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Project Brownian motion modulo ${\mathbb{Z}}^3$ to the compact three-dimensional torus. The projected process has normalized volume as invariant probability measure and is irreducible, so it visits every nonempty open set infinitely often almost surely. The image of $S_3$ contains the radius-$ϵ$ ball about the origin. Hence the original Brownian motion hits $S_3$ at an unbounded set of times.

Let $x$ and $y$ denote the two oriented noncontractible cycles of the torus. Their transport operators intersect once, so the mutual braiding phase gives the Wilson-loop algebra of an Abelian Chern--Simons theoryThe integer charge vectors are identified modulo local electrons: $q$ and $q+K\ell$ have identical braiding with every quasiparticle. The distinct topological charges therefore form the finite quotient group

Choose a ground state that diagonalizes the mutually commuting $x$-cycle operators. Acting with $T_y(q)$ multiplies their eigenvalues by the character $q^{\prime }\mapsto e^{2\pi i{q}^{\prime T}{K}^{-1}q}$. This pairing is nondegenerate on $\mathcal{A}$: if the character is trivial for every $q^{\prime }$, then $K^{-1}q$ is integral and $q$ is the trivial class. Thus the $|\mathcal{A}|$ states obtained from inequivalent $q$ have distinct loop eigenvalues and are orthogonal. The torus ground-state degeneracy of an Abelian Chern--Simons theory is consequently at least $|\det K|$.

The all-ones matrix has eigenvalue $n$ along $t$ and eigenvalue zero on the $(n-1)$-dimensional perpendicular subspace. Hence $K=I_n+pM$ has eigenvalues $1+np,1,\dots ,1$, and

Let $a$ and $b$ be the pullbacks of the standard generator of $H^n(S^n;\mathbb{Z})$ from the two factors. The Künneth theorem and graded commutativity of the cup product give

Each homeomorphism acts invertibly on $H^n\cong {\mathbb{Z}}^2$. To respect ordinary composition, send $h$ to $(h^{-1}{)}^{\ast }$; functoriality of induced map on cohomology then defines a homomorphism

For $n=1$, the space is the torus. Every matrix in $GL(2,\mathbb{Z})$ induces a linear homeomorphism ${\mathbb{R}}^2/{\mathbb{Z}}^2\to {\mathbb{R}}^2/{\mathbb{Z}}^2$, so the image is all of $GL(2,\mathbb{Z})$.

For $n=2$, write $h^{\ast }a=pa+qb$. Since $a^2=0$ and $ab=ba$,so $pq=0$; the same argument applies to $h^{\ast }b$. Invertibility then forces the matrix to be a signed permutation matrix. Every such matrix is realized by swapping the two sphere factors and applying an orientation-reversing homeomorphism to either factor. Thus the image consists exactly of the eight signed permutation matrices.

For each $v\in V(\Gamma )$, the vertex link of a Salvetti complex has two vertices $v^{+}$ and $v^{-}$. If $v$ has degree $d_{\Gamma }(v)$, then each of $v^{+}$ and $v^{-}$ is adjacent to both signed vertices $w^{+},w^{-}$ for every neighbour $w$, and hence has link degree $2d_{\Gamma }(v)$.

Suppose the whole link is a cycle. Every link vertex has degree two, so every vertex of $\Gamma$ has degree one. The link is connected, which forces $\Gamma$ to be connected. A connected one-regular graph consists of one edge. Conversely, if $\Gamma$ is one edge, then $S_{\Gamma }$ is the square torus and its unique vertex has link the four-cycleThus the vertex link is a cycle exactly when $\Gamma$ is an edge.

The hyperplane of a Salvetti complex dual to the loop labelled $v$ is itself the Salvetti complex of the induced subgraph on the neighbours of $v$. If this hyperplane were homeomorphic to a closed surface, its vertex link would be a cycle. Part (c), applied to that induced graph, says that the graph must be one edge. Its Salvetti complex is therefore the two-dimensional torus, a surface of genus one. It cannot be a closed orientable surface of genus two.

