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A smooth surface in ${\mathbb{R}}^3$ is a subset locally parametrized by a smooth map of two variables whose derivative has rank two and which is a homeomorphism onto its image.

Writing $\rho =\sqrt{x^2+y^2}$, the first surface equation becomesIt is the torus of major radius $R=2\sqrt{2}$ and minor radius $r=1$, with smooth parametrizationSince $R>r$, this is an embedding. The Gaussian curvature of a torus is

For the second surface, use the orthogonal transformationIts equation becomes the same torus equation $(\sqrt{X^2+Y^2}-2\sqrt{2}{)}^2+Z^2=1$. Orthogonal transformations are Euclidean isometries and preserve Gaussian curvature. Curvature vanishes where $\cos v=0$, equivalently $Z=\pm 1$. In the original coordinates the zero-curvature points are therefore exactly

False. Excursion coding does not remember the speed of traversal. Let $f$ be any nonzero coding function and let $\varphi :[0,1]\to [0,1]$ be a nonidentity increasing homeomorphism. Set $g=f\circ \varphi$. ThenThus $[s]\mapsto [\varphi (s)]$ induces an isometry $T_g\to T_f$, although generally $g\ne f$.

The finite-dimensional Stinespring dilation theorem states that every quantum channel $\Lambda :\mathcal{B}(H_A)\to \mathcal{B}(H_B)$ has an environment $H_E$ and an isometry $V:H_A\to H_B\otimes H_E$ such thatConversely, every map of this form is completely positive and trace preserving. From Kraus operators $A_j$, one may take $V={\sum }_j{A}_j\otimes |j{⟩}_E$.

Introduce a quantum ancilla with basis $|i⟩$ and defineThe completeness relation gives $V^{†}V=I$, so $V$ is an isometry and extends to a unitary operator on a sufficiently large system-plus-ancilla space. Prepare the ancilla in a fixed state, apply that unitary, and perform the projective measurement $\{I\otimes |i⟩⟨i|\}$. Outcome $i$ has probability $\Vert M_i|\psi ⟩{\Vert }^2$ and leaves the system in the normalized state $M_i|\psi ⟩$. By linearity the same holds for mixed states, implementing the generalized measurement.

Send each orientation-preserving isometry to its isotopy class. This is a group homomorphismFor $g\ge 2$, a hyperbolic isometry isotopic to the identity is the identity. One proof lifts it to the hyperbolic plane: after composing with a deck transformation, its lift commutes with the surface fundamental group. It consequently fixes the endpoints at infinity of every hyperbolic deck transformation. Those endpoints are dense in the circle at infinity, so the lift fixes that circle pointwise and is the identity. The homomorphism is therefore injective, proving the injection of a finite hyperbolic isometry group into a mapping class group.

Both low-genus analogues fail. Every orientation-preserving homeomorphism of the unit sphere is isotopic to the identity, but a round sphere has nontrivial finite rotation groups. On a flat torus, translation by a nonzero torsion point is a finite-order orientation-preserving isometry isotopic to the identity.

The tangent planecuts the hyperboloid inso its intersection is the pair of straight linesTheir accelerations vanish, hence are normal to the surface, so both lines are geodesics. Their tangent vectors at their intersection $(1,0,0)$ are $(0,1,1)$ and $(0,-1,1)$, whose inner product is zero. They therefore intersect at a right angle, giving the second case of plane-section geodesics of the unit one-sheet hyperboloid.

There are also geodesics entirely contained in $z>0$. In the coordinatesthe metric isFor a unit-speed geodesic, the Clairaut first integral for a surface of revolution is $c=(1+z^2)\dot{\varphi }$, and the constant-speed equation becomesChoose $c>1$ and initial height $z_0=\sqrt{c^2-1}$. The geodesic initially tangent to the parallel has $\dot{z}=0$, and the displayed identity prevents it from entering $0<z<z_0$. Since the parallel at $z_0>0$ is not itself a geodesic, the curve turns there and otherwise has $z>z_0$. Hence it remains entirely in $z>0$; this is a geodesic trapped in one half of the unit one-sheet hyperboloid.

Finally, the waistis a geodesic and is preserved setwise by every isometry of the hyperboloid. Indeed, its Gaussian curvature isThe value $K=-1$ occurs exactly at $z=0$. Since Gaussian curvature is intrinsic and therefore preserved by isometries, every isometry preserves the waist. Thus the answer to the final question is also yes, with the isometry-invariant waist geodesic of the unit one-sheet hyperboloid as an example.

Let $X=\{v_1,\dots ,v_m\}$ be the vertices of a regular polygon, indexed cyclically so that $v_i\mapsto v_{i+1}$ is an isometry. We first prove the product lemma needed for the cyclic-symmetry argument.

Suppose $A\subseteq X$ has the property that, for every number of colors, a finite witness forces a copy of $X$ on which the vertices corresponding to $A$ are monochromatic. We claim that for every $N,k$ there is a finite witness forcing an $A$-invariant copy of $X^N$. Proceed by mathematical induction on $N$. The case $N=1$ is the assumption. For the step, choose a finite set $S$ that forces an $A$-invariant $X^{N-1}$ for a coloring with $k^{|X|-|A|+1}$ colors, and choose $T$ that forces an $A$-monochromatic $X$ for a coloring with $k^{|S|}$ colors.

