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The fixed straight line of a nonidentity three-dimensional rotation. For a rotation matrix $R$, its direction vector is an eigenvector with eigenvalue $1$. A unit vector along it also specifies the positive sense of the angle through the right-hand convention.

Choose a primitive cubic lattice with $a=0.400\mathrm{nm}$, one identical point scatterer per primitive unit cell, incident direction ${\mathbf{s}}_{\mathrm{in}}=(0,0,1)$, and the continuous wavelength band $0.080\le \lambda \le 0.160\mathrm{nm}$. This is a hypothetical model, not the zinc blende specimen of the Laue X-ray diffraction experiment of 1912. Write a candidate reciprocal lattice vector as $\mathbf{G}=(2\pi /a)(h,k,-m)$ with $m>0$, and set $S=h^2+k^2+m^2$. The Laue equations and elastic scattering giveA transmission photographic plate at $z=L>0$ intercepts only $s_{\mathrm{out},z}>0$. Its coordinates relative to the direct beam areFor $(h,k,-m)=(2,1,-1)$ and $L=50\mathrm{mm}$, $S=6$, $\lambda =0.13333\dots \mathrm{nm}$ and ${\mathbf{s}}_{\mathrm{out}}=(2/3,1/3,2/3)$, so $(X,Y)=(50,25)\mathrm{mm}$. Its length is one, and each component difference satisfies the corresponding Laue equations. This verifies a direction and a selected wavelength, without predicting an exposure or measured wave intensity.

The first panel has fourfold symmetry because its chosen Bravais lattice, beam axis and square detector share that symmetry. For the tilted panel, apply a rotation matrix $R$ to each index vector, set $v=R(h,k,l)$, and use $\lambda =-2a{v}_z/|v{|}^2$ and ${\mathbf{s}}_{\mathrm{out}}=(0,0,1)+(\lambda /a)v$. All indices with $|(h,k,l)|\le 2a/{\lambda }_{\min }=10$ suffice: elastic scattering bounds $|\mathbf{G}|\le 2|{\mathbf{k}}_{\mathrm{in}}|$. Thus the plotted spectral and angular selection is complete within this ideal point model and detector window. Multiples of an index vector select inversely scaled wavelengths and can land on the same spot; a white-beam spot need not identify a unique diffraction order.

Rotate the standard basis through $30^{\circ }$ using the rotation matrix whose columns are the new basis vectors:Here $u\cdot v=0$ and $u\cdot u=v\cdot v=1$, so $Q$ is an orthogonal matrix with $Q^{T}Q=I$, where $T$ denotes transpose and $I$ is the identity matrix. For the same vector $p=(2,1{)}^T$, its coordinate vector in this Cartesian basis isThus $\alpha \approx 2.232$ and $\beta \approx -0.134$: travel forwards along $u$ and a little backwards along $v$ to reconstruct $p=\alpha u+\beta v$. The orthogonal projections onto the two axes give these signed coordinates because the measuring vectors are both perpendicular and of unit length.

The squared norm remainsThis is useful when choosing axes aligned with a geometric feature: the coordinates change but distances and dot products retain the same sum-of-products formula. The two directions of the change of basis must be kept straight: $Q$ reconstructs the vector in old coordinates, $p=Q[p{]}_{\mathcal{U}}$, and $Q^T$ extracts its new coordinates. Applying $Q$ to the old coordinate vector $p$ instead would actively rotate the vector.

A short proof explains both formulas. If $p=Qc$, multiplication by $Q^T$ gives $c=Q^{T}p$. Then $\Vert p{\Vert }^2=c^T{Q}^{T}Qc=c^{T}c$. Perpendicular unit-length columns make the matrix inverse equal the transpose; this is the computational convenience supplied by a Cartesian basis.

For $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$, the real matrix solutions of $AJ=JA$ are preciselyA nonzero such matrix equals $\sqrt{u^2+v^2}{R}_{\theta }$, where $R_{\theta }$ is a planar rotation matrix with $\cos \theta =u/\sqrt{u^2+v^2}$ and $\sin \theta =v/\sqrt{u^2+v^2}$. It acts like multiplication by the complex number $u+iv$. The zero matrix is the zero scalar multiple of any rotation.

