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The multiplicative norm first rules out zero divisors: if $xy=0$ and $x\ne 0$, then $0=|x||y|$ forces $y=0$. If the dimension is zero the claimed bound already holds, so suppose it is positive.

It is worth checking that the printed description of a ring with an underlying real vector space really permits differentiation of multiplication. This follows from the norm identity even if compatibility with real scalars is not separately stated. For fixed $x$, the map $L_x(y)=xy$ is additive by distributivity, andIt is therefore continuous. Additivity gives rational linearity. Approximating a real number by rationals and using continuity proves $L_x(ty)=t{L}_x(y)$ for every real $t$. The same argument holds in the other variable. This is automatic real bilinearity from a multiplicative norm, so multiplication is a bilinear map between finite-dimensional vector spaces, and hence is smooth.

Let $S=\{x\in A:|x|=1\}$ be the unit sphere of the given positive-definite bilinear form. It is a copy of $S^{n-1}$. The squaring mapis well defined because $|x^2|=|x{|}^2=1$, and is smooth. Commutativity gives its derivativeFor $x\in S$ and $v\ne 0$, the norm identity gives $|D{Q}_x(v)|=2|v|$, so the derivative is injective, including on the tangent space $T_{x}S=x^{\perp }$. It takes tangent vectors into $T_{x^2}S$ since $Q$ maps the sphere to itself. Alternatively, polarization of $|xy{|}^2=|x{|}^2|y{|}^2$ gives $⟨xy,xz⟩=|x{|}^2⟨y,z⟩$, so $⟨x^2,2xv⟩=2⟨x,v⟩=0$ directly.

The only identifications under squaring are antipodal:The factorization uses commutativity, and the last implication uses the absence of zero divisors. Conversely $Q(-x)=Q(x)$. It follows that $Q$ induces an injective smooth mapThe antipodal quotient map is a local diffeomorphism, so the injectivity of $DQ$ implies injectivity of $Dq$. Since Real projective space is compact, the stated compact injective immersion is an embedding theorem makes $q$ a closed embedding.

For $n\ge 2$ its source and target have the same dimension $n-1$. The inverse function theorem makes its image open as well as closed in the connected sphere $S^{n-1}$. It is nonempty, so the image is the whole sphere; consequently $q$ is a diffeomorphism. But for $n\ge 3$ this is impossible:Indeed, the antipodal two-sheeted covering map $S^{n-1}\to {\mathbb{RP}}^{n-1}$ has simply connected total space in this range, and is its universal cover, with deck group $\mathbb{Z}/2$. A sphere of dimension at least two is simply connected. These fundamental groups cannot be those of homeomorphic spaces.

We have obtained the dimension bound for a commutative Euclidean normed algebra:The real numbers and complex numbers with their usual products and Euclidean norms attain dimensions one and two. In dimension two, the induced map ${\mathbb{RP}}^1\to S^1$ is entirely compatible with the argument: the circle squaring map identifies exactly the two antipodal preimages. We neither assume a multiplicative identity nor use a classification of real division algebras; the squaring argument proves the bound from the supplied hypotheses.

For the unit-modulus prime values in this question, the Granville-Soundararajan distance isThe second identity uses $|f(p)|=|g(p)|=1$. Thus it is the Euclidean distance between the finite weighted vectors $(f(p)/\sqrt{2p}{)}_{p\le X}$ and $(g(p)/\sqrt{2p}{)}_{p\le X}$. Nonnegativity, symmetry and the triangle inequality follow from that norm representation, and it vanishes exactly when the two prime-value vectors agree. This proves the prime-restriction metric for pretentious distance.

On whole arithmetic functions at fixed $X$ it is a pseudometric. For example $f=1$ and $g={\mu }^2$ have identical values at every prime and zero distance for every $X$, but different values at four. Thus the literal identity-of-indiscernibles assertion needs the prime-restriction quotient.

