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A plane has both principal curvatures zero, hence $K=H=0$. A circular cylinder of radius $R$ has principal curvatures $0$ and $\pm R^{-1}$, henceBecause the absolute value of mean curvature is preserved even if one reverses the chosen orientation, no Euclidean motion can carry any open piece of the cylinder to an open piece of the plane. Thus the condition $K\equiv 0$ does not characterize planar pieces.

There are analogous examples for both signs of constant Gaussian curvature. Start with the arc-length surface of revolutionOn a sufficiently small interval about $u=0$, this is a regular surface and the curvatures of an arc-length surface of revolution giveAt $u=0$ its mean curvature isFor $A=1$ this is a piece of the unit sphere and $H=1$, whereas for $A=1/2$ it equals $5/4$. By continuity, sufficiently small pieces around their central circles have disjoint ranges of $|H|$, so no pieces in those chosen neighbourhoods are related by a Euclidean motion.

For curvature minus one, useagain on a sufficiently small interval. ThenThe choices $A=1$ and $A=2$ give $H(0)=0$ and $H(0)=-3/4$. Restricting to small enough pieces again separates the ranges of $|H|$. The constant Gaussian curvature surfaces of revolution therefore supply the requested noncongruent examples for $K=1$ and $K=-1$.

For $\mathbf{a},\mathbf{b}\in {\mathbb{R}}^{n+1}$, the triangle inequality and the Cauchy-Schwarz inequality giveThus $S$ is a continuous function. The unit sphere in the finite-dimensional Euclidean normed vector space ${\mathbb{R}}^{n+1}$ is compact. Moreover, $S(\mathbf{a})=0$ implies that the polynomial $T\mathbf{a}$ vanishes identically, so every coefficient $a_r$ is zero. Hence $S$ is strictly positive on the unit sphere. By the extreme value theorem, it has a positive minimum

The supremum norm is homogeneous. For $\mathbf{a}\ne 0$,It follows directly that $\Vert T\mathbf{a}{\Vert }_{\infty }\to \infty$ whenever $\Vert \mathbf{a}{\Vert }_2\to \infty$.

No. On the round sphere, take a latitude circle other than the equator. Rotations about the polar axis preserve the circle and act transitively on its points, so all the isometries ${\varphi }_{x,y}$ exist. The latitude separates the sphere into two caps of unequal area, so those caps are not isometric. It is not a great circle and therefore is not a geodesic; equivalently, its geodesic curvature is nonzero. This is a homogeneous nongeodesic latitude.

If $b$ is positively homogeneous of degree $M$ for large $|\theta |$, then $D_{\theta }^{\beta }b$ is positively homogeneous of degree $M-|\beta |$, while $x$-derivatives preserve the degree. These derivatives are uniformly bounded on the unit sphere when $x$ ranges over a compact subset of $X$. Rescaling $\theta$ givesat large frequency, and smoothness controls the remaining compact region. Therefore

An asymptotically flat spacetime at future null infinity admits a smooth conformal completion $(\overline{\mathcal{M}},\bar{g})$ with the following properties. The physical spacetime $\mathcal{M}$ is the interior of $\overline{\mathcal{M}}$, its metric obeysand the boundary component ${\mathcal{I}}^{+}$ satisfies $\Omega =0$ and $d\Omega \ne 0$. Every future-directed outgoing null geodesic has an endpoint there, the generators of ${\mathcal{I}}^{+}$ are complete, and in four spacetime dimensions ${\mathcal{I}}^{+}\simeq \mathbb{R}\times S^2$. The physical Einstein field equations are vacuum in a neighborhood of this boundary.

Let $n_a={\bar{\nabla }}_a\Omega$. Multiplying the supplied conformal Ricci relation by ${\Omega }^2$, using $R_{ab}=0$, and taking the limit to ${\mathcal{I}}^{+}$ givesThe boundary normal is therefore null. Since a null normal is also tangent to its hypersurface, $n^a$ generates ${\mathcal{I}}^{+}$.

Smoothness makes $s={\Omega }^{-1}{n}^2$ finite at the boundary. Multiplication of the same vacuum equation by $\Omega$ gives thereTaking the trace yields $s=\frac{1}{2}\bar{\square }\Omega$, and substitution givesThe remaining freedom $\Omega \mapsto \omega \Omega$ can be used to impose $\bar{\square }\Omega =0$ on ${\mathcal{I}}^{+}$. In this conformal gauge, ${\bar{\nabla }}_a{n}_b=0$ there, so the generators are affinely parametrized, expansion-free null geodesics of the unphysical metric.

