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Suppose these sets exhaust a manifold, and its Riemannian metric is Euclidean beyond some radius in the end coordinates. A finite-time geodesic has fixed speed and cannot increase its exterior radius faster than that speed. It remains in one compact core enlargement, with velocity in a compact disk bundle. The geodesic equation therefore continues beyond every finite parameter endpoint. The compactness hypothesis is substantive: the open exterior of a closed Euclidean ball, with empty core, is incomplete at its inner boundary despite having a Euclidean metric at infinity.

For a surface of revolution whose induced metric is $g=E(\rho )d{\rho }^2+{\rho }^{2}d{\varphi }^2$, with $\rho >0$ and $E>0$ smooth on a coordinate interval, define $\sigma (\rho )={\int }_{{\rho }_{\ast }}^{\rho }\sqrt{E(r)}dr/r$. Then ${\sigma }^{\prime }>0$ and the conformal factor ${\rho }^{-1}$ gives ${\rho }^{-2}g=d{\sigma }^2+d{\varphi }^2$. This proves the rescaled metric is locally the Euclidean metric without solving a curvature equation. Poles and places where $\rho$ is not a coordinate must be treated in other charts; the angular coordinate is also understood locally.

If $X$ is an infinite set with its cofinite topology and $Y$ is a Hausdorff space, every continuous map $f:X\to Y$ is a constant function. Otherwise two distinct image points would have disjoint open neighbourhoods. Their inverse images would be nonempty disjoint open sets of $X$, contradicting its hyperconnected space property.

For example, $f(n)=n$ from ${\mathbb{N}}_{>0}$ with its cofinite topology into the real line with its usual Euclidean metric is not continuous: the inverse image of the open interval $(1/2,3/2)$ is $\{1\}$, which is not open in the infinite domain. A continuous real-valued readout on that domain can only report one value. Finite cofinite domains are different: their discrete topology allows arbitrary functions into any topological space.

An ellipse is the locus of points whose Euclidean distances from two fixed foci of a conic have constant sum $2a$. If the distance between the foci of a conic is $d<2a$, the semi-major and semi-minor axes are $a$ and $\sqrt{4a^2-d^2}/2$. Writing $c=d/2$ gives the conventional relation $b^2=a^2-c^2$.^[ref]

For an original calculation, choose $a=5$, $b=3$, and hence $c=4$. The ellipsehas foci of a conic $F_{-}=(-4,0)$ and $F_{+}=(4,0)$. The point $P=(3,12/5)$ lies on it since $9/25+16/25=1$. Its focal distances areIts eccentricity of a conic is $e=4/5$. The matching directrix of a conic for $F_{+}$ is $x=a/e=25/4$, and the perpendicular distance from $P$ to that straight line is $13/4$. Therefore the ratio of focal distance to directrix distance is $(13/5)/(13/4)=4/5$, agreeing with $e$.

The constant sum can be turned into a construction. Fix a pin at each focus of a conic and pull a closed string loop taut with a pencil. In this example the segment between the pins has length $8$, so a loop of length $18$ leaves $10$ for the two pencil-to-pin segments. Moving the pencil while keeping that loop taut traces the ellipse. This is a geometric prediction from the length constraint, not an assertion about a measured drawing.

The parametrization $x=5\cos t$, $y=3\sin t$ maps the unit circle onto this ellipse. It makes plotting easy, but is an affine map rather than an isometry: horizontal and vertical lengths have different scale factors. An ellipse is not a circle under the original Euclidean metric merely because this map exists. Its two semiaxes and eccentricity of a conic retain the metric information lost by that equivalence.

A useful proof outline begins with $|X-F_{-}|+|X-F_{+}|=2a$. Isolate one square root, square, and then repeat; the resulting quadratic equation is $x^2/a^2+y^2/(a^2-c^2)=1$. Since $a>c$, the quantities isolated before squaring have the required signs, so the reverse implication introduces no extra branch. The condition $a>c$ is essential: at equality the focal-distance locus becomes the line segment between the foci of a conic.

On Euclidean space, every Killing vector field is a constant translation generator plus a constant infinitesimal rotation generator:Here $A$ is a constant vector and $B$ is an antisymmetric matrix. The second covariant derivative of a Killing vector vanishes for the Euclidean metric, so the components are affine functions of Cartesian coordinates; the Killing equation makes the linear coefficient antisymmetric.