Since $\mu (p)=-1$,For completeness, partial summation and the prime-counting form of the Prime number theorem giveTo verify the constant and the error, replace $\pi$ by $\mathrm{Li}+E$ in the first expression. Differentiating $\mathrm{Li}(u)/u$ shows its two contributions combine to $\log \log X$ plus a constant. The error integral converges absolutely because $|E(u)|/u^2\ll e^{-c\sqrt{\log u}}/u$, and $E(X)/X\to 0$. This proves the Mertens second theorem in the form needed here. In particular the pretentious distance between one and the Möbius function satisfies

For the last request, retain the earlier prime normalization $|f(p)|=1$; the proof also works with the weaker assumption $|f(p)|\le 1$. Without a bound on prime values, the original definition and the absolute convergence of the displayed measure need not apply to an unrestricted multiplicative arithmetic function. For example $f(n)=\mu (n{)}^{2}n$ is multiplicative and squarefree-supported, but for $X>e$ the displayed measure has infinite positive mass: its prime terms alone are ${\sum }_p{p}^{-1/\log X}$, which diverges by the Prime number theorem.

Put $L=\log X$, $\sigma =1+1/L$, and use the Fourier transform convention $\hat{\nu }(t)=\int e^{-itu}d\nu (u)$. Since $f$ is supported on squarefree integers, the defining property of a multiplicative arithmetic function gives the absolutely convergent Euler productThe bound on prime values gives $|f(n)|\le 1$ for squarefree $n$, so the total variation norm of a measure is bounded by $\zeta (\sigma )$; thus $\nu$ is a finite measure. Expanding the logarithm of the modulus, with an absolute $O(1)$ remainder since ${\sum }_p{p}^{-2\sigma }\ll 1$, givesAll constants here are uniform in $t$. Moreoverusing the pole of the Riemann zeta function. The nonnegative deficits $1-\mathrm{Re}(f(p)p^{-it})$ giveIndeed, the replacement of $p^{-\sigma }$ by $1/p$ for $p\le X$ costs at mostby the Mertens first theorem, while discarding the $p>X$ terms costs nothing in this lower bound. Thus the large Fourier coefficient criterion for pretentiousness isIn particular,The comparison function is the Archimedean character $n^{it}$ with this sign because our Fourier transform uses $e^{-itu}$.

A statistical path through $P$ is a family $t\mapsto P_t\in \mathcal{P}$, defined on an open interval containing zero, with $P_0=P$. Write $p_t=d{P}_t/d\mu$ and $p=p_0$. Its differentiability in quadratic mean at zero means that there is a square-integrable function $g\in L^2(P)$ such thatThe function $g$ is the score function of the statistical path. Thus it is the derivative of the square-root probability density function, multiplied by two and divided by $\sqrt{p}$ where $p>0$. The score function is determined only $P$-almost everywhere. The displayed definition also controls any probability mass entering a region where $p=0$; a pointwise derivative of the log probability density function alone would not do that. The required derivative is in the square-root-density $L^2$ norm.

Set $\hslash =1$ and let $N=2I\in {\mathbb{Z}}_{\ge 0}$. A concrete irreducible spin representation is the symmetric powerStart with $N$ copies of the defining SU(2) doublet and restrict their tensor product representation to the completely symmetric subspace. Its orthonormal basis consists of symmetric states with $I+m$ up components and $I-m$ down components, where $m=I,I-1,\dots ,-I$. There are $N+1=2I+1$ such states. The total generators are $J_i={\sum }_{a=1}^N{\sigma }_i^{(a)}/2$, acting on this subspace, where ${\sigma }_i$ are the Pauli matrices. This also constructs the trivial representation when $I=0$.

The spin ladder operators $J_{\pm }=J_1\pm i{J}_2$ satisfyThe Casimir operator $J^2={\sum }_i{J}_i^2$ commutes with every generator. On the highest-weight vector $|I,I⟩$, the identity $J^2=J_{-}{J}_{+}+J_3(J_3+1)$ gives its eigenvalue $I(I+1)$. With phases chosen to make the lowering coefficients positive,The squared norm of the lowered state follows from $J_{+}{J}_{-}=J^2-J_3(J_3-1)$. The ladder stops exactly at $m=-I$. Every weight is connected to every other by these operators, and $J_3$ has distinct eigenvalues. Hence any invariant subspace contains a weight vector and then the whole ladder: the representation is irreducible.

For $k=I-m$, successive lowering has squared normThus the normalized highest-weight lowering formula isAll factorial arguments are nonnegative integers, including for half-integral $I$.

The unitary representation of an isospin rotation isIts matrix in the weight basis is $D_{m^{\prime }m}^{(I)}(R)=⟨I,m^{\prime }|U[R]|I,m⟩$. Insert the completeness relation between two operators to obtainThese verify the group representation and unitarity properties, and the ladder argument gives irreducibility. Here rotations carry their SU(2) lifts: $U(2\pi )=(-1{)}^{2I}I$. Integer $I$ descends to the ordinary SO(3) group; half-integral $I$ is a representation of its double cover, not a single-valued representation of $SO(3)$.

