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The Clebsch-Gordan decomposition for SU2 statesRealize the tensor product as polynomials of bidegree $(m,n)$ in $(x_1,y_1)$ and $(x_2,y_2)$. The determinant $D=x_1{y}_2-y_1{x}_2$ is invariant under the simultaneous SU2 action. For each $r$, the vectorhas torus weight $q=m+n-2r$ and is killed by $E=x_1{\partial }_{y_1}+x_2{\partial }_{y_2}$. Applying $F=y_1{\partial }_{x_1}+y_2{\partial }_{x_2}$ produces $q+1$ nonzero vectors of distinct weights, with $F^{q+1}{v}_r=0$. The relations $[E,F]=H$ and $H{v}_r=q{v}_r$ give $E{F}^j{v}_r=j(q-j+1)F^{j-1}{v}_r$, so their span is an irreducible copy of $V_q$.

These irreducible submodules have distinct highest weights and form a direct sum: a sum of earlier ones cannot contain an irreducible module inequivalent to all its summands. Equivalently use complete reducibility, obtained by averaging an inner product over the compact group. Assuming $m\le n$, their total dimension iswhich equals the dimension of the tensor product. Hence they exhaust it, proving the theorem.

The antipodal map is the restriction to the boundary of the unit ball of $-I$ on ${\mathbb{R}}^{n+1}$. The induced orientation on the sphere is reversed or preserved according to the determinant of that ambient linear map, namely $(-1{)}^{n+1}$. Its degree is therefore $(-1{)}^{n+1}$. With integral coefficients, for $n\ge 1$ the induced top-homology map isAt $n=0$, ordinary $H_0(S^0)={\mathbb{Z}}^2$ instead has its two generators exchanged; multiplication by $-1$ is the statement on reduced $H_0(S^0)\cong \mathbb{Z}$. This distinguishes the small-dimension exception from the usual top-degree formula.

For $\sigma \in S_n$, let $P_{\sigma }$ be its permutation matrix and defineSince a transposition exchanges two columns, it has determinant $-1$. Thus any expression of $\sigma$ as $r$ transpositions has sign $(-1{)}^r$, proving that the parity is independent of the expression. Moreover, $P_{\sigma \tau }=P_{\sigma }{P}_{\tau }$, so the determinant provesHence the sign homomorphism is a surjective homomorphism for $n\ge 2$.

For any homomorphism $\phi :S_n\to \{\pm 1\}$, all transpositions have the same image because they are conjugate. Since transpositions generate $S_n$, image $+1$ would make $\phi$ trivial. Surjectivity therefore forces every transposition to map to $-1$, and hence $\phi =\mathrm{sgn}$.

Along a curve $s\mapsto (x(s),y(s))$, the chain rule givesTogether with the second-order partial differential equation, these form a linear system for $(u_{xx},u_{xy},u_{yy})$ whose coefficient matrix isA characteristic curve is precisely one along which this system is singular, equivalently where the principal symbol vanishes on a normal covector. Taking the determinant yields

Take a Hopf link with its two unknot components labeled by integer surgery coefficients $2,3$.

The slam-dunk move eliminates a meridional component with coefficient $b$ and replaces the other coefficient $a$ by $a-1/b$. Eliminating the component with coefficient three givesThus this Dehn surgery on a framed link becomes $5/3$ Dehn filling on the unknot, with the required orientation. As a check, its surgery linking matrix iswhose determinant is five.

All classes in the nontrivial homological equalities are taken in $H_1(X_{(p,q)};\mathbb{Z})$, where $X_{(p,q)}=M_{S^2}(p^{\ast }/q,-q^{\ast }/p,\ast )$. The two internal gluing torus components still have inclusion maps into this knot exterior. Interpreting the maps instead as maps into the closed $S^3$ would make every displayed class zero.

Before the two Dehn fillings, $H_1(S^1\times \text{pair of pants};\mathbb{Z})$ has generators $f,h_1,h_2,h_3$ and relation $h_1+h_2+h_3=0$. The Dehn fillings add $p^{\ast }f-q{h}_1=0$ and $-q^{\ast }f-p{h}_2=0$. Put $m=-h_3$. Eliminating the relations using $p{p}^{\ast }-q{q}^{\ast }=1$ givesFor completeness, the three relation rows in generators $(f,h_1,h_2)$ are $(p^{\ast },-q,0)$, $(-q^{\ast },0,-p)$, $(0,1,1)$ when computing the quotient by $m$. Their determinant is $p{p}^{\ast }-q{q}^{\ast }=1$, so $m$ really generates the entire first homology group, and no finite torsion or index is hidden in the elimination.