Choose a generator coordinate $u$, the defining function $\Omega$, and angular coordinates $x^A$ whose leading metric $q_{AB}$ is the round metric on the unit sphere. After the conformal and coordinate choices above, the leading unphysical metric iswith the Minkowski term $-{\Omega }^{2}d{u}^2$ entering at the next relevant order. Setting $r={\Omega }^{-1}$ recovers the physical asymptotic formThe displayed leading metric is Minkowski spacetime in outgoing null coordinates. Smooth conformal extendibility controls the lower-order corrections, while the vacuum equations constrain them to the radiative Bondi--Sachs expansion. This is the precise sense in which the permitted spacetimes approach Minkowski spacetime near ${\mathcal{I}}^{+}$ while still allowing outgoing gravitational radiation.

Write $P=P_N+P_{N-1}+\cdots +P_0$ as a sum of homogeneous parts. The operator is elliptic whenContinuity on the unit sphere gives $c={\min }_{|\omega |=1}|P_N(\omega )|>0$. Uniformly in $\omega$,so for sufficiently large $t$, $|P(t\omega )|\ge (c/2)t^N$. Since $t^N\asymp ⟨t\omega {⟩}^N$ at large $t$,for sufficiently large $|\lambda |$.

False. Take $V=\{x^2+y^2+z^2<1\}$, the open unit ball, and $u(x,y,z)=x$. This is a nonzero harmonic function. On the unit sphere, the outward unit normal vector is $\mathbf{n}=(x,y,z)$, soThus the prescribed Robin boundary condition holds. Unlike the preceding case, Green's identities now give ${\int }_V|\nabla u{|}^2={\int }_{\partial V}{u}^2$, which permits nonzero solutions of the differential equation.

Let $v\in S$. Choose a unit vector $u$ perpendicular to $v$, and put $w=v\times u$. The cross product identities show that $(v,u,w)$ is an orthonormal basis with positive orientation: $u\times (v\times u)=v$, soThus the matrix $A_v$ with these columns belongs to the special orthogonal group $G$ and satisfies $A_{v}N=v$. For any $v,z\in S$, the element $A_z{A}_v^{-1}$ sends $v$ to $z$. Therefore the group action on the unit sphere is transitive.

For $X\in {\mathrm{SO}}_3(\mathbb{R})$, the defining identity $X^{T}X=I$ givesThus an orthogonal transformation preserves the Euclidean norm and sends the unit sphere to itself. Restricting the given group action to this invariant set preserves both action axioms.

If $XN=N$, the first column of $X$ is $N=e_1$. The other columns are unit vectors perpendicular to it, so the first row is also $(1,0,0)$. Hence $X=\mathrm{diag}(1,B)$ with $B^{T}B=I_2$ and $\det B=1$. Conversely, every such matrix lies in $G$ and fixes $N$. Since every element of the special orthogonal group in two dimensions has the planar rotation matrix form, the stabilizer subgroup isGeometrically these are exactly the rotations in three dimensions about the first coordinate axis.

Let $F=x+y+z^2$ and $g=x^2+y^2+z^2$. The unit sphere is compact, so the continuous function $F$ has a global maximum and global minimum. Since $\nabla g\ne 0$ on the unit sphere, all such points satisfy the Lagrange multiplier equationsThus $x=y$ and either $z=0$ or $\lambda =1$.

If $z=0$, then $x=y=\pm 1/\sqrt{2}$, giving $F=\pm \sqrt{2}$. If $\lambda =1$, then $x=y=1/2$ and $z=\pm 1/\sqrt{2}$, giving $F=3/2$. These exhaust the candidates, so

The real spectral theorem states that a symmetric matrix has an orthonormal basis of real eigenvectors, and hence is a diagonalisable matrix. Here is a proof in this setting.

For $n\ge 1$, the quadratic form $q(x)=x^{T}Ax$ attains a maximum on the unit sphere, by compactness and the extreme value theorem. Choose a unit maximizer $u$ and put $\lambda =u^{T}Au$. For any $w\perp u$, the Rayleigh quotient along $u+tw$ has a maximum at $t=0$, soThe last equality uses $A=A^T$ and $u^{T}w=0$. Thus $Au$ is orthogonal to every $w\perp u$, so $Au=\lambda u$.