For $I=1$, order the basis as $(|1,1⟩,|1,0⟩,|1,-1⟩)$. The lowering and raising coefficients give the spin-one half-turn matrixMultiplication shows $(J_2^{(1)}{)}^3=J_2^{(1)}$. Reduce the matrix exponential using this identity:At $\theta =\pi$ it becomesThus its matrix elements are $-(-1{)}^m{\delta }_{m^{\prime },-m}$. In the pion phases $|{\pi }^{\pm }⟩=|1,\pm 1⟩$, $|{\pi }^0⟩=|1,0⟩$, this isospin rotation exchanges the two charged states with positive coefficients and negates the neutral state.

Charge conjugation is linear and unitary here. Similarity preserves commutators, soConsequently the conjugated generators obey the same Lie algebra. The specified signs also give $\mathcal{C}{J}_{\pm }{\mathcal{C}}^{-1}=-J_{\mp }$. Since $|{\pi }^{\pm }⟩=J_{\pm }|{\pi }^0⟩/\sqrt{2}$ and the neutral state has positive charge conjugation eigenvalue,The signs depend on the chosen charged-state phases, which have been fixed by the ladder convention.

Conjugation by $R_{\pi }=e^{-i\pi J_2}$ changes $(J_1,J_2,J_3)$ to $(-J_1,J_2,-J_3)$, exactly as charge conjugation does. Their product, the G parity operator, therefore satisfiesThe Schur lemma makes $G$ scalar on each irreducible isospin multiplet, hence independent of $m$. On the neutral pion, $G|{\pi }^0⟩=-\mathcal{C}|{\pi }^0⟩=-|{\pi }^0⟩$. Therefore the pion G-parity is

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

An orthonormal multiresolution analysis consists of closed subspaces of a Hilbert space $(V_j{)}_{j\in \mathbb{Z}}$ of the Hilbert space $L^2(\mathbb{R})$ such thatThere is a scaling function $\varphi$ whose integer function translations form an orthonormal basis of $V_0$. Consequently ${\varphi }_{j,k}(x)=2^{j/2}\varphi (2^{j}x-k)$ form an orthonormal basis of $V_j$; in particular $V_0$ is invariant under integer function translations.

Let $W_j=V_{j+1}\ominus V_j$ be the orthogonal complement of $V_j$ in $V_{j+1}$. Nesting, density and the trivial intersection imply the orthogonal direct sumFor example, telescoping $V_{J+1}=V_L\oplus {⨁}_{j=L}^J{W}_j$ and letting $L\to -\infty$, $J\to \infty$ gives this decomposition; the corresponding orthogonal projections converge strongly to zero and the identity by the intersection and density properties.

To construct the orthonormal wavelet, use $\varphi \in V_1$ to write the scaling refinement equation $\varphi ={\sum }_n{a}_n\varphi (2\cdot -n)$, with $a\in {\ell }^2$ and ${\sum }_n|a_n{|}^2=2$. Put $m(t)=\frac{1}{2}{\sum }_n{a}_n{e}^{-int}$. The orthonormal translates and Fourier periodization criterion proved in part (b), applied to the scaling refinement equation, yieldsThis is the quadrature mirror filter identity. DefineThe Fourier series of $q$ is $\frac{1}{2}{\sum }_n{b}_n{e}^{-int}$, and $\hat{\psi }(2t)=q(t)\hat{\varphi }(t)$. The two rows $(m(t),m(t+\pi ))$ and $(q(t),q(t+\pi ))$ form a unitary $2\times 2$ matrix almost everywhere: both have squared norm one and their inner product is zero. This wavelet completion of a multiresolution filter proves not just orthogonality but completeness. Splitting a finer-scale coefficient Fourier series into its values at $t$ and $t+\pi$, this invertible two-channel matrix resolves it into a coarse-scale channel and its complementary channel. Thus the integer function translations of $\psi$ form an orthonormal basis of $W_0$.

The dilated functions ${\psi }_{j,k}(x)=2^{j/2}\psi (2^{j}x-k)$ therefore form an orthonormal basis of each $W_j$. The orthogonal direct sum proves the conclusion: every orthonormal multiresolution analysis produces an orthonormal wavelet basis $\{{\psi }_{j,k}:j,k\in \mathbb{Z}\}$ of $L^2(\mathbb{R})$.