In the Euclidean plane with metric tensor $g=d{x}^2+d{y}^2$, two useful examples areThe notation ${\partial }_x$ denotes a coordinate basis vector; $dx$ denotes a differential one-form. The corresponding flow of a vector field maps areFor translation, all arrows have the same direction and length. For rotation, arrows are tangent to circles, have length $\sqrt{x^2+y^2}$, and vanish at the origin. The rotation flow keeps $x^2+y^2$ fixed, and its nonstationary integral curves of a vector field are circles. Both flows preserve the Euclidean metric. See Carroll's original notes, section 5, equations (5.45)-(5.46).

The circular integral curves of a vector field are symmetry orbits, whereas geodesics of the Euclidean plane are straight lines. A Killing vector field does not require its own orbits to be geodesics. The geodesic conserved quantity from a Killing vector pairs the field with the velocity of a separate geodesic; it does not require the velocity to equal the Killing vector field.

The first fundamental form of a smooth surface $S\subseteq {\mathbb{R}}^3$ is the restriction of the ambient inner product to each tangent space $T_{p}S$. For a regular parametric surface $X(u,v)$, it is

Here $X_u^2$ means $X_u\cdot X_u$, not a componentwise square. The coefficients form the Gram matrix of the two coordinate tangent vectors. Regularity makes this a positive-definite matrix, so $E>0$ and $EG-F^2>0$.

To calculate the length of a curve $\gamma (t)=X(u(t),v(t))$, apply the chain rule and integrate its speed:The same inner product gives the angle $\alpha$ between nonzero tangent vectors $V,W$ through $\cos \alpha =I(V,W)/\sqrt{I(V,V)I(W,W)}$, and gives the area element $dA=\sqrt{EG-F^2}dudv$ as a positive density. One compact expression therefore supplies rulers, protractors, and surface area measurements.

For the lateral surface of a circular cylinder, take the circumferential arc length $u$ rather than the angle:Direct inner products give $E=G=1,F=0$, hence $I=d{u}^2+d{v}^2$. This is exactly the Euclidean metric on the unrolled plane. For instance, a path with $u=3t,v=4t$, $0\le t\le 1$, has length $5$ on either surface. Every corresponding path retains its length, not merely this example, so sufficiently small neighborhoods retain their intrinsic distances. The local isometry preserves angles and surface areas too. Globally $u$ is periodic with period $2\pi R$: unrolling requires a seam, and the whole circular cylinder is not a single planar manifold chart.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with its quotient Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its volume is the lattice covolume. The dual lattice indexes its complete Fourier series of Laplacian eigenfunctions.

On an oriented Euclidean metric four-space, the Hodge star operator on two-forms is an orthogonal involution. Its two eigenspaces each have dimension three. A self-dual two-form has eigenvalue $+1$, while an anti-self-dual two-form has eigenvalue $-1$. The splitting uses the metric-normalized Riemannian volume form; an arbitrarily rescaled volume in the defining wedge identity would rescale the operator itself.

The Killing equation tests whether a candidate vector field generates a local isometry, without needing to solve for its flow of a vector field. For the Levi-Civita connection of a metric tensor $g$, it isParentheses mean averaging over the two index orders. The equation says the symmetric part of the covariant derivative vanishes, not that the entire covariant derivative vanishes. The Lie derivative of the metric tensor gives the coordinate formThe last two terms account for how tangent vectors move with the field; checking only $K^{\rho }{\partial }_{\rho }{g}_{\mu \nu }$ misses this. See Carroll, section 5, equations (5.33) and (5.41)-(5.42).

For the Euclidean Killing vector field $K=-y{\partial }_x+x{\partial }_y$, the Euclidean metric in Cartesian coordinates makes covariant derivatives ordinary partial derivatives. Its diagonal equations give ${\partial }_x{K}_x={\partial }_y{K}_y=0$, and its mixed equation gives ${\partial }_x{K}_y+{\partial }_y{K}_x=1-1=0$. Thus a varying vector field can preserve a constant metric tensor. In contrast, $K=x{\partial }_x+y{\partial }_y$ gives $({\mathcal{L}}_{K}g{)}_{xx}=2$ and fails the Killing equation, despite the constant metric components.