Take the toruswith the translation-invariant symplectic formThe coordinate vector fields give a global symplectic frame of $TM$, so $TM$ is symplectically trivial. Passing fiberwise to the Lagrangian Grassmannian bundle giveswhich is a smooth trivialization over the compact symplectic manifold $T^{2n}$.

A topological surface is a Hausdorff second-countable space locally homeomorphic to ${\mathbb{R}}^2$. Pairing opposite sides of the hexagon gives neighborhoods modelled on discs, including at edge and vertex classes; compactness follows from the compact polygon quotient. Cutting and rearranging the polygon gives the usual opposite-side square, hence a torus.

The full double cone is not a surface at the origin: deleting the vertex from a small neighborhood leaves two components, unlike a punctured disc. The upper cone with $z>0$ excludes that vertex and has ordinary annular charts, so it is a surface.

The Lefschetz fixed-point theorem says that a continuous self-map $f$ of a compact triangulable space has a fixed point whenever its Lefschetz numberis nonzero. In the smooth nondegenerate case, the Lefschetz-Hopf fixed-point theorem expresses this number as the sum of the local fixed-point indices.

Identify $H_1({\mathbb{R}}^2/{\mathbb{Z}}^2;\mathbb{Z})$ with ${\mathbb{Z}}^2$ by sending $v\in {\mathbb{Z}}^2$ to the class of the loop $t\mapsto tv\mathrm{mod}{\mathbb{Z}}^2$. The image of this loop under $f$ lifts from $0$ to the path $t\mapsto \tilde{f}(tv)$, whose endpoint is $\tilde{f}(v)=A(v)$. Its homology class is therefore $A(v)$. Thus $A=f_{\ast }$ under this identification, and in particular $A$ is a homomorphism represented by an integer matrix.

The expansion inequality implies, after taking a derivative, thatfor every tangent vector. The supplied linear-algebra fact shows that every eigenvalue of $D{\tilde{f}}_x$ has modulus at least $\mu >1$, so $1$ is not an eigenvalue. Every fixed point of $f$ is consequently nondegenerate and contributes local index $+1$ or $-1$.

On the torus, the induced maps on $H_0,H_1,H_2$ have traces $1,\mathrm{tr}A,\det A$. HenceThe original expansion inequality applied to lattice vectors gives $|Av|\ge \mu |v|$ for $v\in {\mathbb{Z}}^2$, and homogeneity and density extend it to all $v\in {\mathbb{R}}^2$. If ${\lambda }_1,{\lambda }_2$ are the possibly complex eigenvalues of $A$, then $|{\lambda }_j|\ge \mu$, soSince every fixed point contributes an index of absolute value one, the Lefschetz-Hopf formula gives

Treat the two squares as a single edge-pairing construction, including the identifications between different squares. Number the corners of the left square counterclockwise from its lower left corner by $A,B,C,D$, and similarly use $E,F,G,H$ on the right. The single arrows identify $D$ with $H$ and $C$ with $E$. The double arrows identify $G$ with $F$ and $H$ with $G$. The triple arrows identify $A$ with $E$ and $B$ with $F$. The quadruple arrows identify $A$ with $C$ and $D$ with $B$.

Thus there are precisely two vertex classes, $\{A,C,E\}$ and $\{B,D,F,G,H\}$. There are four edge classes and two two-dimensional cells, so the resulting closed connected surface has Euler characteristicIt is a nonorientable surface: on the left square the quadruple-arrow sides have the same orientation relative to a traversal of its boundary, giving a twisted band after gluing. Equivalently, no choice of orientations for the two faces makes the induced directions opposite on every paired edge. The vertex links are circles, so these edge identifications produce a closed surface, rather than a singular cell space. The classification of compact surfaces now givesA torus also has Euler characteristic zero, but is excluded by the failed orientation test.

For integers $k,l$,Thus reducing coordinates modulo one gives a well-defined function on the torus ${\mathbb{R}}^2/{\mathbb{Z}}^2$. Its determinant is one and its inverse matrix is the integer matrix $\left(\begin{array}{cc}1& -1\\ -1& 2\end{array}\right)$, so it induces a bijection of the torus. The Jacobian determinant one makes this bijection area-preserving.

