The Projective Plane from a Paired Polygon

From a Disk to the Projective Plane

The projective plane is a surface model built from a simple rule: points on the boundary of a disk are paired with their antipodes, while interior points are left alone. The resulting quotient is not just a picture of a surface; it has a precise topology and a finite cell structure. We will identify the quotient with the standard projective plane, encode the boundary pairing by the polygon word \(aa\), and calculate its Euler characteristic.

Let \(S^2=\{(x,y,z)\in\mathbb{R}^3:x^2+y^2+z^2=1\}\), with the subspace topology from the standard topology on \(\mathbb{R}^3\). Define the real projective plane \(\mathbb{RP}^2\) to be the quotient of \(S^2\) by the equivalence relation \(u\sim -u\), with the quotient topology. Thus a point of \(\mathbb{RP}^2\) is an antipodal pair \(\{u,-u\}\), or equivalently a line through the origin in \(\mathbb{R}^3\).

Let \(D^2=\{(x,y)\in\mathbb{R}^2:x^2+y^2\leq 1\}\), with the subspace topology. On \(D^2\), impose the relation that each interior point is equivalent only to itself, while boundary points satisfy \((x,y)\sim(-x,-y)\). In particular, no two distinct interior points are identified, and no interior point is identified with a boundary point.

Proof. Define \(h:D^2\to S^2\) by \(h(x,y)=(x,y,\sqrt{1-x^2-y^2})\), and let \(q:S^2\to\mathbb{RP}^2\) be the quotient map. The map \(h\) is continuous, and so is \(q\circ h\). Every antipodal class in \(S^2\) has a representative with \(z\geq0\), so \(q\circ h\) is surjective. If \(x^2+y^2<1\), then \(h(x,y)\) has positive third coordinate; its antipode has negative third coordinate, so the class \(q(h(x,y))\) has no other representative in the upper hemisphere. On the boundary, \(h(x,y)=(x,y,0)\), and its antipode is \(h(-x,-y)\). Consequently, two points of \(D^2\) have the same image under \(q\circ h\) exactly when they are equivalent under the stated disk relation. Thus \(q\circ h\) induces a continuous bijection from the disk quotient to \(\mathbb{RP}^2\), by the definition of the quotient topology.

For completeness, \(\mathbb{RP}^2\) is Hausdorff: on antipodal classes define \(d([u],[v])=\min\{\|u-v\|,\|u+v\|\}\), where \(u,v\in S^2\). This is independent of the representatives, is zero exactly when the classes agree, and satisfies the triangle inequality: choose signs realizing the two minima for \([u],[v]\) and \([v],[w]\), then apply the Euclidean triangle inequality to the correspondingly signed representatives of \(u\) and \(w\). This metric gives the quotient topology. Indeed, the quotient map is continuous since \(d([u],[v])\leq\|u-v\|\); and if the preimage of a set is open, a ball around \(u\) contained in that preimage gives a \(d\)-ball around \([u]\) contained in the set, using the antipodal symmetry of the preimage. The sphere is compact, so its quotient \(\mathbb{RP}^2\) is compact. The disk quotient is compact as a continuous image of the compact disk. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism: closed subsets of the compact domain are compact, their images are compact, and compact subsets of a Hausdorff space are closed; hence the bijection is closed and its inverse is continuous. This proves the theorem. \(\square\)

The Boundary Pairing and the Word \(aa\)

Mark two antipodal points \(p\) and \(q\) on the boundary of the disk. The two boundary arcs from \(p\) to \(q\) are the two edges of a topological polygon. Orient each edge in the direction in which the boundary is traversed. The antipodal identification pairs corresponding points of the two arcs, with their boundary-traversal parameters matched. Label both edges \(a\); the resulting boundary word is \(aa\). The word records that the two edge occurrences are paired. Here the pairing sends the start of either edge to the start of the other and the end to the end.

To see why these classes give cells, take the images of the polygon’s vertices, open edges, and interior. The open face is the image of the polygon interior, which has no identifications and is an open disk. Each paired pair of open edges contributes one open 1-cell, and each vertex class contributes one 0-cell. This is the cell structure obtained from a polygon quotient: its \(k\)-cells are the equivalence classes of the polygon’s open \(k\)-dimensional pieces. In this case the counts are \(c_0=1\), \(c_1=1\), and \(c_2=1\).

Proof. The boundary pairing has one vertex class and one edge class, and the polygon interior gives one face. Therefore \(V=1\), \(E=1\), and \(F=1\). Substitution into the definition gives \(\chi=V-E+F=1-1+1=1\). \(\square\)

A Subdivision Gives the Same Count

A cell structure can be refined without changing the underlying quotient space. In this example, subdivide the disk using its diameter from \(p\) to \(q\), and mark the midpoint \(c\) of that diameter. The diameter is now two edges, from \(p\) to \(c\) and from \(c\) to \(q\). It divides the disk into two triangular faces, each bounded by one boundary arc and the two halves of the diameter. After the boundary pairing, \(p\) and \(q\) are still identified, while \(c\) is a separate vertex.

This example illustrates what a cell count can and cannot tell us. It gives a numerical invariant of the displayed cell structure, and the same value appears in the coarse and refined counts here. The Euler characteristic alone is not a construction of the surface: the boundary equivalence relation is what specifies the quotient. One must determine the vertex and edge classes from that relation before using the formula.

A Common Counting Pitfall

A tempting but incorrect count is to pair the two edges but leave their endpoints as two separate vertices. That would give \(V=2\), \(E=1\), and \(F=1\), hence the wrong calculation \(\chi=2-1+1=2\). The endpoint calculation rules this out: matching the edge starts identifies \(v_0\) with \(v_1\), and matching the ends identifies the same two points. There is just one vertex class, so the correct count is \(V=1\), \(E=1\), \(F=1\), and \(\chi=1\). In polygon quotients, edge pairing determines endpoint identifications as well; counting the edges without tracing their endpoints can produce a false cell structure.