The Fundamental Group of the Projective Plane

From the Projective Plane to a Polygon

The projective plane has a compact polygon description that makes its fundamental group accessible. Its key feature is that the boundary of a disk is identified in antipodal pairs. The resulting 1-dimensional part is a circle, and as the disk boundary is read once around, it travels twice around that circle in the same direction. Seifert–van Kampen turns this two-fold traversal into a relation on a generator.

Write \(\mathbb{R}P^2\) for the space of lines through the origin in \(\mathbb{R}^3\), with the quotient topology obtained from the sphere \(S^2\) with its standard subspace topology by identifying each point \(v\) with its antipode \(-v\). Write \(D^2\) for the closed unit disk in \(\mathbb{R}^2\), with its subspace topology, and \(S^1=\partial D^2\), also with its subspace topology. The standard polygon model of \(\mathbb{R}P^2\) is \(D^2\) with each boundary point identified with its antipode; interior points are not identified with other points.

To see why this disk model describes the projective plane, take the closed hemisphere of \(S^2\) given by \(z\geq 0\). It meets every antipodal pair, and meets each pair in exactly one point except for pairs on the equator, where both antipodes lie in the hemisphere. Projecting this hemisphere onto its first two coordinates identifies it with \(D^2\). Thus the only identifications that remain on the disk are antipodal identifications on its boundary. The induced map from the disk quotient to \(\mathbb{R}P^2\) is a continuous bijection. The disk quotient is compact, and \(\mathbb{R}P^2\) is Hausdorff: distinct antipodal orbits on the metric sphere are disjoint finite sets with positive distance, so disjoint small neighborhoods of the orbits give disjoint neighborhoods in the quotient. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism: images of closed sets are compact, hence closed, so the inverse is continuous.

In the disk model the boundary quotient is a circle. It is the 1-skeleton of a cell description of \(\mathbb{R}P^2\): there is one vertex, one edge forming a circle, and one 2-cell attached along the boundary of the disk. The attaching map is the map from the disk boundary to that 1-skeleton induced by the boundary identification. Determining its direction and how many times it goes around the circle is the essential calculation.

Reading the Boundary Identification

Identify both the original boundary circle and the quotient boundary circle with the unit circle in \(\mathbb{C}\). If \(z\in S^1\), its antipode is \(-z\). The map \(z\mapsto z^2\) takes the same value on \(z\) and \(-z\), and its fibers are exactly the antipodal pairs. It therefore gives a parameterization of the quotient boundary. In angular coordinates the attaching map is represented by \(e^{it}\mapsto e^{2it}\).

Applying Seifert–van Kampen

Here \(Y\) is the 1-skeleton, which is a circle with its vertex chosen as basepoint. The circle is path-connected, and its fundamental group is infinite cyclic, generated by the positively oriented loop \(a\). This standard calculation of the fundamental group of the circle is part of the earlier material on the fundamental group of the circle. The attaching map sends the positive generator of \(\pi_1(S^1)\) to \(a^2\), since it makes two positive turns. The theorem therefore says to impose the relator \(a^2\).

Proof. Use the vertex of the polygon model as basepoint. Its 1-skeleton \(Y\) is a circle, so \(\pi_1(Y)\) is generated by the positive edge loop \(a\). The boundary attaching map has degree \(2\), as computed above, and therefore takes the generator of \(\pi_1(S^1)\) to \(a^2\). By the disk-attachment form of Seifert–van Kampen, attaching the 2-cell quotients \(\pi_1(Y)\) by the normal subgroup generated by \(a^2\). This is precisely the presentation \(\langle a\mid a^2=1\rangle\). \(\square\)

Why the Group Has Exactly Two Elements

A presentation describes a group by generators and relations, but it is useful to determine explicitly how many elements the presentation gives. The relation \(a^2=1\) implies \(a^{-1}=a\), because multiplying \(a^2=1\) by \(a^{-1}\) gives \(a=a^{-1}\). Consequently every inverse in a word can be replaced by \(a\). Every word is then a power \(a^n\), and writing \(n=2k+r\), where \(r\) is either \(0\) or \(1\), gives \(a^n=(a^2)^k a^r=a^r\). Thus every element is either the identity or \(a\). This proves there are at most two elements; a homomorphism to a two-element group shows that \(a\) is not the identity.

A Common Orientation Mistake

Common pitfall. It is easy to read the second half of the polygon boundary as \(a^{-1}\) and write the boundary word \(aa^{-1}\). That calculation would make the attaching loop trivial and would impose no new relation. But the boundary parameterization settles the direction: on each half the output angle \(2t\) increases by \(2\pi\). Both halves therefore travel positively, and the correct word is \(aa\).

The orientation here comes from the map between the boundary circles, not from a visual impression that the two boundary arcs lie on opposite sides of the polygon. The algebra depends on that distinction: the word \(aa^{-1}\) is the identity in the free group, whereas \(aa\) gives the nontrivial relator \(a^2\). Reversing one edge without checking the parameterized attaching map changes the computation.

The result can be summarized in three steps: the antipodal boundary quotient gives one circular 1-skeleton; the disk boundary maps around that circle twice in the same direction; and Seifert–van Kampen imposes the square of the edge generator as a relation. The resulting group is cyclic of order two.