Its eigenvalues solve ${\lambda }^2-3\lambda +1=0$, givingThere are therefore expanding and contracting eigendirections and no eigenvalue on the unit circle: the function is a hyperbolic toral automorphism. The Lyapunov exponents per iterate areThey sum to zero, as required by area preservation.

The four vertices of the unit square map to $(0,0),(2,1),(3,2),(1,1)$, the vertices of an area-one parallelogram. To reduce it modulo one, cut it by the integer grid. For each piece in the cell $[j,j+1]\times [k,k+1]$, translate by $(-j,-k)$. Up to boundary identifications these translated pieces partition the unit square, because the inverse integer matrix gives exactly one torus preimage to every point. The colored pieces in the original diagram show the same pieces before and after these translations.

For the Fibonacci identity, at $n=1$ the claimed matrix is $\left(\begin{array}{cc}{F}_2& F_1\\ F_1& F_0\end{array}\right)=A$. If the identity holds at $n$, multiplication on the right by $A$ givesEach equality uses $F_{j+1}=F_j+F_{j-1}$, with the specified indexing $F_0=F_1=1$. This proves the formula for every positive integer $n$.

A fixed point satisfies $(A-I)v\in {\mathbb{Z}}^2$. Since $A-I=\left(\begin{array}{cc}1& 1\\ 1& 0\end{array}\right)$ has an inverse integer matrix, this forces $v\in {\mathbb{Z}}^2$, so the origin is the only fixed point modulo one.

For period dividing two,The second congruence gives $y=-3x(\mathrm{mod}1)$; inserting it into the first gives $5x=0(\mathrm{mod}1)$. Thus the complete list isThe first is the fixed origin. Direct application of $A$ exchanges the other four in exactly two pairs:This both verifies the periodic orbits and excludes additional ones.

Finally, if $v$ is fixed by $F^p$, then $(A^p-I)v=m$ for an integer vector $m$. The eigenvalues of $A^p-I$ are ${\lambda }_{+}^p-1$ and ${\lambda }_{-}^p-1$, neither zero. Hence its inverse exists and has rational entries by the adjugate matrix formula for an integer matrix. Therefore $v=(A^p-I{)}^{-1}m$ has rational coordinates. The derivative of the return map is $A^p$, with eigenvalues ${\lambda }_{+}^p>1$ and $0<{\lambda }_{-}^p<1$. Every periodic point is a hyperbolic saddle fixed point of a map and is unstable, although perturbations in its stable eigendirection contract.

A useful operational definition of chaos for a deterministic flow is bounded, nonperiodic long-time dynamics on an invariant set with sensitive dependence on initial conditions: arbitrarily close starting states can separate by a fixed macroscopic amount. Boundedness distinguishes recurrent irregular dynamics from simple exponential escape; aperiodicity excludes a limit cycle; sensitivity distinguishes chaotic unpredictability from a regular quasiperiodic signal. These are properties of the asymptotic dynamics, not a short irregular transient. One precise topological convention, Devaney chaos on a compact invariant set, instead specifies topological transitivity, dense periodic orbits, and sensitive dependence. Transitivity expresses indecomposability of the recurrent motion, while dense periodic orbits mean every region contains unstable recurrent structure. The convention should be stated because these formulations are not identical in complete generality.

An attractor is an invariant set approached by trajectories from a neighbourhood or positive-measure basin, according to the adopted attractor convention. Calling it strange refers to irregular, typically fractal geometry rather than a smooth point, cycle or torus. Strangeness alone does not force chaos. Strange nonchaotic attractors provide counterexamples: geometrically irregular attracting sets can lack positive exponential separation. Primary examples appear in ^[ref]. This geometric convention does not assert that every such example occurs in the autonomous three-variable Lorenz family.

