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For a connected three-dimensional Riemannian manifold, the Myers-Steenrod theorem bounds the dimension of its isometry group by six. Dimension five cannot occur: isotropy embeds in $O(3)$, whose Lie subalgebras have dimensions zero, one or three. A five-dimensional action would require three-dimensional isotropy and a two-dimensional orbit; full rotational isotropy has no invariant tangent two-plane. The maximal dimension six forces transitivity and isotropy, hence constant sectional curvature. Each of the six listed dimensions is attained by a suitable metric.

A finite group of orientation-preserving isometries of the Euclidean plane or hyperbolic plane is a cyclic group. Take one orbit and its unique minimum-radius enclosing disc. Every group element preserves the orbit and therefore the disc's centre. The orientation-preserving stabilizer of a point acts as planar rotations, so the group embeds in the circle group. In a nontrivial finite subgroup, the least positive rotation angle generates all angles by division with remainder, proving cyclicity. Orientation reversal changes the conclusion: reflection groups need not be cyclic.

For conductivity $\lambda ={\lambda }_{0}T$, no cooling in the hot phase and a zero-temperature boundary at $m=0$, the mass-coordinate equation is linear diffusion. Its similarity profile solves $f^{\prime \prime }+\xi f^{\prime }/2=0$, $f(0)=0$, $f(\infty )=1$, giving the displayed error function. The thermal layer broadens as the square root of time. Conducted energy enters the cold radiating reservoir at rate per area $L={\lambda }_{0}A{T}_m(0)=A{T}_0\sqrt{{\lambda }_0}/\sqrt{\pi \tau }$, hence $L\propto t^{-1/2}$.

For an irreducible complex character of a representation, the indicator is $1$ for real type, $-1$ for quaternionic type and $0$ for complex type, the latter meaning $\chi \ne \overline{\chi }$. It detects the invariant bilinear form on an irreducible representation. For an odd-order group, squaring permutes its elements, so every nontrivial irreducible character has indicator zero by character orthogonality.

Every finite coloring of an IP set has a color class that is an IP set. Here is a reduction to the ordinary Hindman theorem. Given $\mathrm{FS}(x_i)$ in the original set, color an integer $n>0$ by the color of ${\sum }_{i\in {\mathrm{supp}}_2(n)}{x}_i$, where the support consists of positions of its nonzero digits in the binary expansion. The Hindman theorem produces $w_i>0$ with monochromatic $\mathrm{FS}(w_i)$. Replace the $w_i$ by consecutive, disjoint finite block sums $v_j$ with separated binary supports: after a block has been chosen, take $K$ above all previously used positions; two of $2^K+1$ partial sums of a fresh tail agree modulo $2^K$, so their positive difference supplies the next block sum divisible by $2^K$. Now binary addition of distinct $v_j$ has no carries. Therefore $y_j={\sum }_{i\in {\mathrm{supp}}_2(v_j)}{x}_i$ generates a monochromatic finite-sums set in the original IP set.

An IP-star set meets every IP set. Equivalently, its complement is not an IP set: a disjoint IP set is precisely an IP subset of the complement. This is a condition on all infinite finite-sums sets, not merely on a single generating sequence.

Let $E$ be the union of integer intervals $[2^{2j},2^{2j+1})$ for $j\ge 0$. Every nonempty sum of distinct $4^j$ has largest summand $4^J$ and lies in $[4^J,2\cdot 4^J)$, since the sum of all smaller summands is $(4^J-1)/3$. Thus $\mathrm{FS}(4^j)\subseteq E$. Doubling gives $\mathrm{FS}(2\cdot 4^j)\subseteq E^c$. Both sides are IP sets. Each can be adjoined to the IP-star filter while preserving the finite intersection property, yielding two distinct IP ultrafilters.

Consider the Riemann problem on $\mathbb{R}$ with initial condition $u_0(x)=u_L$ for $x<0$ and $u_0(x)=u_R$ for $x>0$. No endpoint boundary conditions are imposed. When $u_L>u_R$, its entropy solution is a shock wave with constant speed $s=(u_L+u_R)/2$. In the original example $u_L=2$, $u_R=0$, the Rankine-Hugoniot condition givesThe left characteristic speed is $2$ and the right characteristic speed is $0$, so $2>1>0$: both sides enter the entropy shock. The Viscous Burgers travelling shock gives an exact smooth family tending to this solution.