Apply $R_1$ to a column coordinate vector, then apply $R_2$ in the same fixed basis. The function composition isThus the rightmost matrix acts first. The product is again a rotation matrix, since $(R_2{R}_1{)}^T(R_2{R}_1)=I$ and $\det (R_2{R}_1)=1$. For planar rotations about the same origin, the sine addition formula and cosine addition formula giveFor a concrete composition of rotation matrices, first rotate $(2,0{)}^T$ by $\pi /3$ radians, reaching $(1,\sqrt{3}{)}^T$, then by $\pi /6$ radians:The total angle is $\pi /2$. Reversing these two planar rotations gives the same answer, but noncommuting rotations about fixed axes show why one cannot assume that property for rotations in three dimensions.^[ref]

The vector space ${\mathbb{R}}^3$ with the cross product as Lie bracket, $[x,y]=x\times y$. Bilinearity and $x\times x=0$ give an Alternating Lie bracket; the vector triple product identity proves the Jacobi identity because the three expressions $x\times (y\times z)=y(x\cdot z)-z(x\cdot y)$ cancel cyclically, using the symmetric dot product.

Its relation to the SO(3) Lie algebra is explicit. For a vector $x=(x_1,x_2,x_3)$, define the skew-symmetric matrixThen $[\hat{x},\hat{y}]=\widehat{x\times y}$: acting on $v$, the left side is $x\times (y\times v)-y\times (x\times v)=(x\times y)\times v$. Every real $3\times 3$ skew-symmetric matrix occurs uniquely, so this hat map is a Lie algebra isomorphism. With the standard basis vectors $e_1,e_2,e_3$,and their matrix commutators are $[X,Y]=Z$, $[Y,Z]=X$, $[Z,X]=Y$. With active rotation matrices acting on column vectors, $\exp (t\hat{x})$ rotates through angle $t\Vert x\Vert$ about the oriented axis $x$ when $x\ne 0$. In particular $\exp (tZ)e_1=(\cos t,\sin t,0)$, fixing the positive-angle convention used in the small rotation commutator.

The map $x\mapsto -(i/2)x\cdot \sigma$ is also a Lie algebra isomorphism to the compact real SU(2) Lie algebra, using the Pauli matrix commutator identity. These two identifications concern Lie algebras; they do not identify the corresponding Lie groups globally.

For an objective material isotropy law, permutation of the principal stretches permutes the labelled principal stresses. This follows by conjugating the left stretch tensor with signed permutation rotation matrices. Each principal stress is symmetric in the other two stretches, but need not be symmetric in all three. The law $\sigma =V^2$ provides the example ${\sigma }_i={\lambda }_i^2$. In an invariant representation $\sigma ={\beta }_{0}I+{\beta }_{1}V+{\beta }_2{V}^2$, the scalar coefficients are symmetric functions of the unordered stretches.

Faraday rotation exhibits optical nonreciprocity: with the bias magnetic field held fixed, a retracing beam gains another rotation in the same laboratory sense. Natural optical activity in a reciprocal medium instead unwinds on retracing the path.^[ref]

To reconcile this with the signed $B_{\parallel }$ formula, retain the laboratory $x,y,z$ axes for comparing physical lines. For the backward ray choose a right-handed ray basis $x^{\prime }=x$, $y^{\prime }=-y$, $z^{\prime }=-z$. Its magnetic flux density component is $B_{\parallel }^{\prime }=-B_z$, so its local rotation increment is $-\theta$. But positive $y^{\prime }$ is negative laboratory $y$: the backward increment is therefore $+\theta$ in the laboratory orientation plot. Reversing the ray reverses both the longitudinal component and the basis used for its signed angle. Reversing only one creates the false prediction of cancellation.

In fixed laboratory transverse components, define the rotation matrixFor an ideal mirror at normal incidence that contributes the same complex amplitude factor $r$ to both transverse components, the round-trip map is $rR(\theta )R(\theta )=rR(2\theta )$. The common mirror factor does not change the oscillation line. A reciprocal natural rotator instead gives $rR(-\alpha )R(\alpha )=rI$. This comparison assumes a retraced straight path and no additional polarization-changing elements.