Given a $k$-coloring $c$ of $S\times T$, color $t\in T$ by its complete vector $(c(s,t){)}_{s\in S}$. In the resulting copy of $X$ in $T$, that vector is the same for every member of $A$. Each $s\in S$ can therefore be colored by the vector consisting of $c(s,a)$ for one $a\in A$ and the colors $c(s,x)$ for $x\in X\setminus A$. Applying the choice of $S$ gives an $A$-invariant copy of $X^{N-1}$, and its product with the chosen copy of $X$ in $T$ is the desired $A$-invariant copy of $X^N$. This proves the lemma rather than assuming it.

We now prove by induction on $r$ that a finite witness forces a copy of $X$ whose first $r$ vertices have one color. The case $r=1$ is trivial. For $r=2$, use that a line segment is a Euclidean Ramsey set; each occurrence of a segment congruent to $v_1{v}_2$ can be extended to a copy of the regular polygon, and only finitely many extensions are needed for a finite witness.

Assume the assertion for $r$, and put $A=\{v_1,\dots ,v_r\}$. The product lemma lets us work inside an $A$-invariantly colored copy of $X^N$, with $N$ as large as required. For an increasing $(m-1)$-element subset $H=\{a_1<\cdots <a_{m-1}\}$ of $[N]$ and $1\le i\le r$, define a word $w_i(H)$ by putting $v_{i+j}$ in coordinate $a_j$, with subscripts read modulo $m$, and putting $v_1$ in every other coordinate. Color $H$ by the $r$-tupleBy the Finite Ramsey theorem, for sufficiently large $N$ there are coordinates $b_1<\cdots <b_m$ on which all $(m-1)$-subsets have the same tuple color.

For $1\le i\le r+1$, let $z_i$ be the word whose entries in coordinates $b_1,\dots ,b_m$ arecyclically, and whose other entries are $v_1$. For $i\le r$, the word $z_i$ differs from $w_i(\{b_2,\dots ,b_m\})$ only in coordinate $b_1$, where the two entries $v_i,v_1$ lie in $A$. Similarly $z_{i+1}$ differs from $w_i(\{b_1,\dots ,b_{m-1}\})$ only in coordinate $b_m$, again by two members of $A$. The $A$-invariance and the homogeneous choice of the $b_j$ therefore giveHence $z_1,\dots ,z_{r+1}$ have one color.

The cyclic words $z_1,\dots ,z_m$ form an isometric copy of $\sqrt{m}X$: in each of the $m$ varying coordinates the cyclic shift is an isometry, so every squared distance is multiplied by $m$. Rescaling the finite witness by $1/\sqrt{m}$ gives a copy of $X$. The induction reaches $r=m$, proving that every regular polygon is a Euclidean Ramsey set.

For three consecutive vertices $p_0,p_1,p_2$ of a regular $M$-gon scaled to have side length one,Since the polygon is Euclidean Ramsey, choosing $M$ large enough proves that $\{0,1,2\}$ is an approximately Euclidean Ramsey set.

In fact every finite $X\subseteq {\mathbb{R}}^d$ is approximately Ramsey. First perturb each coordinate of each point onto a sufficiently fine lattice $\delta \mathbb{Z}$, changing all pairwise distances by less than $\epsilon /2$. For each coordinate, map the finitely many required integers to consecutive vertices of a very large regular polygon, scaled so that one angular step has arc length $\delta$. If the step angle is $\theta$, the chord replacing a difference of $q$ lattice steps has lengthas $\theta \to 0$, uniformly over the finitely many differences involved. Taking the orthogonal Cartesian product of $d$ such polygons therefore embeds the perturbed points with a further distance error below $\epsilon /2$. Each polygon is Euclidean Ramsey, and their product is Euclidean Ramsey by the product theorem for Euclidean Ramsey sets. A monochromatic copy of that product contains the corresponding approximate copy of $X$, completing the proof.

Because $S_1$ and $S_2$ are isometric,The two induced orientations on their common boundary $\gamma$ are opposite. Ifis computed with the boundary orientation from $S_1$, then its value for $S_2$ is $-B$. Applying the Gauss-Bonnet theorem to the two halves givesThe equalities above imply $B=0$. This is the isometric halves give zero total boundary geodesic curvature argument.

Every ambient isometry ${\varphi }_{x,y}$ preserves the magnitude of geodesic curvature and sends $x$ to $y$. Hence $|k_g|$ is constant along the connected curve $\gamma$ by the transitive curve symmetry makes geodesic-curvature magnitude constant. If that constant were positive, continuity would force the signed $k_g$ to have one fixed sign, making its integral nonzero. Therefore $k_g\equiv 0$. A curve has zero geodesic curvature exactly when it is a geodesic, so $\gamma$ is geodesic.