Reversing the states to $u_L=0$, $u_R=2$ still gives a Rankine-Hugoniot condition speed of $1$, but now characteristic curves move away from the discontinuity. That expansion jump is a weak solution and fails entropy admissibility. For general $u_L<u_R$, the selected entropy solution is the continuous self-similar solutionThis rarefaction wave fills the expanding gap between the two exterior characteristic curves. Inside its fan, $u_t=-x/t^2$ and $u{u}_x=x/t^2$, verifying the Inviscid Burgers equation; its values agree continuously with both exterior states. For $0$ to $2$, the fan occupies $0<x<2t$. If $u_L=u_R$, the solution is constant.^[ref]

Before characteristic crossing, the method of characteristics for the Inviscid Burgers equation transports each initial value without changing it:Here $\xi$ labels the starting position. Suppose $u_0$ is a bounded continuously differentiable function with bounded derivative. As long as $1+t{u}_0^{\prime }(\xi )>0$ for every label, the characteristic flow map is invertible and differentiating it gives the displayed formula for $u_x$.^[ref]

If the negative minimum $m={\min }_{\xi }{u}_0^{\prime }(\xi )$ is attained, the first breaking time is $t_{\ast }=-1/m$. Two labels ${\xi }_1<{\xi }_2$ with $u_0({\xi }_1)>u_0({\xi }_2)$ have formal intersection timeTheir neighboring-label limit explains why the most negative initial slope breaks first. As $t\uparrow t_{\ast }$, the amplitude remains bounded but $u_x$ diverges: this is gradient blow-up. Continuing intersecting lines would assign incompatible values at one position; the post-breaking entropy solution instead uses shock waves. Nondecreasing data have no forward characteristic crossing by this mechanism.

For a concrete calculation in dimensionless quantities on the infinite line, choose $u_0(\xi )=1-\tanh \xi$, with limiting states $2$ and $0$. Then $u_0^{\prime }(\xi )=-(1-{\tanh }^2\xi )$ has minimum $-1$ at $\xi =0$, so $t_{\ast }=1$ and the breaking point is $(x,t)=(1,1)$. Along the central characteristic curve $x=t$, $u=1$ and $u_x=-1/(1-t)$ before breaking. The labels $\xi =\pm 1$ give intersecting formal lines at $t=1/\tanh 1\approx 1.313$, after the classical solution has already broken.

For a ring homomorphism $A\to B$, extension of scalars takes an invertible module to an invertible $B$-module. A direct-summand presentation in a finite free module remains a direct-summand presentation after tensoring. At a prime $\mathfrak{q}\subset B$ contracting to $\mathfrak{p}$, localization gives $B_{\mathfrak{q}}{\otimes }_{A_{\mathfrak{p}}}{L}_{\mathfrak{p}}\cong B_{\mathfrak{q}}$. Compatibility with tensor products, identities and composition makes the Picard group of a ring a covariant functor.

In the parametrization with density proportional to $y^{-3/2}\exp [-(\lambda /y+\varphi y)/2]$, the Inverse Gaussian distribution has cumulant-generating function $K(t)=\sqrt{\lambda \varphi }-\sqrt{\lambda (\varphi -2t)}$. For $n$ independent copies, the average has this same family with parameters $(n\varphi ,n\lambda )$, while the sum has parameters $(\varphi ,n^2\lambda )$. These assertions follow by multiplying moment-generating functions and scaling the density. The more usual mean parameter is $\mu =\sqrt{\lambda /\varphi }$, so the sum has mean $n\mu$ and shape $n^2\lambda$.

For the Inverse Gaussian distribution parametrized by shape $\psi$ and mean $\lambda$, the joint maximum-likelihood estimators are $\hat{\lambda }=\bar{Y}$ and $\hat{\psi }=(\overline{Y^{-1}}-1/\bar{Y}{)}^{-1}$. With known $\lambda$, the shape estimator is instead $(\overline{Y^{-1}}-2/\lambda +\bar{Y}/{\lambda }^2{)}^{-1}$. Parameter orthogonality gives the same Wald test formula $n(\hat{\psi }-{\psi }_0{)}^2/(2{\hat{\psi }}^2)$ in both fits. The two estimators differ by $O_p(n^{-1})$ under regular model sampling, so the test statistics agree to first order under the null; they need not be numerically equal in finite samples.

Let $j_i=\mathrm{Im}(\bar{\Phi }{D}_i\Phi )$. The identityand completing the square give, when the current boundary integral vanishes,For flux $2\pi N>0$, the lower bound is $\pi N$, saturated by the two first-order Bogomolny equations. Division by $h$ is interpreted in the finite-energy sense at its zeros.

