Let \(SO(3)\) be the group of orthogonal transformations of \(\mathbb R^3\) of determinant \(1\). Let \[ Q=\{A\in SO(3):A^T=A,\ A\ne I\}, \] and let \(\mathbb{RP}^2\) denote the space of lines through the origin in \(\mathbb R^3\).
Show that \(\mathbb{RP}^2\) and \(SO(3)\) are compact metric spaces in their usual topologies.
Show that \(\mathbb{RP}^2\) and \(Q\) are homeomorphic.
Solution
1. The group \(SO(3)\) is a closed subset of\[ M_3(\RR)\cong\RR^9. \]
One has \[ SO(3) = \{A\in M_3(\RR):A^TA=I,\ \det A=1\}. \] The maps \[ A\longmapsto A^TA \] and \[ A\longmapsto\det A \] are continuous. Therefore \(SO(3)\) is the inverse image of the closed set \[ \{I\}\times\{1\} \] under the continuous map \[ A\longmapsto(A^TA,\det A). \] Hence \(SO(3)\) is closed.
2. The group \(SO(3)\) is bounded in \(M_3(\RR)\).
If \(A=(a_{ij})\in SO(3)\), each column is a Euclidean unit vector because \[ A^TA=I. \] Thus \[ \abs{a_{ij}}\leq1 \] for every \(i,j\). Hence \(SO(3)\) is contained in the bounded cube \[ [-1,1]^9. \]
3. The space \(SO(3)\) is a compact metric space in its usual topology.
By steps <1>1–<1>2, \(SO(3)\) is closed and bounded in the finite dimensional Euclidean space \(M_3(\RR)\cong\RR^9\). The Heine–Borel theorem therefore gives compactness. The Euclidean metric on \(M_3(\RR)\) restricts to a metric inducing the usual topology on \(SO(3)\).
4. The real projective plane is the quotient\[ \mathbb{RP}^2 \cong S^2/(u\sim-u). \]
Every line through the origin intersects the unit sphere in exactly two antipodal points \(u\) and \(-u\). Thus sending a unit vector to the line it spans identifies precisely antipodal pairs, giving the stated quotient.
5. The space \(\mathbb{RP}^2\) is compact.
The sphere \(S^2\) is compact, and the quotient map \[ S^2\longrightarrow S^2/(u\sim-u) \] is continuous and surjective. A continuous image of a compact space is compact. Step 4 identifies the quotient with \(\mathbb{RP}^2\).
6. For lines \([u],[v]\in\mathbb{RP}^2\), represented by unit vectors\(u,v\in S^2\), define\[ d([u],[v]) = \min\{\norm{u-v},\norm{u+v}\}. \]This is a metric inducing the usual quotient topology on\(\mathbb{RP}^2\).
Replacing \(u\) or \(v\) by its negative interchanges the two displayed quantities, so the minimum is well defined on lines.
The formula is the quotient metric for the action of the two-element group \[ \{I,-I\} \] on \(S^2\) by Euclidean isometries. Therefore it is a metric. Its metric balls are exactly the images under the quotient map of antipodally symmetric unions of sufficiently small spherical metric balls, so it induces the quotient topology.
7. Consequently, \(\mathbb{RP}^2\) is a compact metric space in its usualtopology.
Compactness is step 5 and metrizability by the usual projective topology is step 6.
8. If \(A\in Q\), then\[ A^2=I. \]
By definition, \(A\) is orthogonal and symmetric. Orthogonality gives \[ A^TA=I, \] while symmetry gives \(A^T=A\). Hence \[ A^2=I. \]
9. Every \(A\in Q\) has eigenvalues\[ 1,-1,-1, \]counted with multiplicity.
By step 8, every eigenvalue satisfies \[ \lambda^2=1, \] so every eigenvalue is \(\pm1\). Since \(A\in SO(3)\), \[ \det A=1. \] The number of eigenvalues equal to \(-1\) is therefore even. Because \(A\neq I\), this number is not zero. In dimension three it must therefore be exactly two.
10. For \(A\in Q\), the eigenspace\[ L_A=\ker(A-I) \]is a line, and\[ A=2P_{L_A}-I, \]where \(P_{L_A}\) is the orthogonal projection onto \(L_A\).
By step 9, the eigenvalue \(1\) has multiplicity one, so \(L_A\) is one-dimensional. Since \(A\) is symmetric, eigenspaces for distinct eigenvalues are orthogonal. Thus \[ \RR^3=L_A\oplus L_A^\perp, \] and \(A\) acts as \(+I\) on \(L_A\) and as \(-I\) on \(L_A^\perp\).
The operator \(2P_{L_A}-I\) has exactly the same action on these two orthogonal summands, so the operators are equal.
11. For a line \(L=[u]\in\mathbb{RP}^2\), with \(u\) a unit vector, define\[ \Phi(L)=2uu^T-I. \]Then \(\Phi(L)\in Q\).
The matrix \[ uu^T \] is the orthogonal projection onto the line \(\RR u\). Hence \[ \Phi(L)=2P_L-I. \] It is symmetric. It acts as \(+1\) on \(L\) and as \(-1\) on \(L^\perp\), so it is orthogonal, has determinant \[ 1\cdot(-1)^2=1, \] and is not the identity. Therefore \(\Phi(L)\in Q\).
Replacing \(u\) by \(-u\) does not change \(uu^T\), so the formula depends only on the line \(L\).
12. The map\[ \Phi:\mathbb{RP}^2\longrightarrow Q \]is bijective.
For surjectivity, let \(A\in Q\). Step 10 gives \[ A=2P_{L_A}-I=\Phi(L_A). \]
For injectivity, if \[ \Phi(L)=\Phi(L'), \] then their \(+1\) eigenspaces agree. By construction, the \(+1\) eigenspace of \(\Phi(L)\) is exactly \(L\), and similarly for \(L'\). Hence \(L=L'\).
13. The map \(\Phi\) is continuous.
The map \[ S^2\longrightarrow M_3(\RR), \qquad u\longmapsto2uu^T-I, \] is polynomial in the coordinates of \(u\), hence continuous. It is invariant under \(u\mapsto-u\), so by the universal property of the quotient topology it descends to a continuous map \[ \Phi:\mathbb{RP}^2\longrightarrow Q. \]
14. The map \(\Phi\) is a homeomorphism:\[ \boxed{ \mathbb{RP}^2\cong Q. } \]
By step 7, \(\mathbb{RP}^2\) is compact. The space \(Q\) is a subspace of the metric space \(SO(3)\), hence is Hausdorff. Steps <1>12–<1>13 show that \(\Phi\) is a continuous bijection from a compact space to a Hausdorff space. Such a map is a homeomorphism.
15. Q.E.D.
Steps 3 and 7 prove part (1), while step 14 proves part (2).