For example, a chosen single-pass $30^{\circ }$ Faraday rotation produces a returned line at $60^{\circ }$ to the original laboratory $x$ axis. A chosen reciprocal $30^{\circ }$ natural rotator returns it to $0^{\circ }$. The same addition makes an ideal optical isolator possible: polarizers at $0^{\circ }$ and $45^{\circ }$ surround a $45^{\circ }$ Faraday rotator. Forward light follows $0^{\circ }\to 45^{\circ }$ and passes. Returning light selected at $45^{\circ }$ reaches $90^{\circ }$ at the entrance polarizer and is rejected by Malus's law. Real transmission loss and incomplete extinction limit the isolation.

Consider a ball moving horizontally with no net horizontal force in the laboratory. Treat its horizontal motion as that of a free particle: ${\mathbf{r}}_{\mathrm{lab}}=(ut,0)$ with constant $u>0$. Axes rotating counterclockwise at constant angular velocity $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ about the launch point assign it the positionThis is an exact rotation matrix transformation, with the axes coinciding at $t=0$. Taking $u=0.25\mathrm{m}{\mathrm{s}}^{-1}$, $\Omega =0.8\mathrm{rad}{\mathrm{s}}^{-1}$, and $0\le t\le 3\mathrm{s}$ gives the two views below. At $t=3\mathrm{s}$, the laboratory position is $(0.75,0)\mathrm{m}$, but the rotating position is approximately $(-0.5530,-0.5066)\mathrm{m}$. Both assign the same distance $0.75\mathrm{m}$ from the origin.

The entire rotating acceleration must include the centrifugal acceleration as well as the Coriolis acceleration. To see the balance explicitly, write $c=\cos \Omega t$ and $s=\sin \Omega t$. Direct differentiation givesThe two apparent contributions areAt launch $\mathbf{r}=0$, so only the Coriolis acceleration contributes to the initial bending. Later the centrifugal acceleration also contributes; the full curved trajectory cannot be explained by keeping only the Coriolis force. In particular, $|{\mathbf{v}}_{\mathrm{rot}}{|}^2=u^2(1+{\Omega }^2{t}^2)$ grows even though the Coriolis force does no work: the rotating-frame kinetic energy increase is supplied by work from the centrifugal force. The laboratory kinetic energy remains $m{u}^2/2$.

A Givens rotation eliminates one component of a vector by rotating just two coordinates, preserving its Euclidean norm. For example,The length remains five: elimination moves the component into the retained coordinate rather than discarding it. Applied to a matrix, the same operation mixes two rows; QR decomposition by Givens rotations uses these local operations to build an upper triangular matrix.

For indices $i<j$, the matrix $G_{ij}(c,s)$ is the identity matrix except for its block on coordinates $i,j$:This article uses left matrix multiplication on column vectors. If $c=\cos \theta$ and $s=\sin \theta$, the block is a rotation matrix through angle $-\theta$: the minus sign belongs in the lower left. Its transpose reverses the rotation. Direct multiplication gives $G^{T}G=I$ and determinant one, so it is an orthogonal matrix; consequently $\Vert Gx{\Vert }_2^2=x^T{G}^{T}Gx=\Vert x{\Vert }_2^2$. Only two coordinates change, but this Euclidean norm identity holds for the entire vector.

A rotation through an angle of $\pi$. In three dimensions, a half-turn about a straight line with unit vector $\mathbf{n}$ fixes its parallel component and negates its orthogonal complement, giving the rotation matrix $2\mathbf{n}{\mathbf{n}}^T-I$ by orthogonal projection.

Take a proper rotation matrix $R$ with real number entries. Then ${\mathbf{d}}_B=\Delta R\hat{\mathbf{k}}$ hasThis odd linear orbital model has an isotropic gap magnitude. Isotropic gap magnitude does not require s-wave pairing: the spin components can rotate together with the orbital direction. Nor does the magnitude fix the relative spin-orbit orientation represented by $R$.

For both displayed models, $\mathbf{d}\times {\mathbf{d}}^{\ast }=0$. The triplet gap matrix $\hat{\Delta }=(\mathbf{d}\cdot \mathbf{\sigma })i{\sigma }_y$, written with Pauli matrices, satisfiesThus these models give the same gap magnitude for the two spin branches. That last simplification would fail for a general complex triplet vector with a nonzero cross product. The displayed angular curves illustrate these specified matrices; they are not a microscopic derivation of the phase diagram.

For a quarter-turn rotation $T$ of ${\mathbb{R}}^3$ about a unit axis vector $n$, the only real invariant subspaces are $\{0\}$, the axis $\mathbb{R}n$, its perpendicular plane $n^{\perp }$, and all of ${\mathbb{R}}^3$.

