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For maps $A\to X$ and $A\to Y$, the double mapping cylinder $X\bigsqcup (A\times I)\bigsqcup Y$, with each endpoint attached by its given map, models the homotopy pushout. A map out of this model is a pair of maps on $X$ and $Y$ together with a homotopy between their restrictions to $A$. This is why compatible homotopy classes extend over a homotopy pushout. For a CW pair, the inclusion is a cofibration and the ordinary pushout gives the same homotopy type.

For the standard coordinate inclusion of spheres, write points as $(u,v)\in {\mathbb{R}}^{m+1}\times {\mathbb{R}}^{n-m}$, with $v\ne 0$ in the complement. The map $(u,v)\mapsto (u,v/\Vert v\Vert )$ is a homeomorphism to $B_{\mathrm{open}}^{m+1}\times S^{n-m-1}$, with inverse $(u,w)\mapsto (u,\sqrt{1-\Vert u{\Vert }^2}w)$. Contracting the ball factor proves the displayed homotopy equivalence. In codimension one there are two contractible components, so integral $H_0$ is ${\mathbb{Z}}^2$; in higher codimension the only nonzero homology groups are $H_0=\mathbb{Z}$ and $H_{n-m-1}=\mathbb{Z}$.

Give a leaf rank zero, and use Hessenberg natural sum over the immediate rooted subtrees. Every rank is below epsilon zero; conversely repeated exponentiation and finite summation encode every Cantor normal form below it. A rooted-tree homeomorphic embedding implies nondecrease of the rank. When roots match, child subtrees embed into distinct branches and exponentiation and natural sum preserve their inequalities. When the source root maps below the host root, its rank is bounded by that of a proper subtree, strictly below the host rank.

The classical mutual information obtained by any measurement of a quantum ensemble is at most its Holevo quantity. This follows from the data-processing inequality for quantum relative entropy applied to the measurement channel.

For equiprobable qubit states with Bloch vectors $r,s$, their density-matrix difference is $(r-s)\cdot \sigma /2$, with eigenvalues $\pm |r-s|/2$. An effect $0\le E\le I$ declaring the first state yields success $1/2+\mathrm{Tr}[E({\rho }_r-{\rho }_s)]/2$. Selecting its positive-eigenvalue eigenspace maximizes this trace and proves the displayed result. Measure along $r-s$ and attach the corresponding state labels; if the vectors coincide, no measurement improves on guessing.

A decreasing filtration defines a Hodge structure when $H_{\mathbb{C}}=F^p\oplus \overline{F^{w-p+1}}$ for every $p$. The decomposition is recovered as $H^{p,q}=F^p\cap \overline{F^q}$. This filtration is the holomorphic datum that varies in a variation of Hodge structure.

A polarization is a nondegenerate integral bilinear form of symmetry $(-1{)}^w$, orthogonal on Hodge summands except complementary bidegrees, satisfying the displayed positivity for every nonzero $v\in H^{p,q}$. This convention takes the cup-product pairing as the primitive geometric polarization. Omitting the constant sign factor and rescaling $Q$ is an equivalent convention. The full intersection pairing of a K3 surface is not itself a polarization: its real $(1,1)$ space has one positive direction as well as nineteen negative directions.

A pure integral variation consists of a local system of free abelian groups, holomorphic filtration subbundles of its associated flat bundle, and fibrewise Hodge structures of fixed weight and Hodge numbers. Its flat connection satisfies Griffiths transversality. A polarized variation additionally has a flat integral bilinear form polarizing every fibre.

This Poisson equation uses the Hodge Laplacian on differential forms. On a closed Riemannian manifold, the obstruction to solving it is the component of the source along harmonic differential forms. The Green operator of the Hodge Laplacian supplies a solution when that component vanishes.

To adjoin an $n$th root of $g\in G$, let $m$ be the order of a group element $g$ and take a cyclic group $C=⟨a⟩$ of order $nm$ when $m$ is finite, or infinite order when $g$ has infinite order. In the free product $G\ast C$, $⟨a^n⟩$ and $⟨g⟩$ have the same order. The HNN extension with $t^{-1}{a}^{n}t=g$ therefore embeds its base by Britton's lemma, and $h=t^{-1}at$ satisfies $h^n=g$. If $g=1$, take $C$ of order $n$ and identify the two trivial subgroups. This is an existence construction, not an algorithm for finding an unknown element order.

In the HNN extension $⟨P,t\mid t^{-1}\ell t=\ell (\ell \in L)⟩$, the displayed equivalence follows from Britton's lemma. If $p\in L$, the word collapses by the defining isomorphism; otherwise it has stable letters and no pinch, so it is nonidentity. When $P$ is finitely presented and $L$ is finitely generated, imposing commutation with the finitely many subgroup generators gives a finite presentation. An undecidable membership problem for $L$ then becomes an undecidable word problem for a group in the extension.

