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For north-pole stereographic projection $p=(X+iY)/(1-Z)$ on the unit sphere, the antipodal pair point has coordinate $q=-(X+iY)/(1+Z)$. Since $X^2+Y^2=1-Z^2$, this gives $p\overline{q}=-1$ when both coordinates are finite. On the whole Riemann sphere, write $q=-1/\overline{p}$, pairing zero with infinity. The two axis endpoints fixed by a sphere rotation obey this relation.

The Borsuk-Ulam theorem says that recording $n$ continuous real-valued measurements on an $n$-dimensional unit sphere must give identical records somewhere on opposite sides. The measurements can vary wildly; continuity alone forces a coincidence at an antipodal pair.

Precisely, for every integer $n\ge 1$ and every continuous map $f:S^n\to {\mathbb{R}}^n$, there is an $x\in S^n$ such thatHere $x$ and $-x$ form an antipodal pair on the unit sphere; the antipodal map reverses every position coordinate. The target is the Euclidean space ${\mathbb{R}}^n$. Both points have the same entire output, so every one of the $n$ measurements agrees at this common pair. See Hatcher, Algebraic Topology, Corollary 2B.7, p. 176.

Circle example. Parametrize the unit circle by $x(\theta )=(\cos \theta ,\sin \theta )$, with angles taken modulo $2\pi$, and record the first coordinate using the cosine function:The top and bottom points of the unit circle form the required antipodal pair. Their first coordinates agree, while their second coordinates are $1$ and $-1$.

Sphere example. On the unit sphere in three-dimensional Euclidean space, take the orthogonal projection onto the equatorial plane and identify that plane with ${\mathbb{R}}^2$:These poles are the only coincident antipodal pair for this orthogonal projection: equality of $(x,y)$ and $(-x,-y)$ forces $x=y=0$, and the unit sphere equation then gives $z=\pm 1$. The Borsuk-Ulam theorem guarantees an antipodal pair for every continuous map into ${\mathbb{R}}^2$, even when no convenient projection formula is available.

Equal output need not mean equal position coordinates. Keeping all three coordinates gives a continuous map $i:S^2\to {\mathbb{R}}^3$ defined by $i(x,y,z)=(x,y,z)$, the restriction of the identity map of ${\mathbb{R}}^3$. Its outputs at an antipodal pair are always distinct, since $u=-u$ would force $u=0$, which is outside the unit sphere. Thus replacing the target ${\mathbb{R}}^2$ by ${\mathbb{R}}^3$ destroys this guarantee.

Why the circle case follows from the intermediate value theorem. For an arbitrary continuous map $f:S^1\to \mathbb{R}$, write $F(\theta )=f(x(\theta ))$ using the unit circle parametrization above. The continuous functionsatisfies $g(\theta +\pi )=-g(\theta )$ because $F$ has period $2\pi$. If $g(0)=0$, the first antipodal pair already works. Otherwise $g(0)$ and $g(\pi )$ have opposite signs, and the intermediate value theorem supplies a zero between them. At that zero, the two antipodal values of $f$ agree. In the cosine function example, $g(\theta )=2\cos \theta$. This short argument handles only $n=1$. See Hatcher, Algebraic Topology, discussion of Theorem 1.10, p. 32.

An ideal weather interpretation. Model the surface of Earth as a sphere. At one fixed instant, suppose temperature $T$ and pressure $P$ are globally defined continuous scalar measurements on that sphere. Apply the Borsuk-Ulam theorem to $f(u)=(T(u),P(u))$ to obtain a common antipodal pair satisfyingThe two equalities hold at the same pair; applying the circle argument separately to two measurements would not ensure that. This is a conclusion about the ideal continuous model, with both measurements taken at the same instant, rather than an assertion that an exactly matching pair has been empirically measured. Matoušek gives this motivating interpretation and explicitly cautions that physical temperature need not be continuous.^[ref]

