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Suppose each job occupies two distinct resources for one whole slot, and each resource can serve at most one job in that slot. Make resources vertices and jobs edges. Two jobs conflict exactly when their edges share a vertex, so a valid schedule is a proper edge colouring. Each slot is a matching in a graph; the chromatic index is the minimum number of slots. The colour labels specify the slots, rather than physical colours. This model assumes all jobs are available initially, with equal duration, interchangeable slots and no additional conflicts. The school-timetable motivation in Graph Theory by Reinhard Diestel (author-hosted Colouring preview, opening paragraphs, pp. 123–124) uses this resource-pair interpretation.

Here is a complete example. Two technicians $A,B$ must each use three instruments $X,Y,Z$ once. Each job needs its technician and its instrument simultaneously. The six jobs are the edges of the complete bipartite graph $K_{2,3}$:Within each $M_i$, the two jobs use different technicians and different instruments, so $M_i$ is a matching in a graph. The three matchings in a graph together cover every job exactly once. Technician $A$ has three jobs, so fewer than three slots cannot suffice; the displayed schedule uses three. This is an optimal proper edge colouring with chromatic index three, witnessed by both a schedule and a resource-workload lower bound.

Repeated jobs requiring the same resource pair become distinct parallel edges of a multigraph, rather than one edge. Additional restrictions change the problem: prescribed allowed slots lead to list edge coloring, while unequal durations, deadlines, or jobs needing more than two resources require a richer scheduling model. Selecting a largest matching in a graph solves a single-round question; a shortest complete schedule must organize all jobs across rounds.

A projective embedding realizes a variety as a closed subvariety of projective space. A very ample line bundle gives such an embedding through its global sections.

The complete system of quadratic forms maps ${\mathbb{P}}^2$ to a smooth surface of degree four in ${\mathbb{P}}^5$. The six coordinates are $x_0^2,x_0{x}_1,x_0{x}_2,x_1^2,x_1{x}_2,x_2^2$. It occurs as a degree-two image of a K3 surface and, after taking a projective cone, in the plane-quintic exception to the K3 linear system dichotomies.

The pullback of $U(H)\to PU(H)$ along a continuous circle homomorphism is a compact central extension of the circle. Its lifts commute: a scalar commutator is a continuous alternating bicharacter, which vanishes on a dense cyclic subgroup and hence everywhere. Average a rank-one positive operator over this compact extension. The resulting nonzero compact operator has a finite-dimensional invariant eigenspace, and commuting unitaries have a common eigenvector there. Its continuous character has central weight one; dividing the lifts by that character yields an honest continuous circle representation.

Let $M$ be a nonzero finitely generated module over a commutative Noetherian ring. Take a length-$n$ resolution with finite projective terms. If its top Ext functor with coefficients in $A$ vanished, the dual of its last injection would be surjective. The dual of a finite projective module is projective, so that surjection splits. Finite projective modules are isomorphic to their double duals, so dualizing gives a splitting of the original injection. The resolution then shortens, contradicting minimality of $n$. When $n=0$, a nonzero coordinate of a finite free module containing $M$ as a summand gives a nonzero map $M\to A$.

For the line-projectivization of a rank-$n$ complex bundle, $p^{\ast }E=S\oplus S^{\perp }$. The Whitney sum formula for Chern classes gives $c_t(S^{\perp })=p^{\ast }{c}_t(E)/(1+xt)$ as a formal power series. Comparing coefficients proves the formula. Since $S^{\perp }$ has rank $n-1$, its degree-$n$ Chern class vanishes and yields the monic projective-bundle relation.

// Target: geometry-and-topology.bigb

This convention for a projective bundle parametrizes one-dimensional quotients of the fibres of a vector bundle $E$. Its universal quotient is ${\pi }^{\ast }E\twoheadrightarrow {\mathcal{O}}_{\mathbb{P}(E)}(1)$, and ${\pi }_{\ast }\mathcal{O}(1)=E$. Tensoring $E$ by a line bundle on the base leaves the projective bundle unchanged but tensors its universal quotient by the pullback of that line bundle. The alternative convention of holomorphic projectivization by lines replaces $E$ by its dual and reverses the tautological sign.

The profinite topology on $\mathbb{Z}$ has basis the residue classes $a+k\mathbb{Z}$, $k\ge 1$. These classes are both open and closed, since their complements are finite unions of other residue classes. Set $F_n=(4^n-1)/3+4^n\mathbb{Z}$ for $n\ge 1$. Then $F_{n+1}\subseteq F_n$ and every $F_n$ is nonempty. An integer $x$ in their intersection would make $3x+1$ divisible by every $4^n$, forcing $3x+1=0$, impossible. Thus the open complements $\mathbb{Z}\setminus F_n$ form an increasing cover with no finite subcover. The topology is Hausdorff because distinct integers are distinct modulo a sufficiently large integer, but is not compact.

For a continuous function $f:\mathbb{R}\to \mathbb{R}$, $g(t)=t^{2}f(t)$ is differentiable at zero with $g^{\prime }(0)=0$, because $[g(h)-g(0)]/h=hf(h)\to 0$. At any nonzero point $x$, $g$ is differentiable exactly when $f$ is: one direction uses the product rule, and the converse applies it to $f(t)=t^{-2}g(t)$ near $x$.