If $K={\partial }_z$ is a coordinate basis vector, its components are constant and the coordinate form reduces to ${\partial }_z{g}_{\mu \nu }=0$. Hence independence of the metric components from $z$ supplies a Killing vector field. This shortcut applies to that coordinate generator, not to an arbitrary varying vector field.

A Killing vector field describes a direction in which geometry has a continuous symmetry: moving points together along its flow of a vector field preserves the metric tensor, including its pairing of tangent vectors. Such a symmetry gives a geodesic conserved quantity from a Killing vector, often a component of momentum or angular momentum. The direction may vary with position: rotation provides the basic example.

On a Riemannian manifold or Lorentzian manifold with metric tensor $g$, a smooth vector field $K$ is a Killing vector field when its Lie derivative of $g$ vanishes. If ${\Phi }_s$ is its local flow of a vector field, the equivalent statements areThe middle equality uses the pullback of a covariant tensor: the local flow maps are isometries on their domains. The last equality is the Killing equation, using the Levi-Civita connection. The infinitesimal condition holds throughout an open domain and integrates to exact local isometries; it is stronger than making the first derivatives of $g$ vanish at a single point. See Tong, section 4.3.1, equation (4.29).

For example, on the interval $0<x<1$ with Euclidean metric $d{x}^2$, $K={\partial }_x$ translates points by $x\mapsto x+s$. This is a local isometry, but a point reaches the end of the interval in finite flow parameter. A local flow of a vector field can be incomplete; see Killing fields on incomplete manifolds need not generate global isometries. The full-plane translation and rotation examples under Euclidean Killing vector field instead have flows for every real $s$.

In a $2k$-dimensional normed vector space in John position, two orthogonal $k$-dimensional subspaces admit $\Vert x{\Vert }_E\le |x|\le 32\mathrm{vr}(E{)}^2\Vert x{\Vert }_E$. To see this, take $\epsilon$-nets in two orthogonal half-dimensional unit spheres, each of size at most $(1+2/\epsilon {)}^k$, and rotate them uniformly together. The volume ratio bound makes the probability that any net point has norm below $2\epsilon$ at most $2[4\mathrm{vr}(E{)}^2({\epsilon }^2+2\epsilon ){]}^k$. With $\epsilon =(32\mathrm{vr}(E{)}^2{)}^{-1}$ this is less than $1$. Since $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric, the successful nets give the stated inequality everywhere on both spheres. A dimension-independent constant therefore requires a bound on the volume ratio; it does not hold for arbitrary normed vector spaces.

For a real linear change of coordinates $x^{\prime }=Ax$, one has $d{x}^{\prime }=Adx$. The Euclidean metric therefore transforms toIt agrees with $d{x}^{T}dx$ for every displacement precisely when $A^{T}A=I$. For example, equality of the associated quadratic forms determines their symmetric coefficient matrices by polarization identity. The orthogonal group is defined by this condition, soTaking determinants gives $(\det A{)}^2=1$. Thus these matrices are invertible automatically; both signs of the determinant are allowed, since the metric alone does not specify an orientation.

Write the cone as $z=r\cot \alpha$, with $r$ the distance from its axis and $0<\alpha <\pi /2$. Its Euclidean metric givesConsequently a differentiable path expressed as $r=r(\theta )$ has arc lengthFor $L(r,r^{\prime })=\sqrt{r^2+{\csc }^2\alpha (r^{\prime }{)}^2}$, the Euler-Lagrange equation has the Beltrami identity because $L$ has no explicit $\theta$ dependence:For a nonradial minimizing path, $C>0$. Rearranging and integrating givesThe endpoint conditions determine $C$ and ${\theta }_0$ on the selected angular branch.

