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Add respectively the conditions ${\sum }_i{x}_i\equiv 0(\mathrm{mod}4)$ and ${\sum }_i{x}_i-2x_1\equiv 0(\mathrm{mod}4)$ to the displayed common conditions. These define the two integer Euclidean lattices underlying the tetracode length-isospectral lattice construction. They coincide on their all-even vectors and differ on all-odd vectors. Reflecting a coordinate divisible by three preserves the code condition; on odd vectors it changes the coordinate sum by two modulo four, while on even vectors both conditions remain satisfied. This yields the coordinate reflection bijection for tetracode lattices.

Let $M={\mathbb{R}}^d/\Lambda$ be a flat torus, where $\Lambda$ is a full-rank Euclidean lattice, and use the nonnegative Laplace-Beltrami operator $\Delta =-{\sum }_{k=1}^d{\partial }_k^2$. The dual lattice isFor $w\in {\Lambda }^{\ast }$, the function ${\varphi }_w(x)=\mathrm{vol}(M{)}^{-1/2}{e}^{2\pi i⟨w,x⟩}$ is periodic under $\Lambda$, so it defines a function on $M$, and direct differentiation gives $\Delta {\varphi }_w=4{\pi }^2|w{|}^2{\varphi }_w$. Integrating characters on a fundamental parallelepiped proves that these functions are orthonormal. After choosing a lattice basis, the ordinary Fourier series theorem on ${\mathbb{R}}^d/{\mathbb{Z}}^d$ proves completeness, so there are no additional eigenvalues. Thus the spectrum of a flat torus iswith multiplicity equal to the number of dual vectors of the given length; the zero eigenvalue occurs once.

It remains to show that a two-dimensional Euclidean lattice is determined up to an orthogonal transformation by its vector-length multiset. This is the two-dimensional lattice reconstruction from vector lengths. Let $v$ be a shortest nonzero vector, of length $a$. It is primitive: $v=ku$ with integer $|k|>1$ would give a shorter nonzero vector. The lattice vectors on its line are therefore precisely the integer multiples of $v$. Subtract two occurrences of each length $ka$, $k=1,2,\dots$, from the length multiset. This removes exactly one primitive lattice line, even when other vectors happen to have the same lengths. The smallest remaining length $b$ is the length of a shortest vector $w$ independent of $v$.

These two vectors form a basis of the lattice. Otherwise a nonzero coset of $\mathbb{Z}v+\mathbb{Z}w$ has a lattice representative $u=sv+tw$ with $|s|,|t|\le 1/2$. It cannot have $t=0$, since primitivity of $v$ would then force $u=0$. It is therefore independent of $v$, whileThe inequality is strict: if both coefficients are nonzero, the vectors are not parallel and the triangle inequality is strict; if $s=0$, then $|u|\le b/2$. This contradicts the choice of $w$.

The covolume $A$ is also determined by the length multiset. If $N(R)$ counts lattice vectors in a radius-$R$ disk, thenTo justify the estimate, tile the plane by a bounded fundamental parallelogram. Only cells meeting a fixed-width neighbourhood of the disk boundary contribute an error; their total number is $O(R)$. The lattice basis $v,w$ has Gram matrixThus $|c|=\sqrt{a^2{b}^2-A^2}$ is determined. Replacing $w$ by $-w$ makes $c\ge 0$, so the Gram matrix is determined entirely. Equal Gram matrices give an orthogonal map between the lattice bases and hence between the lattices.

Two isospectral two-dimensional flat tori consequently have orthogonally equivalent dual lattices. Taking duals shows that their original lattices are orthogonally equivalent too, and the orthogonal map descends to an isometry of the quotient tori. Two isospectral flat tori of dimension two are isometric.

The central problem of spectral geometry is how much of a Riemannian metric is determined by the eigenvalues, with multiplicities, of its Laplace-Beltrami operator. For a membrane with fixed boundary, the squared vibration frequencies are proportional to the Dirichlet eigenvalues. Kac's 1966 paper brought this inverse problem into focus: the full sequence of frequencies might encode a domain's shape, but recovering a few geometric quantities is much weaker than recovering an isometry class.