We first construct a genuinely compact Besicovitch set; a sequence of unrelated small-area sets would not suffice. The compact Kakeya construction from small projections below gives the limiting compact set together with its segments.

Let $\mathcal{X}$ consist of nonempty compact sets $K\subset [0,1]\times \mathbb{R}$ whose horizontal projection is exactly $[0,1]$. With the Hausdorff metric this is a complete metric space. Indeed a Hausdorff-Cauchy sequence is eventually contained in one bounded closed rectangle and converges there to a nonempty compact set; horizontal projection is Lipschitz for Hausdorff distance, so its value $[0,1]$ is preserved in the limit.

For $t\in [0,1]$ put $L_t(x,y)=y+tx$ and $m_t(K)=|L_t(K)|$. The function $(K,t)\mapsto m_t(K)$ is upper semicontinuous. To verify this, cover the compact set $L_t(K)$ by finitely many open intervals with total length less than $m_t(K)+\epsilon$. Joint continuity of the compact images in the Hausdorff metric puts every sufficiently nearby image inside the same cover. This proves the required upper bound, uniformly over a compact parameter interval when a strict bound holds throughout it.

Fix $q\in [0,1]$ and $\epsilon >0$. The setis open. It is also dense, by the following explicit finite approximation. Given $K$ and tolerance $h>0$, partition $[0,1]$ into $N$ equal subintervals and choose $(j/N,y_j)\in K$ at each left endpoint. Above that subinterval put the segmentEach segment lies within $\sqrt{1+q^2}/N$ of its chosen point in $K$. Add a finite $h/2$-net of points of $K$ to their union, ensuring the reverse Hausdorff inclusion. For large $N$ the resulting compact set $K^{\prime }$ belongs to $\mathcal{X}$ and has Hausdorff distance less than $h$ from $K$. The image of each segment under $L_t$ has length $|t-q|/N$, and the added points contribute zero length. Henceso $K^{\prime }\in U(q,\epsilon )$.

Take a finite grid of $q$'s whose $\epsilon$-neighborhoods cover $[0,1]$. The finite intersection of the open dense sets $U(q,\epsilon )$ is dense and implies ${\sup }_{0\le t\le 1}{m}_t(K)<2\epsilon$. Thusis open and dense. The Baire category theorem gives a compact $K\in {\bigcap }_{r\ge 1}{V}_{1/r}$, and therefore $m_t(K)=0$ for every $t\in [0,1]$.

DefineThis is the continuous image of the compact set $K\times [0,1]$, hence compact. Its section at horizontal coordinate $t$ is $L_t(K)$, of measure zero. The Fubini theorem gives $|B|={\int }_0^1{m}_t(K)dt=0$. For every slope $x\in [0,1]$, horizontal surjectivity supplies $(x,y)\in K$, and $B$ contains the segment $(t,y+tx)$, $0\le t\le 1$, of length $\sqrt{1+x^2}\ge 1$. Four rotated copies, through angles $0,\pi /4,\pi /2,3\pi /4$, cover all unoriented directions. Their finite union $\mathcal{B}$ is compact, has area zero, and contains a unit subsegment in every direction:The point-line dual construction is the mechanism in Körner's construction; the density calculation above supplies the needed proof, rather than assuming that small-area sets automatically admit a suitable nested limit.

For the multiplier discussion, define $T_D$ by $\widehat{T_{D}g}(\xi )=1_{\{|\xi |\le 1\}}\hat{g}(\xi )$. The Plancherel theorem gives boundedness on $L^2$. We explain why a putative $L^p$ bound, $1<p<2$, would contradict Kakeya packing; duality of Lp spaces then treats $p>2$.

First, such a bound would force the vector-valued inequalityfor arbitrary directions $v_j$, where $P_v$ is the half-plane Fourier projection onto the half-plane $\xi \cdot v<0$. To see the important curvature step, let $T_R$ project onto the radius-$R$ disk and modulate by $e^{iR{v}_j\cdot x}$. Its conjugated symbol is $1_{\{|\xi +R{v}_j|\le R\}}$, tending almost everywhere to $1_{\{\xi \cdot v_j<0\}}$. Dilations give the same putative norm bound for every $R$. Apply this common operator to ${\sum }_j{ϵ}_j{e}^{iR{v}_j\cdot x}{g}_j$, average over independent random signs, and use the Khintchine inequality. Letting $R\to \infty$ gives the displayed estimate by $L^2$ convergence and Fatou lemma, first for finitely many smooth function functions and then by approximation. The disk has a boundary normal in every direction; a fixed polygon does not.

