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The Aharonov-Bohm effect is the phase $q\Phi /\hslash$ acquired when a charge $q$ encircles magnetic flux $\Phi$, even when the magnetic field vanishes along the charge's path.

In the magnetic Aharonov-Bohm effect, a coherent electron wave takes two routes around inaccessible magnetic flux. Recombination converts their relative wave phase into displaced interference fringes. In 1959, Yakir Aharonov and David Bohm proposed a shielded solenoid geometry and identified the obstruction to removing the magnetic vector potential throughout the accessible region.^[ref]

Use International System of Units, electric charge $q$, and magnetic minimal coupling $H=(-i\hslash \nabla -q\mathbf{A}{)}^2/(2m)$. Let ${\gamma }_1,{\gamma }_2$ run from the same source $S$ to the same detector point $D$. The oriented closed curve $C$ follows ${\gamma }_1$ and then ${\gamma }_2$ backwards. For a static magnetic vector potential, the magnetic contribution to the relative wave phase isHere ${\Phi }_C={\int }_{\Sigma }\mathbf{B}\cdot d\mathbf{S}$ is the oriented magnetic flux through a spanning surface. Equal or calibrated nonmagnetic propagation phases are assumed. For an electron, $q=-e$, where $e>0$ is the elementary charge; reversing $C$ reverses the sign. The observable is the gauge-invariant holonomy $e^{i\delta }$, rather than a unique local value of the magnetic vector potential.

The accessible region is a multiply connected region: $C$ cannot contract to a point without entering the excluded flux region. Stokes theorem does not set the line integral to zero, because a spanning surface must cross that excluded region, where the magnetic field need not vanish. There is real magnetic flux in the apparatus; only the accessible paths are idealized as free of the magnetic field. The exterior of a finite straight solenoid also permits routes around its ends, so its exact topology and leakage differ from those of an infinite solenoid or an excluded solid torus.

An explicit model makes the obstruction visible. Outside an infinite flux tube of radius $a$, choose cylindrical coordinates $(r,\theta ,z)$ andIts curl vanishes there, but a loop with winding number $n$ has ${\oint }_C\mathbf{A}\cdot d\mathbf{l}=n\Phi$. Locally $\mathbf{A}=\nabla (\Phi \theta /(2\pi ))$; the angular coordinate changes by $2\pi$ on a circuit. For generic $q\Phi /\hslash$, that candidate gauge removal changes the wave function's single-valued boundary condition. On a simply connected region, a smooth static magnetic vector potential with zero curl instead has a globally defined scalar primitive and is a pure gauge potential. Thus local zero magnetic field and globally removable magnetic vector potential are different conditions.

For an original interference example, take mutually coherent scalar amplitudes with constant individual intensities $I_1,I_2$ and an imposed carrier phase $ky$, as in off-axis holography. ThenThe interference fringe spacing is $d=2\pi /k$. An electron enclosing $\Phi =h/(4e)$ has $\delta =-\pi /2$, shifting maxima by $d/4$ for this orientation. Increasing the magnetic flux by $h/e$ changes $\delta$ by $-2\pi$ and restores the same pattern: the electron Aharonov-Bohm effect has magnetic flux quantum $h/e$. This uses $h=2\pi \hslash$, the Planck constant. It does not require the source's magnetic flux to vary continuously during a single exposure.

For $I_1=1$, $I_2=1/4$, the extrema are $I_{\max }=9/4$, $I_{\min }=1/4$, and the interferometric visibility is $4/5$. A fixed Aharonov-Bohm phase translates these fringes without changing their contrast. Averaging unresolved fluctuating phases can reduce the contrast; a darkened pattern alone cannot identify enclosed magnetic flux. These formulas assume coherence, indistinguishable alternatives and stable beam envelopes.

