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For a linear map $\theta :U\to V$, its rank of a linear map is $r(\theta )=\dim \mathrm{im}\theta$, and its nullity is $n(\theta )=\dim \ker \theta$. Choose a basis $e_1,\dots ,e_s$ of its kernel and extend it to a basis $e_1,\dots ,e_d$ of $U$. Then $\theta e_{s+1},\dots ,\theta e_d$ span its image. They are linearly independent: a linear combination mapped to zero would belong to the kernel, contradicting independence of the extended basis unless all its coefficients vanish. Hence $r(\theta )=d-s$, provingFor endomorphisms of $U$, define $(\theta +\varphi )(u)=\theta (u)+\varphi (u)$ and $(\theta \varphi )(u)=\theta (\varphi (u))$. The image of the sum lies in the sum of the two images. ThusTo prove the product inequality, restrict $\varphi$ to $K=\ker (\theta \varphi )$. This map has kernel $\ker \varphi$ and image $\mathrm{im}\varphi \cap \ker \theta$: every vector in the intersection has a preimage in $K$. Applying the rank-nullity theorem to the restricted map givesNow put $d=\dim U$ and $s=r(\theta )+r(\varphi )$. If both equalities hold, $r(\theta +\varphi )=s\le d$, while $n(\theta \varphi )=2d-s\le d$ forces $s\ge d$. Therefore $s=d$, the sum has full rank, and the product has nullity $d$. The sum is an isomorphism and the product is zero.

Conversely, if $\theta \varphi =0$, then $\mathrm{im}\varphi \subseteq \ker \theta$, giving $s\le d$. If the sum is also an isomorphism, its rank is $d\le s$. Thus $s=d$, and the two inequalities become equalities. This proves the equality in the rank-sum and nullity-product inequalities in both directions.

The function is Fréchet differentiable at $(a,b)$ when there is a linear map $L:{\mathbb{R}}^2\to \mathbb{R}$ such thatIf it exists, $L(h,k)=f_x(a,b)h+f_y(a,b)k$, as is seen by taking increments along each coordinate axis.

To prove differentiability from continuous partial derivatives, take a small rectangle in the given neighbourhood and split the increment along its sides. The one-variable mean value theorem givesfor some $0<\theta ,\eta <1$ when the corresponding increment is nonzero. If an increment is zero, its term is simply omitted. The required one-variable continuity follows from existence of the corresponding partial derivative on the segment. Subtracting $L(h,k)$ gives a remainder whose absolute value, by continuity of both partial derivatives at $(a,b)$, is at mostfor sufficiently small increments. Since $ϵ$ is arbitrary, the normalized remainder tends to zero. Thus the function is differentiable at $(a,b)$, with the displayed derivative.

A linear map $\Phi$ is a completely positive map if ${\mathrm{id}}_R\otimes \Phi$ sends every positive operator to a positive operator for every finite-dimensional auxiliary system $R$. Positivity of $\Phi$ alone tests only inputs without an auxiliary system and is weaker.

For the Kraus representation, let $\Omega \ge 0$ be any operator on the auxiliary system and the input system. ThenFor any vector $v$, its expectation is ${\sum }_k⟨(I_R\otimes A_k^{†})v,\Omega (I_R\otimes A_k^{†})v⟩\ge 0$. Every amplification is therefore positive, proving the Kraus-form map is completely positive. No normalization condition on the $A_k$ is needed for this positivity proof. To make it a deterministic quantum channel, one additionally requires ${\sum }_k{A}_k^{†}{A}_k=I$; a physical outcome branch instead obeys ${\sum }_k{A}_k^{†}{A}_k\le I$.

