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All the transformations below fix the origin. A planar rotation preserves lengths and angles and turns every vector through one common oriented angle. Its rotation matrix isAn orthogonal reflection in a line fixes that line and reverses its perpendicular direction; reflection in the horizontal axis has reflection matrix $\mathrm{diag}(1,-1)$. A uniform dilation multiplies every vector by the same positive scale $s$, represented by $sI$; for example $2I$ doubles lengths. A shear mapping fixes one line pointwise and shifts vectors parallel to that line by an amount proportional to their transverse coordinate. A horizontal shear mapping has matrixwith $S_1$ a nontrivial example.

The first given matrix is the clockwise planar rotation through $\pi /4$, namely $A=R_{-\pi /4}$. Direct multiplication givesThus $C$ is a horizontal shear mapping of strength $2$. Because $A^{-1}=R_{\pi /4}$,So $B$ is the composition of that shear mapping followed by an anticlockwise planar rotation through $\pi /4$; order matters. It is not individually one of the four elementary types under these definitions. In particular, its determinant is $1$, but it is not an orthogonal matrix, excluding both a planar rotation and an orthogonal reflection. It is not a scalar matrix, excluding a uniform dilation; its eigenvalues $\sqrt{2}+1$ and $\sqrt{2}-1$ exclude a pure shear mapping, whose eigenvalues are both $1$. As a symmetric positive-definite matrix, $B$ can also be described as stretching two perpendicular principal directions by these reciprocal factors, preserving area.

A Cartesian second-rank tensor is an isotropic tensor if its components are unchanged under every proper rotation matrix $R$:The Kronecker delta is isotropic because $R_{ik}{R}_{jk}={\delta }_{ij}$, the orthogonality relation $R{R}^T=I$.

No symmetry of $T$ needs to be assumed. A half-turn about the first axis has matrix $\mathrm{diag}(1,-1,-1)$; invariance under it forces $T_{12},T_{13},T_{21},T_{31}$ to vanish. A half-turn about the second axis additionally kills $T_{23}$ and $T_{32}$. Thus $T=\mathrm{diag}(d_1,d_2,d_3)$. Each half-turn is the square of an allowed quarter-turn. A quarter-turn about the third axis interchanges $d_1,d_2$, so they are equal. A quarter-turn about the second axis interchanges $d_1,d_3$, so all three are equal. Consequently the complete family of isotropic second-rank tensors isConversely, every member of this family is invariant because $R(\lambda I)R^T=\lambda I$.

Write a member of the special linear group as $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ with $ad-bc=1$. Its Möbius transformation is $z\mapsto (az+b)/(cz+d)$, with the usual values at a pole and at infinity. For nonreal $z$,This follows by multiplying numerator and denominator by $c\bar{z}+d$. Thus the sign of the imaginary part is preserved, while the extended real line is preserved as a set.

Fixing zero requires $b=0$. Therefore its stabilizer subgroup and group orbit areTranslations send zero to each finite real point, and $\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ sends it to infinity.

For $gi=i$, comparing real and imaginary parts gives $a=d$ and $b=-c$, with $a^2+c^2=1$. Exactly the same conditions follow from $g(-i)=-i$. HenceFor any $x\in \mathbb{R}$ and $y>0$, the upper triangular matrixsends $i$ to $x+iy$ and $-i$ to $x-iy$. Therefore the two group orbits are the open complex upper half-plane and the open lower half-plane respectively. Along with the extended real line they exhaust the Riemann sphere, so there are exactly three orbits, the real determinant-one Möbius orbits.

Every $h\in H$ has the form $\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)$ and acts by $z\mapsto a^{2}z+ab$. Thus $Hi$ is the entire complex upper half-plane, by the same explicit matrices $h_{x,y}$. Given $g\in G$, choose $h\in H$ such that $hi=gi$. Then $k=h^{-1}g$ fixes $i$, so $k\in SO(2)$ and $g=hk$. This proves the triangular-rotation factorization of real determinant-one matrices.

For uniqueness, if $h_1{k}_1=h_2{k}_2$, then $h_2^{-1}{h}_1=k_2{k}_1^{-1}\in H\cap SO(2)$. An upper triangular rotation matrix must have $\sin \theta =0$, soThere are consequently exactly two matrix factorizations, $(h,k)$ and $(-h,-k)$. Restricting the first diagonal entry of $h$ to be positive makes the factorization unique. For example, if $g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$ and $q=\sqrt{c^2+d^2}$, that unique branch isAngles are understood modulo $2\pi$ when counting matrices; allowing unrestricted real angle representatives gives infinitely many labels for those same two factorizations.