The first major line of development extracts geometric information from spectral sums. On a smooth closed $d$-dimensional manifold, the heat trace has the short-time expansionwhere ${\lambda }_j$ are the nonnegative eigenvalues of $P=-\Delta$ and $R$ is scalar curvature. The coefficients are heat invariants, formed by integrating local curvature expressions. They recover dimension, Riemannian volume and total scalar curvature. For smooth planar domains with the Dirichlet boundary condition,Thus area and boundary length are audible. The constant term also contains topological information through Gauss-Bonnet theorem, with corner corrections when the boundary is polygonal. The Weyl law is the corresponding leading eigenvalue-counting result. Neither the leading term nor the whole list of local heat coefficients automatically reconstructs global geometry.

In the 1970s, the wave trace introduced a complementary link with global closed geodesics. The distributionis spectral, while its singularities are governed by periodic geodesic motion. The 1975 Duistermaat-Guillemin work made this relationship precise. Its trace formula places singularities at geodesic lengths and computes their coefficients under suitable clean or nondegenerate hypotheses. When a coefficient is nonzero, the corresponding length is audible. Possible cancellation must be addressed before identifying the entire length spectrum with the singular support. This approach led to spectral rigidity questions, rather than just the recovery of integral curvature quantities.

The second main line constructs counterexamples to metric determination. A precedent was the 1964 flat-torus example: different sixteen-dimensional Euclidean lattices could give the same Laplacian eigenvalues. Subsequent lattice constructions lowered the dimension; Conway and Sloane's 1992 work exhibited four-dimensional Euclidean lattices with matching lattice theta series. For a flat torus ${\mathbb{R}}^d/\Lambda$, the spectrum is the multiset $4{\pi }^2|\xi {|}^2$ for $\xi \in {\Lambda }^{\ast }$, so equality of lattice theta series of the dual lattices gives isospectrality even when the Euclidean lattices are not related by an orthogonal transformation.

A systematic geometric construction arrived with the Sunada theorem in 1985. For a common Riemannian cover with finite isometry group $T$, almost conjugate subgroups have the same number of invariant vectors in each Laplacian eigenspace. Their freely acting quotients are therefore isospectral manifolds. This reduces an analytic equality to a finite-group representation identity. It produces many examples on surfaces and in higher dimensions, but nonconjugacy of the subgroups must be supplemented by an argument excluding extra quotient isometries. Transplantation theorem provides an explicit version: piecewise eigenfunctions are recombined by a fixed invertible linear transformation that respects the gluing and boundary conditions.

The planar problem itself was answered negatively by Gordon, Webb and Wolpert in 1992: noncongruent simply connected polygonal plane domains have equal Dirichlet spectra. This is stronger than counterexamples among abstract manifolds or flat tori, since these are genuine two-dimensional planar membranes. It does not settle every restricted version, such as asking for two domains within a specified smooth or convex class.

By the 1990s, isospectrality was also known to coexist with continuous metric variation and changes in local geometry. For example, a 1997 construction gives continuous isospectral families on products of spheres and tori whose members need not be locally isometric; in some examples the maximum scalar curvature changes. Conversely, rigidity and compactness results show that an isospectral class is not unrestricted. The 1988 surface compactness theorem makes fixed-spectrum sets of closed orientable surface metrics compact modulo diffeomorphisms. Compactness does not mean there is only one metric, or even finitely many metrics. By 2005, the subject therefore combined local heat invariants, global geodesic information, group-theoretic constructions, explicit transplantation and geometric rigidity. The spectrum determines substantial geometry, while generally failing to determine the full metric.

The given two-dimensional ternary subspace is the tetracode. Its nonzero words form four one-dimensional subspaces, represented byEvery nonzero word has exactly three nonzero coordinates; in particular every codeword has a zero coordinate. We use this zero coordinate to construct a length-preserving reflection, but then choose unequal coordinate weights to rule out a global isometry.

The parity congruence description of tetracode lattices is especially convenient. Define integer Euclidean latticesThese congruences define additive subgroups of ${\mathbb{Z}}^4$ containing $12{\mathbb{Z}}^4$, so they are discrete and have full rank. For all-even vectors the last conditions coincide. For all-odd vectors they require respectively coordinate sum zero or two modulo four.

For an explicit basis, setTheir columns reduce to $T$ modulo three and satisfy the stated parity congruences, so $B_{\pm }{\mathbb{Z}}^4\subseteq {\Gamma }_{\pm }$. Since $S^2=-3I$, $|\det B_{\pm }|=144$. Conversely, the modulo-three condition has index nine, while each allowed set of residues modulo four has sixteen elements out of $4^4$: eight all-even and eight all-odd residues. The Chinese remainder theorem therefore gives $[{\mathbb{Z}}^4:{\Gamma }_{\pm }]=9\cdot 16=144$. The inclusions have equal index and are equalities.

For any $x\in {\Gamma }_{+}$, choose the first coordinate $j$ for which $x_j\equiv 0(\mathrm{mod}3)$, and reflect that coordinate:Such a coordinate always exists by the codeword description. Its reflection preserves $x\mathrm{mod}3\in T$ and the common coordinate parity. If $x$ is odd, its sum changes by two modulo four, taking the plus condition to the minus condition. If $x$ is even, the sum changes by zero modulo four, and the two conditions already coincide. The set of coordinates divisible by three is unchanged, so applying the same rule again recovers $x$. This proves a bijection $F:{\Gamma }_{+}\to {\Gamma }_{-}$ preserving each absolute coordinate, the coordinate reflection bijection for tetracode lattices.

Now choose positive weights linearly independent over $\mathbb{Q}$, for instanceThe Euclidean lattices $L_{+}$ and $L_{-}$ have the same multiset of vector lengths, because $\Vert Dx{\Vert }^2={\sum }_j{a}_j{x}_j^2$ is preserved by $F$. A closed geodesic of a flat torus is the projection of a straight line with displacement a lattice vector. Thus the two tori have identical length spectra, including multiplicities of lattice displacements. Primitive lengths can also be recovered by subtracting proper iterates successively, so they agree as well.

Suppose a torus isometry existed. Its lift to ${\mathbb{R}}^4$ is affine with orthogonal linear part $R$ taking $L_{+}$ onto $L_{-}$. Put $A=D^{-1}RD$, so $A{\Gamma }_{+}={\Gamma }_{-}$. Since $12e_i\in {\Gamma }_{+}$, its image is an integer vector $y$ satisfyingRational independence of the weights forces $y_j^2=1441_{j=i}$. Hence $A{e}_i=\pm e_i$ for every $i$: the rationally independent axis weights obstruct a lattice isometry by forcing it to be a diagonal sign change.

Such a sign change would preserve $T$ modulo three. The code has a unique nonzero projective word supported on each triple of coordinates, so preserving each of its four support sets forces the three signs on each triple to agree. All four signs therefore agree, leaving only $A=I$ or $A=-I$. Both preserve ${\Gamma }_{+}$ rather than exchanging it with ${\Gamma }_{-}$. These lattices are distinct: $(3,-1,-1,-1)\in {\Gamma }_{+}$ is odd and has sum zero, so is not in ${\Gamma }_{-}$. This contradiction provesThe choice of weights is essential; setting all four equal would make the displayed bases orthogonal with equal lengths and would not prove nonisometry.

Use the given integer $u$ satisfying $u^2\equiv -1(\mathrm{mod}p)$. The Euclidean lattice under consideration has basis $(1,u),(0,p)$, because every vector in it is uniquely $(a,ua+bp)$ with $a,b\in \mathbb{Z}$. Its fundamental parallelogram therefore has areaWe use the following two-dimensional form of the Minkowski convex body theorem: a convex, centrally symmetric measurable subset of ${\mathbb{R}}^2$ of area strictly greater than four times the covolume of a full-rank Euclidean lattice contains a nonzero vector of that lattice. Apply it to the closed disk of squared radius $3p/2$. Its area is $3\pi p/2>4p$, so it contains a nonzero lattice vector $(a,b)$ satisfyingOn the other hand the defining modular congruence and $u^2\equiv -1(\mathrm{mod}p)$ implyThere is exactly one positive multiple of $p$ strictly below $2p$, so $\boxed{p=a^2+b^2}$. Neither coordinate can be zero, since a prime is not the square of an integer. This congruence lattice proof of the prime sum of two squares proves the required case of the sum of two squares theorem without assuming any conclusion at the critical area $4p$.