Equivalent formulations. A continuous map $h:S^n\to {\mathbb{R}}^n$ is odd when $h(-x)=-h(x)$, extending the symmetry of an odd function. The Borsuk-Ulam theorem equivalently says every such map has a zero. For a general $f$, the difference $h(x)=f(x)-f(-x)$ has this symmetry; conversely, equal antipodal outputs of an odd map must be zero. Another equivalent formulation is that no odd map between spheres $S^n\to S^{n-1}$ exists. Normalizing a nonvanishing odd map by its Euclidean norm would produce such an odd map between spheres. These formulations are useful starting points for algebraic topology; the general proof is beyond the circle argument here.^[ref]

For further study, Jiří Matoušek's Using the Borsuk-Ulam Theorem (2003) develops the theorem and its applications to combinatorics and Euclidean geometry. An early publication is Karol Borsuk's 1933 paper Drei Sätze über die n-dimensionale euklidische Sphäre.

For north-pole stereographic projection $z=(X+iY)/(1-Z)$ on the unit sphere, substitute $|z{|}^2=(1+Z)/(1-Z)$ into a planar circle equation. The displayed plane equation follows, and its intersection with the sphere is a circle. The plane contains the sphere's centre exactly when $r^2=1+|c{|}^2$, giving a great circle.

A real bilinear form $B$ on a normed vector space $X$ is coercive when there is one constant $\alpha >0$ such thatThe lower bound uniformly controls every direction in the chosen norm. Unlike the upper bound defining a bounded bilinear form, it forces positive diagonal values at every nonzero vector. No symmetry is assumed. For complex Hilbert spaces, the usual analogue uses a sesquilinear form and $\mathrm{Re}B(v,v)\ge \alpha \Vert v{\Vert }^2$; it is not the same assertion for a complex bilinear form.

A concrete bounded and coercive example on ${\mathbb{R}}^2$, with the Euclidean norm, isIts quadratic form isThe best lower constant is $\alpha =2$, attained at $(1,0)$. Nevertheless, $B(e_1,e_2)=1$ and $B(e_2,e_1)=-1$, so coercivity does not imply a symmetric bilinear form. The symmetric part of a matrix $S=(A+A^T)/2=\mathrm{diag}(2,3)$ determines $q$, while the skew-symmetric matrix part changes mixed pairings without changing $q$. This is the symmetric and alternating parts of a bilinear form decomposition.^[ref]

For the upper constant, direct multiplication gives $A^{T}A=\left(\begin{array}{cc}5& -1\\ -1& 10\end{array}\right)$. Its eigenvalues are $(15\pm \sqrt{29})/2$, so the optimal upper constant, the spectral norm, is $C=\sqrt{(15+\sqrt{29})/2}$. The two constants answer different questions: how large a mixed pairing can be, and how small a self-pairing can be relative to squared norm.

The plotted quadratic forms show the mechanism: $x^2-y^2$ descends in one direction, $x^2$ leaves one direction uncontrolled, and $2x^2+3y^2$ controls both. These are self-pairings; their level sets cannot reveal the skew-symmetric matrix part. Calling $q$ a quadratic energy here refers to this diagnostic quantity, not to a claim that a nonsymmetric equation comes from minimizing an energy functional.

On a real finite-dimensional vector space with any norm, strict positivity $B(v,v)>0$ for every nonzero $v$ does imply coercivity. Indeed the quadratic form is continuous, and its positive minimum on the compact unit sphere supplies $\alpha$. If $B$ is also a symmetric bilinear form, this is positive definiteness. For general $B$, it is the symmetric and alternating parts of a bilinear form decomposition that isolates the positive part. Switching between equivalent norms preserves coercivity, but usually changes its constant.

The infinite-dimensional vector space distinction is explicit on the real l2 sequence space:The Cauchy-Schwarz inequality gives boundedness with $C=1$, and $B(v,v)>0$ for every nonzero vector. But $B(e_n,e_n)=1/n$ with $\Vert e_n{\Vert }_2=1$, so no positive $\alpha$ works. On the span of $e_1,\dots ,e_N$, the best lower constant is $1/N$: each finite restriction is coercive, but the control disappears as $N$ grows. Strict positivity is therefore weaker than a uniform lower bound.