The appeal of the Aharonov-Bohm effect is that a local Lorentz force calculation along either ideal path misses information carried by their global arrangement. It gives a physical example of holonomy and supplies the encircling phase used in Chern--Simons flux attachment. This alone does not establish intrinsic topological order. The static comparison also matters: changing magnetic flux can induce an electric field, so a time-dependent experiment requires the scalar-potential contribution as well as the displayed spatial line integral.

On a compact oriented Riemannian manifold without boundary, a smooth $(n-p)$-form $\chi$ annihilates all harmonic $p$-forms under wedge integration exactly when its harmonic projection vanishes. Indeed ${\int }_M\phi \wedge \chi =⟨\ast \phi ,\chi {⟩}_{L^2}$, and the Hodge star maps ${\mathcal{H}}^p$ isomorphically onto ${\mathcal{H}}^{n-p}$. By the Hodge decomposition theorem, this annihilator isThus both exact and coexact additions are allowed; a harmonic representative is unique. The potentials of these additions need not be unique, and out-of-range degree spaces are zero. For a concrete coexact ambiguity, take the torus $(\mathbb{R}/2\pi \mathbb{Z}{)}^2$ with metric $d{x}^2+d{y}^2$. The form $\chi =\delta (\sin ydx\wedge dy)=\cos ydx$ annihilates every harmonic one-form by orthogonality, yet $d\chi =\sin ydx\wedge dy$ is nonzero. Thus an annihilating representative need not even be closed, let alone exact.

An Anosov surface homeomorphism has nonsingular invariant expanding and contracting foliations. In the closed orientable surface setting it occurs on the torus, with a hyperbolic integral linear model.

The standard four-torus has trivial complex tangent bundle, so its symplectic canonical class is zero. The symplectic adjunction formula gives $F^2=2g-2$ for a smooth genus-$g$ fiber of a symplectic Lefschetz pencil. Two generic fibers meet once positively at each base point and nowhere else, giving the displayed count.

For a compact torus $T\cong U(1{)}^r$, its continuous one-dimensional characters form a free abelian group of rank $r$. They are precisely $z\mapsto z_1^{m_1}\cdots z_r^{m_r}$ with integer exponents. This follows from the classification of representations of the circle group, applied to each factor. These characters are the integral group weights occurring upon restricting a group representation to a maximal torus; global integrality must be distinguished from integrality against coroots for a simply connected semisimple covering group.

A Chern insulator is a two-dimensional band insulator whose occupied energy bands have a nonzero total Chern number. It has a bulk band gap, yet an interface with a topologically trivial band insulator carries a net excess of quantum Hall edge states propagating in one direction. Time-reversal symmetry must be broken. The Haldane model shows that a nonzero average magnetic flux is unnecessary.

Why distinguish two solids that both have a band gap? The energy eigenvalues tell us whether an excitation costs energy; the energy eigenstates also describe how the state changes around the Brillouin zone. Their Berry curvature can integrate to a nonzero integer. A small deformation changes the Berry curvature continuously, but cannot continuously change an integer. A Chern insulator and a trivial band insulator therefore cannot be connected through a path of isolated occupied energy bands with a common bulk band gap. The distinction survives sufficiently weak local perturbations. It does not make every microscopic quantity or every boundary state immune to every disturbance.

For a single filled energy band, set the Berry connection to ${\mathcal{A}}_j=i⟨u_{-}|{\partial }_{k_j}{u}_{-}⟩$, and orient the momentum plane by $(k_x,k_y)$. ThenThis is the First Chern class evaluated on the Brillouin zone, a torus. Smooth eigenvectors may require multiple patches; their transition phases have integer winding numbers. For charge $-e$ and conventional electric current $j_i={\sigma }_{ij}{E}_j$, these conventions give ${\sigma }_{yx}=C{e}^2/h$ and ${\sigma }_{xy}=-C{e}^2/h$. The measurable magnitude agrees with Chern number and integer Hall conductivity. Reversing the momentum orientation or the sign of the Berry connection reverses the displayed $C$.

