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Manifolds 11 | Projective Space is a Manifold [dark version] thumbnail

Manifolds 11 | Projective Space is a Manifold [dark version]

5 min read

Based on The Bright Side of Mathematics's video on YouTube. If you like this content, support the original creators by watching, liking and subscribing to their content.

TL;DR

Real projective space P^n\mathbb{R} is the quotient of S^n by the identification x \sim -x, with topology induced by the quotient map Q.

Briefing

Real projective space P^n\mathbb{R} is shown to be a well-defined n-dimensional manifold by building an explicit atlas from the sphere S^n and the standard identification x \sim -x. The construction starts with the quotient map Q: S^n \to P^n\mathbb{R}, sending each point to its equivalence class. Two points on the sphere represent the same projective point exactly when they are antipodal, so the topology on P^n\mathbb{R} is the quotient topology induced by Q.

To prove manifold structure, the key move is to cover P^n\mathbb{R} with n+1 open sets that correspond to where a chosen homogeneous coordinate is nonzero. For each i = 1,\dots,n+1, define V_i \subset P^n\mathbb{R} as the set of equivalence classes [x] for which the i-th coordinate of x is not zero. Because Q^{-1}(V_i) consists of points on S^n where x_i \neq 0, it becomes a union of two open hemispheres on the sphere (the regions where x_i is positive or negative). Since those preimages are open in S^n and the topology is quotient, each V_i is open in P^n\mathbb{R}. Together, the V_i cover all of P^n\mathbb{R} because every point on the sphere has at least one nonzero coordinate.

The atlas then comes from charts H_i that turn each V_i into Euclidean space. The transcript illustrates the idea first for n=1, where P^1\mathbb{R} can be visualized as lines through the origin in the plane. On the circle S^1, the chart domain V_1 corresponds to lines excluding the one where x_1=0; the chart uses slope as the coordinate. For a point on the sphere with x_1 \neq 0, the slope is x_2/x_1, and this ratio is well-defined on equivalence classes because flipping x \mapsto -x cancels the sign. The chart map is bijective onto an open subset of \mathbb{R}, and its inverse is built by taking the vector (1, m), scaling it to lie on the unit circle, and then passing to its equivalence class. Continuity of both directions yields a homeomorphism, establishing local Euclidean behavior.

For general n, the same mechanism scales up: each chart H_i maps an equivalence class [x] (with x_i \neq 0) to a vector in \mathbb{R}^n by dividing every coordinate by x_i and omitting the i-th slot. Concretely, the image is (x_1/x_i, \dots, x_{i-1}/x_i, x_{i+1}/x_i, \dots, x_{n+1}/x_i), producing an n-component vector. The normalization ensures the representative lies on S^n, and the inverse is constructed by inserting a 1 in the i-th position and scaling back to the sphere before taking the equivalence class. Each (V_i, H_i) therefore provides a homeomorphism with \mathbb{R}^n, so P^n\mathbb{R} is locally Euclidean of dimension n.

With this atlas in place—and using standard background properties like second countability and the Hausdorff condition—the result is that real projective space is a bona fide n-dimensional manifold. The next step, hinted for a later installment, is adding differentiability structure to these manifolds.

Cornell Notes

Real projective space P^n\mathbb{R} is built from the sphere S^n by identifying antipodal points x \sim -x. The quotient map Q: S^n \to P^n\mathbb{R} induces the quotient topology, and the space is covered by n+1 open sets V_i consisting of classes where the i-th homogeneous coordinate is nonzero. On each V_i, a chart H_i sends a class [x] to an n-tuple obtained by dividing all coordinates by x_i and omitting the i-th entry; this produces a homeomorphism from V_i to \mathbb{R}^n. The n=1 case uses slope of lines through the origin as the coordinate, with an inverse built by scaling (1,m) back to the unit circle. Together, these charts show P^n\mathbb{R} is an n-dimensional manifold.

How are points in real projective space identified, and what role does the quotient map play?