Transposition is positive on one system: for any positive $\rho$ and any $v$, $v^{†}{\rho }^{T}v$ is the complex conjugate of ${\bar{v}}^{†}\rho \bar{v}$, hence is real and nonnegative. It fails complete positivity already for a qubit. Apply partial transposition to the second subsystem of the Bell state $|{\Phi }^{+}⟩=(|00⟩+|11⟩)/\sqrt{2}$. The resulting operator isOn the antisymmetric vector $(|01⟩-|10⟩)/\sqrt{2}$, its eigenvalue is $-1/2$. Thus the partial transpose is not positive on this entangled input, andThe same witness embeds into any larger input dimension; in the exceptional one-dimensional case transposition is simply the identity.

For $X\in L$, define the Adjoint representation of a Lie algebra by ${\mathrm{ad}}_X(Y)=[X,Y]$. The Jacobi identity gives, for every $Z$,Thus $\mathrm{ad}$ preserves brackets and is a Lie algebra representation. Its Killing form isUsing the representation identity and the cyclic property of the trace,The Killing form is also symmetric by cyclicity.

In a basis $T_a$, write $[T_a,T_b]=c_{ab}{}^d{T}_d$. The lowered structure constants satisfyThe first expression is antisymmetric in $a,b$ and the second in $b,c$. These transpositions generate all permutations, so $c_{abc}$ is totally antisymmetric.

An invariant subspace of $L$ under its Adjoint representation of a Lie algebra is exactly an ideal of a Lie algebra: $[L,I]\subseteq I$. A simple Lie algebra has no nonzero proper ideal, proving the adjoint representation is irreducible.

Put $D_X=d(X)$. The same trace calculation proves invariance of the Trace form of a Lie algebra representation:Because the $D_X$ are anti-Hermitian, $H$ is a real symmetric bilinear form. The Killing form of a simple compact Lie algebra is negative definite. Define a real linear map $T$ by $H(X,Y)=\kappa (TX,Y)$. Invariance of both forms implies that $T$ commutes with every ${\mathrm{ad}}_Z$, while symmetry makes $T$ self-adjoint for the positive inner product $-\kappa$. By the spectral theorem, its real eigenspaces are invariant under the Adjoint representation of a Lie algebra. Irreducibility forces a single eigenvalue, so $T=\mu I$ and $H=\mu \kappa$. This also avoids the extra care needed when applying the complex form of the Schur lemma to a real vector space.

The kernel of $d$ is an ideal of a Lie algebra. It cannot be all of $L$: a trivial representation of dimension greater than one is reducible. Simplicity therefore makes $d$ faithful. For every nonzero $X$,Hence the positive trace index for a compact simple Lie algebra givesin the adapted basis.

For the cubic trace tensor of a Lie algebra representation, the identityexpands toSet $Y=T_d$, $X=T_a$, $Z=T_b$, $W=T_c$. Cyclicity of $B$ converts $B_{\ell bc}$ to $B_{bc\ell }$ and $B_{a\ell c}$ to $B_{ca\ell }$. Thus the invariance identity for a cubic trace tensor is

For the final contraction, it is important to keep the specified index-raising convention. In the adapted basis write $C_{ab}{}^c=c_{ab}{}^c$. Then $c_{abc}=-C_{ab}{}^c$, while raising the last two indices with ${\kappa }^{ab}=-{\delta }^{ab}$ gives $c_a{}^{mn}=-C_{am}{}^n$. Total antisymmetry and the definition of the Killing form implyOnly the antisymmetric part of $B$ in $m,n$ contributes, soConsequently the Killing-normalized contraction of a cubic trace tensor isThe negative sign printed in the last requested identity is incompatible with raising indices by the inverse Killing form. The opposite sign is obtained if one instead defines the lowered structure constants using the positive Euclidean metric, a different convention from the one specified.

For a concrete check, take the two-dimensional representation of $\mathfrak{su}(2)$ with $T_a=-i{\sigma }_a/(2\sqrt{2})$, where the ${\sigma }_a$ are Pauli matrices. ThenTheir contraction is $+{\delta }_{ab}/8$, confirming the positive sign with $\mu =1/4$ and providing a counterexample to the printed sign.