Use the orthonormal right-handed basisDirect multiplication gives $B(r)u=u$, $B(r)v=rv+w/\sqrt{2}$ and $B(r)w=-v/\sqrt{2}+rw$. Thus the matrix in this basis isThe last two columns are orthogonal and both have squared norm $r^2+1/2$. Therefore $B(r)$ is an orthogonal matrix precisely when $r^2+1/2=1$, giving the unique positive valueIts determinant is $r_0^2+1/2=1$, so it is a proper rotation matrix, not a reflection. The transverse block has cosine and sine both $1/\sqrt{2}$. ConsequentlyWith the oriented axis $u$, the angle is positive by the right-hand rule, since $u\times v=w$. The axis also follows from the unit-eigenvalue eigenspace. As a check, $\mathrm{tr}B(r_0)=1+\sqrt{2}=1+2\cos (\pi /4)$.

If either vector vanishes, both arguments are zero. Otherwise choose a rotation matrix $R$ sending $x/|x|$ to $y/|y|$. The surface measure on a sphere is rotation invariant. Changing variables $\sigma =R\omega$ givesThus the two spherical integrals are equal. The property extends to complex continuous $\varphi$ by treating real and imaginary parts separately, as needed for the Fourier exponential.

Under a proper rotation matrix $R$, invariance of a Cartesian second-rank tensor reads $T=RT{R}^T$, equivalently $TR=RT$. The half-turn $R_z(\pi )=\mathrm{diag}(-1,-1,1)$ forces all entries mixing the $z$ direction with the $xy$ plane to vanish. Thus $T$ consists of a planar $2\times 2$ block $B$ and the entry $T_{33}=c$.

Commutation with the planar quarter-turn $J=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)$ gives $B=aI+bJ$. This block already commutes with every planar rotation. Now use the half-turn $R_x(\pi )=\mathrm{diag}(1,-1,-1)$: on the planar block its conjugation fixes $I$ and sends $J$ to $-J$, so invariance forces $b=0$. ThereforeThe formula uses the Kronecker delta and is sufficient as well as necessary: rotations preserving the unoriented $z$-axis fix both $I$ and $e_3{e}_3^T$. This is an axially invariant second-rank tensor with the planar antisymmetric part removed by the horizontal half-turn.

A counterclockwise planar rotation matrix has the formTo recover this form from preservation of the Euclidean norm, let $\mathbf{u},\mathbf{v}$ be the columns of $R$. Applying the norm condition to the two coordinate unit vectors gives $|\mathbf{u}|=|\mathbf{v}|=1$. Applying it to their sum givesso $\mathbf{u}\cdot \mathbf{v}=0$. The columns are an orthonormal basis, and $R^{T}R=I$: $R$ is an orthogonal matrix.

Write $\mathbf{u}=(\cos \theta ,\sin \theta {)}^T$. There are two perpendicular unit vectors available for $\mathbf{v}$, namely $\pm (-\sin \theta ,\cos \theta {)}^T$. The first gives determinant $1$, the second $-1$. The positive determinant hypothesis selects the first. A norm-preserving planar map with positive determinant is a rotation.

For the commutation condition, writeDirect matrix multiplication givesEquality forces $c=-b$ and $d=a$, so, with $u=a$ and $v=-b$,If $u=v=0$, take $\lambda =0$ and $R=I$. Otherwise set $\lambda =\sqrt{u^2+v^2}>0$ and choose $\cos \theta =u/\lambda$, $\sin \theta =v/\lambda$. ThenThis describes the centralizer of a planar quarter-turn. Its nonzero elements act as a rotation followed by a uniform scaling, just as multiplication by a nonzero complex number does.

Let $R$ be an orthogonal matrix describing a Cartesian change of coordinates, with transformed vector components $v_i^{\prime }=R_{ia}{v}_a$ and $w_j^{\prime }=R_{jb}{w}_b$. Their outer product transforms aswhich is the transformation law of a second-order Cartesian tensor. Here “rank two” counts tensor indices, not the rank of a matrix; a nonzero outer product has matrix rank one.

For an isotropic second-rank tensor, invariance means $RT{R}^T=T$ for every proper rotation matrix. The half-turn matrices $\mathrm{diag}(1,-1,-1)$, $\mathrm{diag}(-1,1,-1)$ and $\mathrm{diag}(-1,-1,1)$ force all off-diagonal entries to vanish. Quarter-turns around coordinate axes then force the three diagonal entries to be equal. Conversely $RI{R}^T=I$, so the most general second-rank isotropic tensor isSince $v{w}^T$ has matrix rank at most one, it cannot equal a nonzero scalar multiple of the three-dimensional identity. Thus its isotropy forces $v{w}^T=0$. If $v\ne 0$, a nonzero component $v_i$ makes the entire $i$th row vanish only when $w=0$; the converse is immediate. The required choices are exactlyFor the Levi-Civita symbol, multilinearity and antisymmetry of the determinant givefor every $R\in SO(3)$. This proves the rotation invariance of the Levi-Civita symbol, hence its third-rank isotropy. The convention is invariance under proper rotations: under an orthogonal reflection $\det R=-1$, the sign reverses. Thus it is a pseudotensor if transformations of both orientations are included.