Orient each exceptional-fiber longitude by ${\mu }_i\cdot {\lambda }_i=1$. In the boundary basis $(m_i,{\tilde{f}}_i)=(-{\tilde{h}}_i,{\tilde{f}}_i)$, write ${\mu }_i={\alpha }_i{m}_i+{\beta }_i{\tilde{f}}_i$ and ${\lambda }_i=u_i{m}_i+v_i{\tilde{f}}_i$, where ${\alpha }_i{v}_i-{\beta }_i{u}_i=1$. ThenThe filled meridian of a solid torus maps to zero, and $({\alpha }_1,{\alpha }_2)=(q,p)$, yieldingAdding multiples of ${\mu }_i$ to ${\lambda }_i$ has no effect. The word “any” therefore means any longitude with this compatible orientation; negating a longitude would negate the corresponding equality.

To compute Turaev torsion, use the general product formula $\tau (S^1\times F)\doteq (1-[f]{)}^{-\chi (F)}$ and multiplicativity under gluing along torus components. For a pair of pants, $\chi (F)=-1$, so the unfilled piece contributes $1-[f]$. The two filling solid torus pieces contribute $(1-[{\lambda }_1]{)}^{-1}$ and $(1-[{\lambda }_2]{)}^{-1}$. With $t=[m]$, the equalities above give $[f]=t^{pq}$, $[{\lambda }_1]=t^p$, and $[{\lambda }_2]=t^q$. ThereforeHere $\doteq$ allows the unit $\pm t^n$: a fully refined Turaev torsion requires an Euler structure and a homology orientation, neither of which the statement specifies. The displayed rational function is the representative whose expansion at $t=0$ starts with $1$, equivalently the standard nonnegative-exponent normalization.

Let $Z_{ir}=1_{\{\text{observation }i\text{ belongs to rat }r\}}$. After integrating out the random intercepts,The fixed-effect design matrix has four independent columns, so the given $A$ has dimensions $176\times 172$, with $A^{T}A=I_{172}$ and $A^{T}X=0$. Consequently the error contrasts $U=A^{T}Y$ have a multivariate normal distribution with mean zero and covariance matrix $W=A^{T}VA$.

The restricted likelihood from orthogonal error contrasts is their density. Equivalently, the estimates maximizeThis is restricted maximum likelihood, not maximization over individual rat effects. The unknown $\beta$ disappears because $A^{T}X=0$. Replacing $A$ by $AQ$ for an orthogonal $Q$ rotates $U$ and conjugates $W$, preserving the determinant and quadratic form; hence the objective is independent of the chosen orthonormal basis. The software's integrated REML log-likelihood can differ by an additive constant depending only on $X$, which has no effect on these variance estimates.

The canonical momentum equation is $\dot{X}=eP+u{X}^{\prime }$. Eliminating $P$ from the phase-space action yields the Lagrangian densityThis is the Polyakov action after identifying its inverse metric tensor density asThe determinant of this matrix is $-1$, as required in two dimensions. The Weyl transformation leaves the matrix unchanged, so $e,u$ encode the metric modulo its conformal factor.

Define the induced worldsheet metric by $g_{\mu \nu }={\eta }_{mn}{\partial }_{\mu }{X}^m{\partial }_{\nu }{X}^n$. Varying the independent inverse metric tensor in the Polyakov action givesConsequently $g_{\mu \nu }=f{\gamma }_{\mu \nu }$, where $f=\frac{1}{2}{\gamma }^{\rho \sigma }{g}_{\rho \sigma }$. For a nondegenerate Lorentzian string worldsheet with the usual compatible time orientation, take $f>0$, or ${\gamma }_{\mu \nu }=e^{2\omega }{g}_{\mu \nu }$. The undetermined function is precisely Weyl invariance. Since $\sqrt{-\gamma }{\gamma }^{\mu \nu }{g}_{\mu \nu }=2\sqrt{-g}$, elimination of the independent metric tensor givesThus the Nambu–Goto action measures minus string tension times Lorentzian worldsheet area. The metric elimination holds in the nondegenerate interior; null endpoints are understood as boundary limits.

Write $I_a={\bar{I}}_{aa}$ and use the right-handed rotationIn the integral defining the second mass moment tensor, change variables to $\mathbf{y}=L^{-1}\mathbf{x}$. The determinant is one and $x_i=L_{ik}{y}_k$, givingSet $C=(I_1+I_2)/2$ and $\Delta =I_1-I_2$. Multiplication givesIn particular, $I_{kk}=I_1+I_2+I_3$ is constant. These expressions also show how the rotating anisotropy enters the mass quadrupole moment.

Put $h_{ab}=\delta g_{ab}$, $h=g^{ab}{h}_{ab}$, and $h^{ab}=g^{ac}{g}^{bd}{h}_{cd}$. Throughout this part use the Riemann curvature tensor convention printed in the paper. Varying $g^{ac}{g}_{cb}={\delta }^a{}_b$ gives the inverse-metric variation. The determinant identity $\delta \log |\det g|=g^{ab}{h}_{ab}$ gives the volume form variation. Together these areIn Lorentzian coordinates the latter reads $\delta \sqrt{-g}=\frac{1}{2}\sqrt{-g}h$.