Use ${\beta }_T=1/(k_{B}T)$ for inverse temperature, to distinguish it from the order-parameter critical exponent $\beta$. In a translation-invariant pure thermodynamic phase, the connected two-point correlation function isFor finite correlation length, at $|r|\gg \xi$ it decays exponentially, possibly with an algebraic prefactor; ${\xi }^{-1}$ is the asymptotic exponential decay rate. Below an ordered transition the pure phase is selected by taking the thermodynamic limit before $h\to 0^{+}$ or $h\to 0^{-}$. A zero-field symmetric mixture of the two ordered phases has a nondecaying connected term and does not describe the usual pure-phase magnetic susceptibility.

Let $\mathcal{F}=-k_{B}T\log Z$ denote extensive physical free energy and $S={\sum }_r{\sigma }_r$. Since the field contributes $-hS$ to the statistical Hamiltonian, differentiation givesExpanding the variance and using translation invariance proves the correlation-function susceptibility sum rule:The sum includes $r=0$. If the source is instead defined as the dimensionless quantity ${\beta }_{T}h$, the corresponding derivative omits the displayed inverse-temperature factor.

A real-space renormalization group step of length factor $b>1$ uses a local blocking kernel $B_b({\sigma }^{\prime },\sigma )\ge 0$ satisfying ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )=1$. For example, a deterministic majority-spin rule assigns one coarse Ising spin to each block; tied blocks can be resolved with equal probabilities. Define the blocked statistical Hamiltonian byThe spacing becomes $a^{\prime }=ba$. All generated interactions, including multispin terms and the field-independent constant, must be retained for an exact transformation. Summing over ${\sigma }^{\prime }$ uses normalization to prove $Z(u^{\prime },C^{\prime },N^{\prime })=Z(u,C,N)$. Iteration gives $N_p=b^{-pD}N$ and $Z(u_p,C_p,N_p)=Z(u_0,C_0,N)$.

The coarse probability law is the exact marginal law of these block variables. Indeed, for any coarse observable $A$, its expected value is the microscopic expected value of ${\sum }_{{\sigma }^{\prime }}{B}_b({\sigma }^{\prime },\sigma )A({\sigma }^{\prime })$. A blocking rule that retains the long-wavelength order-parameter mode, with source and observable transformations tracked, therefore retains long-distance response and correlation functions. Their field normalization may change, while their physical length scales and critical singularities do not. Partition-function equality alone would not justify this assertion for an arbitrary kernel that discards all slow variables. Truncating the generated operator family is likewise an approximation.

For the free-energy recursion use dimensionless free energy per site,in the thermodynamic limit. Exact partition-function invariance gives $F(u_0,C_0)=b^{-pD}F(u_p,C_p)$. If one step sends $C$ to $C^{\prime }=b^{D}C+c(u)$, the coefficient $c(u)$ records the identity-operator contribution from eliminated fluctuations and normalization. ConsequentlyRepeated substitution derives the additive free-energy recursion under blocking:In physical energy units, divide this convention by ${\beta }_T$. The additive term is essential for absolute free energies even when it does not affect the leading critical powers.

A renormalization-group fixed point satisfies $R_b(u_{\ast })=u_{\ast }$. Linearize in scaling coordinates $v_i$ so that $v_i^{\prime }=b^{y_i}{v}_i+O(v^2)$. A relevant operator has $y_i>0$, an irrelevant operator has $y_i<0$, and a marginal operator has $y_i=0$ with behaviour decided by nonlinear terms. Thus the eigenvalues of the discrete linear map are $b^{y_i}$; the sign criterion applies to the generator exponents $y_i$, not to the discrete multipliers. The critical surface is the stable manifold on which every relevant scaling field has been tuned away. Irrelevant coordinates flow to the renormalization-group fixed point along it. A repulsive renormalization-group trajectory leaves the renormalization-group fixed point along an unstable relevant direction, ultimately entering a noncritical phase or another scaling regime.