A connected compact surface is determined by its orientability, number $b$ of boundary circles and genus. An orientable surface of genus $g$ has Euler characteristic $2-2g-b$; a surface which is not orientable with $r$ crosscaps has Euler characteristic $2-r-b$. In particular, a closed connected surface with vanishing first mod-two homology is a sphere, and a connected closed orientable surface of Euler characteristic zero is a torus.

There are exactly four types: the two-sphere, real projective plane, torus, and Klein bottle. Four sides admit three pairings and four direction choices for each pairing. Up to cyclic relabelling, labels are adjacent, $a{a}^{s}b{b}^t$, or alternating, $ab{a}^s{b}^t$. For adjacent labels, the vertex counts for $(s,t)=(-1,-1)$, a mixed-sign pair, and $(1,1)$ are $3,2,1$. For alternating labels they are $1,1,2$. There are two edge classes and one face, so Euler characteristic is $V-1$. Orientability occurs exactly when $s=t=-1$. The classification theorem for surfaces identifies every case, leaving no additional closed surfaces from a square. Examples are $a{a}^{-1}b{b}^{-1}$ for the sphere, $aab{b}^{-1}$ for the projective plane, $ab{a}^{-1}{b}^{-1}$ for the torus and $ab{a}^{-1}b$ for the Klein bottle.

For a Diophantine frequency vector, the equation $\omega \cdot {\partial }_{x}f=g$ on a periodic torus has an analytic solution on each smaller complex strip exactly when $g$ has zero average. Its nonzero Fourier coefficients are $f_k=g_k/(ik\cdot \omega )$, and its zero coefficient is arbitrary. Thus the normalized zero-mean solution is unique.

The torus and the wedge sum $S^1\vee S^1\vee S^2$ both have integral cohomology $\mathbb{Z},{\mathbb{Z}}^2,\mathbb{Z}$ in degrees zero, one and two. Their fundamental groups differ: the torus has ${\mathbb{Z}}^2$, while the wedge has the nonabelian free group on two generators. Thus even compact finite cell complexes can have the same additive cohomology without being homotopy equivalent.

Attach two one-handles to a four-dimensional zero-handle, then a two-handle along the commutator of their generators with the product zero framing. The result is $T^2\times D^2$. Its spine is the usual cell structure of the torus: one vertex, two one-cells, and a two-cell attached by $xy{x}^{-1}{y}^{-1}$. An embedded representative of its $H_2$ generator is obtained by taking a once-punctured torus carried by the two one-handles and capping its boundary by the core disk of the two-handle. It meets the relative co-core once and has self-intersection zero. The boundary is $T^3$, whose integral homology has ranks $1,3,3,1$.

Remove small disks around a nondegenerate maximum of a smooth function on one closed surface and a nondegenerate minimum on another. Rescale their values so the boundary values agree with the two ends of a neck. Their regular level collars allow a strictly monotone function on the neck, with nonzero derivative, matching smoothly to both sides. The resulting connected sum has all the old critical points except the deleted maximum and minimum. This permits one degenerate torus saddle to be retained while adding ordinary handles for higher genus.

Although topological degree classifies maps $S^n\to S^n$ up to homotopy, it is not a complete invariant for other manifolds. On the torus, the identity and the integer shear $\left(\begin{array}{cc}1& 1\\ 0& 1\end{array}\right)$ both have degree one, but induce distinct maps on $H_1(T^2;\mathbb{Z})$ and cannot be homotopic.

Suppose a nonconstant holomorphic map $f:X\to {\mathbb{P}}^1$ from a connected compact Riemann surface has degree of a holomorphic map two and exactly four distinct branch values of a holomorphic map. The valency theorem forces each of their fibres to contain one ramification point with $e_p=2$, a simple ramification point of a holomorphic map. HenceThis determines the genus without recognizing a torus in a drawing.

