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Exercise 1343Topology › Topology of Mn(ℝ) › Applications and the case n = 2L3Added 6 Nov 2026

Rank-one orthogonal projections form the projective space

#projective space#projections#compact Hausdorff#homeomorphism

Statement

Show that the space P1={P∈Mn(R): P=PT=P2, rank⁡P=1}\mathcal P_1=\{P\in M_n(\mathbb{R}):\ P=P^T=P^2,\ \operatorname{rank}P=1\} of rank-one orthogonal projections is homeomorphic to the real projective space RPn−1\mathbb{RP}^{n-1}.

Hint

Map a unit vector xx to xxTxx^T.

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Solution

Idea. A line through the origin is determined by a unit vector up to sign, and the orthogonal projection onto it is xxTxx^T, which does not see the sign.

Step 1: description of P1\mathcal P_1. If P∈P1P\in\mathcal P_1, its image is a line Rx\mathbb{R}x with ∣x∣=1|x|=1, and PP is the orthogonal projection onto it: Py=(x⋅y)x=xxTyPy=(x\cdot y)x=xx^Ty. So P=xxTP=xx^T. Conversely, for a unit vector xx, xxTxx^T is symmetric, satisfies (xxT)2=x(xTx)xT=xxT(xx^T)^2=x(x^Tx)x^T=xx^T, and has rank 11.

Step 2: the map. ϕ:Sn−1→P1\phi:S^{n-1}\to\mathcal P_1, ϕ(x)=xxT\phi(x)=xx^T, is continuous and onto by Step 1. Moreover xxT=yyTxx^T=yy^T iff x=±yx=\pm y: apply both sides to xx to get x=(y⋅x)yx=(y\cdot x)y, so xx is a multiple of yy, and both are unit vectors.

Step 3: homeomorphism. ϕ\phi induces a continuous bijection ϕˉ:RPn−1=Sn−1/{±1}→P1\bar\phi:\mathbb{RP}^{n-1}=S^{n-1}/\{\pm1\}\to\mathcal P_1. The source is compact (a quotient of the sphere) and P1⊂Mn(R)\mathcal P_1\subset M_n(\mathbb{R}) is Hausdorff. A continuous bijection from a compact space to a Hausdorff space is a homeomorphism.

Remark. The same construction with x∈Cnx\in\mathbb{C}^n and xx∗xx^* gives CPn−1\mathbb{CP}^{n-1} as the space of rank-one Hermitian projections. For n=2n=2, P1\mathcal P_1 is a circle: RP1≅S1\mathbb{RP}^1\cong S^1.

Lesson. Compact-to-Hausdorff continuous bijections are homeomorphisms: use them to identify quotient spaces.