To separate the singular free-energy density from its regular background, suppose the eliminated-mode contribution $g$ is analytic near the renormalization-group fixed point. Look for an analytic background $a(u)$ satisfying $a(u)=b^{-D}a(R_{b}u)+g(u)$. Then $F_s=f-a$ satisfies $F_s(u)=b^{-D}{F}_s(R_{b}u)$. In linear thermal and field coordinates, an analytic monomial $g_{mn}{t}^m{h}^n$ has background coefficientprovided the denominator is nonzero. This explicitly explains when analytic subtraction of an inhomogeneous renormalization recursion works. The printed inhomogeneous equation may describe the free-energy function before this subtraction, or a singular part defined only up to analytic terms. The inhomogeneous sum need not be small compared with the full free energy: an analytic constant can dominate a vanishing singular contribution. It can be omitted in deriving leading singular powers after regular subtraction, provided no renormalization-group free-energy resonance, marginal logarithm or dangerously irrelevant coupling obstructs the homogeneous critical asymptotics. At a renormalization-group free-energy resonance $m{\lambda }_t+n{\lambda }_h=D$, additive logarithms can remain, so simply dropping the term would miss them.

Suppose there are exactly two relevant scaling fields, thermal $t$ and magnetic $h$, with positive exponents ${\lambda }_t,{\lambda }_h$, and the remaining fields approach a regular limit. Physical reduced temperature and magnetic field agree with these coordinates to leading order, up to nonuniversal metric factors. The homogeneous scaling hypothesis for critical phenomena isas an asymptotic scaling equation. Choose $\ell =|t{|}^{-1/{\lambda }_t}$ to make the thermal argument $+1$ or $-1$. Defining $f_{\pm }(z)=F_s(\pm 1,z)$ givesThe two functions describe respectively $t>0$ and $t<0$: approaching the transition from disordered and ordered sides need not give the same function. In particular the ordered zero-field function has one-sided derivatives associated with spontaneous magnetization. Metric factors multiply this formula in different microscopic normalizations.

The correlation length in Bravais lattice units obeys $\xi (t,h)=\ell \xi ({\ell }^{{\lambda }_t}t,{\ell }^{{\lambda }_h}h)$. At $h=0$, the same choice of $\ell$ gives $\nu =1/{\lambda }_t$. Two thermal derivatives of $F_s(t,0)$ give $2-\alpha =D/{\lambda }_t$. One and two field derivatives give, respectively, $\beta =(D-{\lambda }_h)/{\lambda }_t$ and $\gamma =(2{\lambda }_h-D)/{\lambda }_t$, assuming the relevant branch amplitudes are nonzero. At $t=0$, choose $\ell =|h{|}^{-1/{\lambda }_h}$; then $M\propto \mathrm{sgn}(h)|h{|}^{(D-{\lambda }_h)/{\lambda }_h}$, so $\delta ={\lambda }_h/(D-{\lambda }_h)$. HenceSimple algebra now gives the Rushbrooke scaling relation $\alpha +2\beta +\gamma =2$ and the hyperscaling relation $\alpha =2-D\nu$. Also $\beta \delta ={\lambda }_h/{\lambda }_t=\beta +\gamma$, giving the Widom scaling relation. These statements refer to leading singular contributions with the homogeneous two-field scaling assumptions. Above the upper critical dimension, the quartic coupling is dangerously irrelevant because it is still needed to stabilize the ordered phase; ordinary mean-field critical exponents then satisfy Rushbrooke and Widom but generally not hyperscaling. Marginal logarithms also require qualifications at the upper critical dimension.

