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A filter on a set equipped with a topology converges to $x$ if it contains every neighbourhood of $x$. A Hausdorff space gives every convergent filter at most one limit: disjoint neighbourhoods of distinct purported limits would have empty intersection in the filter.

For a function $f:X\to Y$, the pushforward of a filter on a set $X$ consists of the subsets $B\subseteq Y$ with $f^{-1}(B)\in \mathcal{F}$. Preimages preserve inclusions and finite intersections, so this is a proper filter. Equivalently it is generated by $f(A)$ for $A\in \mathcal{F}$. Because preimages preserve complements too, the pushforward of an ultrafilter is an ultrafilter.

For a source in pure plume balance, $Q(z)=C{F}_s^{1/3}(z+z_v{)}^{5/3}$ and $Q_s=Q(0)$. A single ideal exhaust passes the source volume flux. Above the exhaust height, it removes unprocessed lower fluid and the filling-box first front descends at rate $-\pi Q/A_c$. Below it, processed fluid leaves and only entrained volume grows the processed region. The floor is reached in finite time if the opening is at the floor; otherwise the final approach is exponential. A finite-height aperture supporting bidirectional exchange requires a different model.

An acknowledged file-system transaction remains recoverable after a crash under the stated storage assumptions. In the full-data timeline, durable journal payload and a durable journal commit record suffice even before the journal checkpoint. A crash before durable commit can lose the append while still preserving file-system crash consistency. A successful ordinary write may only have reached a buffer cache; an application requiring persistence must use its documented interface and check errors. On Linux kernel-based systems, fsync on a file descriptor persists the associated file contents and file metadata, but persisting a newly created name also requires fsync on the containing file-system directory.^[ref]

In $L={\mathbb{Q}}_p(\theta )$ with ${\theta }^2=p$, the Eisenstein polynomial $X^2-p$ gives $v_L(\theta )=1$ and $v_L(p)=2$. The element $x=\theta /p$ has field trace zero, but valuation $-1$, so is not in the valuation ring. Cancellation in a field trace contrasts with the multiplicative test in unit norm detects units in a local extension.

For an integral element of a finite extension of non-Archimedean local fields, reduce the multiplication endomorphism on ${\mathcal{O}}_L$ modulo a base uniformizer. Its filtration by the powers of the maximal ideal has $e$ successive quotients, each the residue field ${\kappa }_L$. Multiplication by $a$ induces multiplication by $\bar{a}$ on each quotient. Additivity of the matrix trace through the filtration proves the formula. If the residue characteristic divides the ramification index, the trace of every integral element lies in the base maximal ideal; a Galois extension is not required.

For a finite extension of local fields with normalized discrete valuations, the field norm satisfies the displayed equality. The extended absolute values on a field of all conjugates agree, so $|N(x){|}_K=|x{|}_L^{[L:K]}$. Using $[L:K]=e(L/K)f(L/K)$ and $v_L{|}_K=e(L/K)v_K$ gives the equality. Thus the field norm is a unit precisely when the original element is a unit.

Let $A$ be an integrally closed domain with fraction field $K$, and let $L/K$ be finite. If $b\in L$ is an integral element over $A$, its monic minimal polynomial lies in $A[T]$: its roots are integral, so its coefficients are integral elements of $K$, hence belong to $A$. If that polynomial has degree $r$ and constant coefficient $a_0$, then $N_{L/K}(b)=((-1{)}^r{a}_0{)}^{[L:K(b)]}\in A$. This remains true for inseparable extensions. An infinite algebraic extension has no usual determinant field norm.

A field isomorphism is a bijective field homomorphism; its inverse is also a field homomorphism. For fields extending a specified field $K$, an isomorphism may be required to fix $K$ pointwise, a stronger requirement than an abstract field isomorphism.

More generally, two field extensions inside a common overfield are linearly disjoint field extensions when the multiplication map from their tensor product of fields into that overfield is injective. The following basis criterion is the finite-degree case.

Finite extensions $L_1/K$ and $L_2/K$ are linearly disjoint when products of basis elements form a basis of their compositum. In particular,Extensions of coprime finite degrees are linearly disjoint.

The field automorphism group fixing $K$ consists of all field automorphisms of $L$ fixing $K$ pointwise, with composition. It is defined without assuming that $L/K$ is Galois. For $L=K(\alpha )$ with algebraic degree $d$, the action on the roots of the minimal polynomial inside $L$ is faithful, so this group embeds in $S_d$ and is finite.

For all finite-dimensional density operators, including pure states and mixed states, the unsquared quantum fidelity and trace distance obeyIn the squared quantum fidelity convention this is $1-\sqrt{F_{\mathrm{sq}}}\le D\le \sqrt{1-F_{\mathrm{sq}}}$. The bounds translate a high quantum fidelity into a small operationally meaningful trace distance. If $F\ge 1-\epsilon$ with $0\le \epsilon \le 1$, then $D\le \sqrt{2\epsilon -{\epsilon }^2}$; if $D\le \delta$, then $F\ge 1-\delta$. The upper bound is attained by the pure-state trace distance and fidelity identity. For the fidelity of commuting mixed states example, the bounds read $0.0652\lesssim 0.3\lesssim 0.3551$. ^[ref]

A POVM $\{E_i\}$ gives outcome probability distributions $p_i=\mathrm{Tr}(\rho E_i)$ and $q_i=\mathrm{Tr}(\sigma E_i)$ by the Born rule. Their classical overlap, the Bhattacharyya coefficient, is $B(p,q)={\sum }_i\sqrt{p_i{q}_i}$. Fuchs and Caves showed that minimizing this overlap over all POVMs gives the unsquared quantum fidelity:The same POVM is applied to either possible input quantum state. Each measurement in quantum mechanics has $B(p,q)\ge F$; some measurement attains equality, including when the density operators have zero eigenvalues. Thus quantum fidelity quantifies the smallest classical overlap that observation can reveal. For fidelity of commuting mixed states, measurement in a common eigenbasis attains the minimum. This operational interpretation compares statistics without assigning a probability that two mixed states are equal. ^{[ref] [ref]}

