Plane Embeddings on the Sphere

Graphs, Drawings and Embeddings

A graph records which vertices are joined, not where those vertices lie in space. To study planarity, we place the vertices and edges in a surface and require that edges meet only where the graph says they meet. The plane and the sphere may look like different settings, but removing one point from the sphere leaves a surface equivalent to the plane. This equivalence lets us move between plane drawings and sphere drawings without changing the graph or introducing crossings.

Throughout, graphs are finite and simple. A drawing places each vertex at a distinct point and represents each edge by an arc joining its endpoints. An arc is a continuous, non-self-intersecting curve. A drawing is an embedding if the interior of each edge arc contains no vertex and edge arcs meet only at a shared endpoint. A plane embedding is an embedding in the plane; a sphere embedding is an embedding on the sphere.

The distinction is important. Planarity is a property of the abstract graph: it asks whether some embedding exists. A particular drawing is a proposed realization of the graph and may fail to be an embedding if two edge interiors cross. A graph can therefore be planar even when a particular drawing of it is not an embedding.

Stereographic Projection

Let \(S^2=\{(X,Y,Z)\in\mathbb{R}^3:X^2+Y^2+Z^2=1\}\) be the unit sphere, and let \(N=(0,0,1)\) be its north pole. Stereographic projection gives a one-to-one correspondence between the plane and \(S^2\setminus\{N\}\). We use the following direction: map a point \((x,y)\) in the plane to the point where the line from \(N\) through \((x,y,0)\) meets the sphere.

Proof. Write \(r^2=x^2+y^2\). The sum of the squares of the three coordinates of \(\Phi(x,y)\) is \[ \frac{4x^2+4y^2+(r^2-1)^2}{(r^2+1)^2} =\frac{4r^2+r^4-2r^2+1}{(r^2+1)^2} =\frac{(r^2+1)^2}{(r^2+1)^2}=1. \] Thus \(\Phi(x,y)\in S^2\). Its third coordinate is less than \(1\), because \[ 1-\frac{r^2-1}{r^2+1}=\frac{2}{r^2+1}>0, \] so its image omits \(N\).

For a point \((X,Y,Z)\in S^2\setminus\{N\}\), define \[ \Psi(X,Y,Z)=\left(\frac{X}{1-Z},\frac{Y}{1-Z}\right). \] The denominator is nonzero because \(Z\ne1\). Substituting the coordinates of \(\Phi(x,y)\) gives \(X/(1-Z)=x\) and \(Y/(1-Z)=y\). Conversely, using \(X^2+Y^2=1-Z^2\), one obtains \[ \left(\frac{X}{1-Z}\right)^2+\left(\frac{Y}{1-Z}\right)^2 =\frac{1-Z^2}{(1-Z)^2} =\frac{1+Z}{1-Z}. \] Substitution into the formula for \(\Phi\) recovers \((X,Y,Z)\). Both maps are continuous on their domains, so they are inverse homeomorphisms. A homeomorphism is a continuous bijection with continuous inverse; it preserves which points and arcs meet, and which do not. Therefore it takes an embedding to an embedding. \(\square\)

The omitted point matters. A plane drawing maps to a sphere drawing that avoids \(N\). Conversely, to transfer a sphere embedding to the plane, we must remove a point not used by the embedded graph, then use a rotated version of \(\Psi\). A finite embedded graph is a finite union of vertices and arcs and cannot cover the sphere: each arc is closed and has empty interior, and a finite union of such sets cannot cover a surface. Thus an unused point can be chosen.

Proof. Suppose first that \(G\) has a plane embedding. Apply \(\Phi\) to every vertex and every edge arc. Since \(\Phi\) is a homeomorphism onto \(S^2\setminus\{N\}\), distinct vertices remain distinct, each edge still joins its original endpoints, and edge interiors still meet only at shared endpoints. This gives a sphere embedding.

Now suppose that \(G\) has a sphere embedding. Choose a point \(q\) on the sphere outside the embedded graph. Rotate the sphere so that \(q\) becomes \(N\). Rotations preserve incidences and intersections. The rotated embedding avoids \(N\), so apply \(\Psi\) to its vertices and arcs. The result is a plane embedding. If \(G\) has no vertices, the empty drawing embeds in both surfaces, so the equivalence also holds in that case. \(\square\)

Moving an Embedding Between Surfaces

One Abstract Graph, Different Drawings

An abstract graph consists of its vertex set and edge set. A drawing adds geometric information: vertex positions and the chosen edge arcs. Two drawings can represent the same abstract graph even if the positions and curves differ. Some drawings are embeddings and some are not; the graph is planar if at least one is an embedding.

What the Correspondence Does—and Does Not—Say

The plane and sphere descriptions are equivalent because they are related by a homeomorphism after one point is removed. This does not mean that every drawing of a planar graph is a plane embedding, nor does it mean that a fixed drawing has a unique spherical counterpart independent of the chosen projection point. The graph is the same; the placement is additional data.

When comparing drawings, keep three questions separate:

• Which abstract graph is represented? Read the vertex set and edge set.
• Is this particular drawing an embedding? Check that edge interiors have no crossings and contain no vertices.
• Is the abstract graph planar? Determine whether at least one embedding exists, in the plane or equivalently on the sphere.

A common mistake is to see a crossing in one drawing and conclude that the abstract graph is nonplanar. The \(K_4\) example gives a concrete correction: the crossing diagonals show only that the second drawing fails to be an embedding. The first drawing is an embedding of exactly the same graph, so the graph is planar. Conversely, a drawing with no visible crossings must still represent the intended edges correctly; omitting an edge or identifying two vertices would change the graph rather than establish an embedding of it.

These distinctions provide the language needed for studying regions and their boundaries in a particular embedding. Those are properties of the embedded drawing, not merely of the abstract edge list.