For the connected-correlation scaling form $G(r)=|r{|}^{-(D-2+\eta )}{f}_G(|r|/\xi )$, the function $f_G(x)$ tends to a finite nonzero value as $x\to 0$ and decays exponentially at large $x$, up to powers of $x$, in a massive short-range phase. The scaling ansatz alone does not fix that prefactor. If the usual massive Gaussian field correlation tail $G(r)\propto r^{-(D-1)/2}{e}^{-r/\xi }$ applies, matching the powers gives $f_G(x)\sim B{x}^{(D-3)/2+\eta }{e}^{-x}$, with a correlation-length-dependent amplitude in $G$ fixed by matching the critical scaling form.

In physical coordinates the lattice sum has site-density factor $a^{-D}$. Its long-distance singular part therefore giveswhere ${\Omega }_{D-1}$ is the area of the unit sphere in $D$ dimensions. For $\eta <2$, finite $f_G(0)$, an integrable exponential tail and a nonzero integral, the last factor tends to a finite nonzero constant as $\xi \to \infty$. Microscopic distances contribute a regular background. Thus ${\chi }_s\propto {\xi }^{2-\eta }\propto |t{|}^{-(2-\eta )\nu }$, proving the Fisher scaling relationEquivalently the field has scaling dimension $(D-2+\eta )/2$, so its conjugate field has ${\lambda }_h=(D+2-\eta )/2$, which gives the same identity from the exponent formulas above.

First scale the stated unit-ball approximation. For every nonzero residual $r\in F$, apply the hypothesis to $r/\Vert r\Vert$ and multiply the resulting vector by $\Vert r\Vert$. This gives $v\in E$ withFor zero residual use $v=0$. Start with $r_0=y$, choose $v_j$ by this rule for $r_j$, and set $r_{j+1}=r_j-T{v}_j$. Inductively,The Banach space $E$ therefore contains the sum $x={\sum }_{j\ge 0}{v}_j$, withThe partial sums satisfy $T{\sum }_{j=0}^{n-1}{v}_j=y-r_n$. Boundedness of $T$ gives convergence of the left side to $Tx$, while $r_n\to 0$ gives $Tx=y$. Thus $T$ is surjective with the stated lifting bound. This geometric correction for approximate surjectivity used completeness of $E$, not any unproved completeness of $F$.

Put $C=R/(1-k)$. Given a Cauchy sequence $(y_n)$ in $F$, choose a subsequence $(y_{n_j})$ such that $\Vert y_{n_{j+1}}-y_{n_j}\Vert \le 2^{-j}$ for $j\ge 1$. Lift each difference to $u_j\in E$ with $T{u}_j=y_{n_{j+1}}-y_{n_j}$ and $\Vert u_j\Vert \le C{2}^{-j}$. Lift $y_{n_1}$ to $u_0$. The sum $u=u_0+{\sum }_{j\ge 1}{u}_j$ exists in $E$, and its image is the limit of the subsequence. A Cauchy sequence with a convergent subsequence converges to the same limit: use the triangle inequality between an arbitrary late term, a later subsequence term, and the limit. This proves $F$ is complete, the completeness forced by uniformly bounded lifting principle.

The open mapping theorem states that a bounded surjective linear operator between Banach spaces maps open sets to open sets. In particular a bounded linear bijection between Banach spaces has a bounded inverse. To deduce the closed graph theorem, let $S:E_1\to F_1$ be an everywhere-defined linear map between Banach spaces with closed graph. Its graph is a closed subspace of the Banach product $E_1\times F_1$, with norm $\Vert (x,y)\Vert =\Vert x\Vert +\Vert y\Vert$, so it is itself Banach. The projectionis a bounded linear bijection. The open mapping theorem makes its inverse bounded. Composing that inverse with the bounded second projection proves that $S$ is bounded. This is the required closed graph conclusion, rather than a continuity assumption on $S$.