The geometric input for this overview is the standard finite Perron tree packing used in Fefferman's original proof: for arbitrarily small $\eta >0$ one obtains long thin rectangles $R_j$ with $A={\sum }_j|R_j|$ and $|{\bigcup }_j{R}_j|\le \eta A$, while fixed longitudinal translates $R_j^{\prime }$ are pairwise disjoint. This is a quantitative finite arrangement, stronger than merely having a compact null set. The splitting/sliding procedure subdivides an angular triangle into narrow triangles and overlaps their bases; the extensions along their axes separate in reverse order. Its area estimate has the form ${\alpha }^{2m}+2(1-\alpha )$ relative to the original total triangle area: the merged central part shrinks by ${\alpha }^2$ at each generation, and the accumulated caps contribute at most $2(1-\alpha )$. Taking $\alpha$ close to one and then $m$ large makes the ratio arbitrarily small. Inscribed rectangular portions retain a fixed proportion of the total area. The reflected extensions explain why the translates can be disjoint even though the original rectangles overlap strongly.

Here is the analytic contradiction, with the relevant powers explicit. Take $g_j=1_{R_j}$. In coordinates parallel and perpendicular to the long axis $v_j$, the directional Hilbert transform integrates only along that axis. If the longitudinal interval is $[0,\ell ]$, then outside itOn a fixed translate such as $s\in [5\ell ,6\ell ]$, this has magnitude at least ${\pi }^{-1}\log (6/5)$. Since $P_v=(I-i{H}_v)/2$ up to the harmless choice of half-plane sign, $|P_{v_j}{g}_j|\ge c_0>0$ on $R_j^{\prime }$. Their disjointness gives a lower bound $c_0{A}^{1/p}$ for the vector-valued output norm.

Writing $h={\sum }_j{1}_{R_j}$, Hölder's inequality for $p<2$ givesThe putative vector-valued inequality would implywhich is impossible as $\eta \downarrow 0$. The disk multiplier is self-adjoint, so a bound for $p>2$ would give one for its conjugate exponent below two. Thus the disk multiplier is bounded only at $p=2$ among $1<p<\infty$. This explains both the random-sign step and the precise geometric amplification; it is not an assertion that frequency rectangles by themselves produce a logarithmic norm blow-up.

When $\mathbf{x}\cdot \mathbf{y}=\cos (2\pi /3)=-1/2$, the squared norm of their sum is $2+2(-1/2)=1$. ThereforeThe minimizing vector is indeed a unit vector. All three pairwise scalar products then equal $-1/2$, so these three unit vectors lie in a plane, separated by $120$ degrees.

Two nonzero vectors are linearly independent when $s\mathbf{x}+t\mathbf{y}=0$ forces $s=t=0$. Their linear span then has dimension two. If they are linearly dependent, one is a nonzero scalar multiple of the other, and their span has dimension one. Thus the two requested dimensions are $\boxed{2\text{ and }1}$.

The Euclidean scalar product and its norm are $\mathbf{x}\cdot \mathbf{y}={\sum }_{j=1}^n{x}_j{y}_j$ and $\Vert \mathbf{x}\Vert =\sqrt{\mathbf{x}\cdot \mathbf{x}}$. To prove the Cauchy-Schwarz inequality, first handle $\mathbf{y}=0$ trivially. Otherwise the squared normgivesEquality holds exactly when the residual vector is zero, meaning the vectors are linearly dependent; this includes the zero-vector cases. Expanding $\Vert \mathbf{x}+\mathbf{y}{\Vert }^2$ and applying Cauchy-Schwarz then givesTaking nonnegative square roots proves the triangle inequality.

For the unit-vector optimization, let $\mathbf{w}=\mathbf{x}+\mathbf{y}$. Independence ensures $\mathbf{w}\ne 0$. The variable part of $S$ is $\mathbf{z}\cdot \mathbf{w}$, which Cauchy-Schwarz bounds below by $-\Vert \mathbf{w}\Vert$. Equality occurs only for the antiparallel unit vector. ThusThis is minimizing a linear functional on a sphere. The two consequences follow below.

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, $\mathrm{tr}B(r_0)=1+\sqrt{2}=1+2\cos (\pi /4)$.