Two complementary numerical diagnostics are to estimate the largest Lyapunov exponent and to construct a return map. For the first, integrate the variational equation $\dot{\eta }=DF(x(t))\eta$ alongside the Lorenz trajectory, repeatedly normalize $\eta$, and average the logarithms of the accumulated stretch factors after discarding a transient. A positive converged exponent suggests sensitive exponential separation; the full spectrum should include a zero flow exponent and have sum $-(\sigma +1+b)$, providing checks. For the second, record transverse crossings of a Poincaré section or successive maxima of a state variable, then look for the stretching and folding of an aperiodic return set rather than one point or a finite cycle. Nearby-orbit separation or a broadband power spectrum can add evidence, but plots alone do not prove chaos. Repeat with smaller integration step, longer runs and several initial conditions to distinguish numerical error, long transients and coexistence of attractors from persistent dynamics.

The Lorenz vector field has constant divergence $\nabla \cdot F=-\sigma -1-b$. Liouville formula for a fundamental matrix for the derivative of the flow therefore givesThis concerns transported three-dimensional phase volume, not the Euclidean distance between each pair of trajectories.

A quasiperiodic orbit with the usual smooth two-frequency torus closure would lie on an invariant embedded two-torus. That torus bounds a solid torus of positive finite volume in ${\mathbb{R}}^3$. Since the vector field is tangent to its boundary, its outward flux is zero, contradicting the divergence theorem, whose volume integral is $-(\sigma +1+b)V<0$. This negative-divergence obstruction to invariant tori excludes such quasiperiodic tori. The argument uses global negative divergence and the enclosed volume; saying only that an attracting torus has zero volume would not exclude it.

At an equilibrium the sum of eigenvalues is $-(\sigma +1+b)<0$, so they cannot all have positive real parts: no equilibrium is a fully repelling source. For a periodic orbit of period $T$, one Floquet multiplier equals one in the flow direction. The product of the other two is $e^{-(\sigma +1+b)T}<1$, so they cannot both have modulus greater than one: no periodic orbit is fully repelling either. Saddle equilibria and saddle periodic orbits remain possible; contraction of volume does not imply contraction in every direction.

On the torus with basepoint $(1,1)$ take $p(z,w)=(z^2,w)$ and $p^{\prime }(z,w)=(z,w^2)$. Both are connected two-sheeted coverings. Their induced subgroups of ${\pi }_1(S^1\times S^1)={\mathbb{Z}}^2$ are $2\mathbb{Z}\times \mathbb{Z}$ and $\mathbb{Z}\times 2\mathbb{Z}$, which are different. Equivalence over the fixed base would force equal based subgroups. Thus the two degree-two covers are inequivalent.

For an open cover $X=U\cup V$, the Mayer–Vietoris theorem gives the long exact sequenceIt ends in $H_0(U\cap V)\to H_0(U)\oplus H_0(V)\to H_0(X)\to 0$. To prove it, let $C_{\ast }^{U,V}(X)$ be the subcomplex generated by singular simplices lying wholly in $U$ or wholly in $V$. There is a short exact sequence of chain complexesExactness follows from the singular-simplex bases: the intersection of the two chain subgroups consists exactly of simplices lying in both sets. The small-chain locality theorem for singular homology says the inclusion of this small-chain complex into $C_{\ast }(X)$ induces an isomorphism on homology; barycentric subdivision is its standard justification. The long exact sequence of a short exact sequence of chain complexes now proves the theorem. For a homological cycle represented as $a+b$, the connecting map is $[a+b]\mapsto [\partial a]$, since $\partial a=-\partial b$ lies in the intersection. This also identifies the displayed maps and their signs.

For the glued solid tori, use open collar enlargements of the two halves. Each retracts to a circle; the overlap retracts to their boundary torus. Thus the halves have $H_0=H_1=\mathbb{Z}$ and higher homology zero, while the overlap has $H_0=H_2=\mathbb{Z}$ and $H_1={\mathbb{Z}}^2$. In the overlap choose longitude and meridian generators. The meridian bounds a disk in each solid torus and the longitude generates each first homology, so the map on $H_1$ isIts kernel is the meridian copy of $\mathbb{Z}$ and its cokernel is $\mathbb{Z}$. The map on $H_0$ is the injective map $z\mapsto (z,-z)$. Exactness therefore gives $H_2(X)=\mathbb{Z}$ and $H_1(X)=\mathbb{Z}$. The preceding segment is $0\to H_3(X)\to H_2(U\cap V)\to 0$, giving $H_3(X)=\mathbb{Z}$. Connectedness gives $H_0(X)=\mathbb{Z}$, and all higher groups vanish. HenceAs a geometric check, the identity boundary gluing identifies the product with $S^1\times (D^2{\cup }_{S^1}{D}^2)=S^1\times S^2$.

