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Use a fixed right-handed Cartesian basis in three dimensions and active rotations in three dimensions through $\pi /2$ radians. The right-hand rule gives the following rotation matrices about the positive $x$ and $z$ axes:For the vector $v=(1,1,0{)}^T$, the two orders giveBoth outputs have Euclidean norm $\sqrt{2}$, but they point in different directions. Each arrow below is a vector from the origin; the curved paths track its endpoint during each rotation in three dimensions.

The axes in this example remain fixed throughout. A command to turn about an object's own current axis describes a different operation because that axis can move after the first turn. With an object's orthonormal basis represented by columns of $Q$, a fixed-axis rotation matrix $R$ acts by $RQ$, whereas the corresponding rotation expressed in the object's own basis acts by $QR$.^[ref]

The dihedral group here has order $2n$. Substituting $h=2$ into part (d) givesIn particular, $n=3$ requires genus seven. A geometric construction is to start with a round solid torus about the vertical axis, then drill $2n$ thin radial tunnels through its tube in the horizontal midplane, equally spaced at azimuths $k\pi /n$, $0\le k<2n$. The tunnels run from the central opening to the exterior; their directions are horizontal and radial. Round all their rims in the same way, preserving the dihedral symmetry, and let $X$ be the boundary of the resulting solid.

Here is why the construction has exactly the required properties. Every nonidentity power of $\beta$ is a rotation in three dimensions fixing the vertical axis, which lies in the central opening and is disjoint from $X$. Each element $\alpha {\beta }^j$ is a half-turn about a horizontal line at azimuth $j\pi /n$, modulo $\pi$. The $n$ such lines meet the original solid torus in $2n$ short radial segments. Each segment lies strictly inside one drilled tunnel, so these axes are also disjoint from $X$. A nonidentity rotation in three dimensions fixes only its axis. Thus no nonidentity element fixes a point of $X$, proving a free action of a group.

Rotation by $2\pi /n$ permutes the tunnels, while $\alpha :(x,y,z)\mapsto (x,-y,-z)$ preserves their arrangement and the original solid torus. Hence $X$ is preserved by the two generators and therefore by $G$. The tunnels can be chosen sufficiently thin to be pairwise disjoint, and the rounding can be made group-invariant in small disjoint neighbourhoods of their rims. It neither changes the genus nor introduces an intersection with any axis.

Each tunnel removes two discs from the existing boundary and inserts the cylindrical wall of the bore. Its contribution to Euler characteristic is $-2$: the two removed discs contribute $-2$, and the inserted annulus and its boundary circles have zero Euler characteristic. The boundary remains connected and orientable, so each tunnel increases genus by one. Starting from a torus and drilling $2n$ tunnels therefore gives $g=1+2n$. The construction also supplies a Riemann surface: use the complex structure associated with the induced oriented Riemannian metric. The ambient generators are isometries preserving the solid and ambient orientation of a vector space, so they preserve the induced boundary orientation and act by biholomorphisms. Part (d) then confirms that the quotient space has genus two. This is a free dihedral action on a drilled torus.

With determinant-one normalization, the necessary and sufficient condition for sphere rotations as special-unitary Möbius transformations isEquivalently, the matrix $M=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ belongs to the special unitary group $SU(2)$. The two determinant-one representatives $M$ and $-M$ induce the same Möbius transformation, and both satisfy this condition.

Use north-pole stereographic projection in the direction from the plane to the sphere:Its chord-distance identity is $|\phi (z)-\phi (w)|=2\chi (z,w)$, where $\chi$ is the chordal metric. Direct subtraction, using $ad-bc=1$, givesThis is first calculated away from poles and then extended continuously to the Riemann sphere. A sphere rotation preserves chord distances, so for distinct finite nonpole pointsUsing three distinct points shows all these positive ratios equal one. Hence $H(z)=1+|z{|}^2$ everywhere. Comparing its coefficients givesThus $M$ is a unitary matrix. Comparing $M^{-1}=M^{\ast }$ with $M^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$ proves the boxed relations.

Conversely, these relations give $H(z)=1+|z{|}^2$, so the induced sphere map $T=\phi g{\phi }^{-1}$ preserves all Euclidean chord distances. Such a bijection of the unit sphere is the restriction of an orthogonal matrix: let $v_j=T(e_j)$; preservation of distances makes these an orthonormal basis, and $T(p)\cdot v_j=p\cdot e_j$ forces $T(p)={\sum }_j{p}_j{v}_j$. Finally the Möbius transformation is holomorphic with positive real Jacobian at every ordinary coordinate point. Conjugating by stereographic projection preserves orientation, so this orthogonal matrix has determinant $+1$. It is therefore a genuine rotation in three dimensions, proving sufficiency rather than just spherical isometry.