Finally consider the identity map $I:(C(X),\Vert \cdot \Vert )\to (C(X),\Vert \cdot {\Vert }_{\infty })$. If $f_n\to f$ in the new norm and $f_n\to g$ uniformly, each assumed continuous point evaluation functional gives $f_n(x)\to f(x)$, while uniform convergence gives $f_n(x)\to g(x)$. Hence $f=g$. Since both spaces are metric, this sequential argument proves the graph is closed. Both norms are complete, so the closed graph theorem makes $I$ bounded. Its inverse is bounded by the open mapping theorem. Consequently there are constants $c,C^{\prime }>0$ such thatThis proves Banach norm rigidity from continuous point evaluations. Pointwise continuity alone would not give a common bound on all evaluations; completeness supplies that through the closed graph argument.

A principal bundle $\pi :P\to B$ with structure group $G$ has a smooth free right $G$-action, each fibre is a single orbit, and it is locally equivariantly isomorphic to $U\times G\to U$. A representation $\rho :G\to GL(V)$ defines the associated vector bundleLocal sections of $P$ identify $E$ with $U\times V$, and changes of section act by $\rho$ on the fibre coordinates.

A principal bundle is trivial if and only if it has a global smooth section. Given $s:B\to P$, the map $(b,g)\mapsto s(b)g$ is an equivariant isomorphism $B\times G\to P$: freeness and transitivity give its unique inverse on each fibre, and local trivializations make that inverse smooth. Conversely a product bundle has section $b\mapsto (b,e)$. This proves principal bundle trivialization by a global section. A contractible paracompact base is a sufficient condition for triviality of such a bundle; local sections alone are insufficient on a general base.

A manifold is parallelizable when its tangent bundle is trivial, equivalently when it possesses a global smooth frame. The circle $S^1$ is an example, with its nowhere-vanishing unit tangent field. The sphere $S^3$ is another: regard it as the unit quaternions. At a unit quaternion $q$, the three tangent vectors $qi,qj,qk$ are independent and smoothly depend on $q$. Quaternion multiplication preserves the norm, so these form an orthonormal frame. This gives the quaternionic left-invariant frame on the three-sphere.

For any Lie group $G$ of dimension $n$, choose a basis $E_1,\dots ,E_n$ of $T_{e}G$. Left translations defineThese fields are smooth; at every $g$, the differential of the diffeomorphism $L_g$ is invertible, so they are a basis of $T_{g}G$. Hence every Lie group manifold is parallelizable, as in parallelization of a Lie group by left translations.

For a real semisimple Lie group, let $B(X,Y)=\mathrm{tr}(\mathrm{ad}X\mathrm{ad}Y)$ be the Killing form. It is nondegenerate and invariant under the adjoint action. Left translating $B$ gives a bi-invariant pseudo-Riemannian metric. For left-invariant fields, the Koszul formula reduces toThus ${\nabla }_{X}Y=[X,Y]/2$. The curvature convention of Question 1 givesTo obtain the Ricci tensor, trace the linear map $X\mapsto R(X,Y)Z$. Since $[[X,Y],Z]=\mathrm{ad}Z\mathrm{ad}YX$,This proves the Killing-form Einstein metric construction for every real semisimple group. If $G$ is compact semisimple, $-B$ is positive definite and its Ricci tensor is $-\frac{1}{4}B=\frac{1}{4}g$, giving a Riemannian Einstein metric. For a noncompact semisimple group the general Killing-form construction is indefinite. Thus the unrestricted assertion is understood in the pseudo-Riemannian sense appropriate here; positive definiteness is not obtained by simply taking $-B$ in the noncompact case.

The two unheaded requests are addressed here; the individual group-cover calculations are in the scoped parts below. A quaternion $q=a+bi+cj+dk$ has norm $|q{|}^2=a^2+b^2+c^2+d^2$. Thus the unit quaternions form exactly the sphere $S^3\subset {\mathbb{R}}^4$, with quaternion multiplication giving its smooth group structure. The explicit identification with the special unitary group $SU(2)$ isThe matrices representing $i,j,k$ satisfy the quaternion multiplication relations. The displayed matrix obeys $U^{†}U=|q{|}^{2}I$ and $\det U=|q{|}^2$; conversely every $SU(2)$ matrix has that form with $|q|=1$. Hence $S^3$ is the manifold of unit quaternions, and its group is $SU(2)$.