The orthogonal complement $u^{\perp }$ is an invariant subspace, because for $x\perp u$,The restriction of $A$ to $u^{\perp }$ is again symmetric with respect to the induced inner product. By mathematical induction on dimension, it has an orthonormal basis of eigenvectors; adjoining $u$ gives one for ${\mathbb{R}}^n$. If $Q$ has those basis vectors as columns, then $Q^{T}AQ$ is a real diagonal matrix and $Q^{-1}=Q^T$.

For $a=b=1$, the stationary points found in parts (a) and (b) have temperatures $0$, $1$, $1/4$ and $1/5$. The unit sphere is compact and $T$ is a continuous function, so the extreme value theorem ensures that a hottest point exists; it must be one of those stationary points. ThusFor a direct bound, put $s=x^2\in [0,1]$. The identity $(y^2-z^2{)}^2\ge 0$ gives $y^2{z}^2\le (y^2+z^2{)}^2/4=(1-s{)}^2/4$, andsince $1-s^2-(1-s{)}^2/4=(1-s)(5s+3)/4\ge 0$. Equality forces $s=1$, giving exactly the two points above.

A sufficient regular-level condition is $\nabla f(x)\ne 0$ at every point with $f(x)=0$. The implicit function theorem then makes the zero level a two-dimensional continuously differentiable submanifold. Its tangent space isFor a minimizing pair $(a,b)$ and any tangent vector $v\in T_a{S}_f$, take a differentiable surface curve $x(s)$ with $x(0)=a$, $x^{\prime }(0)=v$. The function $\Vert x(s)-b{\Vert }^2$ has a minimum at zero, so its derivative $2(a-b)\cdot v$ vanishes. Varying the other point similarly proves orthogonality to $T_b{S}_g$.

For the particular surfaces, minimize the distance from the origin to the second surface first. Put $u=x^2$, $v=y^2$, so $u,v\ge 0$ and $z=(u-v-6)/2$. ThenLet $s=u-v$. Since $u+v\ge |s|$, this is at least $|s|+(s-6{)}^2/4$, with equality when one of $u,v$ is zero. For $s\ge 0$, this quadratic has its minimum at $s=4$, with value 5. For $s\le 0$, it decreases up to zero and has minimum9 on that half-line. The global minimum is therefore5, attained atAll points of this surface lie outside the unit sphere. For any such $b$, the distance to the sphere is $\Vert b\Vert -1$, attained at $a=b/\Vert b\Vert$ by the triangle inequality. ConsequentlyThis is a global minimization, not merely a stationary-point calculation.

Measure single-particle energy from the lowest state and write ${\epsilon }_k=c{k}^{\alpha }$, with $c>0$. In a homogeneous system, the number of wavevector states in a thin spherical shell is proportional to $k^{d-1}dk$. Changing variables from $k$ to $\epsilon$ gives the density of states per unit spatial volume,where $S_{d-1}$ is the area of the unit sphere in $d$ dimensions; an additional internal state degeneracy would multiply this expression by a constant.

The Bose-Einstein distribution has $\mu \le 0$. At saturation, $\mu \to 0^{-}$, the maximum excited-state number density isOnly the lower endpoint can cause divergence. There the integrand is proportional toThe improper integral converges precisely when its power exceeds $-1$. ThusWhen $d=\alpha$, the divergence is logarithmic; when $d<\alpha$, it is a power divergence. In either case the excited states can accommodate any fixed finite density at every positive temperature, so there is no bulk Bose-Einstein condensation at nonzero temperature.

For completeness, putting $s=d/\alpha >1$ and using the Gamma function and Riemann zeta function givesAt fixed density $n$, the right-hand side reaches $n$ at $T_c$; below it the excess particles occupy the lowest state, with condensate fraction $1-(T/T_c{)}^s$. This is Bose condensation with power-law dispersion in the homogeneous thermodynamic limit. A finite vessel or a trapping potential changes the low-energy state count and must not be silently substituted for that setting.

Interpret the norms as Euclidean norms and set $B=A^{\mathsf{T}}A$. This is a real symmetric matrix with ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}=\Vert A\mathbf{x}{\Vert }^2\ge 0$, hence a positive semidefinite matrix. Squaring the nonnegative objective does not change its argmin. Let ${\lambda }_{\min }$ be the smallest eigenvalue of $B$ and $E_{\min }$ its eigenspace.