Use the Einstein summation convention. Substitute the given basis transformation into the expansion of the vector:By linear independence of the basis vectors, $v_j=R_{ij}{v}_i^{\prime }$. In matrix notation $v=R^T{v}^{\prime }$. A rotation matrix is orthogonal, so multiplication by $R$ givesThe placement of the transpose follows from the specified convention $e_i^{\prime }=R_{ij}{e}_j$. It must not be guessed from a different convention for an active rotation of a vector.

Centre the cube at the origin with faces having normals $\pm e_1,\pm e_2,\pm e_3$. Every rotational symmetry fixes the centre and permutes these normals, so its rotation matrix is a signed permutation matrix of determinant one. Conversely any such matrix permutes the cube’s coordinates up to signs and preserves the cube. For each of $3!$ coordinate permutations, four of the eight choices of signs give determinant one. Consequently the rotational symmetry group of a cube hasThis counts the rotations directly, without an unverified stabilizer calculation. The TeX’s $2^k$ is damaged transcription: the PDF prints 24.

Let $X$ be the $3^6$ face colourings. A colouring fixed by a rotation must be constant on each permutation cycle of faces, giving $3^c$ fixed colourings if the rotation has $c$ face cycles. The rotation types are:

There are three face axes, each with two quarter-turns and one half-turn; six edge axes, each with one half-turn; and four vertex axes, each with two nonidentity third-turns. These rotations are distinct and, with the identity, total 24, so the table is exhaustive. Face-axis rotations fix their two axial faces; the remaining four faces cycle or pair. Edge half-turns pair all faces, while vertex third-turns form two triples.

Burnside lemma now givesThe three colours are labelled, orthogonal reflections are excluded, and no condition requires every colour to appear.

Assume first $0<k<N$. Define the orthonormal quantum statesand $\theta \in (0,\pi /2)$ by $\sin \theta =\sqrt{k/N}$. The relevant invariant subspace is the two-dimensional span of these quantum states, not the entire Hilbert space. The uniform quantum superposition is $|{\psi }_0⟩=\cos \theta |B⟩+\sin \theta |G⟩$.

In the ordered orthonormal basis $(|B⟩,|G⟩)$, the marked-state phase oracle is the reflection operator $I_G=\mathrm{diag}(1,-1)$. Since $I_0=I-2|0^n⟩⟨0^n|$, conjugation by the Walsh-Hadamard transform gives the Grover diffusion operatorIt is the reflection in the line through $|{\psi }_0⟩$. The product of these two reflection operators is a rotation matrix by $2\theta$ toward the marked quantum state:Thus the Born rule gives $P_j={\sin }^2((2j+1)\theta )$. Choose a nonnegative integer $j_{\ast }$ nearest to $\pi /(4\theta )-1/2$. Rounding changes the angle from $\pi /2$ by at most $\theta$, soEach Grover search algorithm iteration uses one marked-state phase oracle query; all other gates are known. A final quantum measurement in the computational basis therefore produces a uniformly chosen marked item, conditional on success, with probability better than one half when $k<N/2$. This uses the supplied marked count $k$ to select the stopping time; an unknown $k$ needs an additional search schedule or counting procedure. If $k=N$, no oracle query is needed; if $k=0$, no marked item exists, so a successful search is impossible.

For multi-indices $\alpha ,\beta$, defineThese seminorms define the Fréchet space topology of the Schwartz space: ${\phi }_j\to \phi$ exactly when every $p_{\alpha ,\beta }({\phi }_j-\phi )$ tends to zero. An equivalent increasing family isA tempered distribution is a continuous linear functional on this space. Equivalently, $|⟨T,\phi ⟩|\le C{q}_m(\phi )$ for some $C,m$. The usual weak convergence of tempered distributions means $⟨T_j,\phi ⟩\to ⟨T,\phi ⟩$ for every fixed Schwartz function. The strong dual topology instead requires uniform convergence on every subset of the Schwartz space that is a bounded set in a topological vector space; the Fourier maps below are continuous in both topologies.

Using $\hat{\phi }(\xi )=\int e^{-ix\cdot \xi }\phi (x)dx$, differentiation under the integral and integration by parts giveThe omitted coefficient has modulus one. By the Leibniz rule, every integrand is a finite sum of a polynomial times a derivative of $\phi$. Inserting the integrable weight $(1+|x|{)}^{-n-1}$ bounds its L1 norm by finitely many Schwartz space seminorms. In particular $q_m(\hat{\phi })\le C_m{q}_{m+n+1}(\phi )$. Thus the Fourier transform maps $\mathcal{S}$ continuously into itself.