For ordered simplicial cochains, define the Alexander-Whitney cup product byusing the front and back faces of each characteristic simplex. Write $\delta$ for the coboundary, $\delta \varphi =\varphi \partial$. Expanding the face-boundary formula gives the graded Leibniz identityIn this expansion, face omissions on the front and back portions supply the two displayed terms, and their extra terms at the shared vertex cancel with opposite signs. Hence the cup of cocycles is a cocycle. If $\varphi$ changes by $\delta a$, its product with a cocycle changes by $\delta (a⌣\psi )$; changing $\psi$ by $\delta b$ changes it by $(-1{)}^i\delta (\varphi ⌣b)$. The product therefore descends to cohomology. It is associative by the three consecutive-face formula and has the constant degree-zero cochain as unit.

In the wedge, every positive-dimensional simplex belongs to exactly one factor. Restricting positive-degree cochains therefore gives a direct sum of the two cochain groups and splits the coboundaries in degrees above one. In degree one, the same splitting of coboundaries holds: any pair of degree-zero cochains on the factors can have their values at the glued vertices made equal by adding a constant to one factor, which does not change its coboundary. Thus restriction inducesRepresent classes from the two factors by cochains extended by zero on the other factor. If both degrees are positive, a simplex lies wholly in one factor, so one of the front or back evaluations is zero. Their cup product is already zero as a cochain. This proves all the stated mixed products vanish.

For the indicated wedge there is one unit in degree zero, two generators $a,b$ in degree one, and one generator $c$ in degree two; higher groups vanish. Every product of positive-degree elements is zero. Its ring is thereforeThe torus has the same additive groups, but its degree-one generators have nonzero cup product generating degree two. The two graded rings are not isomorphic. Since a homotopy equivalence induces a cohomology-ring isomorphism, the torus and this wedge are not homotopy equivalent.

There are exactly four homeomorphism types: the two-sphere, real projective plane, torus, and Klein bottle.

For completeness, enumerate all pairings. Four sides have three pairings; each pair can be matched in either direction, giving twelve signed pairings before symmetries. Up to cyclic relabelling, the repeated labels are either adjacent, of form $a{a}^{s}b{b}^t$, or alternating, of form $ab{a}^s{b}^t$, where $s,t\in \{1,-1\}$. The second adjacent pattern $abba$ rotates to the first. Endpoint identification gives the following table:Here the quotient always has two edge classes and one face, so its Euler characteristic is $V-2+1=V-1$. To reproduce the vertex count, label consecutive square vertices $0,1,2,3$. For paired sides $i,j$, a same-direction pairing identifies $i$ with $j$ and $i+1$ with $j+1$; a reversed-direction pairing identifies $i$ with $j+1$ and $i+1$ with $j$, with indices modulo four. Taking the generated equivalence classes gives exactly the displayed values of $V$.

The quotient is orientable precisely when each label is traversed once positively and once negatively around the boundary: then the face orientation extends across every paired edge. A same-direction pair obstructs such an orientation. Thus the two $\chi =0$ possibilities are distinguished as the orientable torus and nonorientable Klein bottle; the other rows are fixed by Euler characteristic and the classification theorem for surfaces.

Explicit square representatives areAll edge interior points have disk neighbourhoods by joining two half-disks, and each identified vertex has a cyclic sequence of corner sectors, again a disk neighbourhood. There are no leftover boundary edges. The enumeration therefore describes closed surfaces, and the twelve signed pairings leave no additional cases.

False: the result is again a torus with one open disc removed. Label the hexagon polygon vertices clockwise from the upper left asWith the same arrow labels as in part (ii),There are two vertex equivalence classes, $P=\{v_0,v_2,v_4\}$ and $Q=\{v_1,v_3,v_5\}$. The free diagonal polygon edges are $v_1{v}_2$ and $v_4{v}_5$. Each joins $P$ to $Q$, so together they form one boundary component of a surface, not two. At each vertex equivalence class the three corner sectors join in a single chain between these two free polygon edges. Thus each vertex link in a polygonal schema is a closed interval, and the quotient space is an orientable surface with boundary.