For the connection, vary metric compatibility ${\nabla }_a{g}_{bc}=0$:Add the equations with derivatives $a,b$, subtract that with derivative $c$, and use symmetry of the lower indices of the Levi-Civita connection. This givesThe difference of two connections is a tensor, so this formula is covariant.

Varying the coordinate curvature formula yields the Palatini identityThe contractions of the connection variation areFinally, $\delta R=(\delta g^{ab})R_{ab}+g^{ab}\delta R_{ab}$ gives the metric variation of scalar curvature:The double divergence may equivalently be written ${\nabla }_a{\nabla }_b{h}^{ab}$: relabeling the contracted dummy indices and using symmetry of $h^{ab}$ gives the same expression. The derivative terms are a covariant divergence, useful when varying a curvature-dependent action.

For an axisymmetric vertical mode of a shearing sheet in this Boussinesq approximation, the perturbations have no horizontal gradients, so background-shear advection vanishes. The linear Coriolis acceleration and perturbation advection of the background shear givewhile thermal conduction and incompressibility giveHere $\beta$ is the thermal diffusion rate of a Fourier mode, not the cooling exponent in Question 2. Since $k\ne 0$, incompressibility forces $u_z^{\prime }=0$, and the vertical momentum equation then forces $P^{\prime }=0$ for this nonzero-wavenumber amplitude.

The three remaining amplitudes satisfyA nontrivial solution requires the determinant to vanish, giving the convective overstability dispersion relationNo division by $s$ or $s+\beta$ was needed, so the thermal energy mode of a shearing sheet is retained. The notation $N^2$ denotes a signed buoyancy frequency squared and can be negative for an adverse stratification; it must not be restricted to the square of a real positive frequency.

Let $q=\alpha h$, $d=\alpha (1-h)$, and use the four regional amplitudes above. Continuity of $\hat{w}$ at the two jumps givesPressure continuity, using vorticity-jump matching for an inviscid shear flow, gives the other two equations:These are the required homogeneous four equations. As a useful reduction, put $X=\tanh q$, $Y=\tanh d$, $H=q(1+XY)$ and $J=q(X+Y)$. Eliminating $A,D$ leavesThe vanishing determinant gives $c^2=(H-Y)(J-XY)/(HJ)$, which expands to the printed dispersion relation.

Label the two atoms in primitive cell $j$ by $u_{2j}$ and $u_{2j+1}$, with $j$ understood modulo $N$ because of the periodic boundary condition. Choose the spring between them to have constant $\lambda$. Newton's second law gives

For a longitudinal normal mode, putPeriodicity requires $e^{2iNka}=1$, so $k=\pi r/(Na)$. Since the primitive-cell length is $2a$, wavevectors differing by $\pi /a$ describe the same mode, and the first Brillouin zone may be chosen asThe amplitudes obeySetting its determinant to zero gives the alternating-spring chain dispersion relationThe lower sign is the acoustic phonon branch and the upper sign is the optical phonon branch.

At the edge of the Brillouin zone,so the frequency band gap isIt vanishes at $\alpha =1$, when the apparent two-atom primitive cell can be reduced to one atom.

Near the centre of the Brillouin zone, a Taylor expansion givesandThe group velocity of the acoustic branch at long wavelength is the speed of sound,

The Leibniz formula for determinants isEquivalently, the determinant is the unique alternating function of the rows that is multilinear and takes value one on the identity matrix. Multiplying one row by $\lambda$ therefore gives

Let $M(t_1,\dots ,t_n)=\mathrm{diag}(t_1,\dots ,t_n)-A$. Expanding the determinant along row $i$ shows thatwhere $C_{ii}$ is the corresponding cofactor. Since a diagonal entry of the adjugate matrix is that same cofactor, the multivariable chain rule along the diagonal $t_1=\cdots =t_n=t$ yieldsTherefore

At a point with $xy\ne 0$, the determinant of the derivative isThe inverse function theorem therefore supplies neighbourhoods on which $f$ has a continuously differentiable, hence differentiable, inverse. Concretely, if $(u,v)=f(x,y)$, the local inverse is the restriction ofwhich is continuously differentiable near the image whenever $y\ne 0$.

Put $w_k=C_k^{-1}$, $u_i=H_i^{-1}$, $A={\sum }_k{w}_k$, $B={\sum }_i{u}_i$, and form the weighted meansThe two equations obtained from the score function areConsequently the maximum-likelihood estimators areThe Hessian matrix of the log likelihood isIts first leading principal minor is negative and its determinant is $AB>0$, so it is a negative-definite matrix. Thus the stationary point is the unique global maximum.