The practical payoff is a stability estimate for a weak formulation. Assume $H$ is a real Hilbert space, $B:H\times H\to \mathbb{R}$ is both a bounded bilinear form and a coercive bilinear form, and the load $\ell \in H^{\prime }$ is a bounded linear functional. The Lax-Milgram theorem gives a unique $u\in H$ satisfying $B(u,v)=\ell (v)$ for all $v\in H$, including nonsymmetric $B$.^[ref] The dual norm $\Vert \ell {\Vert }_{H^{\prime }}={\sup }_{v\ne 0}|\ell (v)|/\Vert v\Vert$ measures load size. Testing with $v=u$ gives the short energy estimateFor $u=0$ the bound holds directly; otherwise divide by $\Vert u\Vert$. Subtracting the equations for two loads and using the same argument gives $\Vert u_1-u_2\Vert \le \Vert {\ell }_1-{\ell }_2{\Vert }_{H^{\prime }}/\alpha$. Thus a positive lower constant limits the response to load errors. The upper bound and completeness of the Hilbert space enter the existence theorem; the estimate itself uses the lower bound and the load's dual norm.

For the displayed matrix $A$, take $\ell (v)=f^{T}v$. Then $B(u,v)=\ell (v)$ for every $v$ means $A^{T}u=f$, not $Au=f$, with this ordering of the bilinear form. Since $\det A=7$,For $f=t(1,0{)}^T$, the exact response is $u=t(3,-1{)}^T/7$ and $\Vert u{\Vert }_2=\sqrt{10}|t|/7<|t|/2$. For $f=t(0,1{)}^T$, it is $u=t(1,2{)}^T/7$, with norm $\sqrt{5}|t|/7$. The actual maximal amplification is the operator norm $\Vert (A^T{)}^{-1}{\Vert }_2=\sqrt{2/(15-\sqrt{29})}<1/2$: the coercivity estimate is valid but need not be sharp for a nonsymmetric bilinear form.

A finite-dimensional commutative real algebra with a Euclidean multiplicative norm has dimension at most two. Automatic real bilinearity from a multiplicative norm supplies bilinearity if it was not part of the initial description. On its unit sphere, squaring has derivative $v\mapsto 2xv$, injective because there are no zero divisors. The only equal squares are antipodal pairs, so the map induces an injective immersion ${\mathbb{RP}}^{n-1}\to S^{n-1}$. Compactness and the inverse function theorem make it a diffeomorphism when $n\ge 2$. For $n\ge 3$, the fundamental groups $\mathbb{Z}/2$ and zero contradict this. The real and complex number algebras attain dimensions one and two.

For two inertial frames with equal speed in a laboratory, a photon is measured at equal angular frequency by their comoving observers exactly when its laboratory propagation unit vector $\mathbf{n}$ obeys the displayed condition. Indeed, measured angular frequency is $-k\cdot U=\gamma {\omega }_{\mathrm{lab}}(1-\mathbf{n}\cdot \mathbf{v}/c)$, and equal laboratory Lorentz factors cancel. Thus the permitted directions form a great circle on the unit sphere. In the plane of the two laboratory velocities, only the internal angle bisector line remains. The velocity information alone fixes neither direction along that line nor coplanarity of the ray.

If $r\ge \sqrt{2}$ and a convex polytope satisfies $r^{-1}{B}^n\subseteq P\subseteq B^n$, then it has at least $(1-r^{-2}{)}^{-n/2}\ge e^{n/(2r^2)}$ facets. Its outward facet normals define spherical caps covering the unit sphere; the spherical cap area upper bound supplies the estimate.

