ideal triangle graph whose complement consists of two discs. A pure homeomorphism fixes all three endpoints and carries each arc to an isotopic arc. The simultaneous-isotopy lemma makes it fix the three arcs, and the Alexander trick on each complementary disc makes it isotopic to the identity. Therefore

Fix punctures $p,q,r$. There is one isotopy class ${\ell }_p$ of essential arc from $p$ back to $p$: by the Jordan curve theorem, it separates $q$ from $r$, and any two such arcs are isotopic relative to the punctures.

Altogether $\mathcal{A}(S_{0,3})$ has six vertices:The joining arcs $a_{pq},a_{pr},a_{qr}$ span a triangle. For each puncture $p$, the three verticesspan another triangle. These are all the maximal simplices: ${\ell }_p$ must cross $a_{qr}$, and returning arcs based at distinct punctures cannot be made disjoint.

Every orientation-preserving permutation of the three punctures is realizable, while a homeomorphism acting trivially on the punctures is isotopic to the identity. ThusIt permutes the labels in the displayed description. There are two vertex orbits, the three joining arcs and the three returning arcs, as summarized by the arc complex of the three-punctured sphere.

Suppose that two closed geodesics ${\gamma }_1$ and ${\gamma }_2$ were disjoint. Under the usual convention that a closed geodesic curve is simple, the Jordan curve theorem on the sphere shows that they bound an annulus $A\subset S$. Its Euler characteristic is zero. Give $A$ the induced orientation and apply the Gauss-Bonnet theorem:Both boundary components are geodesics, so their geodesic curvature vanishes. It follows thatThis is impossible because $K>0$ everywhere and $A$ has positive area. Hence the two closed geodesics must intersect, as summarized by the intersection of closed geodesics on a positively curved sphere.