Indeed $T^2$ fixes the axis and negates the perpendicular plane. Hence $P=(I+T^2)/2$ is the linear projection onto the axis, while $I-P$ is the linear projection onto the perpendicular plane. An invariant subspace $L$ is preserved by these polynomials in $T$, soThe first intersection is either zero or the axis. If the second contains a nonzero $v$, it also contains $Tv$. These are perpendicular nonzero vectors, so they span the perpendicular plane. This proves the classification.

Over the complex numbers the situation changes: the rotation matrix has eigenvalues $1,i,-i$, giving three one-dimensional eigenspaces. The absence of real eigenvalues on the perpendicular plane explains why it has no real one-dimensional invariant subspace.

A second-rank Cartesian tensor invariant under every proper rotation matrix is a scalar multiple of the Kronecker delta. Half-turns about the coordinate axes kill its off-diagonal entries, and quarter-turns make its diagonal entries equal. A tensor of the form $v_i{w}_j$ is an outer product with matrix rank at most one, so it is isotropic in three dimensions only when it vanishes, equivalently when $v=0$ or $w=0$.

A rank-three Cartesian tensor invariant under every proper rotation matrix in three dimensions is a multiple of the Levi-Civita symbol. Half-turns about coordinate axes force every nonzero component to contain each coordinate label once, and quarter-turns make these components alternating. Conversely, the determinant formula $R_{ia}{R}_{jb}{R}_{kc}{ϵ}_{abc}=(\det R){ϵ}_{ijk}$ proves invariance when $\det R=1$. Requiring invariance under reflections as an ordinary tensor forces $b=0$; invariance under proper rotations alone permits nonzero $b$.

A Lie group is a group that is also a finite-dimensional smooth manifold, with multiplication $G\times G\to G$ and inversion $G\to G$ both smooth maps. Its tangent space at the identity element carries a natural Lie algebra structure. Smoothness lets us differentiate the group operation; being a group means that every transformation can be composed and undone. The dimension of a manifold counts independent local parameters, rather than the number of elements of the group. See MIT 18.745, Lecture 2, Definition 2.1 and Example 2.5.

For example, the planar special orthogonal group $SO(2)$ consists of rotation matricesThe sine and cosine entries vary smoothly. Four entries appear, but they are constrained by $R_{\theta }^T{R}_{\theta }=I$ and $\det R_{\theta }=1$: locally, one angle specifies the whole matrix. Multiplication adds angles, with angles differing by $2\pi$ representing the same rotation. Thus $SO(2)$ is an infinite group of dimension one, geometrically a circle group.

The positive-dimensional examples motivate the language of continuous symmetry, but the definition also admits finite groups with the discrete topology as zero-dimensional Lie groups. Conversely, being an infinite group alone says nothing about having a continuous angle parameter. The compatibility between smooth maps and the group operation is what makes differentiation useful.

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]

In a right-handed reference frame, the rotation matrix from an orbital plane frame whose first axis points along the ascending node is $R_z(\Omega )R_x(I)$. Its columns are the orbital frame's orthonormal basis expressed in reference coordinates. The unit vector normal is $(\sin \Omega \sin I,-\cos \Omega \sin I,\cos I)$; its dot product with another such normal gives the cosine of their mutual inclination.

For a real basis matrix $E$ with basis vectors as rows, the reciprocal basis has row matrix $E^{-T}$. If either basis is orthonormal, both are, and $E^{-T}=E$. Thus $E$ is an orthogonal matrix, with determinant either $1$ or $-1$. It is a rotation matrix precisely when the original ordered basis is positively oriented relative to the reference coordinates. For example, $\mathrm{diag}(1,1,-1)$ is its own orthonormal reciprocal basis matrix but is an improper orthogonal transformation.

A real skew-symmetric matrix is an orthogonal matrix exactly when its square is minus the identity matrix. Indeed, $B^{\mathsf{T}}B=-B^2$. Such a matrix can exist only in even dimension: taking determinants of $B^2=-I$ would otherwise give $(\det B{)}^2=-1$. In two dimensions every example isThese are rotation matrices through opposite right angles. Their orthogonal similarity transforms preserve eigenvalues, but do not in general produce a diagonal matrix.