Again, no. On the unit round sphere and for a sufficiently small nonzero $\epsilon$, considerThis is a smooth embedded closed curve and may be reparametrized by arc length. It obeysso the antipodal isometry preserves the curve and exchanges the two complementary discs. Its restriction therefore gives an isometry $S_1\to S_2$.

The curve is not contained in any plane through the origin: its third-harmonic height cannot satisfy a nontrivial linear relation with $\cos \theta$ and $\sin \theta$. It is therefore not a great circle. Since the closed geodesics of the round sphere are precisely the great circles, $\gamma$ is not geodesic. This antipodally symmetric nongeodesic spherical curve shows that the point-transitivity assumption cannot be omitted.

An open-boundary matrix product state with physical dimension $q$ and bond dimension $\chi$ iswhere the $A^i$ are $\chi \times \chi$ matrices and $|\ell ⟩,|r⟩$ are boundary vectors. For periodic boundary conditions, replace the boundary contraction by the matrix trace $\mathrm{Tr}(A^{i_1}\cdots A^{i_N})$.

The same matrices define the matrix product state transfer mapWhen ${\sum }_i(A^i{)}^{†}{A}^i=I$, the Stinespring dilationis an isometry. Repeatedly applying $V$ stores each Kraus label in a fresh physical register:Contracting the remaining virtual system with $⟨\ell |$ gives an MPS, while tracing over all recorded labels gives repeated application of the completely positive map $\mathcal{E}$. The MPS is therefore a coherent unravelling of the channel. Equivalently, retaining the Kraus-label registers realizes a purification of its output. A non-normalized MPS tensor gives the same construction with a general completely positive map; an appropriate canonical gauge normalizes the transfer map on its support.

The map $F$ is a measure-preserving transformation whenfor every Borel set $B$, equivalently $F_{\ast }\omega =\omega$. In this case its Koopman operator is an isometry on $L^2(\omega )$.

Write $X=\{x_1,\dots ,x_s\}$. Since $X$ is not a spherical point set, there are real coefficients ${\lambda }_i$, not all zero, such thatIndeed, take a minimal nonspherical subset; its points are affinely dependent, and centering the proper spherical subset shows that the corresponding quadratic sum is nonzero. The three relations are invariant under isometries, and rescaling the ${\lambda }_i$ lets us assume $b=1/2$.

Choose $\delta <1/(2s)$ and color every $y$ by the $s$ intervals of length $\delta$ containing the fractional parts of ${\lambda }_i\Vert y{\Vert }^2$. This uses finitely many colors. If $y_1,\dots ,y_s$ were a monochromatic isometric copy of $X$, then eachwould lie within $\delta$ of an integer. Their sum is within $s\delta <1/2$ of an integer, but because ${\sum }_i{\lambda }_i=0$ it equalsa contradiction. Hence $X$ is not a Euclidean Ramsey set.

Put ${\rho }^2=28-28\sqrt{5}/5$ and, for $j\in \mathbb{Z}/30\mathbb{Z}$, defineThe shift $v_j\mapsto v_{j+1}$ is an isometry acting transitively on the finite set $V=\{v_j\}$, so $V$ is a cyclic transitive point set and hence a Euclidean Ramsey set by the Kriz theorem for cyclic transitive point sets.

For every $j,k$,For $k=1,5,11,15,4,14$, these squared distances are respectivelyConsequently $v_0,v_1,v_{26},v_{15}$, in that order, have consecutive side lengths $9,5,9,10$ and equal diagonals $\sqrt{131}$. This is an isometric copy of the required isosceles trapezium. Since every monochromatic copy of $V$ contains this four-point subset, the trapezium is Euclidean Ramsey.

For $x,y\in X$, the triangle inequality and the fact that $\varphi$ is an isometry giveThus the displacement function is $2$-Lipschitz, hence continuous.

For $|a|<1$, the Möbius transformation $f_a(z)=(z-a)/(1-\overline{a}z)$ is an isometry of the Poincare disc model sending $a$ to zero. Rotations combined with these transformations describe its orientation-preserving isometries.

A geodesic hexagon in the hyperbolic plane with all six interior angles $\pi /2$. Any three positive alternating side lengths determine it up to isometry. If those sides are ${\ell }_i/2$, the opposite seam satisfies $\cosh s_i=(\cosh ({\ell }_i/2)+\cosh ({\ell }_j/2)\cosh ({\ell }_k/2))/(\sinh ({\ell }_j/2)\sinh ({\ell }_k/2))$. Two congruent such hexagons glued along the other three sides form a pair of pants with boundary lengths ${\ell }_1,{\ell }_2,{\ell }_3$.

A Riemannian metric on a Lie group is right-invariant when every right translation is an isometry. It is determined by a positive-definite inner product at the identity, extended with a coframe of right-invariant differential forms. Its translation-generated Killing vector fields are left-invariant, rather than the right-invariant vector fields dual to that coframe.

A class of metrics is spectrally rigid if equality of their specified spectra forces the metrics to be isometric, or, in a deformation version, if continuous isospectral deformations are trivial. The spectrum of a flat torus determines every two-dimensional flat torus up to isometry. The Wolpert generic spectral rigidity theorem is a generic, rather than universal, uniqueness statement.