On the unit sphere, use $\zeta =(x+iy)/(1-z)$ away from the north pole and $\eta =(x-iy)/(1+z)$ away from the south pole. Their overlap relation $\eta =1/\zeta$ is a holomorphic map, giving a holomorphic atlas for the Riemann sphere. The opposite sign of $iy$ in the second manifold chart ensures a holomorphic rather than antiholomorphic transition.

Writing the sphere coordinates as $(X,Y,Z)$ avoids confusing their height with the complex coordinate $z$. The two manifold charts cover the whole unit sphere:On their overlap, $X^2+Y^2=1-Z^2$ givesThe coordinate transformation and its inverse are holomorphic, so these manifold charts make the sphere a Riemann surface. The sign in $X-iY$ is essential. Geometric projection from the south pole with the same planar axis labels would give $v=(X+iY)/(1+Z)=1/\overline{z}$. The complex conjugation in that transition prevents it from being holomorphic. For $z=i$, the correct second coordinate is $w=-i$: positive $Y$ corresponds to negative imaginary part in this manifold chart.

A vector norm on a finite-dimensional real or complex vector space induces the displayed matrix norm on its linear operators. It is the corresponding operator norm, and the supremum can equivalently be taken over the unit sphere. Applying the length bound twice proves submultiplicativity: $\Vert BCx\Vert \le \Vert B\Vert \Vert C\Vert \Vert x\Vert$, hence $\Vert BC\Vert \le \Vert B\Vert \Vert C\Vert$.

The interior consists exactly of strictly copositive matrices. If $x^{T}Ax$ has positive minimum $\delta$ on the compact nonnegative unit sphere, perturbations of operator norm less than $\delta$ preserve this positive lower bound there. Conversely, if a unit nonnegative vector has zero quadratic form, $A-\epsilon I$ leaves the copositive cone for every $\epsilon >0$. Thus such a matrix is not interior. The identity matrix is a convenient interior point.

An epsilon-net in a metric space is a set within distance $\epsilon$ of every point. A maximal $\epsilon$-separated set is an epsilon-net: any uncovered point could be added. On the unit sphere of ${\mathbb{R}}^k$, disjoint Euclidean balls of radius $\epsilon /2$ around separated points lie in the ball of radius $1+\epsilon /2$. Comparing Lebesgue measures gives a net of cardinality at most $(1+2/\epsilon {)}^k$.

For a real matrix $A$, let $B=A^{\mathsf{T}}A$. It is positive semidefinite, and the spectral theorem for real symmetric matrices gives $x^{\mathsf{T}}Bx\ge {\lambda }_{\min }\Vert x{\Vert }^2$. If ${\lambda }_{\min }>0$, rescaling any longer vector to unit length decreases its image norm, so minimizers lie on the unit sphere. Lagrange multipliers applied to $x^{\mathsf{T}}Bx$ with $\Vert x\Vert =1$ give $Bx=\lambda x$. Thus the minimizers are exactly the unit vectors in the least eigenspace.

If ${\lambda }_{\min }=0$, the minimum is zero and the full argmin consists of all $x\in \ker A$ with $\Vert x\Vert \ge 1$. Indeed, $\ker (A^{\mathsf{T}}A)=\ker A$ because $x^{\mathsf{T}}{A}^{\mathsf{T}}Ax=\Vert Ax{\Vert }^2$. Restricting the minimizers to the sphere would omit longer kernel vectors in this case. For a ball constraint $\Vert x\Vert \le 1$, the argmin is always $\ker A$ intersected with the ball.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.

The retraction of a Hilbert ball onto its sphere exists. In the real Hilbert space ${\ell }^2$, let $S(x_0,x_1,\dots )=(0,x_0,x_1,\dots )$ be the unilateral shift operator, and let $e_0=(1,0,\dots )$. For $x$ in the closed unit ball $\mathbb{B}$, defineThe orthogonality of $e_0$ and $Sx$ gives $\Vert F(x){\Vert }^2=1$. Thus $F$ is a continuous map into the unit sphere. It has no fixed point: $F(x)=x$ would imply $x_{j+1}=x_j$ for every $j\ge 0$. A constant sequence in ${\ell }^2$ must be zero, whereas $F(0)=e_0\ne 0$.