A Lie group is a group that is also a smooth manifold, with differentiable multiplication and inversion. A Lie algebra is a vector space with a bilinear alternating Lie bracket satisfying the Jacobi identity.

For a Matrix Lie group $G$, put $\mathfrak{g}=T_{I}G$. If $X,Y\in \mathfrak{g}$, the matrix commutator $[X,Y]=XY-YX$ again lies in $\mathfrak{g}$: the group commutator $e^{tX}{e}^{sY}{e}^{-tX}{e}^{-sY}$ lies in $G$, and the coefficient of $ts$ in its matrix logarithm is $[X,Y]$. Bilinearity and antisymmetry are immediate, while associativity of matrix multiplication gives the Jacobi identity. Thus $T_{I}G$, with the commutator bracket, is the Lie algebra of a matrix Lie group $\mathcal{L}(G)$.

The special linear group $SL(2,\mathbb{R})$ is the inverse image of the regular value $1$ under the smooth determinant map, and multiplication and inversion are smooth. Differentiating $\det (I+tA)=1+t\mathrm{tr}A+O(t^2)$ shows thatThe Cayley-Hamilton theorem applied to a trace-zero two-by-two matrix gives

The exponential map of a matrix Lie group is the matrix exponentialSince $\det (e^A)=e^{\mathrm{tr}A}=1$, its image lies in $SL(2,\mathbb{R})$. Put $d=\det A$. The identity $A^2=-dI$ sums the series explicitly. If $d>0$, with $r=\sqrt{d}$,If $d=0$, the trace is $2$, while if $d<0$, with $r=\sqrt{-d}$,HenceBut $\mathrm{diag}(-2,-1/2)\in SL(2,\mathbb{R})$ has trace $-5/2$. It is therefore outside the image, so the exponential map is not surjective.

The horizontally invariant magnetized shearing-sheet equations are linear in the horizontal fields, so perturbations about the equilibrium satisfy the same equations. The fixed surface boundary conditions require $\delta B_x=\delta B_y=0$ at $z=\pm H$. Choose a normal mode withwhere the vertical wavenumber is $k=m\pi /(2H)$, $m=1,2,\dots$. With the Alfvén frequencythe four amplitude equations becomeEliminating the velocities leavesA nonzero amplitude requires the determinant of this system to vanish, yielding the ideal magnetorotational dispersion relationEquivalently, with ${\kappa }_r^2=2(2-q){\Omega }^2$,As a quadratic equation for $s^2$, its discriminant is ${\kappa }_r^4+16{\Omega }^2{\omega }_a^2>0$. If $0<{\omega }_a^2<2q{\Omega }^2$, its constant term is negative, so one root $s^2$ is positive and there is an exponentially growing mode. If ${\omega }_a^2>2q{\Omega }^2$, both the constant term and the coefficient of $s^2$ are positive, giving two negative roots and only oscillatory modes. Equality is marginal.

The lowest allowed vertical wavenumber, $k=\pi /(2H)$, is the last to be stabilized as $|B_z|$ increases. Therefore the finite-thickness magnetorotational instability criterion isThere is also a vertically uniform velocity mode with zero magnetic perturbation; for the usual orbitally stable regime $q<2$, it is just stable epicyclic motion. For $q>2$, that uniform mode is already hydrodynamically unstable, independently of the magnetic criterion. At $q=2$ it is marginal. The criterion above concerns the magnetic modes.

At each fixed time, the maximum over nonzero initial conditions is the optimal energy amplification of a linear system:Write $a=e^{{\lambda }_{1}t}$, $d=e^{{\lambda }_{2}t}$ and $b=(a-d)/({\lambda }_1-{\lambda }_2)$. The symmetric matrix $A^{T}A$ has trace $\tau =a^2+b^2+d^2$ and determinant $a^2{d}^2$. Its largest eigenvalue isAn optimal initial condition is a right singular vector of $A(t)$ associated with its largest singular value.

If ${\lambda }_2>{\lambda }_1$, let $\alpha =1/({\lambda }_2-{\lambda }_1)$. Then $A(t)/e^{{\lambda }_{2}t}\to \left(\begin{array}{cc}0& 0\\ \alpha & 1\end{array}\right)$, so the precise long-time asymptotic statement isThe Cauchy-Schwarz inequality shows that the unit initial conditions achieving this leading factor areThis is an adjoint eigenvector, since $L^T(\alpha ,1{)}^T={\lambda }_2(\alpha ,1{)}^T$. It differs from the right eigenvector $v_2=(0,1{)}^T$: the initial condition maximizes the projection onto the slow mode, whereas the eventual state aligns with that right eigenvector. For ${\lambda }_2<0$, even the optimal energy ultimately decays to zero; the prefactor describes enhanced excitation of the slow mode.