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With $\Sigma =I_d$, constrained likelihood maximization minimizes $\Vert \theta -\overline{X}{\Vert }^2$. On the unit sphere this is $1+\Vert \overline{X}{\Vert }^2-2{\theta }^T\overline{X}$. By the Cauchy-Schwarz inequality its minimizer isIf $\overline{X}=0$, every unit vector is a maximum-likelihood estimator; a specified measurable tie-breaking choice makes an estimator defined everywhere. This event has probability zero in the nonsingular normal model, but the optimization still has that exceptional case.

Two norms $N_1,N_2$ are equivalent norms when constants $c,C>0$ satisfyfor every $v$.

For $x,y\in {\mathbb{R}}^n$, the reverse triangle inequality givesby the Cauchy-Schwarz inequality. Hence $g$ is Lipschitz continuous, and therefore continuous.

On the compact Euclidean unit sphere, $g$ is positive and continuous, so the extreme value theorem gives $0<m\le g\le M<\infty$. Homogeneity then yieldsThus every norm on a finite-dimensional vector space is equivalent to the Euclidean norm; comparing two such bounds proves any two norms on $V$ are Lipschitz equivalent.

Let $q$ denote the quotient map and write $[X]$ for a point of Real projective space. For $i=1,\dots ,n+1$, take the open setIt is open in the quotient topology because its inverse image under $q$ is the open set $\{X:X_i\ne 0\}$. These sets cover, since a nonzero vector has at least one nonzero coordinate. Define the manifold chartThe hat means that the $i$th coordinate is omitted. Ratios are unchanged by multiplying $X$ by any nonzero real number, so the chart is well defined. Its inverse inserts $1$ into position $i$ and takes the resulting equivalence class. The ratios and this inverse are continuous, giving a homeomorphism $U_i\to {\mathbb{R}}^n$.

For a compact description of the smooth transition maps, write $a_i=1$ and let $a_k=X_k/X_i$ for $k\ne i$. On $U_i\cap U_j$ one has $a_j\ne 0$, andEach coordinate is a rational smooth function on the open domain $a_j\ne 0$; reversing $i,j$ gives a smooth inverse. Thus these $n+1$ charts form the standard affine atlas of real projective space.

For completeness, the underlying space is Hausdorff: normalizing $X$ identifies it with the antipodal quotient of the unit sphere, and the continuous injective map $[X]\mapsto X{X}^T/(X^{T}X)$ into the space of symmetric matrices separates any two distinct classes by disjoint open neighborhoods. It is also a second-countable space: pull back the countable bases of rational balls in these finitely many charts, and take their union. Therefore

The great circle containing the pointing and celestial pole is the intersection of the unit sphere with the meridian plane $y=0$. The orthogonal projection of $\mathbf{s}$ onto that plane, normalized to unit length, isWithin the local meridian chart $\cos \Delta >0$, its declination is therefore $q$. Outside that chart, the same construction gives an oriented great circle parameter; the foot can lie on the opposite right-ascension half of the meridian, and $q$ need not be its ordinary declination.

To verify the right angle geometrically, the tangent to the great circle from $L$ toward $S$ is proportional to $\mathbf{s}-(\mathbf{s}\cdot \mathbf{\ell })\mathbf{\ell }$. Since $\mathbf{s}\cdot \mathbf{\ell }=\rho$, that tangent has only a $y$ component. The tangent from $L$ toward the pole, $\mathbf{P}-(\mathbf{P}\cdot \mathbf{\ell })\mathbf{\ell }$, lies in $y=0$. Their orthogonality proves the claimed spherical right angle. Thus within the local meridian chart, $q$ is the declination of the perpendicular foot on the pointing meridian.

There are antipodal perpendicular feet on the complete great circle. On the local meridian chart, the telescope field selects the foot near the pointing; the cotangent equation alone determines $q$ only modulo $\pi$. At a coincident foot or pole, the corresponding angle is understood by continuity rather than by a nonzero tangent vector.

Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

Uniform orientation on the unit sphere means that probability is proportional to the solid-angle elementBecause the minor axis is unoriented, it is enough to take $0\le i\le \pi /2$. After integration over the azimuth and normalization,Equivalently, the random variable $\mu =\cos i$ has the uniform distribution on $[0,1]$.