Put $d=x-F(x)$, which is nonzero everywhere, and defineThis is a continuous map on the whole closed unit ball, including zero. To check its range, observe that $⟨F(x),d⟩=⟨F(x),x⟩-1$, soIf $\Vert x\Vert =1$, then $\Vert x-F(x){\Vert }^2=2(1-⟨F(x),x⟩)$, hence $\alpha (x)=1$ and $r(x)=x$. Therefore $r:\mathbb{B}\to \partial \mathbb{B}$ is a retraction.

Geometrically this is the second intersection of the line from $F(x)$ through $x$ with the unit sphere. In factThe construction illustrates why the finite-dimensional hypothesis in the no-retraction theorem matters. The closed unit ball of ${\ell }^2$ is not a compact set in the norm topology.

Let $S_{+}^{n-1}$ be the nonnegative portion of the unit sphere specified in the paper. If $Ax=0$ for some $x\in S_{+}^{n-1}$, then this nonzero vector is an eigenvector with eigenvalue zero, which already proves the assertion.

Otherwise $Ax\ne 0$ for every $x\in S_{+}^{n-1}$, andis a continuous map from $S_{+}^{n-1}$ to itself. Indeed, every entry of $Ax$ is nonnegative because $A$ is a nonnegative matrix and every entry of $x$ is nonnegative. By the supplied homeomorphism $h:{\mathbb{B}}^{n-1}\to S_{+}^{n-1}$, the continuous map $h^{-1}Th$ maps the finite-dimensional closed unit ball into itself. The Brouwer fixed-point theorem supplies a fixed point, hence $T(x_{\ast })=x_{\ast }$ for some $x_{\ast }\in S_{+}^{n-1}$. ThereforeTogether with the zero case this proves that $A$ has a nonnegative eigenvalue, with a nonnegative eigenvector. For $n=1$, the auxiliary closed unit ball is the one-point space ${\mathbb{B}}^0$ and the argument remains valid. No claim of strict positivity of every entry of the eigenvector is needed.

Put $d{\Omega }_2^2=d{\theta }^2+{\sin }^2\theta d{\phi }^2$, the standard metric tensor on the unit sphere. From $t=v+m/\lambda$ one has $dt=dv-m{\lambda }^{-2}d\lambda$. The apparently divergent terms cancel:The $(v,\lambda )$ metric tensor block has determinant $-m^2\ne 0$, independently of $\lambda$. Its coefficients are real analytic, so this formula defines a regular Lorentzian metric onThe polar-coordinate degeneracies at the poles are handled by ordinary two-sphere charts. Thus $\lambda =0$ is a coordinate singularity of the original chart and is crossed by this spacetime extension.

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.

Two norms are Lipschitz equivalent norms if constants $c,C>0$ satisfy $c\Vert u{\Vert }_1\le \Vert u{\Vert }_2\le C\Vert u{\Vert }_1$ for every $u$. On the vector space of finitely supported sequences, the norms $\Vert x{\Vert }_1=\sum |x_i|$ and $\Vert x{\Vert }_{\infty }=\max |x_i|$ are not equivalent: a vector with $N$ entries equal to one has the two norms $N$ and one.

For dimension $d<\infty$, choose a basis and write $u=\sum x_i{e}_i$. Any norm satisfies $\Vert u\Vert \le \sum |x_i|\Vert e_i\Vert \le C|x{|}_2$. This also implies its continuity in the Euclidean coordinates, by $|\Vert u\Vert -\Vert v\Vert |\le \Vert u-v\Vert \le C|x-y{|}_2$. Its positive continuous values on the compact Euclidean unit sphere have a positive minimum $c$. Homogeneity gives $c|x{|}_2\le \Vert u\Vert$, proving equivalence with the Euclidean norm and hence between any two norms.

