Real projective plane

Consider a sphere, and let the great circles of the sphere be "lines", and let pairs of antipodal points be "points". It is easy to check that this system obeys the axioms required of a projective plane:

• any pair of distinct great circles meet at a pair of antipodal points; and
• any two distinct pairs of antipodal points lie on a single great circle.

If we identify each point on the sphere with its antipodal point, then we get a representation of the real projective plane in which the "points" of the projective plane really are points. This means that the projective plane is the quotient space of the sphere obtained by partitioning the sphere into equivalence classes under the equivalence relation ~, where x ~ y if y = x or y = −x. This quotient space of the sphere is homeomorphic with the collection of all lines passing through the origin in R³.

The quotient map from the sphere onto the real projective plane is in fact a two sheeted (i.e. two-to-one) covering map. It follows that the fundamental group of the real projective plane is the cyclic group of order 2; i.e., integers modulo 2. One can take the loop AB from the figure above to be the generator.

A hemisphere can represent a real projective plane by attaching opposite points on the equator together.

Because the sphere covers the real projective plane twice, the plane may be represented as a closed hemisphere around whose rim opposite points are similarly identified.[2]

The projective plane can be immersed (local neighbourhoods of the source space do not have self-intersections) in 3-space. Boy's surface is an example of an immersion.

Polyhedral examples must have at least nine faces.[3]

An animation of the Roman Surface

Steiner's Roman surface is a more degenerate map of the projective plane into 3-space, containing a cross-cap.

The tetrahemihexahedron is a polyhedral representation of the real projective plane.

A polyhedral representation is the tetrahemihexahedron,[4] which has the same general form as Steiner's Roman Surface, shown here.

Looking in the opposite direction, certain abstract regular polytopes – hemi-cube, hemi-dodecahedron, and hemi-icosahedron – can be constructed as regular figures in the projective plane; see also projective polyhedra.

Various planar (flat) projections or mappings of the projective plane have been described. In 1874 Klein described the mapping:[1]

${\displaystyle k(x,y)=\left(1+x^{2}+y^{2}\right)^{\frac {1}{2}}{\binom {x}{y}}}$

Central projection of the projective hemisphere onto a plane yields the usual infinite projective plane, described below.

A closed surface is obtained by gluing a disk to a cross-cap. This surface can be represented parametrically by the following equations:

${\displaystyle {\begin{aligned}X(u,v)&=r\,(1+\cos v)\,\cos u,\\Y(u,v)&=r\,(1+\cos v)\,\sin u,\\Z(u,v)&=-\operatorname {tanh} \left(u-\pi \right)\,r\,\sin v,\end{aligned}}}$

where both u and v range from 0 to 2π.

These equations are similar to those of a torus. Figure 1 shows a closed cross-capped disk.

Figure 1. Two views of a cross-capped disk.

A cross-capped disk has a plane of symmetry which passes through its line segment of double points. In Figure 1 the cross-capped disk is seen from above its plane of symmetry z = 0, but it would look the same if seen from below.

A cross-capped disk can be sliced open along its plane of symmetry, while making sure not to cut along any of its double points. The result is shown in Figure 2.

Figure 2. Two views of a cross-capped disk which has been sliced open.

Once this exception is made, it will be seen that the sliced cross-capped disk is homeomorphic to a self-intersecting disk, as shown in Figure 3.

Figure 3. Two alternative views of a self-intersecting disk.

The self-intersecting disk is homeomorphic to an ordinary disk. The parametric equations of the self-intersecting disk are:

${\displaystyle {\begin{aligned}X(u,v)&=r\,v\,\cos 2u,\\Y(u,v)&=r\,v\,\sin 2u,\\Z(u,v)&=r\,v\,\cos u,\end{aligned}}}$

where u ranges from 0 to 2π and v ranges from 0 to 1.

