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    Stabilizers and a coset of half-turns in $SO(3)$

    P-BKF80-4

    Area
    Prelim
    Seen at
    BERKELEY
    Years
    1980
    Status
    draft

    Let \(G=SO(3)\). Fix a unit vector \(v\in\RR^3\) and set \[ H_v=\{T\in G:T v=v\}. \]

    1. Show that \(H_v\) is a subgroup of \(G\).
    2. Let \(S_v\) be the set of rotations through \(180^\circ\) about lines orthogonal to \(v\). Show that \(S_v\) is a coset of \(H_v\) in \(G\).
    Solution

    Choose a unit vector \(w\in v^\perp\), and let \(R\in SO(3)\) be the rotation through \(180^\circ\) about the line \(\RR w\).

    1. \(H_v\) is a subgroup of \(G\).

    The identity fixes \(v\), so \(I\in H_v\). If \(T,U\in H_v\), then \[ (TU)v=T(Uv)=Tv=v, \] so \(TU\in H_v\). Finally, if \(T\in H_v\), then applying \(T^{-1}\) to \(Tv=v\) gives \[ T^{-1}v=v, \] so \(T^{-1}\in H_v\). Hence \(H_v\leq G\).

    2. One has\[ S_v=\{T\in SO(3):Tv=-v\}. \]

    If \(T\in S_v\), its rotation axis is a line \(L\subset v^\perp\). A half-turn fixes \(L\) pointwise and acts as \(-I\) on \(L^\perp\). Since \(v\perp L\), one has \(Tv=-v\).

    Conversely, suppose \(T\in SO(3)\) and \(Tv=-v\). Because \(T\) is orthogonal, the plane \(v^\perp\) is \(T\)-invariant. Moreover, \[ \det(T|_{v^\perp})=\frac{\det T}{-1}=-1. \] An orthogonal operator on a two-dimensional real inner-product space with determinant \(-1\) has eigenvalues \(1\) and \(-1\). Hence there is a unit vector \(u\in v^\perp\) with \(Tu=u\). If \(z\in v^\perp\) is a unit vector orthogonal to \(u\), then the determinant condition forces \(Tz=-z\). Together with \(Tv=-v\), this shows that \(T\) fixes the axis \(\RR u\) and negates its orthogonal complement. Thus \(T\) is the rotation through \(180^\circ\) about the line \(\RR u\subset v^\perp\), so \(T\in S_v\).

    3. \(S_v=\boxed{R H_v}\).

    Since the axis of \(R\) lies in \(v^\perp\), step 2 gives \(Rv=-v\).

    If \(H\in H_v\), then \[ (RH)v=R(Hv)=Rv=-v, \] so step 2 gives \(RH\in S_v\). Hence \(RH_v\subseteq S_v\).

    Conversely, let \(T\in S_v\). Since a half-turn satisfies \(R^{-1}=R\), step 2 gives \[ (R^{-1}T)v=R(-v)=v. \] Thus \(R^{-1}T\in H_v\), so \(T\in RH_v\). Therefore \(S_v=RH_v\).

    4. Q.E.D.

    Step 1 proves part 1, and step 3 proves that \(S_v\) is a left coset of \(H_v\) in \(G\).