If a finite-dimensional subspace $U$ of a normed space contains a sequence converging in the ambient norm, its coefficient vectors are Cauchy by this equivalence. The coefficients converge in ${\mathbb{R}}^d$ or ${\mathbb{C}}^d$, and the reconstructed vector lies in $U$ and is the ambient limit. Thus finite-dimensional subspaces are closed.

A finite-dimensional normed space has nonempty open balls with compact closures by equivalence and Heine-Borel. Conversely let an open set $O$ have compact closure. Choose $w\in O$ and $r>0$ such that $B(w,r)\subset O$. Its closed ball of radius $r/2$ is a closed subset of the compact closure of $O$ and is therefore compact. Translation and scaling make the closed unit ball compact.

If the space were infinite-dimensional, construct unit vectors $x_j$ with $\mathrm{dist}(x_j,\mathrm{span}(x_1,\dots ,x_{j-1}))>1/2$. Here is the needed Riesz lemma: for a proper closed subspace $M$, choose $z\notin M$, put $d=\mathrm{dist}(z,M)>0$, and choose $m\in M$ with $\Vert z-m\Vert <2d$; then $(z-m)/\Vert z-m\Vert$ has distance greater than $1/2$ from $M$. The finite spans are closed by the earlier argument, so this constructs the sequence. Its members are separated by more than $1/2$, and have no convergent subsequence, contradicting compactness. Therefore the space is finite-dimensional exactly when it has a nonempty open set with compact closure.

For a $d$-dimensional normed vector space, with unit ball $K$ and John ellipsoid $\mathcal{E}$, its volume ratio isIt is at least $1$ and invariant under invertible linear coordinate changes.

For ${\ell }_1^n$, the unit ball has volume $2^n/n!$: its intersection with each orthant is the standard simplex of volume $1/n!$. Its John ellipsoid is $n^{-1/2}{B}_2^n$. To verify maximum volume, let $A{B}_2^n$ be any centered inscribed ellipsoid. For each sign vector $\epsilon \in \{-1,1{\}}^n$, containment gives $|A^T\epsilon |\le 1$. Averaging over independent Rademacher random variables yields $\mathrm{tr}(A{A}^T)\le 1$. The arithmetic-geometric mean inequality for its squared singular values gives $|\det A|\le n^{-n/2}$. The scalar matrix $n^{-1/2}I$ attains this bound because $\Vert x{\Vert }_1\le \sqrt{n}|x|$. Centered ellipsoids suffice: if $a+A{B}_2^n$ is contained in the symmetric convex unit ball, averaging it with $-a+A{B}_2^n$ gives the centered copy with the same volume.

Using the stated Euclidean ball volume and $n=2k$,since $(2k)!/k!={\prod }_{j=k+1}^{2k}j\ge k^k$. Therefore $\mathrm{vr}({\ell }_1^{2k})\le \sqrt{8/\pi }$ uniformly in $k$.

The subsequent assertion for an arbitrary $2k$-dimensional space is false as printed. Here is a counterexample independent of any asymptotic section theorem. Take $E={\ell }_{\infty }^{2k}$. Its John ellipsoid is $B_2^{2k}$: an inscribed $A{B}_2^{2k}$ has every row of $A$ of Euclidean length at most $1$, so the Hadamard inequality gives $|\det A|\le 1$, attained by $A=I$. Let $F$ be any $k$-dimensional subspace and $f_1,\dots ,f_k$ an orthonormal basis of it. The random unit vectorhas $|x|=1$. Since ${\sum }_j{f}_j(i{)}^2\le 1$ for each coordinate $i$, the moment-generating function bound for independent Rademacher random variables givesChoose $r=\sqrt{2\log (8k)/k}$; the latter probability is at most $1/2$, so some unit vector in $F$ has $\Vert x{\Vert }_{\infty }\le r$. The proposed section inequality would therefore forceThus even one half-dimensional section cannot have the printed universal constant.