For the circle assertion, a nondegenerate generalized circle has an equationFor $A\ne 0$ this is an ordinary circle, and for $A=0$ it is a line with infinity included. On the sphere write coordinates $(p_1,p_2,p_3)$, with $z=(p_1+i{p}_2)/(1-p_3)$ and $|z{|}^2=(1+p_3)/(1-p_3)$. Multiplication by $1-p_3$ gives the planeIf $n=(2\mathrm{Re}B,-2\mathrm{Im}B,A-C)$ and $h=-(A+C)$, then $|n{|}^2-h^2=4(|B{|}^2-AC)>0$. Thus the plane intersects the unit sphere in a proper circle. It contains the north pole exactly when $A=0$, accounting for the point at infinity on a line.

Conversely, a circle on the sphere is the intersection with a plane $n\cdot p=h$, $|h|<|n|$. Set $A=(n_3-h)/2$, $C=(-n_3-h)/2$ and $B=(n_1-i{n}_2)/2$. The preceding equation is recovered and has $|B{|}^2-AC=(|n{|}^2-h^2)/4>0$. Its inverse stereographic image is an ordinary circle or a line with infinity attached. This proves both directions of the circle-plane relation under stereographic projection.

Write $\mathbf{x}=s\mathbf{n}+\mathbf{v}$, with $\mathbf{v}\cdot \mathbf{n}=0$. The linear map sends $s\mathbf{n}$ to $as\mathbf{n}$ and sends $\mathbf{v}$ to $a\mathbf{v}+b(\mathbf{n}\times \mathbf{v})$. On the perpendicular plane, $\mathbf{v}$ and $\mathbf{n}\times \mathbf{v}$ are perpendicular vectors of equal length. ThusThe expression in parentheses is a rotation in three dimensions by $\vartheta$ about the oriented $\mathbf{n}$ axis, with the positive direction fixed by the cross product. Therefore the map scales the axial component by $a$, scales the perpendicular component by $\sqrt{a^2+b^2}$, and rotates the perpendicular plane by $\arctan (b/a)$. Since $a,b>0$, the angle lies between zero and $\pi /2$. This is an axis-angle decomposition, with different axial and transverse scale factors.

Write $G=\mathrm{Rot}(D)$ for the rotational symmetry group of a regular dodecahedron. Centre the solid at the origin. A rotation in three dimensions is determined by the image of a chosen face and one of its vertices, so there are at most $12\cdot 5=60$ rotations. All these choices occur. The outward unit normal to a face and the direction from its centre to the chosen vertex determine an oriented orthonormal frame. The proper rotation mapping one such frame to another maps the whole regular pentagonal face. At each edge, the common dihedral angle and convexity then determine the adjacent face uniquely; propagating across the connected face-adjacency graph shows that the rotation maps the entire solid to itself. Hence $|G|=60$, and the action is transitive on faces, vertices and edges.

Its rotations can be counted geometrically. There are six axes joining opposite face centres, each allowing four nonidentity rotations of order five; ten axes joining opposite vertices, each allowing two of order three; and fifteen axes joining opposite edge midpoints, each allowing a half-turn. The orbit-stabilizer theorem also verifies these possibilities: vertex stabilizers have order $60/20=3$ and edge stabilizers have order $60/30=2$, and their nonidentity elements rotate about the corresponding vertex or edge-midpoint axis. Together with the identity this accounts for all $1+24+20+15=60$ elements. In particular there are no elements of order four.

We now determine the conjugacy classes. The group is transitive on edges, so all fifteen edge half-turns are conjugate. For a rotation whose angle is neither zero nor $\pi$, a commuting rotation must preserve its oriented axis: conjugation carries the axis and its rotation angle to their images, and reversing the axis would reverse a non-half-turn angle. Thus its centralizer consists of rotations about the same axis. For a face rotation this centralizer has order five, and for a vertex rotation it has order three. The orbit-stabilizer theorem for conjugation therefore gives class sizes $60/5=12$ and $60/3=20$. The twenty vertex rotations form one class, and the twenty-four face rotations form two classes. The class sizes are $1,15,20,12,12$.