To verify that this stationary path really minimizes arc length, use the local isometry from a circular cone to the plane. Set $s=r/\sin \alpha$ and $\varphi =\theta \sin \alpha$; then $d{s}^2+s^{2}d{\varphi }^2$ is the plane metric. The displayed path is a straight line $s\cos (\varphi -{\varphi }_0)=C/\sin \alpha$, so its segment realizes the plane distance. Choose endpoint angular lifts with $|\Delta \theta |\le \pi$; then $|\Delta \varphi |<\pi$, and the joining segment avoids the apex. Its length isOther lifts with developed angular difference less than $\pi$ give at least this distance; an angular change of at least $\pi$ has infimum at least $s_1+s_2$, attained only if passage through the apex is allowed. This proves the global choice above. Endpoints on a common generator have the radial segment as shortest path, which is the limiting case omitted by the $r(\theta )$ parametrization. If an endpoint is the apex, its shortest path is a generator. Thus shortest paths are straight segments after developing the cone.

The pullback of the flat Euclidean metric under $x^{\prime }{}^{\mu }=x^{\mu }+ϵ{\omega }^{\mu }(x)$ isFor a conformal transformation it must be a scalar multiple of ${\delta }_{\mu \nu }$. Taking the trace determines that multiple. With $\rho ={\partial }_{\alpha }{\omega }^{\alpha }$, the necessary and sufficient infinitesimal condition is the Conformal Killing equationDifferentiate its $(\alpha ,\beta )$ instance with respect to $x^{\gamma }$, its $(\alpha ,\gamma )$ instance with respect to $x^{\beta }$, and its $(\beta ,\gamma )$ instance with respect to $x^{\alpha }$. Adding the first two and subtracting the last, commuting partial derivatives, leaves $2{\partial }_{\beta }{\partial }_{\gamma }{\omega }_{\alpha }$. Hence flat conformal Killing integrability givesContracting $\beta$ with $\gamma$ gives $\Delta {\omega }_{\alpha }=(2-d){\partial }_{\alpha }\rho /d$. Its divergence is $\Delta \rho =(2-d)\Delta \rho /d$, so $2(d-1)\Delta \rho =0$: thus $\boxed{\Delta \rho =0\text{ for }d>1}$. Taking instead the $\alpha$ divergence of the displayed second-derivative identity givesConsequently $\boxed{{\partial }_{\beta }{\partial }_{\gamma }\rho =0\text{ for }d>2}$.

On a connected flat domain, write $\rho =d\lambda +2db\cdot x$, with constant $\lambda$ and $b$. Integrating the second-derivative identity, and imposing the Conformal Killing equation on its remaining affine part, gives the complete Conformal Killing vector fieldIndeed the quadratic part has precisely the required Hessian matrix, while the symmetric part of the remaining linear term must be $\lambda {\delta }_{\mu \nu }$. The constants generate translations, rotations, dilations and special conformal transformations, respectively. There are $d$, $d(d-1)/2$, $1$ and $d$ independent parameters, so the full conformal group has dimensionHere “full” means local transformations, or global transformations of the conformal compactification of Euclidean space; its connected group is locally $SO(d+1,1)$. Requiring every transformation to be nonsingular on all of uncompactified ${\mathbb{R}}^d$ would exclude the special conformal transformations with finite poles.

Let $F_{+}=(F+\ast F)/2$ and $F_{-}=(F-\ast F)/2$, using the oriented Euclidean metric. For a self-dual two-form $\omega$,Apply this componentwise in the Lie algebra. Since the three ${\omega }_i$ span the self-dual subspace, their vanishing wedges force $F_{+}=0$. Thus $\ast F=-F$. The gauge-theory Bianchi identity is $D_{A}F=0$, so $D_A\ast F=-D_{A}F=0$, the Euclidean Yang-Mills equations. This proves the implication locally; finite action would additionally be needed to call the solution an instanton.

Use the complex orientation for $w=x^1+i{x}^2$ and $z=x^3+i{x}^4$. The real and imaginary parts of $dw\wedge dz$, and $i(dw\wedge d\bar{w}+dz\wedge d\bar{z})$, span the self-dual two-forms. Accordingly the three gauge curvature equations areFor a spectral parameter $\lambda \in {\mathbb{CP}}^1$, consider the Lax pairTheir commutator isIt vanishes for every $\lambda$ exactly when all three coefficients vanish. The derivative directions ${\partial }_w-\lambda {\partial }_{\bar{z}}$ and ${\partial }_z+\lambda {\partial }_{\bar{w}}$ are mutually orthogonal null vectors for the complexified metric. Flatness on the corresponding two-planes makes the linear parallel-transport equations locally compatible. Conversely compatibility for an invertible fundamental matrix $\Psi$ forces the commutator to vanish; a single zero solution would not establish it. This is the Euclidean complex-coordinate anti-self-dual Lax pair.

For an Abelian U(1) connection, all commutators of the connection components vanish, and the gauge curvature equations therefore readLocally the first two equations say that $A^{1,0}$ is $\partial$-closed and $A^{0,1}$ is $\bar{\partial }$-closed. The Dolbeault-Poincaré lemma, and its complex conjugate, give scalar functions $u,v$ withSubstituting in the third equation givesIn a complex gauge transformation, use $g=e^{-u}$ and the derivative-plus-connection convention $A\mapsto A+g^{-1}dg$. ThenThis is the scalar Laplace equation, since for the stated metric $\Delta =2({\partial }_z{\partial }_{\bar{z}}+{\partial }_w{\partial }_{\bar{w}})$. The complex gauge transformation need not remain in $U(1)$.

A reduction using only a unitary gauge transformation is also available. Since $A$ is anti-Hermitian, $A_{\bar{w}}=-\overline{A_w}$ and $A_{\bar{z}}=-\overline{A_z}$, so the potential for the barred components can be chosen as $v=-\bar{u}$. Write $u=p+iq$ with real $p,q$. Then $A=\partial p-\bar{\partial }p+idq$, and $g=e^{-iq}$ removes the last term. With $f=-2p$ this givesIts gauge curvature is $\partial \bar{\partial }f$. Thus the Abelian anti-self-dual connections from harmonic scalar potentials reduction is local and does not require silently enlarging the real gauge group.

The Wick rotation gives the radial Euclidean metric $d{s}_E^2=Vd{\tau }^2+d{r}^2/V$. For a simple root, write $b=V^{\prime }(r_0)\ne 0$ and choose the side with V positive. There $r-r_0=b{\rho }^2/4$ for $\rho \ge 0$, so $V=b^2{\rho }^2/4+O({\rho }^4)$ andThe leading geometry is a plane in polar coordinates with angle $\phi =|b|\tau /2$. For a time identification of period beta, the circumference/proper-radius ratio approaches $|b|\beta /2$. Unless it is exactly $2\pi$, the origin is a conical singularity, with deficit $2\pi -|b|\beta /2$ or an excess if this is negative. This is the Euclidean metric near a simple static horizon, rather than a divergent Lorentzian curvature invariant at the horizon. The side with V negative does not give a positive-definite metric under this real Wick rotation.

In orthonormal Cartesian coordinates, a second-rank tensor has components transforming under a change of axes $Q$ as $M_{ij}^{\prime }=Q_{ik}{Q}_{j\ell }{M}_{k\ell }$, where $Q{Q}^T=I$. Its contraction is invariant:Thus $M_{ii}$ is a scalar. This is tensor contraction with the Euclidean metric. In general nonorthonormal coordinates the invariant expression is $g^{ij}{M}_{ij}$, not an unweighted sum of diagonal covariant components.

For the planar plate, every mass element has $z=0$. The inertia tensor is consequentlyIt follows directly that ${\mathbf{e}}_z$ is an eigenvector andThe real symmetric matrix has three real eigenvalues. Its trace is $2{\int }_D(x^2+y^2)\rho dS=2M_{\perp }$, so $M_1+M_2+M_{\perp }=2M_{\perp }$. Hence the perpendicular axis theorem gives $M_{\perp }=M_1+M_2$.

For a disc of constant areal density ${\rho }_0$, symmetry gives ${\int }_{D}xydS=0$, and polar integration gives ${\int }_D{x}^{2}dS={\int }_D{y}^{2}dS=\pi a^4/4$. Thereforewhere the scalar $M$ in the last expression denotes the total mass rather than the tensor.

For a $d$-dimensional normed vector space, with unit ball $K$ and John ellipsoid $\mathcal{E}$, its volume ratio isIt is at least $1$ and invariant under invertible linear coordinate changes.

For ${\ell }_1^n$, the unit ball has volume $2^n/n!$: its intersection with each orthant is the standard simplex of volume $1/n!$. Its John ellipsoid is $n^{-1/2}{B}_2^n$. To verify maximum volume, let $A{B}_2^n$ be any centered inscribed ellipsoid. For each sign vector $\epsilon \in \{-1,1{\}}^n$, containment gives $|A^T\epsilon |\le 1$. Averaging over independent Rademacher random variables yields $\mathrm{tr}(A{A}^T)\le 1$. The arithmetic-geometric mean inequality for its squared singular values gives $|\det A|\le n^{-n/2}$. The scalar matrix $n^{-1/2}I$ attains this bound because $\Vert x{\Vert }_1\le \sqrt{n}|x|$. Centered ellipsoids suffice: if $a+A{B}_2^n$ is contained in the symmetric convex unit ball, averaging it with $-a+A{B}_2^n$ gives the centered copy with the same volume.

Using the stated Euclidean ball volume and $n=2k$,since $(2k)!/k!={\prod }_{j=k+1}^{2k}j\ge k^k$. Therefore $\mathrm{vr}({\ell }_1^{2k})\le \sqrt{8/\pi }$ uniformly in $k$.

The subsequent assertion for an arbitrary $2k$-dimensional space is false as printed. Here is a counterexample independent of any asymptotic section theorem. Take $E={\ell }_{\infty }^{2k}$. Its John ellipsoid is $B_2^{2k}$: an inscribed $A{B}_2^{2k}$ has every row of $A$ of Euclidean length at most $1$, so the Hadamard inequality gives $|\det A|\le 1$, attained by $A=I$. Let $F$ be any $k$-dimensional subspace and $f_1,\dots ,f_k$ an orthonormal basis of it. The random unit vectorhas $|x|=1$. Since ${\sum }_j{f}_j(i{)}^2\le 1$ for each coordinate $i$, the moment-generating function bound for independent Rademacher random variables givesChoose $r=\sqrt{2\log (8k)/k}$; the latter probability is at most $1/2$, so some unit vector in $F$ has $\Vert x{\Vert }_{\infty }\le r$. The proposed section inequality would therefore forceThus even one half-dimensional section cannot have the printed universal constant.

The valid orthogonal splitting with bounded volume ratio is that $L$ may depend on $V=\mathrm{vr}(E)$, and $L=32V^2$ suffices. This supplies a dimension-independent constant for the preceding ${\ell }_1^{2k}$ case, or whenever $V$ is bounded uniformly. We prove the complete qualified assertion, including its epsilon-net ingredient.

In John position, $\Vert x{\Vert }_E\le |x|$. Polar integration of the radial function $\rho (u)=1/\Vert u{\Vert }_E$ giveswhere $\sigma$ is normalized spherical measure.

A maximal $\epsilon$-separated subset of $S^{k-1}$ is an epsilon-net. Balls of radius $\epsilon /2$ about its points have disjoint interiors and lie in the ball of radius $1+\epsilon /2$. Comparing volumes bounds its cardinality by $(1+2/\epsilon {)}^k$. Finite existence follows from this packing bound, and maximality proves the net property without assuming a separate covering theorem.

Choose two orthogonal $k$-dimensional subspaces $F_0,F_0^{\perp }$ and one such net on each unit sphere. Rotate both by a common uniformly distributed orthogonal matrix $U$. This distribution can be obtained from Question 1 applied to the compact orthogonal group with its Frobenius metric; left and right multiplication are transitive isometries. Consequently $Uv$ is uniformly distributed on $S^{2k-1}$ for every fixed unit vector $v$. The union bound shows that the probability that some rotated net point has norm less than $2\epsilon$ is at mostSet $\epsilon =(32V^2{)}^{-1}$. Because $V\ge 1$, the bracket is at most $1/4+1/256$, and twice its $k$th power is less than $1$ for every $k\ge 1$. Some rotation therefore makes every point of both nets have norm at least $2\epsilon$.

The triangle inequality and $\Vert x{\Vert }_E\le |x|$ imply that $\Vert \cdot {\Vert }_E$ is $1$-Lipschitz for the Euclidean metric. Every unit vector of $U{F}_0$ or $U{F}_0^{\perp }$ is within $\epsilon$ of a good net point and thus has norm at least $\epsilon$. By homogeneity, on both orthogonal subspaces,For ${\ell }_1^{2k}$ the first computation gives the absolute choice $L=256/\pi$. The missing bounded-volume ratio restriction is essential; the literal unrestricted assertion cannot be proved.