The valid orthogonal splitting with bounded volume ratio is that $L$ may depend on $V=\mathrm{vr}(E)$, and $L=32V^2$ suffices. This supplies a dimension-independent constant for the preceding ${\ell }_1^{2k}$ case, or whenever $V$ is bounded uniformly. We prove the complete qualified assertion, including its epsilon-net ingredient.

In John position, $\Vert x{\Vert }_E\le |x|$. Polar integration of the radial function $\rho (u)=1/\Vert u{\Vert }_E$ giveswhere $\sigma$ is normalized spherical measure.

A maximal $\epsilon$-separated subset of $S^{k-1}$ is an epsilon-net. Balls of radius $\epsilon /2$ about its points have disjoint interiors and lie in the ball of radius $1+\epsilon /2$. Comparing volumes bounds its cardinality by $(1+2/\epsilon {)}^k$. Finite existence follows from this packing bound, and maximality proves the net property without assuming a separate covering theorem.

Choose two orthogonal $k$-dimensional subspaces $F_0,F_0^{\perp }$ and one such net on each unit sphere. Rotate both by a common uniformly distributed orthogonal matrix $U$. This distribution can be obtained from Question 1 applied to the compact orthogonal group with its Frobenius metric; left and right multiplication are transitive isometries. Consequently $Uv$ is uniformly distributed on $S^{2k-1}$ for every fixed unit vector $v$. The union bound shows that the probability that some rotated net point has norm less than $2\epsilon$ is at mostSet $\epsilon =(32V^2{)}^{-1}$. Because $V\ge 1$, the bracket is at most $1/4+1/256$, and twice its $k$th power is less than $1$ for every $k\ge 1$. Some rotation therefore makes every point of both nets have norm at least $2\epsilon$.

The triangle inequality and $\Vert x{\Vert }_E\le |x|$ imply that $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric. Every unit vector of $U{F}_0$ or $U{F}_0^{\perp }$ is within $\epsilon$ of a good net point and thus has norm at least $\epsilon$. By homogeneity, on both orthogonal subspaces,For ${\ell }_1^{2k}$ the first computation gives the absolute choice $L=256/\pi$. The missing bounded-volume ratio restriction is essential; the literal unrestricted assertion cannot be proved.

Two norms $\Vert \cdot {\Vert }_a,\Vert \cdot {\Vert }_b$ are Lipschitz equivalent norms when constants $c,C>0$ satisfy $c\Vert x{\Vert }_a\le \Vert x{\Vert }_b\le C\Vert x{\Vert }_a$ for every vector $x$. Applying these inequalities to $x_j-x_k$ shows that a sequence is a Cauchy sequence for one norm exactly when it is so for the other; applying them to $x_j-x$ likewise equates convergence. If one norm is complete, a Cauchy sequence for the other converges using the first norm and hence also the second. Interchanging the norms proves completeness is preserved in both directions.

Write $x={\sum }_i{x}_i{e}_i$ and set $C={\sum }_i\Vert e_i\Vert$. If $\Vert x{\Vert }_2=1$, then $|x_i|\le 1$, and the triangle inequality givesScaling a nonzero $x$ gives $\Vert x\Vert \le C\Vert x{\Vert }_2$ for every $x$. The reverse triangle inequality now impliesso $x\mapsto \Vert x\Vert$ is continuous, indeed Lipschitz, in the Euclidean norm.

Suppose there were no positive lower bound on the Euclidean unit sphere. We could choose $x_j$ with $\Vert x_j{\Vert }_2=1$ but $\Vert x_j\Vert <1/j$. The Bolzano-Weierstrass theorem supplies a subsequence converging in the Euclidean norm to $x$. Continuity of both norms gives $\Vert x{\Vert }_2=1$ and $\Vert x\Vert =0$, contradicting definiteness. Therefore some $m>0$ satisfies $\Vert x\Vert \ge m$ on the unit sphere. Scaling again proves the finite-dimensional equivalence of norms: