The Real Projective Plane

Consider in the sphere \mathbb{S}_2 \subset \mathbb{R}^3, the relation induced by identification of antipodal points; that is, given z_1, z_2 \in \mathbb{S}_2, set z_1 \sim z_2 if and only z_1 = \pm z_1. The corresponding quotient space \mathbb{P} = \big( \mathbb{S}_2 / \sim \big) is what we call real projective space.

Since we are interested in the topological properties of this space, we actually define a real projective space to be any homeomorphic set to \mathbb{P}.  Among those, we are interested in one that can be realized from a square, by identification of its sides (in a similar manner as we did with the torus).  We proceed as follows:

Assume that we start from the unit sphere \{ (x_1,x_2,x_3) \in \mathbb{R}^3 : x_1^2+x_2^2+x_3^2=1\}, and note that the upper hemisphere H=\{ (x_1,x_2,x_3) \in \mathbb{R}^3 : x_1^2+x_2^2+x_3^2=1, x_3 \geq 0 \} contains at least one of each pair of antipodal points.  If both antipodal points occur in H, they will necessarily lie over the circle \{ (x_1, x_2, 0) \in \mathbb{R}^3 : x_1^2+x_2^2=1\}.  The hemisphere H is obviously homeomorphic to a disk (by a simple vertical projection onto the plane \{x_3=0\}, for example).  And the disk is homeomorphic to a square, so we may use a composition of both to realize a homomorphism from H to \square_2.

Define in H an equivalence relation that identifies two antipodal points on the border, and notice that the homeomorphism just computed takes that identification to the following: Given (x_1,x_2), (y_1,y_2) \in \square_2, it is (x_1,x_2) \sim (y_,y_2) if

  • x_1=y_1 and x_2=y_2, or
  • x_1y_1 = -1 and  x_2=1-y_2, or
  • x_2y_2=-1 and x_1=1-y_1.

A diagram representing the quotient space \big( \square_2 / \sim \big) is presented below:

The punch-line is, of course, to construct a homeomorphism from the real projective plane \mathbb{P} as defined above, to the quotient space \big( \square_2 / \sim \big). The reader should not have much trouble to give an analytic expression of such a map following the steps above.

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