The musical isomorphisms are the pointwise maps induced by the Riemannian metric:In local coordinates they are $(X^{♭}{)}_i=g_{ij}{X}^j$ and $({\alpha }^{♯}{)}^i=g^{ij}{\alpha }_j$. The inverse metric makes these inverse bundle maps, hence inverse maps between smooth vector fields and smooth differential one-forms.

For a covariant two-tensor $\beta$, its metric trace is ${\mathrm{tr}}_g\beta =g^{ij}{\beta }_{ij}={\sum }_i\beta (e_i,e_i)$ in a Riemannian orthonormal frame. The contraction is independent of the chosen frame. There is a terminology distinction here: if $\beta$ is literally an alternating differential two-form, this trace is zero. The object needed below is the covariant two-tensor $D(X^{♭})$, generally not alternating. Interpreting that tensor as an alternating form would make the stated divergence construction impossible.

Let $D$ be the Levi-Civita connection. The Riemannian gradient, divergence and positive Laplace-Beltrami operator areThus $g({\mathrm{grad}}_{g}f,Y)=Yf$, and in coordinatesThis choice makes the Laplacian eigenvalues nonnegative on a closed Riemannian manifold. The alternative convention $+\mathrm{div}\mathrm{grad}$ reverses all eigenvalue signs, but none of the comparisons below.

For the Riemannian submersion, let $F=f\circ \pi$ and choose orthonormal horizontal lifts $E_i$ of a base frame and an orthonormal vertical frame $V_{\alpha }$. The horizontal part of $D_{E_i}{E}_i$ projects to the base Levi-Civita connection; this follows directly from its Koszul formula applied to basic horizontal fields, because their inner products and projected brackets are pulled back from the base. Therefore the horizontal trace of the Riemannian Hessian of $F$ is the pulled-back base Hessian trace. Since $V_{\alpha }F=0$, its vertical terms areWriting $H={\sum }_{\alpha }(D_{V_{\alpha }}{V}_{\alpha }{)}^{\mathrm{hor}}$ for the unnormalised fibre mean curvature vector givesThe totally geodesic submanifold condition on the fibres makes $H=0$. Consequently eigenfunction pullback by a Riemannian submersion givesHere the Riemannian submersion is onto, so the map is injective. Its image is exactly the members of the upstairs eigenspace constant on each whole fibre: such a function descends smoothly by local sections of the submersion, and the displayed identity forces the descended function to have the same eigenvalue. In particular, whenever these eigenspaces are finite dimensional, the downstairs multiplicity is no larger than the upstairs multiplicity. The upstairs eigenspace may also contain functions varying along fibres.

A flat torus is ${\mathbb{R}}^d/\Lambda$ with the metric induced by the Euclidean metric, where $\Lambda$ is a full-rank Euclidean lattice. Its dual lattice is ${\Lambda }^{\ast }=\{w:⟨w,v⟩\in \mathbb{Z}\text{ for all }v\in \Lambda \}$. The characters $e_w(x)=e^{2\pi i⟨w,x⟩}$ descend to the quotient and form an orthogonal Fourier basis of $L^2$ on it. Direct differentiation gives the spectrum of a flat torus:with multiplicity equal to the number of dual lattice vectors of each length. In particular zero has multiplicity one.

Let the given four-dimensional isospectral manifolds be ${\mathbb{R}}^4/{\Lambda }_1$ and ${\mathbb{R}}^4/{\Lambda }_2$. For $d=4+k$, take their products with the same $k$ circles of length $\epsilon$, equivalently the lattices ${\Lambda }_i\oplus \epsilon {\mathbb{Z}}^k$. The dual lattices are ${\Lambda }_i^{\ast }\oplus {\epsilon }^{-1}{\mathbb{Z}}^k$, so their Laplacian eigenvalues areThe given equality of base spectra with multiplicities makes these product spectra identical.