When two density operators are commuting elements, choose a common orthonormal eigenbasis and write $\rho =\mathrm{diag}(p_i)$, $\sigma =\mathrm{diag}(q_i)$. Their quantum fidelity becomes the classical overlap of the two probability distributions, while their trace distance becomes total variation distance:This follows by computing the positive square root of an operator entry by entry in the common eigenbasis. For the qubit mixed states $\rho =\mathrm{diag}(0.9,0.1)$ and $\sigma =\mathrm{diag}(0.6,0.4)$,The two contributions to quantum fidelity are about $0.734847$ and $0.2$. In contrast, the chance that two independent computational basis measurements, one on each preparation, print the same label is $0.9(0.6)+0.1(0.4)=0.58$. That particular event has a specified probability; neither $F$ nor $F_{\mathrm{sq}}$ is that probability.

Here trace distance $0.3$ means an equal-prior quantum state discrimination success of $0.65$ by the Holevo–Helstrom theorem. Measuring in the common computational basis and guessing $\rho$ on outcome zero and $\sigma$ on outcome one achieves $[0.9+0.4]/2=0.65$. Also $\sqrt{1-F^2}\approx 0.3551>D$: the pure-state trace distance and fidelity identity is an equality for pure states, rather than a formula for arbitrary mixed states.

For any unitary $U$, the unique positive square root of an operator obeys $\sqrt{UA{U}^{†}}=U\sqrt{A}{U}^{†}$. Conjugating the positive operators appearing in trace distance and quantum fidelity therefore conjugates their square roots. Cyclicity of the trace removes the unitary factors, proving invariance of both quantities under simultaneous unitary transformations of the states.

// Target: quantum-information-theory.bigb

For normalized pure states $\rho =|\psi ⟩⟨\psi |$ and $\sigma =|\phi ⟩⟨\phi |$, the trace distance and quantum fidelity satisfyThe last two expressions distinguish the unsquared quantum fidelity from the squared quantum fidelity. Identical pure states give $D=0$, and orthogonal pure states give $D=1$. The equality need not hold for mixed states; the Fuchs-van de Graaf inequalities give bounds in that case.

Brief proof: for two distinct pure states, their two-dimensional span contains the nonzero part of $\rho -\sigma$. This restriction has trace zero and determinant $-(1-F^2)$, hence eigenvalues $\pm \sqrt{1-F^2}$. Its trace norm is twice the positive eigenvalue, giving trace distance $D=\sqrt{1-F^2}$. For identical pure states, $\rho -\sigma =0$ directly.

This is the square of the unsquared quantum fidelity, another common convention for fidelity of quantum states. If one state is the normalized pure state $|\psi ⟩⟨\psi |$, thenBy the Born rule, the squared quantum fidelity here is the probability that a measurement in quantum mechanics with acceptance orthogonal projector $|\psi ⟩⟨\psi |$ accepts $\sigma$. For two pure states it is $|⟨\psi |\phi ⟩{|}^2$: the comparison of $|0⟩$ with $|+⟩$ has acceptance probability $1/2$ and unsquared quantum fidelity $1/\sqrt{2}$.

For two general mixed states, neither quantum fidelity nor squared quantum fidelity is a probability that the states are equal. Equality of two specified density operators is a definite mathematical condition. The measurement characterization of quantum fidelity instead describes overlap between outcome probability distributions. The ensemble average used in reliable quantum source compression is normally the average squared quantum fidelity with the emitted pure state; the convention must be stated when comparing formulas.

For a connected oriented topological surface of finite type with negative Euler characteristic, every orientation-preserving mapping class has a periodic representative, preserves an essential multicurve, or has a pseudo-Anosov map representative. In the usual disjoint classification, the reducible case means a nonperiodic class preserving such a multicurve. The assertion concerns representatives up to isotopy, rather than requiring an arbitrary homeomorphism in a periodic class to have finite order.

For a torus with one puncture, the action on first homology identifies its mapping class group with $S{L}_2(\mathbb{Z})$. For the corresponding matrix $M$, the periodic classes have $|\mathrm{tr}M|<2$ or $M=\pm I$. Nonperiodic reducible classes have $|\mathrm{tr}M|=2$ and $M\ne \pm I$; pseudo-Anosov map classes have $|\mathrm{tr}M|>2$. For example, $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ is a Dehn twist, while $I$ is periodic despite having the same trace.^[ref]

A puncture is distinguished from a boundary circle fixed pointwise: boundary twists give additional mapping classes invisible to the displayed $S{L}_2(\mathbb{Z})$ action. For the closed torus, the hyperbolic case has an Anosov homeomorphism representative.

The curvature form of a connection on a vector bundle is $F=dA+A\wedge A$ in a local frame. A principal connection has the corresponding Lie algebra-valued form $F=dA+\frac{1}{2}[A\wedge A]$. It measures the failure of horizontal directions to be integrable.

The curvature is $F(A)=d_A^2$ and in a local frame satisfies $F(A)=dA+A\wedge A$.

A section of a fiber bundle $\pi :E\to B$ is a continuous map $s:B\to E$ with $\pi \circ s={\mathrm{id}}_B$. It chooses a point of each fibre continuously. A section of a vector bundle is the special case for vector fibres; a projective bundle section chooses a line in each fibre of the underlying vector bundle.