Identify the pointed torus with ${\mathbb{R}}^2/{\mathbb{Z}}^2$ with $p$ at the origin. Its fundamental group is ${\mathbb{Z}}^2$, generated by the two coordinate loops. A homeomorphism fixing $p$ induces an automorphism of this group, givingFunctoriality makes it a homomorphism. It is nontrivial, indeed onto: every integer invertible matrix defines a pointed torus homeomorphism by $x\mapsto Ax$ modulo the euclidean lattice.

If $f_{\ast }=I$, lift $f$ to a continuous $F:{\mathbb{R}}^2\to {\mathbb{R}}^2$ with $F(0)=0$. Its deck-transformation law is $F(x+n)=F(x)+n$ for every $n\in {\mathbb{Z}}^2$, because the induced map on coordinate loops is identity. Thenis well defined: adding $n$ changes the bracketed expression by exactly $n$. It is a continuous pointed homotopy from $f$ to identity. Hence every kernel element is homotopic to the identity, although the intermediate maps need not be homeomorphisms.

A plane crystallographic group is a discrete subgroup of the Euclidean plane's isometry group with compact quotient; equivalently a compact set has translates covering the plane. We prove the crystallographic restriction theorem without assuming its conclusion. The orientation-preserving subgroup $G^{+}$ has index at most two and still has a cocompact group action. Its subgroup $T$ of translations is discrete as an additive subgroup of ${\mathbb{R}}^2$.

Suppose $T$ were trivial. Linear parts of orientation-preserving plane isometries commute, so every group commutator in $G^{+}$ is a translation. Thus $G^{+}$ would be an abelian group. Every member would then commute with the given nontrivial rotation $g$ and preserve its unique fixed point. A discrete subgroup of the compact rotation group about that point is finite. Then $G$ would be finite, contradicting cocompactness on the unbounded plane. Hence $T$ contains a nonzero vector $v$.

If $g$ is a half-turn its order is already two. Otherwise its linear part $R$ rotates $v$ to a linearly independent vector $Rv$, and conjugating translations shows that $Rv\in T$. Thus $T$ spans the plane and is a euclidean lattice. A discrete additive subgroup meets every bounded closed set in finitely many points: an accumulating sequence would give nonzero differences tending to zero, contradicting isolation of the identity. To see the euclidean lattice assertion directly, choose a shortest nonzero translation $v$. Translations on its line are integer multiples of $v$. Reducing all parallel components modulo $v$ shows that nonzero perpendicular components are bounded away from zero, since otherwise infinitely many translation vectors would accumulate in a compact rectangle. Choose a vector $w$ with the smallest positive perpendicular component; reducing by $w$ and then $v$ proves $T=\mathbb{Z}v+\mathbb{Z}w$.

Conjugation by $g$ preserves this euclidean lattice. In its euclidean lattice basis, $R$ is therefore an invertible integer matrix with determinant one. Its trace is an integer and is also $2\cos \theta$, where $\theta$ is the rotation angle. Since $g$ is nontrivial,These values give angles of order $2,3,4,6$ respectively. Hence

Use the following skew-symmetric matrix, whose columns reduce modulo three to four of the specified nonzero tetracode words:Set $B_{+}=3I+S$, $B_{-}=-3I+S$, and define the full Euclidean lattices $L_{\pm }^0=B_{\pm }{\mathbb{Z}}^4$. Their determinants have absolute value $144$, sinceFor arbitrary positive real numbers $a_1,\dots ,a_4$, put $D=\mathrm{diag}(\sqrt{a_1},\dots ,\sqrt{a_4})$ and $L_{\pm }=D{L}_{\pm }^0$. This is the tetracode length-isospectral lattice construction. We prove equality of the length multisets by an explicit bijection, not just equality of the shortest lengths.

Modulo three, both lattices have image $\mathrm{im}S=T$. The matrix $S$ has rank two over ${\mathbb{F}}_3$; its four column lines are precisely the four lines in $T$. Also $S^2=0$ modulo three, so $\ker S=\mathrm{im}S=T$. Each nonzero tetracode word has exactly one zero coordinate. Every $v\in L_{\pm }^0$ therefore has some coordinate divisible by three. Moreover, all four coordinates of $v$ have the same parity, because every entry of either $B_{\pm }$ is odd.