A normal subgroup is a union of these classes and contains the identity. Its possible orders areBy Lagrange's theorem its order must divide $60$. Only $1$ and $60$ in this list do so. Therefore $\boxed{G\text{ is simple of order }60}$.

Let $Q\in \mathrm{SO}(3,\mathbb{R})$. Since $Q^{T}Q=I$ and $\det Q=1$,so $\det (Q-I)=0$. Choose a unit vector $e$ with $Qe=e$. Its orthogonal complement is an invariant subspace: for $v\perp e$, $e\cdot Qv=(Q^{T}e)\cdot v=e\cdot v=0$. In an orthonormal basis $(e,e_1,e_2)$, $Q$ therefore has a scalar block $1$ and a two-dimensional orthogonal matrix block of determinant one. Such a block has columns $(\cos \theta ,\sin \theta {)}^T$ and $(-\sin \theta ,\cos \theta {)}^T$. Consequently $Q$ is a rotation in three dimensions about the axis $\mathbb{R}e$ through angle $\theta$, including the identity when $\theta =0$.

To construct the Householder reflections, let $S_0$ reflect in the plane spanned by $e,e_1$ and let $S_{\psi }$ reflect in the plane spanned by $e,\cos \psi e_1+\sin \psi e_2$. They fix $e$, while on $e^{\perp }$ their matrices areTheir product is the planar rotation matrix through $2\psi$. Taking $\psi =\theta /2$ proves $\boxed{Q=S_{\theta /2}{S}_0}$. This gives the composition of two plane reflections explicitly; each is a Householder reflection $I-2n{n}^T$ for the corresponding unit plane normal $n$.

TakeIts defining matrix has nonzero determinant eight, so it is a Möbius transformation. Direct substitution shows that it fixes $p=1$ and $q=-1$, which satisfy $p\overline{q}=-1$. However, $M^{\prime }(1)=8/(3-1{)}^2=2$. Differentiating the inverse projection formula gives the round spherical line element $ds=2|dz|/(1+|z{|}^2)$. At a fixed finite point, its local length multiplier under a Möbius transformation is exactly $|M^{\prime }(p)|$, since the metric factors in numerator and denominator agree. Here that multiplier is two, so this map is not an isometry and cannot be a rotation in three dimensions of the sphere. It demonstrates that antipodal fixed points do not characterize sphere rotations.

For stereographic projection, choose the north pole $N=(0,0,1)$ of the unit sphere and identify the plane $Z=0$ with the complex plane by $z=X+iY$. Project a point $P=(X,Y,Z)\ne N$ along the line joining it to $N$; adjoin $z=\infty$ for the north pole to identify the whole sphere with the Riemann sphere. A point of that line is $N+t(P-N)$. Its last coordinate vanishes when $t=1/(1-Z)$, givingUsing $X^2+Y^2+Z^2=1$, or solving this line's second intersection with the sphere, gives the inverseA great circle lies in a plane $h_{1}X+h_{2}Y+h_{3}Z=0$ through the origin. Its projected equation isIt is a straight line only if $h_3=0$, meaning the plane contains the north pole and the south pole. Thus if all three sides of a nondegenerate spherical triangle projected to straight sides of a Euclidean triangle, all three distinct side planes would contain both poles. Each pair of side planes could then meet on the sphere only at those same two poles, which cannot supply three distinct triangle vertices. A nondegenerate spherical triangle cannot project to a Euclidean triangle. If a side passes through the projection pole, its image is unbounded and also cannot be a finite triangle side.

A nonidentity rotation in three dimensions has an axis through the origin: an orthogonal three-by-three matrix of determinant one has an eigenvalue one, since nonreal eigenvalues pair as conjugates, real eigenvalues are $\pm 1$, and their product is one. Its axis meets the unit sphere in an antipodal pair, which is fixed. In the identity case every point is fixed, so an antipodal pair can again be selected. Projecting $P$ and $-P$ giveswhenever both points are finite. Thus the associated Möbius transformation fixes an antipodal pair satisfying the claimed relation. The universally valid relation is $q=-1/\overline{p}$ on the Riemann sphere, with $0$ and $\infty$ paired. A rotation about the projection axis fixes exactly that pair, so the PDF's literal finite product needs this extended-coordinate convention. This is the antipodal stereographic coordinate relation.