First minimize on the unit sphere. The Lagrange multiplier function $L(\mathbf{x},\lambda )={\mathbf{x}}^{\mathsf{T}}B\mathbf{x}-\lambda ({\mathbf{x}}^{\mathsf{T}}\mathbf{x}-1)$ has stationarity condition $B\mathbf{x}=\lambda \mathbf{x}$. Multiplication by ${\mathbf{x}}^{\mathsf{T}}$ shows that the squared objective at such a point is $\lambda$. The spectral theorem for real symmetric matrices also gives, for every unit vector, ${\mathbf{x}}^{\mathsf{T}}B\mathbf{x}\ge {\lambda }_{\min }$, with equality exactly in $E_{\min }$. This proves global minimality, rather than merely finding stationary candidates.

If ${\lambda }_{\min }>0$, every point with $\Vert \mathbf{x}\Vert >1$ has a strictly smaller objective after rescaling to unit length. ThereforeIf ${\lambda }_{\min }=0$, the kernel of a linear map satisfies $\ker B=\ker A$, since $B\mathbf{x}=0$ implies $\Vert A\mathbf{x}{\Vert }^2=0$, and the converse is immediate. The minimum is then zero, attained precisely atIn this case minimizers need not have unit length. This distinction is essential for a rank-deficient $A$, including any real $m\times n$ matrix with $m<n$. The result is the minimum image norm outside a Euclidean unit ball, governed by the least singular value.

The null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

For a symmetric matrix $H$ and an $\eta$-metric net $U$ of the unit sphere, with $0<\eta <1/2$, approximate a maximizing unit vector $x$ by $u\in U$. Expanding $x^{\top }Hx-u^{\top }Hu=(x-u{)}^{\top }Hx+u^{\top }H(x-u)$ bounds its absolute value by $2\eta \Vert H{\Vert }_{\mathrm{op}}$. Rearrangement proves the bound. The volumetric bound for Euclidean metric nets limits the number of needed directions, allowing a union bound to control a random matrix.

Removing any interior point $a$ from the Euclidean closed unit ball allows a retraction onto its unit sphere. For $v=x-a\ne 0$, the positive root of $|a+tv{|}^2=1$ isThen $r(x)=a+\tau (x)v$ is a continuous map on the punctured closed unit ball, has norm one, and fixes its boundary because $\tau (x)=1$ there. The quadratic equation has one positive and one negative root since $|a|<1$, so the chosen root is unambiguous.

The closed unit ball $B$ in the infinite-dimensional real Hilbert space ${\ell }^2$ admits a retraction onto its unit sphere. Write $S$ for the forward unilateral shift operator and set $F(x)=\sqrt{1-\Vert x{\Vert }^2}{e}_0+Sx$. Then $\Vert F(x)\Vert =1$ and $F(x)\ne x$: equality would force all coordinates of $x$ to coincide, hence vanish, contradicting $F(0)=e_0$.

For $d=x-F(x)$, setThe nonvanishing denominator makes $r$ a continuous map. Expanding its squared norm gives $\Vert r(x)\Vert =1$. On $\Vert x\Vert =1$, the fraction is one, so $r(x)=x$. Thus the finite-dimensional no-retraction theorem does not extend to this closed unit ball.

The Riemann sphere is $\mathbb{C}\cup \{\infty \}$ with two charts related by reciprocal coordinates.

The extra point completes the complex plane into a compact Riemann surface. It is one point shared by every direction of escape: $n$, $-n$ and $in$ all tend to $\infty$ as $n\to \infty$. A basis of neighbourhoods at $\infty$ isThis is the one-point compactification of the complex plane; convergence to $\infty$ means that the modulus tends to infinity.^[ref]

For a concrete picture, use the unit sphereand identify $z=x+iy$ with $(x,y,0)$ in its equatorial plane. The north pole $N=(0,0,1)$ represents $\infty$ and the south pole $S=(0,0,-1)$ represents $0$. The stereographic projection below gives the identification. This convention fixes the scale: the equator corresponds to $|z|=1$.^[ref]

Why add this point? The Möbius transformation $z\mapsto 1/z$ then exchanges $0$ and $\infty$ and becomes a holomorphic map everywhere on the Riemann sphere. Its apparent blow-up at zero disappears when its output is expressed in the reciprocal chart at infinity: $w=1/(1/z)=z$. This is a concrete instance of a meromorphic function as a holomorphic map to the Riemann sphere. The additional point makes the complex plane compact, while its second manifold chart lets us do ordinary complex analysis near infinity.