Projecting the self-intersecting disk onto the plane of symmetry (z = 0 in the parametrization given earlier) which passes only through the double points, the result is an ordinary disk which repeats itself (doubles up on itself).

The plane z = 0 cuts the self-intersecting disk into a pair of disks which are mirror reflections of each other. The disks have centers at the origin.

Now consider the rims of the disks (with v = 1). The points on the rim of the self-intersecting disk come in pairs which are reflections of each other with respect to the plane z = 0.

A cross-capped disk is formed by identifying these pairs of points, making them equivalent to each other. This means that a point with parameters (u, 1) and coordinates ${\displaystyle (r\,\cos 2u,r\,\sin 2u,r\,\cos u)}$ is identified with the point (u + π, 1) whose coordinates are ${\displaystyle (r\,\cos 2u,r\,\sin 2u,-r\,\cos u)}$. But this means that pairs of opposite points on the rim of the (equivalent) ordinary disk are identified with each other; this is how a real projective plane is formed out of a disk. Therefore, the surface shown in Figure 1 (cross-cap with disk) is topologically equivalent to the real projective plane RP².

A line in P² can be represented by the equation ax + by + cz = 0. If we treat a, b, and c as the column vector ℓ and x, y, z as the column vector x then the equation above can be written in matrix form as:

x^Tℓ = 0 or ℓ^Tx = 0.

Using vector notation we may instead write x ⋅ ℓ = 0 or ℓ ⋅ x = 0.

The equation k(x^Tℓ) = 0 (which k is a non-zero scalar) sweeps out a plane that goes through zero in R³ and k(x) sweeps out a line, again going through zero. The plane and line are linear subspaces in R³, which always go through zero.

In P² the equation of a line is ax + by + cz = 0 and this equation can represent a line on any plane parallel to the x, y plane by multiplying the equation by k.

If z = 1 we have a normalized homogeneous coordinate. All points that have z = 1 create a plane. Let's pretend we are looking at that plane (from a position further out along the z axis and looking back towards the origin) and there are two parallel lines drawn on the plane. From where we are standing (given our visual capabilities) we can see only so much of the plane, which we represent as the area outlined in red in the diagram. If we walk away from the plane along the z axis, (still looking backwards towards the origin), we can see more of the plane. In our field of view original points have moved. We can reflect this movement by dividing the homogeneous coordinate by a constant. In the adjacent image we have divided by 2 so the z value now becomes 0.5. If we walk far enough away what we are looking at becomes a point in the distance. As we walk away we see more and more of the parallel lines. The lines will meet at a line at infinity (a line that goes through zero on the plane at z = 0). Lines on the plane when z = 0 are ideal points. The plane at z = 0 is the line at infinity.

The homogeneous point (0, 0, 0) is where all the real points go when you're looking at the plane from an infinite distance, a line on the z = 0 plane is where parallel lines intersect.

In the equation x^Tℓ = 0 there are two column vectors. You can keep either constant and vary the other. If we keep the point x constant and vary the coefficients ℓ we create new lines that go through the point. If we keep the coefficients constant and vary the points that satisfy the equation we create a line. We look upon x as a point, because the axes we are using are x, y, and z. If we instead plotted the coefficients using axis marked a, b, c points would become lines and lines would become points. If you prove something with the data plotted on axis marked x, y, and z the same argument can be used for the data plotted on axis marked a, b, and c. That is duality.

The equation x^Tℓ = 0 calculates the inner product of two column vectors. The inner product of two vectors is zero if the vectors are orthogonal. In P², the line between the points x₁ and x₂ may be represented as a column vector ℓ that satisfies the equations x₁^Tℓ = 0 and x₂^Tℓ = 0, or in other words a column vector ℓ that is orthogonal to x₁ and x₂. The cross product will find such a vector: the line joining two points has homogeneous coordinates given by the equation x₁ × x₂. The intersection of two lines may be found in the same way, using duality, as the cross product of the vectors representing the lines, ℓ₁ × ℓ₂.