A maximal external point of a finite set has no other point strictly larger in every coordinate. This weak nondominance differs from a maximal element in the product order when coordinate ties exist: for example $(1,0)$ passes the weak test against $(1,1)$ but is not maximal in the product order. In an independent continuous-coordinate random sample, ties have probability zero, so the two notions coincide almost surely.

Two isospectral manifolds remain isospectral after taking a Riemannian product with the same closed factor: eigenvalues add, with multiplicities given by the product Laplacian eigenspace decomposition. For nonisometric closed connected factors $M_1,M_2$, choose $\epsilon <\delta <{\min }_i\mathrm{inj}(M_i)$. Every closed product geodesic of length less than $\delta$ has constant projection to $M_i$, since a nonconstant closed geodesic shorter than the injectivity radius is impossible. The short coordinate geodesics of the small flat torus span exactly its tangent factor at every point. Any product Riemannian isometry must preserve that intrinsically defined smooth distribution and its orthogonal complement, hence restrict to an isometry of the original factors. Thus these products are still nonisometric. Taking $M_i$ to be four-dimensional tori proves the higher-dimensional stabilization statement without assuming a cancellation theorem for arbitrary products.

On the probability simplex, ${\sum }_i{x}_i=1$. The Cauchy-Schwarz inequality gives $1\le \sqrt{n}\Vert x\Vert$, with equality only when all coordinates are equal. Thus the displayed centroid is the unique closest point to the origin. It is also equidistant from all coordinate vertices, since $\Vert x-e_i{\Vert }^2=\Vert x{\Vert }^2-2x_i+1$.

Every abstract subgroup of finite index in a finitely generated pro-p group is open. Fixed-generator commutator collection in nilpotent groups makes the abstract commutator subgroup closed. The resulting abelian quotient is a quotient of a finite power of the p-adic integers, so its subgroup of $p$th powers is closed and of finite index. Thus the algebraic $G^p[G,G]$ is open and every abstract index-$p$ subgroup is open. An arbitrary finite quotient is a finite p-group; induction on its order, using a normal subgroup of index $p$ and the Schreier lemma, gives the assertion. The normal core reduces nonnormal finite-index subgroups to this case.

A finitely generated powerful p-group in the category of pro-p groups is uniform if all successive indices in its lower p-series are equal. Equivalently it is torsion-free. If its generator number is $d$, the successive indices are $p^d$, the power map carries $G^{p^n}$ bijectively onto $G^{p^{n+1}}$, and an ordered minimal generating set gives unique coordinates $g_1^{{\lambda }_1}\cdots g_d^{{\lambda }_d}$ with ${\lambda }_i$ in the p-adic integers. For $p=2$, powerfulness means $[G,G]\le G^4$.

For a finitely generated pro-p group, its profinite Frattini subgroup has finite elementary abelian quotient. Continuous group homomorphisms to ${\mathbb{F}}_p$ are precisely the linear functionals on this quotient. The topological generation criterion identifies a minimal generating set with a basis, giving the displayed equality. Finite generation is essential when generator number means the least cardinality of an unrestricted dense generating family.

For finite $X$, this is the pro-p completion of the free group on $X$. Every function from $X$ to a pro-p group extends uniquely to a continuous group homomorphism. More generally one can require a basis tending to the identity and impose the universal property for maps with that convergence condition; alternatively the pro-$p$ completion of the abstract free group represents unrestricted maps from an arbitrary set. These infinite-rank conventions should not be conflated when interpreting generator cardinalities.

The inverse limit of the quotients of an abstract group by its normal subgroups of finite $p$-power index. Maps to pro-p groups factor uniquely through the canonical dense image in this completion. Unlike profinite completion, only finite p-groups are used.

A pro-p group whose closed commutator subgroup lies in its closed power subgroup $G^{p^{ϵ}}$, with $ϵ=1$ for odd $p$ and $ϵ=2$ for $p=2$. This is the profinite analogue of a powerful p-group. Finitely generated torsion-free examples are exactly the uniform pro-p groups. The additive group of a finite product of p-adic integers is an abelian example.

If a pro-$p$ presentation on $d$ generators and possibly infinitely many relators $r_i$ satisfieswith the sum convergent and $D$ the Magnus degree of a free pro-p group element, then the presented pro-p group is infinite. The Golod-Shafarevich inequality forces its completed group algebra to be infinite-dimensional, which would be impossible for a finite group. This formulation applies to relators whose degrees tend to infinity and is suitable for enumerated power-relation constructions.

Over $k={\mathbb{F}}_p$, the Nottingham group is $\mathcal{N}(k)=\{f(t)=t+{\sum }_{i\ge 2}{a}_i{t}^i:a_i\in k\}$ with composition and the $t$-adic topology. Equivalently it is the group of continuous $k$-automorphisms of the Laurent series field $k((t))$ that act trivially on $(t)/(t^2)$. With ordinary series composition, the isomorphism sends $f$ to the automorphism $h(t)\mapsto h(f^{-1}(t))$; this inverse avoids reversing the group product. It is a pro-p group, recovered from its finite truncations, and contains elements of arbitrarily large finite $p$-power order.