For $x\ge 2$, the harmonic number $H_{\lfloor x\rfloor }$ exceeds $\log x$. The finite Euler product over the primes at most $x$ contains every summand of this harmonic number. Taking logarithms and applying the geometric-series bound for the logarithmic remainder bounds the total remainder by $\frac{1}{2}{\sum }_{n=2}^{\infty }[n(n-1){]}^{-1}=1/2$, proving the displayed bound.

For directions spaced in the same rotational sense, Euler formula for normal curvature gives $k_n(\theta )=H+(k_1-k_2)\cos (2\theta )/2$. The sum of the $m$ equally spaced complex exponentials $e^{2i\theta +2\pi ij/m}$ is zero when $m\ge 2$, proving the formula. Consecutive unsigned angles alone do not imply this arrangement: they permit directions to double back.

A finite set of distinct great circles not all passing through a common antipodal pair has a vertex lying on exactly two circles. Its cells are spherical polygons with at least three edges: a digon would have antipodal endpoints, and every circle's defining hemisphere would then have to contain both endpoints, forcing a common axis. With $V-E+F=2$ and $3F\le 2E$, the graph has a vertex of valence less than six. Valences are even and at least four, so one is exactly four. Central projection identifies this with an ordinary intersection in a real projective line arrangement. If affine line intersections are counted only at finite points, parallel-line configurations require a separate qualification.

Poincare duality gives the displayed formula for connected closed oriented four-manifolds. Every integer occurs. A connected sum of $r$ copies of the Complex projective plane and $s$ copies of $S^1\times S^3$ has Euler characteristic $2+r-2s$, because the summand characteristics are three and zero. Given any integer $h$, choose $s$ large enough that $r=h-2+2s\ge 0$; when both counts are zero use $S^4$. If simple connectivity is additionally required, the possible values instead begin at two, since $b_1=0$; connected sums of projective planes realize every such value.

Use Poincare duality with coefficients in ${\mathbb{F}}_2$, valid also without orientability. The ranks in complementary degrees agree, and their signs in the alternating sum are opposite. They therefore cancel in pairs. Euler characteristic is independent of the coefficient field, giving the stated integer equality.

For a finite CW complex, $b_0$ counts connected components: its connected components are its path components, as recorded by the zeroth homology group. For a finite graph regarded as a one-dimensional CW complex, no higher homology groups occur, so $b_1=E-V+b_0$ counts independent graph cycles, and $\chi =b_0-b_1$. For a connected planar surface obtained by removing $h$ disjoint open disks from the interior of a closed disk, $(b_0,b_1,b_2)=(1,h,0)$ and $\chi =1-h$; the annulus is $h=1$.

The word "hole" needs its dimension. A two-sphere encloses a cavity but has $b_1=0$ and $b_2=1$. A closed orientable surface of genus $g$ instead has $(b_0,b_1,b_2)=(1,2g,1)$ and $\chi =2-2g$: increasing the genus by one adds two independent one-dimensional homology classes. In higher dimensions, further homology groups enter with alternating signs; there is no universal "components minus handles" rule.

A homeomorphism preserves every homology group, hence the Euler characteristic wherever the finite rational Betti number sum is defined. In fact, homotopy invariance of homology makes Euler characteristic invariant under homotopy equivalence. Thus different finite CW complex structures on the same space have equal alternating cell counts.^[ref]

In the illustrated simplicial subdivision, inserting the three edge midpoints turns one filled triangle into four. The counts change from $(V,E,F)=(3,3,1)$ to $(6,9,4)$, but $6-9+4=1$. Splitting a single edge adds one vertex and one edge, so its net contribution is $1-1=0$. Adding a diagonal inside a polygonal face adds one edge and one face, giving $-1+1=0$.

These local accounts explain common refinements. For arbitrary finite simplicial subdivisions, including barycentric subdivision in any dimension, the underlying geometric realization is unchanged. The topological invariance of Euler characteristic therefore guarantees the same alternating count, even when the individual counts change substantially.

Identify opposite sides of a square by translations to obtain a torus. All four corners become one vertex; the horizontal sides give one topological cell $a$ and the vertical sides one topological cell $b$. The open square supplies one two-dimensional topological cell. Thus $(c_0,c_1,c_2)=(1,2,1)$ and $\chi =0$. The cellular boundary of each one-dimensional topological cell is zero because its endpoints agree. Traversing the face boundary gives the group commutator $ab{a}^{-1}{b}^{-1}$, whose signed counts of both $a$ and $b$ vanish. Hence its cellular boundary is also zero, and $(b_0,b_1,b_2)=(1,2,1)$.^[ref]

The boundary of a tetrahedron is a two-sphere, with four vertices, six edges and four triangular topological cells: $\chi =4-6+4=2$. Its rational Betti numbers are $(1,0,1)$ by homology of a sphere. The torus's two independent one-dimensional homology classes cancel its degree-zero and degree-two contributions. The torus and annulus both have Euler characteristic zero, yet different homology groups; one integer cannot recover all of homology.

Build a closed annulus from two nested $n$-sided polygons, $n\ge 3$, with $n$ radial edges joining corresponding vertices. There are $2n$ vertices, $3n$ edges and $n$ quadrilateral topological cells, giving $\chi =2n-3n+n=0$. The central missing region and the exterior are not faces of the annulus. A deformation retraction onto a circle gives $(b_0,b_1,b_2)=(1,1,0)$, agreeing with the cell count. This is a useful reduction: the annulus is two-dimensional, yet its homology is that of a one-dimensional cycle graph.