Write $\Phi =e^{u/2+i\chi }$ away from its zeros and use the oriented planar Hodge star with $\ast d{x}^1=d{x}^2$. The first Bogomolny equation gives $A=d\chi -\frac{1}{2}\ast du$. A zero of multiplicity $n_j$ contributes $u=2n_j\log |x-p_j|+\text{smooth}$ and phase winding $2\pi n_j$. Combining the smooth magnetic field with these delta sources gives the displayed equation and the topological boundary condition $u\to 0$. Prescribed zero positions and multiplicities reduce multi-vortex construction to this scalar elliptic problem; a regular solution reconstructs both smooth fields.

In coordinates $(\theta ,z)$ on a circular cylinder of radius $a$, the induced metric is $a^{2}d{\theta }^2+d{z}^2$. Its coefficients are constant, so the Christoffel symbols vanish and the affinely parametrized geodesic equation is ${\theta }^{\prime \prime }=z^{\prime \prime }=0$. The nonconstant solutions are generators when $c=0$, circular parallels when $d=0$, and helices when both are nonzero. Unrolling the cylinder with coordinate $u=a\theta$ makes all these curves straight lines in the Euclidean metric. Unit-speed parametrization additionally requires $a^2{c}^2+d^2=1$.

For a worked helix on a circular cylinder of radius $1$, choose $\theta (t)=2\pi t,z(t)=2t$ for $0\le t\le 1$. Unrolling gives the straight segment $(u,z)=(2\pi t,2t)$, so its arc length is $\sqrt{4{\pi }^2+4}$. Its acceleration in space points normally to the surface, consistent with the geodesic equation.

This geodesic is not globally shortest between its endpoints: they share the same angular position, and the vertical generator has length $2$. The helix minimizes length among paths with the same endpoints and one full winding, because their unrolled lifts join $(0,0)$ to $(2\pi ,2)$. In general, geodesics minimize sufficiently short subarcs; changing winding can produce a shorter route. Keeping this local/global distinction makes unrolling useful without mistaking every straight lift for an unrestricted shortest path.

For $c,r>0$, an intersection inside a disk requires at least two lines to meet that disk. The latter count is Poisson with mean $2\pi cr$, giving the non-strict bound first. It is strict because two lines with nearby normal angles but separated perpendicular distances can both hit the disk while intersecting outside. Choose two positive-measure parameter boxes forcing this configuration; the event of exactly one line in each box and no other disk-hitting line has positive probability. This also yields a void probability too large to equal that of a Poisson point process with the intersection process's intensity measure.

In the product $(S^{2d}{)}^{2\ell }$, for $d,\ell \ge 1$, vary the factors indexed by an $\ell$-element subset $I$ and fix points in the other factors. Orient the resulting submanifold $S_I$ in increasing factor order. These homology classes form an integral middle-dimensional basis by the Künneth theorem. Complementary subsets give a transverse intersection at one point, with positive sign since all factor dimensions are even. For noncomplementary subsets some factor is fixed in both representatives; moving that point makes them disjoint. In particular every basis class has zero self-intersection number. Complementary pairs give hyperbolic blocks in the intersection form.

The intersection matrix records an intersection pairing on chosen classes. Its nonsingularity proves linear independence: a relation between the classes, paired with each class in turn, gives a vector in the matrix's kernel.

For an isolated singular point of an $N$-dimensional oriented pseudomanifold, the unshifted middle intersection complex restricts to $\mathbb{Q}$ on the regular stratum and has no negative cohomology. Its support condition is $H^j(i_x^{\ast }I{C}_{\bar{p}})=0$ for $j>\bar{p}(N)$; its cosupport condition is $H^j(i_x^{!}I{C}_{\bar{p}})=0$ for $j\le \bar{p}(N)+1$. These conditions, together with constructibility and normalization, characterize it. On a Witt space they are exchanged by $IC\cong D(IC)[-N]$.

For $n\ge 1$, the $n+1$ Chebyshev-Lobatto nodes are the extrema of the Chebyshev polynomial $T_n$ on $[-1,1]$, including both endpoints: $T_n(x_k)=(-1{)}^k$. Imagine $n+1$ equally spaced angles on a semicircle and project their points onto the horizontal axis. The interpolation nodes crowd near $-1$ and $1$ because equal angular steps have smaller horizontal projections there. Reversing the order gives the same set. These are distinct from the roots of a polynomial $\cos ((2k+1)\pi /(2n+2))$ of $T_{n+1}$, which omit the endpoints. Driscoll and Braun call this endpoint-including family Chebyshev points of the second kind or Chebyshev extreme points.^[ref]

On an interval $[a,b]$, use $x_k=(a+b)/2+(b-a)\cos (k\pi /n)/2$. The Runge phenomenon example compares polynomial interpolation at eleven Chebyshev-Lobatto nodes with eleven equally spaced interpolation nodes. The distinction matters whenever sampling locations are a choice rather than fixed observations.