For the exterior-square double cover of SO(3,3), fix a basis $e_1,\dots ,e_4$ and volume element $v=e_1\wedge e_2\wedge e_3\wedge e_4$. Define the symmetric bilinear form $Q$ on the six-dimensional exterior square byIt is symmetric because two-forms commute under the wedge product. PutThen $Q(a_i,a_j)=Q(b_i,b_j)=0$ and $Q(a_i,b_j)={\delta }_{ij}$. The Gram matrix is $\left(\begin{array}{cc}0& I_3\\ I_3& 0\end{array}\right)$: the vectors $(a_i+b_i)/\sqrt{2}$ have norm $+1$, while $(a_i-b_i)/\sqrt{2}$ have norm $-1$. Therefore the quadratic form $\omega \mapsto \omega \wedge \omega$ has signature $(3,3)$.

For $A\in SL(4,\mathbb{R})$,Thus $\rho (A)={\Lambda }^{2}A$ preserves $Q$. The special linear group is connected: polar decomposition of an invertible real matrix writes $A=OP$ with $O\in SO(4)$ and $P$ positive definite of determinant one, and both factors can be joined to the identity. The image consequently lies in $S{O}_0(3,3)$.

If ${\Lambda }^{2}A=I$, then $Au\wedge Aw=u\wedge w$ for every pair, so $A$ preserves every two-plane. A line is an intersection of two such planes, hence every line is preserved. A linear map preserving every line is scalar: apply it to basis vectors and then to their pairwise sums. Thus $A=\lambda I$, and ${\Lambda }^{2}A=I$ forces ${\lambda }^2=1$. The kernel is exactly $\{\pm I\}$. It is discrete, so the derivative of $\rho$ is injective. Both Lie algebras have dimension $15$, making the image open; an open subgroup of a connected group is the whole group. HenceThe identity-component restriction is essential. The same principle of an explicit form-preserving representation, kernel calculation and dimension argument supplies the five remaining covers.

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.

Differentiability at $p$ means there is a linear map $A=Df{|}_p$ such thatThis is the Fréchet derivative definition; the same linear approximation must work in every direction, not merely along coordinate axes.

The chain rule says that if $f$ is differentiable at $p$ and $g$ is differentiable at $f(p)$, then $g\circ f$ is differentiable at $p$ andFor its proof, put $A=Df{|}_p$, $B=Dg{|}_{f(p)}$, and $k=f(p+h)-f(p)=Ah+r(h)$. Since a linear map between these finite-dimensional normed spaces is bounded, $\Vert k\Vert =O(\Vert h\Vert )$. Differentiability of $g$ gives $g(f(p)+k)=g(f(p))+Bk+s(k)$ with $s(k)=o(\Vert k\Vert )$. ThereforeIf $k=0$, take $s(0)=0$; otherwise $\Vert s(k)\Vert /\Vert h\Vert =(\Vert s(k)\Vert /\Vert k\Vert )(\Vert k\Vert /\Vert h\Vert )\to 0$. This proves the required full linear approximation.

For the Euclidean norm, at every $x\ne 0$ the ordinary one-coordinate differentiation givesAt the origin its difference quotient in direction $e_i$ is $|t|/t$, whose right and left limits are $1$ and $-1$. Thus none of its coordinate partial derivatives exists at the origin.