Its CW complex has CW cells comprising two vertices, four edges and one face, givingIn contrast, a torus with two disjoint open discs removed has two boundary components of a surface and Euler characteristic $2-2-2=-2$. Either topological invariant rules out the proposed homeomorphism.

True. Label the pentagon polygon vertices clockwise from the upper left asUse $a$ for the upward single arrows and $b$ for the rightward double arrows in the original diagram. Their endpoint identifications areAll five polygon vertices are therefore identified. The remaining diagonal polygon edge, from $v_3$ to $v_4$, has its endpoints identified and is the only boundary component of a surface.

The paired polygon edges join half-disc neighbourhoods into disc neighbourhoods. At the common vertex, the five corner sectors join in one chain whose two free ends lie on the unpaired polygon edge: its vertex link in a polygonal schema is a closed interval. Hence that point has a half-disc neighbourhood. This verifies that the quotient space is a surface with boundary, rather than merely counting its CW cells. The paired polygon edges are oppositely traversed around the polygon, so this surface is orientable.

Its CW cells comprise one vertex, three edges (the $a$ pair, the $b$ pair, and the free polygon edge), and one face. ThereforeThe classification of compact surfaces identifies it as a torus with an open disc removed.

Label the hexagon polygon vertices $v_0,\dots ,v_5$ in the direction in which its boundary word is $abc{a}^{-1}{b}^{-1}{c}^{-1}$, with the first side running from $v_0$ to $v_1$. Matching the endpoints according to the arrows givesThus the two vertex equivalence classes are $\{v_0,v_2,v_4\}$ and $\{v_1,v_3,v_5\}$.

The polygonal schema gives a compact connected surface with empty manifold boundary. Interior points have disc neighbourhoods; a paired polygon edge glues two half-disc neighbourhoods into a disc. At each of the two vertex equivalence classes, its three corner sectors join cyclically, so the vertex link in a polygonal schema is a circle, again giving a disc neighbourhood. Pairing every label once positively and once negatively makes these gluings consistent with an orientation, so the resulting surface is orientable.

The resulting CW complex has CW cells comprising two vertices, three edges and one face. Its Euler characteristic isBy the classification of compact surfaces, a connected closed orientable surface with $\chi =0$ has genus $g$ satisfying $2-2g=0$, hence $g=1$. The quotient space is a torus.

Every complete polygon edge pairing gives a compact connected surface with empty manifold boundary. Here is the local reason at the vertices. Replace each polygon corner by a small link interval. The polygon edge pairings identify all endpoints of these link intervals in pairs. In a given vertex equivalence class, the resulting graph is connected: the endpoint identifications generating that vertex equivalence class also connect its corner intervals. Every link graph vertex has degree of a vertex two, so the graph is a circle. Its topological cone is a disc, giving the required local surface model. Edge and face CW cell interiors have the usual disc neighbourhoods as well. A finite triangulation compatible with the polygon edge pairings realizes the result as a finite simplicial complex, giving a Hausdorff, second-countable space.

For every such surface, $\chi =V-2\ge -1$. If it is orientable, the classification of compact surfaces gives $\chi =2-2g$, so $g=0$ or $1$. If it is a nonorientable surface, write $k\ge 1$ for its nonorientable genus; then $\chi =2-k\ge -1$ gives $k\le 3$. The exhaustive list is therefore

To prove that every listed possibility actually occurs, the following six-letter boundary words specify complete pairings of a hexagon:The vertex counts follow by matching starts and ends of equally directed arrows, just as in part (b). The first and third words pair every letter with its inverse, giving orientable surfaces. Each other word contains a letter twice with the same sign, giving a nonorientable surface: an orientation on the face would induce equal, rather than opposite, boundary directions along that glued pair. With the displayed Euler characteristics, the classification of compact surfaces identifies the examples exactly as listed. All five types occur, and no others occur.