Both maxima exist by continuity on the compact set $[a,b]$. Their sum is nonnegative and vanishes only for $f=0$. The identities $\Vert cf{\Vert }_1=|c|\Vert f{\Vert }_1$ and the triangle inequality follow by applying the corresponding supremum norm facts to $f$ and $f^{\prime }$. This proves that $\Vert \cdot {\Vert }_1$ is a norm.

For $m=(a+b)/2$, $|\Phi (f)|=|f^{\prime }(m)|\le \max |f^{\prime }|\le \Vert f{\Vert }_1$. Hence the first unit ball has image contained in $[-1,1]$: derivative evaluation is a bounded linear functional for this norm.

For the supremum norm alone, take $f_N(x)=\sin (N(x-m))$. These smooth functions satisfy $\Vert f_N{\Vert }_0\le 1$, but $\Phi (f_N)=N$. Thus the second image is unbounded. The distinction is the unboundedness of derivative evaluation in the supremum norm, not a failure of differentiability of the individual functions.

A Riemannian metric is a smooth field of positive-definite symmetric bilinear forms $g_x$ on the tangent spaces, equivalently a smooth section of ${\mathrm{Sym}}^2{T}^{\ast }M$ with that positivity. Its musical isomorphism isNondegeneracy makes every fiber map an isomorphism. In coordinates the map has matrix $g_{ij}$, and its inverse ${♯}_g$ has matrix $g^{ij}$. Smoothness of the inverse follows from the inverse-matrix formula and the nonvanishing determinant. Both maps cover the identity on $M$, giving a smooth vector bundle isomorphism.

For local flattening of a Riemannian metric, choose a manifold chart neighborhood $V$ around $p$ with closure contained in $U$, and let $h={\varphi }^{\ast }\delta$ on $V$. Choose a smooth bump function $0\le \chi \le 1$, with compact support in $V$ and equal to one on a smaller neighborhood $W$ of $p$. SetThe formulas glue smoothly because the modification is supported strictly inside $V$. A convex combination of positive-definite forms is positive definite. On $W$ the metric is exactly $h$, so $\varphi$ gives an isometry to a Euclidean open subset. Outside $U$ the original metric remains unchanged.

Geodesic completeness means that every geodesic with arbitrary initial point and tangent vector extends for every real affine parameter. On each connected component, the Hopf-Rinow theorem equates this with completeness of the Riemannian distance.

The Euclidean-end claim needs a compact-core interpretation that is absent from its literal hypotheses. Indeed, $M=\{x\in {\mathbb{R}}^n:|x|>1\}$ with $n\ge 2$, $X=\varnothing$ and $g=\delta$ satisfy those hypotheses. The geodesic $\gamma (t)=(2-t)e_1$ reaches the missing unit sphere at $t=1$ and cannot continue within $M$. Thus the printed assumptions alone do not prove completeness.

Here is the intended compact-core completeness for a Euclidean end. Write $\Phi :M\setminus X\to \{|x|>1\}$ for the end coordinates and assume additionally thatis compact for every sufficiently large $R$. Choose such an $R$ beyond the metric transition and large enough to include the initial point of a geodesic. Its speed $s$ is constant. On every segment outside $K_R$, the Euclidean radius changes at a rate at most $s$, so over a finite parameter interval it cannot exceed $R+sT$. The full segment therefore stays in the compact set $K_{R+sT}$. Its velocities also stay in a compact subset of $TM$, since their metric norm is fixed and the base set is compact. The smooth geodesic ordinary differential equation consequently extends past any supposed finite endpoint. The reversed-time argument is identical. This proves completeness under the compact-core condition, and explains exactly what the exterior-of-a-ball counterexample lacks.

For upward stability of Riemannian completeness, lengths of all curves satisfy $L_{\tilde{g}}\ge L_g$, hence $d_{\tilde{g}}\ge d_g$. A $d_{\tilde{g}}$-Cauchy sequence is therefore $d_g$-Cauchy and has a limit $p$ because $g$ is complete. Smooth positive-definite metrics induce the manifold topology. More explicitly, on a small coordinate ball around $p$, $\tilde{g}$ has a bounded largest matrix eigenvalue, so the coordinate straight segment gives $d_{\tilde{g}}(p,q)\le C|\varphi (q)-\varphi (p)|$. Thus convergence to $p$ is also in $d_{\tilde{g}}$. That distance is complete, and Hopf-Rinow yieldsFor disconnected $M$, apply this argument on each component.