A concrete realization is the compact Riemann surface obtained by adding the two points at infinity to the even-degree hyperelliptic model given by the algebraic curveProjection to $z$ has degree of a holomorphic map two. At a simple zero $a$ of $P$, $w$ is a local coordinate and $z-a=w^2/P^{\prime }(a)+O(w^4)$, so $e_p=2$. At infinity set $u=1/z$ and $s=w/z^2$: then $s^2=(1-4u^2)(1-u^2)$. There are two points with $u=0$, $s=\pm 1$, each unramified since $u$ is a local coordinate. The four finite roots are therefore all the branch values of a holomorphic map.^[ref]

Choose a triangulation of a surface on the Riemann sphere with these four branch values of a holomorphic map as vertices, combinatorially the boundary of a tetrahedron. Its counts are $V=4,E=6,F=4$. Upstairs, each vertex has one preimage and each open edge and face has two lifts: $V_X=4,E_X=12,F_X=8$, giving Euler characteristic $4-12+8=0$. This is a cell schematic, not a geometric immersion of $X$. The Triangulation proof of the Riemann-Hurwitz formula applies the same counting idea after refining a general triangulation of a surface around the branch values of a holomorphic map.^[ref]

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.

Two identical planar simple pendulums of length $l$ and bob mass $m$, with pivots separated horizontally by $r>0$, are coupled by a straight spring of natural length $r$ and spring constant $k$. In the ideal model without contact restrictions, their configuration space is the torus $S^1\times S^1$. Measure both angles from the downward vertical towards the same horizontal direction. Their bob separation isThe exact Lagrangian is the expression in the title. To derive it, use bob positions $(l\sin {\theta }_1,-l\cos {\theta }_1)$ and $(r+l\sin {\theta }_2,-l\cos {\theta }_2)$. Kinetic energy from Cartesian position vectors gives $T=m{l}^2({\dot{\theta }}_1^2+{\dot{\theta }}_2^2)/2$. Gravitational potential energy is $-mgl(\cos {\theta }_1+\cos {\theta }_2)$ and spring potential energy is $k(d-r{)}^2/2$; subtracting these from $T$ proves the formula. Near the downward equilibrium, $d-r=l({\theta }_2-{\theta }_1)+O(|\mathbf{\theta }{|}^3)$ for fixed $r$, giving the familiar normal modes of two spring-coupled equal pendulums.

In the orientable case an excellent compact three-manifold is an irreducible three-manifold and a boundary-irreducible three-manifold, is not a ball, contains a two-sided properly embedded incompressible surface, and has every properly embedded incompressible zero-Euler characteristic surface a boundary-parallel surface. In particular it has no essential annulus or torus. These properties give lower bounds on the Thurston norm of classes with essential boundary.

The torus and Klein bottle are not homeomorphic because only the former is orientable. Adding a projective plane to either by connected sum of surfaces makes a nonorientable surface with Euler characteristic minus one. The classification of compact surfaces identifies both with three projective planes. Thus connected-sum cancellation can fail even for closed connected surfaces.

In a product symplectic manifold $S^2(a)\times T^{2n-2}$, every compatible almost complex structure admits a J-holomorphic sphere in class $F$ through each point. Here $S^2(a)$ has area $a$, and the torus carries a constant symplectic form. All spherical classes are integer multiples of $F$, so a curve in class $F$ cannot split into positive-area nonconstant spheres. For the product complex structure the fibre through a point is unique and regular. Its one-point count modulo two is one; generic transversality and the compact one-dimensional parameterized moduli spaces preserve this count. Approximation and Gromov compactness for spheres remove the genericity requirement. This gives the area-$a$ curve used in the proof of the Gromov non-squeezing theorem.

Removing an embedded open disk from the torus leaves a topological surface with one boundary circle. In the square model, radial expansion away from the removed disk gives a deformation retraction onto the square boundary with opposite sides identified, a wedge of two circles. Its fundamental group is therefore the free group on two group generators.