Every $Q\in O(3)$ with $\det Q=1$ is a rotation in three dimensions, hence is the product of two reflections in planes through its axis whose angle is half the rotation angle.

If $\det Q=-1$, its real eigenvalue is $-1$: the other two eigenvalues are either a complex-conjugate pair or two real signs, and their product is positive. Let $v$ be a corresponding unit eigenvector and let $H$ be reflection in $v^{\perp }$. Then $HQ$ fixes $v$ and has determinant $1$, so it is a rotation and hence a product of two reflections. Since $Q=H(HQ)$, every member of $O(3)$ is a product of at most three reflections.

Two reflections have determinant $1$, whereas one reflection has eigenvalues $(-1,1,1)$. The map $-I\in O(3)$ has determinant $-1$ but is not a reflection, so it cannot be a product of at most two reflections. Thus three are sometimes necessary.

The unit sphere and the kernel $e^{-c|y-x{|}^2}$ are invariant under every rotation in three dimensions fixing $y$. Therefore $T(y)$ has the same axisymmetry, so part (a) gives

Choose spherical polar coordinates with polar axis $y$ and put $u=\cos \theta =x\cdot y$. Since $|x|=|y|=1$,The tensor contraction giving the trace isSimilarly,At $c=0$ these expressions have the continuous limiting values $4\pi$ and $4\pi /3$. Writing them as $A=T_{kk}=3\alpha +\beta$ and $B=T_{ij}{y}_i{y}_j=\alpha +\beta$ gives

The orthogonal group isThe special orthogonal group is its determinant-one subgroup

Every eigenvalue of an orthogonal matrix has modulus one. A real three-by-three matrix has at least one real eigenvalue, and nonreal eigenvalues occur in a complex conjugate pair. If $Q\in SO(3)$ has such a pair $e^{\pm i\theta }$, their product is one, so the remaining eigenvalue is $\det Q=1$. If all three eigenvalues are real, each is $\pm 1$ and their product is one; an odd number of three signs with product one must include $+1$. Hence every element of $SO(3)$ has an eigenvector of eigenvalue one and represents a rotation in three dimensions about its span.

It is false that every element of $O(3)$ is either a rotation or a plane reflection. For example,is an improper orthogonal transformation combining a rotation with a reflection. Its eigenvalues are $e^{\pm i\theta }$ and $-1$, whereas a plane reflection has eigenvalues $1,1,-1$ and a rotation has determinant one.

If $T^2=I$, every $v\in V$ has the decompositionwhose first term belongs to $E_1$ and second to $E_{-1}$. Their intersection is zero by ii, so $V=E_1\oplus E_{-1}$ is a direct sum. Choosing bases of the two eigenspaces gives a diagonal matrix with entries $1$ and $-1$.

Since $T^2\ne T$ while $T^2=I$, we have $T\ne I$. Thus $\dim E_{-1}$ is $1$, $2$, or $3$. Reordering the basis gives exactlyFor $-I_3$, $T$ is point inversion through the origin. For $\mathrm{diag}(1,1,-1)$, it fixes the plane $E_1$ pointwise and reverses the complementary line $E_{-1}$: if $v=p+\ell$ in this direct sum, then $Tv=p-\ell$. For $\mathrm{diag}(1,-1,-1)$, it fixes the line $E_1$ pointwise and reverses the complementary plane $E_{-1}$.

These are instances of oblique symmetry of complementary subspaces, allowing oblique eigenspaces. They are a Euclidean reflection in a hyperplane and a rotation in three dimensions through $\pi$ about a line, respectively, when the two eigenspaces are perpendicular. The hypotheses do not require an orthogonal transformation or an orthonormal basis, so that additional metric property cannot be inferred just from the displayed diagonal forms.

A screw motion combines a rotation in three dimensions about a fixed line with a translation parallel to that line. With unit axis direction $e$, axial speed $v$, and rotation matrix $R_t$ fixing $e$, each map ${\Phi }_t(x)=R_{t}x+tve$ is a proper Euclidean motion of Euclidean three-space, and together these maps form a one-parameter subgroup. Each off-axis group orbit is a circular helix when both rotational speed and axial speed are nonzero. A pure rotation or translation is a limiting case.

The SO(3) group consists of real three-by-three orthogonal matrices with determinant one. It acts by rotation in three dimensions. Its group manifold is ${\mathbb{RP}}^3$, since the Adjoint double cover from SU(2) to SO(3) identifies antipodal points of the three-sphere.