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An aleph-one-like linear order has size ${\aleph }_1$ and every point has only countably many predecessors. Consequently every proper initial segment is countable, since it is bounded by any point outside it. This differs from requiring every nonempty interval to be uncountable.

A total order is dense when between every two distinct ordered points lies a third. Endpoints are allowed; “without endpoints” is an additional condition. A dense order with two points $a<b$ cannot be a well-order, since $\{x:a<x\}$ is nonempty and has no least member. Its first-order axioms are the total-order axioms plus $\forall x,y(x<y\Rightarrow \exists z(x<z<y))$. Countability and uncountability are not first-order axiomatizable restrictions, by the upward and downward witness constructions of the Löwenheim-Skolem theorem.

A totally ordered set is a set equipped with a total order. Internally in a topos, weak totality does not imply decidable equality. The empty ordered set is allowed unless an inhabitation axiom is explicitly imposed.

A maximum of a subset $S$ of a total order is an element $m\in S$ with $s\le m$ for every $s\in S$. If it exists it is unique. Every nonempty finite subset has a maximum, by repeated pairwise comparison. Unlike a supremum, a maximum must belong to the subset. The completion time of jobs processed on finitely many parallel machines is the maximum of the machine completion times.

A minimum of $A$ is a member of $A$ less than or equal to every member of $A$. It equals the infimum when that lower bound is attained. A nonempty finite subset of a totally ordered set has a minimum; a nonempty open real interval does not.

Every closed regular space curve has total curvature at least $2\pi$. Equality holds exactly for convex plane curves traversed once. Here curvature is unsigned and integration is with respect to arc length.

For coprime p,q with absolute values at least two, the knot group is $⟨a,b\mid a^p=b^q⟩$. Its nontrivial central element $z=a^p=b^q$ maps nontrivially to the abelianization. A meridian $a^u{b}^v$, where $qu+pv=1$, has infinite order in the quotient $C_{|p|}\ast C_{|q|}$. A composite knot group would be an amalgamated free product of proper factors over its meridian subgroup, whose center lies in that subgroup. Then z would be a meridian power, contradicting the infinite order of the meridian modulo z.

// Target: geometric-group-theory.bigb

Collapsing an essential meridian circle of a torus to one point gives the pinched torus. In the torus's CW structure, the two-cell attaches to two circles by the commutator. Collapsing one circle leaves one one-cell and a two-cell attached by a null-homotopic word. Thus the quotient is homotopy equivalent to $S^1\vee S^2$. Its cellular chain groups are $\mathbb{Z}$ in degrees zero, one and two, with zero differentials, so its integral homology groups are $H_0=H_1=H_2=\mathbb{Z}$ and vanish in all higher degrees. This illustrates that collapsing a circle can preserve a top-dimensional homology class even while reducing the first homology rank.

Take $u,v$ modulo $\pi$; shifting either coordinate by $\pi$ leaves $h$ unchanged. Its partial derivatives are $h_u=\sin v\sin (2u-v)$ and $h_v=\sin u\sin (u-2v)$. The only critical points are $(0,0)$, $(\pi /3,2\pi /3)$ and $(2\pi /3,\pi /3)$. The latter are a nondegenerate minimum and maximum, with values $\mp 3\sqrt{3}/8$. At the origin the leading cubic is $uv(u-v)$, linearly equivalent to a monkey saddle; its critical point is isolated but degenerate. Thus three smooth critical points coexist with the Morse lower bound of four for the torus.

For an abelian group $G$ and an integer $m\ge 1$, its $m$-torsion subgroup is $G[m]=\{x\in G:mx=0\}$, the kernel of multiplication by $m$. In a finite abelian group, every nonempty fiber of this homomorphism has cardinality $|G[m]|$, so $|G:mG|=|G[m]|$. This follows directly: if $mx=my$, then $x-y\in G[m]$, and conversely translating by an element of $G[m]$ leaves the image unchanged. On the cyclic group ${\mathbb{Z}}_N$, the solutions of $mx=0$ number $\gcd (m,N)$; divide both integers by their greatest common divisor to see that $x$ must be a multiple of $N/\gcd (m,N)$.

Enumerate all words $w_i$ on two generators in a free pro-p group. Choose powers $p^{e_i}$ so large that ${\sum }_i{\tau }^{p^{e_i}}<2\tau -1$ for some $1/2<\tau <1$, and impose the closed normal relations $w_i^{p^{e_i}}=1$. The Golod-Shafarevich infinitude criterion for pro-p groups makes the resulting pro-$p$ group infinite. Its abstract subgroup generated by the two generator images is dense and infinite; each element is represented by an enumerated word and has finite order. The ambient finite quotients separate its elements, so it is also a residually finite group. The abstract subgroup, rather than the whole topological group, is the desired finitely generated torsion group.

Every nontrivial countable torsion-free group embeds in a countable torsion-free group with precisely two conjugacy classes, the identity and all nonidentity elements. At each stage, enumerate all ordered pairs of nonidentity elements of the current group and successively adjoin HNN extensions identifying their infinite cyclic subgroups. Britton's lemma preserves embeddings and torsion in an HNN extension preserves torsion-freeness. Repeat on the resulting countable union, so pairs involving newly introduced elements are eventually included. A final increasing union is countable and torsion-free, and any two nonidentity elements become conjugate at a later stage. No effective enumeration of nonidentity elements is claimed.