For completeness, the Fourier inversion theorem follows here by Gaussian regularization. The inverse transform of $e^{-t|\xi {|}^2}$ is $g_t(x)=(4\pi t{)}^{-n/2}{e}^{-|x{|}^2/(4t)}$. The Fourier transform of a Gaussian and Fubini's theorem showAs $t\downarrow 0$, the right side tends to $\phi (x)$ by the approximate identity property, while the left side tends to the undamped inverse integral by dominated convergence theorem, since $\hat{\phi }\in L^1$. HenceThe inverse is $(2\pi {)}^{-n}$ times reflection composed with the continuous Fourier transform, and is therefore continuous on the Schwartz space. This proves a continuous linear isomorphism with continuous inverse.

Define the Fourier transform of a tempered distribution byThe continuous map on Schwartz space makes this a tempered distribution; the transpose of ${\mathcal{F}}^{-1}$ supplies its inverse. Pointwise convergence of pairings proves weak continuity. For the strong dual topology, the transform of a bounded set of Schwartz functions is bounded, so uniform convergence of pairings on bounded sets proves continuity of both Fourier maps there too.

Now let $R\in SO(n)$ be a rotation matrix in the special orthogonal group and write ${\rho }_{R}T=T\circ R$. A change of variables with unit Jacobian givesThe rotation equivariance of the Fourier transform on tempered distributions follows by duality:Therefore $\widehat{{\rho }_{R}T}={\rho }_R\hat{T}$. Applying this identity and the inverse Fourier transform gives the two directions:This is precisely preservation of radial tempered distributions. For $n>1$, $SO(n)$ acts transitively on spheres, so an invariant smooth function is an ordinary radial function.

A nematic director is an unoriented axis: $\mathbf{n}$ and $-\mathbf{n}$ describe the same molecular alignment. A vector average cannot represent this head-tail symmetry. The orientational second moment $⟨n_i{n}_j⟩$ is a symmetric second-rank tensor, and removing its isotropic part gives the traceless second-rank tensor $Q_{ij}\propto ⟨n_i{n}_j-{\delta }_{ij}/2⟩$. It vanishes for an isotropic angular distribution and transforms correctly under rotation matrices.

Every nonzero real symmetric traceless $2\times 2$ matrix has eigenvalues $\lambda /2,-\lambda /2$ and can be written $Q=\lambda (\mathbf{n}{\mathbf{n}}^{\mathsf{T}}-I/2)$ by choosing the eigenvector of the positive eigenvalue as director and $\lambda \ge 0$. Consequently $\mathrm{Tr}{Q}^2={\lambda }^2/2$. The bulk Landau-de Gennes free energy becomesFor $a<0$, its minimum occurs at ${\lambda }^2=-a/b$. The elastic density $K|\nabla \cdot Q{|}^2/2$ is nonnegative and vanishes for a uniform field. Thus, with no boundary condition imposing a texture, the pointwise lower bound is attained byThe negative-amplitude representation is equivalent in two dimensions to rotating the director by $\pi /2$, so the positive choice loses no tensor minima. This is minimization of the specified coarse-grained functional, not a claim of true thermodynamic long-range nematic order at every temperature in an infinite two-dimensional system. The PDF explicitly uses the divergence elastic term; the TeX's $|\nabla Q{|}^2$ loses the dot and would give different elastic coefficients.

The characteristic polynomial of the displayed rotation matrix is $(1-\lambda )({\lambda }^2-2\lambda \cos \theta +1)$, so its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, all of modulus one. The real eigenvector $(0,0,1{)}^T$ spans the rotation axis, while the arguments $\pm \theta$ of the conjugate eigenvalues record the rotation angle on its perpendicular plane.

Applying first the $z$-rotation and then the $x$-rotation givesIts $1$-eigenspace is spanned by $(1,-1,1{)}^T$, so $\mathbf{n}=(1,-1,1{)}^T/\sqrt{3}$. The matrix trace identity $\mathrm{tr}R=1+2\cos \varphi$ gives $0=1+2\cos \varphi$. Thus $\varphi =2\pi /3$.

The basic Galilean transformations include spacetime translations ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{a}$, $t^{\prime }=t-b$; spatial rotations ${\mathbf{x}}^{\prime }=R\mathbf{x}$, $t^{\prime }=t$; and boosts ${\mathbf{x}}^{\prime }=\mathbf{x}-\mathbf{v}t$, $t^{\prime }=t$, where $R$ is a rotation matrix and $\mathbf{v}$ is constant.