Points on the sphere S^n are identified by the equivalence relation x \sim -x. The quotient space under this relation is P^n\mathbb{R}, equipped with the quotient topology. The canonical projection (quotient map) Q sends each point x on S^n to its equivalence class [x]. Continuity of Q and the quotient topology are what make the sets defined via preimages on S^n automatically open in P^n\mathbb{R}.

Why do the sets V_i form an open cover of P^n\mathbb{R}?

For each i = 1,\dots,n+1, V_i consists of all equivalence classes [x] where the i-th coordinate of x is not zero. The preimage Q^{-1}(V_i) is exactly the subset of S^n where x_i \neq 0, which splits into two open regions on the sphere: where x_i>0 and where x_i<0. Those are open in S^n, so V_i is open in the quotient space. Every point on S^n has at least one nonzero coordinate, so the V_i collectively cover all of P^n\mathbb{R}.

In the n=1 case, how does the chart coordinate arise, and why is it well-defined on equivalence classes?

When n=1, P^1\mathbb{R} corresponds to lines through the origin in the plane. On S^1, a point has coordinates (x_1,x_2) with x_1^2+x_2^2=1. For classes in V_1 (where x_1 \neq 0), the chart uses slope m = x_2/x_1. This ratio is well-defined under x \mapsto -x because both numerator and denominator change sign, canceling out. The inverse map takes a slope m, forms the vector (1,m), scales it to unit length to land on S^1, and then passes to its equivalence class.

What is the general formula for the chart H_i on V_i when n is arbitrary?

For a class [x] with x_i \neq 0, H_i maps it to an n-dimensional vector by dividing by x_i and omitting the i-th coordinate: H_i([x]) = (x_1/x_i, \dots, x_{i-1}/x_i, x_{i+1}/x_i, \dots, x_{n+1}/x_i). The resulting vector has n components and lies in \mathbb{R}^n. The construction is designed so that antipodal representatives give the same output, since dividing by x_i removes the sign ambiguity.

How is the inverse of H_i constructed in the general case?

Given a vector in \mathbb{R}^n, the inverse inserts a 1 into the i-th position (to reconstruct an (n+1)-component homogeneous coordinate vector), then scales the whole vector back onto the unit sphere S^n. After that normalization, the map takes the resulting point to its equivalence class in P^n\mathbb{R}. This mirrors the n=1 inverse built from (1,m) and scaling to the unit circle.

Review Questions

  1. What are the definitions of V_i and H_i, and how do they depend on the condition x_i \neq 0?
  2. Why does the ratio x_j/x_i define a coordinate on equivalence classes under x \sim -x?
  3. How does the inverse chart construction ensure the representative lies on S^n before passing to the quotient?

Key Points

  1. 1

    Real projective space P^n\mathbb{R} is the quotient of S^n by the identification x \sim -x, with topology induced by the quotient map Q.

  2. 2

    Open sets V_i are defined by requiring the i-th homogeneous coordinate to be nonzero; their preimages on S^n split into two open hemispheres (x_i>0 and x_i<0).

  3. 3

    The sets V_i cover P^n\mathbb{R} because every point on S^n has at least one nonzero coordinate.

  4. 4

    Each chart H_i maps V_i to \mathbb{R}^n by dividing all coordinates by x_i and omitting the i-th entry, producing an n-tuple of ratios.

  5. 5

    The coordinate functions are well-defined on equivalence classes because the sign flip x \mapsto -x cancels in the ratios.

  6. 6

    For n=1, the chart coordinate becomes the slope x_2/x_1 of a line through the origin, with an inverse built by scaling (1,m) back to S^1.

  7. 7

    Homeomorphisms from each V_i to \mathbb{R}^n establish that P^n\mathbb{R} is locally Euclidean of dimension n, hence an n-dimensional manifold.

Highlights

The atlas for P^n\mathbb{R} comes from the simple rule: pick a coordinate x_i \neq 0 and use ratios x_j/x_i as local coordinates.
In the n=1 picture, projective space becomes lines through the origin, and the chart coordinate is exactly the line’s slope.
The sign ambiguity from x \sim -x disappears automatically because every coordinate is divided by the same nonzero x_i.
The inverse charts rebuild a point on S^n by inserting a 1 in the i-th slot and scaling to unit length before taking the equivalence class.

Topics