For axial stretch $\lambda$ and twist $\alpha$ per reference length, an incompressible neo-Hookean solid has current radius $a=R/\sqrt{\lambda }$. In local cylindrical orthonormal bases the deformation gradient is $A=\left(\begin{array}{ccc}{\lambda }^{-1/2}& 0& 0\\ 0& {\lambda }^{-1/2}& \alpha r\\ 0& 0& \lambda \end{array}\right)$. The Cauchy stress tensor is $\mu A{A}^T+qI$. Radial equilibrium gives $q^{\prime }=\mu {\alpha }^{2}r$, and a traction-free curved surface gives $q=-\mu /\lambda +\mu {\alpha }^2(r^2-a^2)/2$. The nominal end traction is the displayed vector, measured per reference area. Its moment about the axis is $\int (x_1{t}_2-x_2{t}_1)d{S}_0=\mu \alpha \int r^{2}d{S}_0=\mu \pi R^4\alpha /(2\lambda )$. This calculation uses current moment arms with reference-area tractions.

A pair of equal and opposite forces acting along distinct lines has zero resultant force but a nonzero torque. If $\mathbf{F}$ acts at ${\mathbf{x}}_1$ and $-\mathbf{F}$ at ${\mathbf{x}}_2$, the total torque about any origin is $({\mathbf{x}}_1-{\mathbf{x}}_2)\times \mathbf{F}$. Shifting that origin changes the moment by the cross product of the shift with the resultant force, which is zero. A prescribed couple therefore specifies a pure turning moment independently of its reference point. In Stokes flow it can drive a rotating sphere in Stokes flow without a directly imposed translational force.

The lever arm of a nonzero force about an origin is the shortest distance from that origin to the line through the application point parallel to the force. Its length is $\ell =|\mathbf{r}\times \hat{\mathbf{F}}|$, with unit vector $\hat{\mathbf{F}}=\mathbf{F}/|\mathbf{F}|$. Thus torque magnitude is $F\ell$, maximal at a right angle for fixed $r$ and $F$, and zero if the line passes through the origin. Sliding the application point along that line leaves torque unchanged, since $(\mathbf{r}+s\hat{\mathbf{F}})\times \mathbf{F}=\mathbf{r}\times \mathbf{F}$. The same perpendicular-distance construction gives angular momentum magnitude $p\ell$ for a line parallel to nonzero momentum through the particle.

For circle-independent fields, the metric gives the type-IIA string-frame metric, dilaton and RR one-form. Decomposing $A_3=C_3+B_2\wedge dy$ gives the RR three-form and Kalb–Ramond field. With $H_3=d{B}_2$, $F_2=d{C}_1$, the horizontal four-form is ${\tilde{F}}_4=d{C}_3-C_1\wedge H_3$. The Einstein term reduces, modulo a total derivative, to $\sqrt{-g}[e^{-2\Phi }(R+4(\nabla \Phi {)}^2)-\frac{1}{4}{F}_{2,\mu \nu }{F}_2^{\mu \nu }]$. The eleven-dimensional form norm gives $\sqrt{-g}[e^{-2\Phi }|H_3{|}^2+|{\tilde{F}}_4{|}^2]$, where $|F_k{|}^2=F^2/k!$. Ordinary reduction generates massless type IIA supergravity, not Romans mass.

For constant-density incompressible ideal magnetohydrodynamics about $\mathbf{B}=Jr{\mathbf{e}}_{\varphi }$, the cylindrical basis contributions make the linearized Lorentz force equal to $imJ\mathbf{b}+2J{\mathbf{e}}_z\times \mathbf{b}$, apart from a magnetic pressure gradient. The induction equation reduces to $i\omega \mathbf{b}=imJ\mathbf{u}$. Taking the curl after eliminating velocity gives the displayed relation, with $\tilde{J}=J/\sqrt{{\mu }_0\rho }$. A second curl gives a vector Helmholtz equation and squared frequencies ${\omega }^2={\tilde{J}}^2[m^2\pm 2|mk|/({\alpha }^2+k^2{)}^{1/2}]$. The lower $|m|=1$ branch grows when $3k^2>{\alpha }^2$.

For constant-density cylindrical rotation $r\Omega (r){\mathbf{e}}_{\varphi }$ and a purely toroidal magnetic field, an axisymmetric radial displacement gives $b_{\varphi }=(B_{\varphi }/r-B_{\varphi }^{\prime }){\xi }_r$. Combining its curvature force with conservation of angular momentum gives the restoring coefficient $\mathcal{D}$ above. A local incompressible radial-vertical mode has ${\omega }^2=(k_z^2/k^2)\mathcal{D}$. Positive $\mathcal{D}$ gives stability in this ideal interchange setting; negative $\mathcal{D}$ permits growth. This is distinct from weak-field magnetorotational instability: a steeply increasing toroidal field can release current energy even without differential rotation.

In axisymmetric ideal magnetohydrodynamics, $\Omega =u_{\varphi }/R$ is the material angular velocity. Cylindrical basis-vector derivatives and mass conservation give the displayed winding equation. Thus $B_{\varphi }/(\rho R)$ is an advected scalar only when ${\mathbf{B}}_p\cdot \nabla \Omega =0$. Differential rotation along a field line otherwise produces toroidal field. This does not violate magnetic flux freezing, which concerns flux through a material surface rather than a cylindrical vector component.

For a lattice polytope $\Delta$, form the semigroup of lattice points $(m,l)$ with $l\ge 0$ and $m\in l\Delta$, using only $(0,0)$ in degree zero. Its semigroup algebra is graded by $l$, and its Proj construction gives ${\mathbb{P}}_{\Delta }$. The chart at a lattice vertex $v$ is $\mathrm{Spec}k[\mathrm{cone}(\Delta -v)\cap M]$; these charts give the normal fan of a polytope construction. For a polytope of smaller dimension, use its intrinsic affine lattice after translation.

// Target: geometry-and-topology.bigb