Projective Space as the Space of Lines

From Vectors to Lines

A nonzero vector in Euclidean space determines a line through the origin, but many different vectors determine the same line: multiplying a vector by any nonzero real number changes its length and may reverse its direction without changing the line. Real projective space records these lines as points. Its topology is chosen so that a set of lines is open exactly when the collection of nonzero vectors lying on those lines is open.

Fix an integer \(n\geq 0\). Give \(\mathbb{R}^{n+1}\) its Euclidean topology and give \(\mathbb{R}^{n+1}\setminus\{0\}\) the subspace topology. In particular, the zero vector is not part of the space being quotiented.

The relation is an equivalence relation: \(x=1x\) gives reflexivity; if \(y=\lambda x\) with \(\lambda\ne0\), then \(x=\lambda^{-1}y\), giving symmetry; and if \(y=\lambda x\) and \(z=\mu y\), then \(z=(\mu\lambda)x\) with \(\mu\lambda\ne0\), giving transitivity. Each class is exactly one line through the origin with its zero vector removed. Indeed, all its vectors are scalar multiples of \(x\), and every nonzero vector on that line is such a multiple.

The quotient topology is not an additional choice of a metric or a notion of distance between lines. It is a precise openness rule. For any \(A\subseteq\mathbb{RP}^n\), its preimage is the union of all the equivalence classes represented by points of \(A\). Such a union is called saturated: whenever it contains a vector, it contains every nonzero scalar multiple of that vector. Thus, to test whether a set of lines is open, test whether its full preimage of vectors is open.

The Quotient Map and Its Fibers

The map \(q\) is surjective because every point of \(\mathbb{RP}^n\) is, by definition, an equivalence class represented by some nonzero vector. Its fibers describe exactly which vectors have been identified.

Proof. If \(q(x)=q(y)\), then \(x\) and \(y\) represent the same equivalence class. By the definition of the relation, \(y=\lambda x\) for some \(\lambda\ne0\). Conversely, if \(y=\lambda x\) for some \(\lambda\ne0\), then \(x\sim y\), so they belong to the same equivalence class and \(q(x)=q(y)\). The fiber containing \(x\) is therefore \(\{\lambda x:\lambda\in\mathbb{R}\setminus\{0\}\}\), which is the line spanned by \(x\) with the origin removed. \(\square\)

The quotient map is continuous: if \(A\) is open in \(\mathbb{RP}^n\), the definition of the quotient topology says precisely that \(q^{-1}(A)\) is open in the domain. The same definition gives a useful test in the other direction: a subset of \(\mathbb{RP}^n\) is open whenever its preimage under \(q\) is open. This test only applies to the full preimage, which includes every representative of every line in the set.

Representing Projective Space by the Unit Sphere

Every line through the origin meets the unit sphere \(S^n=\{u\in\mathbb{R}^{n+1}:\|u\|=1\}\) in exactly two points, which are negatives of one another. This gives another construction of projective space. In this section \(S^n\) has the subspace topology from Euclidean space, and its antipodal quotient means the quotient by the relation \(u\sim v\) if \(v=u\) or \(v=-u\).

Proof. Define \(r:\mathbb{R}^{n+1}\setminus\{0\}\to S^n\) by \(r(x)=x/\|x\|\). This map is continuous because the norm is continuous and is nonzero on its domain. It is surjective since \(r(u)=u\) for \(u\in S^n\).

We first show that \(r\) is open. Let \(O\) be open in \(\mathbb{R}^{n+1}\setminus\{0\}\), and take \(u\in r(O)\). There is \(x\in O\) with \(r(x)=u\); write \(x=tu\), where \(t=\|x\|>0\). Because the punctured Euclidean space is open in \(\mathbb{R}^{n+1}\), the subspace-open set \(O\) is also open in \(\mathbb{R}^{n+1}\). There is therefore \(\delta>0\) such that the Euclidean ball of radius \(\delta\) about \(x\) lies in \(O\). If \(v\in S^n\) and \(\|v-u\|<\delta/t\), then \(\|tv-x\|=t\|v-u\|<\delta\), so \(tv\in O\) and \(r(tv)=v\). Thus the neighborhood \(\{v\in S^n:\|v-u\|<\delta/t\}\) of \(u\) lies in \(r(O)\). Each point of \(r(O)\) has such a neighborhood, so \(r(O)\) is open in \(S^n\).

A continuous, open, surjective map is a quotient map: for a subset \(D\) of its target, if \(r^{-1}(D)\) is open, then \(D=r(r^{-1}(D))\) is open because \(r\) is open; the reverse implication follows from continuity. Thus \(r\) is quotient. Now \(q=p\circ r\). For any \(A\subseteq\mathbb{RP}^n\), this identity gives \(q^{-1}(A)=r^{-1}(p^{-1}(A))\). Since \(r\) is quotient, \(p^{-1}(A)\) is open exactly when \(q^{-1}(A)\) is open. Since \(q\) is quotient, the latter holds exactly when \(A\) is open. This proves that \(p\) is quotient.

Finally, \(p(u)=p(v)\) exactly when \(v=\lambda u\) for some \(\lambda\ne0\). If \(u,v\) are both unit vectors, taking norms gives \(1=|\lambda|\), so \(\lambda=1\) or \(\lambda=-1\). Therefore the fibers are precisely the antipodal pairs. The map from the antipodal quotient \(S^n/{\sim}\) to \(\mathbb{RP}^n\) sending the class of \(u\) to \([u]\) is consequently a well-defined bijection. A set in either quotient is open exactly when its preimage in \(S^n\) is open: for \(\mathbb{RP}^n\) this follows from the quotient criterion for \(p\), and for \(S^n/{\sim}\) it is the definition of its quotient topology. The bijection and its inverse are therefore continuous, so it is a homeomorphism. \(\square\)

A Closed Set and a Common Pitfall

Common pitfall. Identifying vectors only when one is a positive multiple of the other would keep opposite directions separate. That does not describe lines through the origin. For example, \((1,2)\) and \((-1,-2)\) span the same line because \((-1,-2)=(-1)(1,2)\), so they represent the same point of \(\mathbb{RP}^1\). On the unit circle they are antipodal points and must be identified. A condition such as \(x_0>0\) also cannot by itself describe a set of lines: \((1,2)\) satisfies it, but its representative \((-1,-2)\) does not. By contrast, \(x_0\ne0\) is unchanged under every nonzero rescaling and can describe a saturated preimage.

The quotient description makes the topology practical: openness and closedness can be checked using representatives, as long as the preimage includes every representative of each selected line. The sphere description removes the varying lengths of vectors and leaves only the identification of antipodal points. Both constructions encode the same space, but neither says that distinct nearby-looking vectors are distinct projective points; all nonzero multiples of a vector represent just one point.