Choose $\epsilon$ smaller than the shortest nonzero vector length in either original Euclidean lattice. By short-vector cancellation for flat tori, the shortest nonzero vectors in the product lattice are exactly $\pm \epsilon e_j$ in its new factor. To spell out the nonisometry argument, any isometry of the product flat tori lifts to an isometry of their universal Euclidean covers. This is affine, and its orthogonal linear part maps one product lattice onto the other. It must map the span of their shortest vectors, namely the new $k$-dimensional factor, onto itself; orthogonality then preserves the original four-dimensional factor and maps ${\Lambda }_1$ onto ${\Lambda }_2$. That would give an isometry of the original flat tori, contrary to the assumption. Isospectral nonisometric flat tori therefore exist in every dimension $d\ge 4$.

Use the positive Laplace-Beltrami operator, $\Delta =-\mathrm{div}\nabla$. A flat torus is the quotient ${\mathbb{R}}^d/\Lambda$, where $\Lambda$ is a full-rank Euclidean lattice and the metric descends from the Euclidean metric. Its dual lattice isIf $V$ is the covolume of $\Lambda$, the functions ${\varphi }_{\xi }(x)=V^{-1/2}{e}^{2\pi i⟨\xi ,x⟩}$ descend to the torus. Direct differentiation gives the spectrum of a flat torus:The multiset includes one entry for each dual vector. In particular zero has multiplicity one. Writing $\Lambda =B{\mathbb{Z}}^d$ identifies the dual vectors with $B^{-T}m$, $m\in {\mathbb{Z}}^d$. Standard Fourier series on the unit cube then prove that these functions are an orthonormal basis of $L^2$ on the torus. If $\Delta f=\lambda f$, its Fourier coefficients satisfy $(4{\pi }^2|\xi {|}^2-\lambda )\hat{f}(\xi )=0$. Thus every eigenfunction is a linear combination of the indicated shell of frequencies, with no additional eigenvalues. Compactness gives the discrete self-adjoint realization, whose eigenfunctions are smooth. For real functions the two characters at $\xi ,-\xi$ become cosine and sine, giving the same real multiplicities. The opposite Laplacian sign would negate all displayed eigenvalues without changing the rigidity conclusion.

To prove two-dimensional spectral rigidity, let $L={\Lambda }^{\ast }$ and recover its Gram matrix from its vector-length multiset. Let $a$ be the shortest nonzero length and choose $v\in L$ with $|v|=a$. This vector is primitive: a proper integer multiple would have a shorter lattice vector. Hence $L\cap \mathbb{R}v=\mathbb{Z}v$. From the spectral multiset subtract exactly two occurrences of each length $ka$, $k=1,2,\dots$. The remaining multiset consists precisely of the vectors not on this line, including correct multiplicities at coincident lengths. Its least length $b$ is the length of a shortest vector $w$ independent of $v$.

The pair $(v,w)$ is a basis of $L$. Otherwise take a lattice point in a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ and reduce its two coefficients into $[-1/2,1/2]$. The resulting nonzero lattice vector $z=\alpha v+\beta w$ is not on $\mathbb{R}v$, by primitivity of $v$. Butwith strict inequality: when both coefficients are nonzero the triangle inequality is strict for independent vectors, and when one vanishes its length is at most $b/2$. This contradicts the definition of $b$. Subtract an integer multiple of $v$ from $w$ and, if necessary, change its sign to arrange $0\le c=⟨v,w⟩\le a^2/2$; minimality of $b$ makes this reduction possible without decreasing its length.

The covolume $A$ of $L$ is also spectral. The count of lattice vectors with length at most $R$ obeysFor example, translate a bounded fundamental parallelogram of diameter bound $D$: the union of tiles with centres in $B_R$ contains $B_{R-D}$ and lies inside $B_{R+D}$. Comparing their areas proves the leading coefficient. The spectrum supplies this count. Since $(v,w)$ is a lattice basis, its parallelogram area satisfies $A^2=a^2{b}^2-c^2$. ThereforeAll three quantities are determined by the spectrum. Equal Gram matrices give an orthogonal map between the dual lattices; taking duals gives an orthogonal map between the original lattices, descending to a Riemannian isometry of the tori. Isospectral flat two-tori are isometric. This is the two-dimensional lattice reconstruction from vector lengths; it uses lengths with multiplicities, not merely the set of distinct lengths.