A filled triangle is a disk with three vertices, three edges and one two-dimensional topological cell, so $\chi =3-3+1=1$. Its simplicial homology has $(b_0,b_1,b_2)=(1,0,0)$: the filled triangle is contractible, and the three-edge chain cycle is the chain boundary of its face. Keeping only the cycle graph on those three vertices gives $\chi =3-3=0$ and $(b_0,b_1)=(1,1)$, since that chain cycle no longer bounds a face. The surrounding plane's exterior is not part of either simplicial complex; do not add it to $F$.

Let $X$ be a finite CW complex, with $c_i$ topological cells of dimension $i$ and none above $d$. Its Euler characteristic is the displayed alternating sum. For a finite simplicial complex, $c_i$ counts its $i$-dimensional simplexes, each counted once rather than once for each containing simplex. In dimension two this becomes $V-E+F$. A topological cell's interior is counted; identified parts of its boundary are counted in their own lower-dimensional topological cells.^[ref]

The cellular homology theorem identifies cellular homology with singular homology, so the count equals the alternating rational Betti numbers. These are the free ranks of integral homology; torsion subgroups do not contribute to them. Over any field, the alternating homology dimensions still give the same Euler characteristic, although individual Betti numbers can depend on the field.

The spectral radius of the matrix exponential gives the lower bound through the spectral abscissa $s(A)$. Differentiating the squared Euclidean norm along $q^{\prime }=Aq$ and bounding its Hermitian quadratic form gives the upper bound through the Euclidean logarithmic norm ${\mu }_2(A)$. A non-normal matrix can have ${\mu }_2(A)>0$ despite $s(A)<0$, allowing transient growth before its ultimate decay.

A frieze group is a discrete subgroup of the isometry group of the Euclidean plane whose translation subgroup is infinite cyclic and has finite index. It describes a pattern periodic in one planar direction. The planar hypothesis is part of the definition; analogous groups acting in three-dimensional Euclidean space need not be frieze groups.

Translations along a fixed line give $\mathbb{Z}$. Adding reflections in perpendicular parallel lines gives the infinite dihedral group; adding a reflection in the translation axis instead gives $\mathbb{Z}\times C_2$. A glide reflection supplies another way to combine the single periodic direction with orientation reversal.

The translation subgroup of a group of Euclidean isometries consists of its translations. It is a normal subgroup, because conjugation by $x\mapsto Ax+b$ takes translation by $v$ to translation by $Av$. Its vectors form an additive subgroup of Euclidean space; for a plane crystallographic group they form a rank-two translation lattice.

In Euclidean geometry, a rotation about the origin is an orientation-preserving orthogonal linear map, represented by an element of the special orthogonal group. About another center $a$, the map has form $x\mapsto a+R(x-a)$. In three dimensions a nonidentity rotation fixes an axis.

The vertex set of a triangle with three distinct side lengths has trivial setwise stabilizer subgroup in the plane isometry group. Each vertex is characterized by the unordered pair of lengths of its incident edges, so a setwise-preserving Euclidean isometry fixes every vertex. An isometry fixing three noncollinear plane points is the identity: squared Euclidean distances to them determine every other point, since subtracting two of the Euclidean distance equations gives two independent linear equations. For example, $(0,0),(1,0),(0,2)$ have side lengths $1,2,\sqrt{5}$.

The ring of integers $\mathbb{Z}[(1+\sqrt{-7})/2]$ is a Euclidean domain for $N(z)=z\overline{z}$. In its lattice choose the nearest row, whose imaginary-coordinate error is at most $\sqrt{7}/4$, then the nearest point on that row, whose real-coordinate error is at most $1/2$. Every complex number is thus within squared distance $11/16<1$ of a lattice point, giving Euclidean division. The suborder $\mathbb{Z}[\sqrt{-7}]$ has the nonprincipal ideal $(2,1+\sqrt{-7})$ and consequently is not a Euclidean domain for any Euclidean function.

For a Euclidean radial-time metric $C(r)(f(r)d{t}_E^2+d{r}^2/f(r))$, assume $f$ has a simple horizon zero and $C$ is smooth, finite and positive there. A proper radial coordinate $\rho =2\sqrt{C(r_H)(r-r_H)/f^{\prime }(r_H)}$ puts the leading metric in the form $d{\rho }^2+f^{\prime }(r_H{)}^2{\rho }^{2}d{t}_E^2/4$. Smoothness fixes the displayed period independently of $C(r_H)$. When the Lorentzian time is normalized at infinity, the period is the inverse Hawking temperature in natural units. A conformal factor singular at the horizon need not preserve this argument.

// Target: general-relativity.bigb

For a normalized static metric $d{s}^2=-Vd{t}^2+d{r}^2/V+r^{2}d{\Omega }^2$ with a simple outer zero at $r_0$, set $b=V^{\prime }(r_0)>0$. The Wick rotation and $r-r_0=b{\rho }^2/4$ give $d{\rho }^2+(b\rho /2{)}^{2}d{t}_E^2$ to leading order, so smoothness fixes period $4\pi /b$. In an ingoing null chart, ${\nabla }_{k}k=(b/2)k$ on the future horizon; the outgoing chart gives the opposite sign on the past branch. The Hawking temperature is $b/(4\pi )=|\kappa |/(2\pi )$ and uses the time generator normalized at infinity.