Every finite-order element of an HNN extension is conjugate to a finite-order element of its base. Choose a conjugate minimizing the number of stable letters. If this number is positive, no pinch is possible either internally or across the end-to-start join, since a boundary pinch would allow a conjugation shortening it. Every positive power is then reduced with positive stable-letter length, and Britton's lemma makes it nonidentity. In particular, an HNN extension of a torsion-free group is torsion-free.

Near a flat quartic maximum, $V\simeq \alpha {\sigma }^4-3\alpha {\varphi }^4$. With reduced Planck mass $M$, the slow-roll approximation gives $N\simeq {\sigma }^4/(24M^2{\varphi }_N^2)$, ${\eta }_V\simeq -3/(2N)$ and ${ϵ}_V\simeq {\sigma }^4/(192M^4{N}^3)$. Hence $n_s\simeq 1-3/N$, $n_T\simeq -2{ϵ}_V$, and ${\mathcal{P}}_{\mathcal{R}}\simeq 8\alpha N^3/{\pi }^2$. The result is dominated by the near-maximum part of the trajectory; extending slow roll to a distant minimum is unnecessary.

// Target: cosmic-inflation.bigb

For a strongly continuous unitary representation of a compact group, this Bochner integral exists in the Hilbert-Schmidt norm and is a positive compact operator commuting with the representation. It is nonzero if $\xi \ne 0$, because $⟨K_{\xi }\xi ,\xi ⟩={\int }_G|⟨U_g\xi ,\xi ⟩{|}^{2}dg>0$. A nonzero eigenspace is finite-dimensional and invariant. Applying this on orthogonal complements yields an orthogonal direct sum of finite-dimensional irreducible representations, with no separability assumption on the represented Hilbert space.

Take $H_0=0$, $H_1={\sigma }_y/2$, ${\rho }_d=(I+{\sigma }_z)/2$ and $\rho =(I+{\sigma }_x)/2$. The overlap is $f=1/2$, but $C=\mathrm{Tr}({\rho }_d[-i{H}_1,\rho ])=-1/2$, so the Hilbert-Schmidt Lyapunov feedback identity gives $\dot{V}=1/2>0$. A target overlap and the derivative of that overlap along the control direction are different quantities.

For an unramified extension of non-Archimedean local fields, let $\varphi$ generate its cyclic Galois group. The ordinary Hilbert theorem 90 supplies $b\in L^{\times }$ with $x=\varphi (b)/b$. A common uniformizer $\pi \in K$ gives $y={\pi }^{-v_L(b)}b\in {\mathcal{O}}_L^{\times }$, retaining the same quotient because $\varphi$ fixes $\pi$. The reverse implication follows by telescoping the field norm.

For a Finite Galois extension $E/K(T)$, take a primitive element and exclude all ramification and denominator problems. A Hilbert subset keeping its minimal polynomial irreducible has residue degree $[E:K(T)]$. The residue group is an unramified decomposition subgroup of the generic Galois group with the same order, so it is the whole group. This specializes a regular Galois realization over $K(T)$ to realizations over $K$ when $K$ is Hilbertian.

A first-order embedded deformation of a closed subscheme $Z\subseteq X$ is a closed subscheme $Z_{\epsilon }\subseteq X{\times }_k\mathrm{Spec}(k[\epsilon ]/({\epsilon }^2))$ flat over the dual numbers and with special fibre equal to the given embedded $Z$. It is a tangent vector of the Hilbert scheme, not a deformation modulo arbitrary automorphisms of the ambient scheme. These deformations are naturally the global sections of the normal sheaf.

A finite abelian extension $E/K$ of number fields lies in the ordinary Hilbert class field exactly when it is unramified at every finite prime and every real place of $K$ splits completely in $E$. Its Galois group is then a quotient of the ordinary ideal class group through the Artin map.

This gives a useful recognition test: construct such an extension and show that its degree equals the class number $h_K$. The inclusion $E\subseteq H_K$ and equality of degrees then force $E=H_K$. Finite-prime unramifiedness alone gives the corresponding narrow-class-field condition when real places may become complex.

The ideal class group of $\mathbb{Q}(\sqrt{-30})$ is $C_2\times C_2$. Its four reduced binary quadratic forms have coefficients $(1,0,30)$, $(2,0,15)$, $(3,0,10)$ and $(5,0,6)$. The displayed multiquadratic field has degree four over the base. The seven quadratic-subfield discriminants are $8,-3,5,-24,40,-15,-120$, whose product in absolute value is $120^4$. The conductor-discriminant formula and discriminant tower formula imply that the relative discriminant over $\mathbb{Q}(\sqrt{-30})$ is the unit ideal. This unramified abelian degree-four extension is therefore its Hilbert class field.