A minimal surface can have a planar point, where $k_1=k_2=0$. There $dN=0$, which fails the stipulated conformality condition $\lambda >0$. Thus minimality need not make the Gauss map conformal everywhere; a plane is the simplest counterexample.

Conversely, conformality gives $|k_1|=|k_2|\ne 0$. It permits either $k_1=-k_2$, which is minimal, or $k_1=k_2\ne 0$, which is umbilical and nonminimal. On a round sphere, the Gauss map is conformal while both principal curvatures are equal and nonzero. Thus conformality alone does not imply minimality.

DefineThis is continuous and takes values on the unit sphere. It agrees at $(0,y)$ and $(1,y)$, is constant on the bottom edge, and is constant on the top edge. Thus it is constant on every $R$-class, so the universal property of the quotient topology gives a continuous map

For $0<y<1$, the third coordinate $\cos (\pi y)$ determines $y$, and the first two coordinates determine $x$ modulo one. Consequently the only equal values of $f$ in the open strip arise from $x=0$ and $x=1$. At $y=0$ the whole edge maps to the north pole, and at $y=1$ the whole edge maps to the south pole. These are exactly the identifications defining $R$, so $\overline{f}$ is injective. The spherical-coordinate formula also shows that it is surjective.

The square is compact, hence its quotient $X/R$ is compact by part (ii), while $S^2$ is Hausdorff as a subspace of ${\mathbb{R}}^3$. The compact-to-Hausdorff continuous bijection theorem now makes $\overline{f}$ a homeomorphism. Geometrically, identifying the vertical sides produces a cylinder and collapsing each boundary circle to a point produces the suspension of a topological space $S^1$, which is $S^2$. This proves the square quotient model of the two-sphere.

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 $|\lambda |$.

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.

For independent Bernoulli distribution upper-triangular entries of a symmetric zero-diagonal adjacency matrix of a graph, put $W=A-\mathbb{E}A$. For a unit vector $x$, $x^{\top }Wx=2{\sum }_{i<j}{x}_i{x}_j{W}_{ij}$ has sub-Gaussian variance proxy at most ${\sum }_{i<j}{x}_i^2{x}_j^2\le 1/2$, by the Hoeffding lemma. Hence $\mathbb{P}(|x^{\top }Wx|>u)\le 2e^{-u^2}$. The volumetric bound for Euclidean metric nets gives a $1/4$-metric net of the unit sphere with at most $9^n$ points. The quadratic form net bound gives $\Vert W{\Vert }_{\mathrm{op}}\le 2{\max }_x|x^{\top }Wx|$ on that net. The union bound gives $\mathbb{P}(\Vert W{\Vert }_{\mathrm{op}}>2\sqrt{n\log 9+\log 2+s})\le e^{-s}$; integrating this tail proves the displayed expectation bound.

On the unit sphere $S^{n-1}$, let $D_{\rho }$ be a spherical cap of angular radius $0<\rho <\pi /2$. If $\sin \rho \ge 1/\sqrt{n}$, its surface area is at most $s_{n-1}{\sin }^n\rho$. This includes $\rho \ge \pi /4$ for every $n\ge 2$. The bound follows by comparing the cap-area derivative with the derivative of ${\sin }^n\rho$ and checking the two endpoint values.

A real symmetric matrix whose quadratic form is positive at every nonzero vector in the nonnegative orthant. By compactness of the nonnegative unit sphere, this is equivalent to a positive uniform lower bound $x^{T}Ax\ge \delta \Vert x{\Vert }_2^2$ on that orthant. It need not be a positive-definite matrix.

A cover of the unit ball by balls of radius $\delta /2$ gives a radius-$\delta$ metric net on the unit sphere: discard balls missing the sphere and choose a sphere point in each remaining ball. Every sphere point lies within $\delta$ of the selected point in its ball. The argument keeps the number of balls and ensures that the selected points have unit length, regardless of the initial covering centers.

A subset of the radius-$R$ Euclidean ball admits an $\epsilon$-metric net of size at most $(1+2R/\epsilon {)}^n$. Take a maximal separated set: its disjoint radius-$\epsilon /2$ Euclidean balls lie in the radius-$(R+\epsilon /2)$ ball. Comparing Lebesgue measures gives the estimate. In particular, the unit sphere admits a net of radius one with at most $3^n$ points.