Regard $B$ as a linear map ${\mathbb{R}}^n\to {\mathbb{R}}^p$ and $A$ as a linear map ${\mathbb{R}}^p\to {\mathbb{R}}^m$ (the same argument holds over any field). ThenIts dimension cannot exceed either $\dim \mathrm{im}A$ or $\dim \mathrm{im}B$, since applying a linear map cannot increase dimension. The rank bound for a matrix product is thereforeThe bound $p$ is attained whenTheir product has an $I_p$ block and zeros elsewhere, so its rank of a matrix is $p$. Thus the bound depending only on the given matrix sizes is sharp.

The row rank of a matrix is the dimension of the linear span of its row vectors. Its column rank is the dimension of the linear span of its column vectors, equivalently the dimension of the image of the corresponding linear map. The equality of row rank and column rank states that these two numbers are equal for every matrix. Their common value is the rank of a matrix.

The dual map reverses the direction of the linear map:Here the dual spaces consist of complex-linear functionals. Composition with $\alpha$ is linear both as a functional on $V$ and as an operation on $W^{\ast }$, so this does define a linear map between the dual spaces.

Let $(v_1,\dots ,v_n)$ and $(w_1,\dots ,w_m)$ be the chosen bases, and let $(v^1,\dots ,v^n)$ and $(w^1,\dots ,w^m)$ be their dual bases. If $A$ represents $\alpha$, then $\alpha (v_i)={\sum }_j{A}_{ji}{w}_j$. ThereforeThe coefficient in row $i$, column $j$ of the matrix of ${\alpha }^{\ast }$ is thus $A_{ji}$, giving the matrix transpose $A^{\mathsf{T}}$. This calculation introduces no complex conjugation: the algebraic dual map uses complex-linear functionals, whereas a Hermitian transpose belongs to the inner-product adjoint construction.

A weight-$2k$ modular form gives an invariant tensor differentialbecause $d(\gamma \tau )=(c\tau +d{)}^{-2}d\tau$ cancels its automorphy factor. Assume first $k\ge 0$. At an ordinary quotient point this differential is holomorphic. At an elliptic point of order $e$, use $u$ and $w=u^e$ from part (a), and write the pullback as $A(u)(du{)}^k$, with $A$ holomorphic. Its invariance says $A(\zeta u){\zeta }^k=A(u)$, so a nonzero term $a_r{u}^r$ has $r+k\equiv 0(\mathrm{mod}e)$. Sincethe exponent in the descended coefficient is the integer $(r-k(e-1))/e$. With $r\ge 0$, it is at least $-\lfloor k(e-1)/e\rfloor$. At the modular cusp, $d\tau =dq/(2\pi iq)$ and $f$ has a power series with nonnegative exponents, so ${\omega }_f$ has a pole of order at most $k$.

Consequently the linear map $f\mapsto {\omega }_f$ injects $M_{2k}$ into the meromorphic sections of $K^{\otimes k}$ with poles bounded byThis section space is $H^0(X(1),K^{\otimes k}(D))$. It is finite-dimensional on a compact Riemann surface by the Riemann-Roch theorem. Concretely choose one nonzero meromorphic section of the line bundle; dividing by it identifies these sections with a Riemann-Roch space with a fixed finite divisor bound. This proves the requested finite-dimensionality by compact-surface theory.

In fact the injection is an isomorphism. If a descended coefficient has order at least $-\lfloor k(e-1)/e\rfloor$, its pulled-back exponent $ed+k(e-1)$ is nonnegative; the modular cusp pole bound likewise gives a holomorphic Fourier expansion of a modular form. These pullbacks therefore recover modular forms. Since $X(1)$ is the Riemann sphere, $\mathrm{deg}K=-2$, and a line bundle of degree $d$ on the sphere has $\max (0,d+1)$ sections. HenceFor negative $k$, the valence formula excludes a nonzero holomorphic modular form because all its orders would be nonnegative but their sum would be $k/6<0$; the space is zero. The mechanism behind finite-dimensionality is modular forms as meromorphic differentials on X(1), with explicit elliptic and modular cusp pole bounds.