For $v\in L_{+}^0$, choose the least index $i$ with $3\mid v_i$ and set $w=v-2v_i{e}_i$. The coordinate reflection bijection for tetracode lattices will show that $w\in L_{-}^0$. First, $w\equiv v(\mathrm{mod}3)$, so $Sw\equiv 0(\mathrm{mod}3)$. Second, positive-lattice membership implies $Sv\equiv 3v\equiv -v(\mathrm{mod}4)$. HenceThe expression in parentheses has $i$th coordinate zero and every other coordinate $v_j+v_i{S}_{ji}$ even, by the common parity and $S_{ji}=\pm 1$. Thus $(I-S)w\equiv 0(\mathrm{mod}4)$. Together the congruences modulo three and four say $(-3I-S)w\equiv 0(\mathrm{mod}12)$, which is precisely $B_{-}^{-1}w\in {\mathbb{Z}}^4$.

Conversely, if $v\in L_{-}^0$, the same reflection has unchanged reduction modulo three, and now $Sv\equiv v(\mathrm{mod}4)$. ThusIts $i$th coordinate is $4v_i$ and its other coordinates are even multiples of two, so it vanishes modulo four; it vanishes modulo three as well. The reflected vector is in $L_{+}^0$. The chosen index is unchanged by reflection, including for vectors reducing to zero, so this map is an involution between the two lattices. It preserves every $|v_j|$ and therefore preserves $\Vert Dv{\Vert }^2={\sum }_j{a}_j{v}_j^2$. The two lattices have the same length spectrum, including every multiplicity, for every positive choice of weights.

To prove a genuine failure of isometry, chooseThese four numbers are linearly independent over $\mathbb{Q}$: changing the signs of the independent square roots in the multiquadratic field $\mathbb{Q}(\sqrt{2},\sqrt{3},\sqrt{5})$ isolates each coefficient of a rational relation. If an orthogonal transformation $O$ maps $L_{+}$ to $L_{-}$, thenand $P^T\mathrm{diag}(a_j)P=\mathrm{diag}(a_j)$. Comparing rational coefficients of the independent $a_j$ in the diagonal entries gives $P_{ji}^2={\delta }_{ji}$. Consequently $P$ is diagonal with entries $\pm 1$.

Such a coordinate sign change must preserve $T$ modulo three. For each zero coordinate $i$, there are exactly two opposite nonzero tetracode words. Preserving their line requires the other three signs to be equal. Taking two different zero coordinates forces all four signs to be equal. Thus $P=I$ or $P=-I$, either of which would imply $L_{+}^0=L_{-}^0$. ButThis contradiction proves $L_{+}$ and $L_{-}$ are not isometric. Allowing affine isometries changes nothing: translating the image of zero back to zero reduces them to the orthogonal case.

For the flat tori, take period lattices equal to the dual lattices:The standard spectrum of a flat torus lemma states that its characters of a real torus are indexed by the dual of its period lattice. Here they are $e^{2\pi i⟨v,x⟩}$ for $v\in L_{\pm }$, with Laplacian eigenvalue $4{\pi }^2\Vert v{\Vert }^2$. These characters give a complete orthogonal basis, so our length-preserving bijection proves that $X_{+}$ and $X_{-}$ are isospectral manifolds. The second standard lemma is that an isometry of flat tori lifts to an affine isometry of their Euclidean universal covers. Its orthogonal part maps the period lattices to one another, and therefore maps their dual lattices to one another. This would give the impossible isometry $L_{+}\cong L_{-}$. Hence $X_{+}$ and $X_{-}$ are isospectral non-isometric flat four-dimensional tori.

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 $d\ge 4$.

A plane plane crystallographic group is a discrete subgroup of the Euclidean isometry group with compact quotient. Its translations form a normal rank-two Euclidean lattice $\Lambda$; this is the standard translation-lattice structure of a plane crystallographic group. Its point group is the finite group of linear parts, equivalently the quotient by those translations. Conjugating a translation by an isometry sends its translation vector to its image under the linear part, so every point-group element preserves $\Lambda$.

Choose a Euclidean lattice basis. A rotational linear part is then represented by an integer matrix $A$, since it sends the basis to Euclidean lattice vectors. In an orthonormal basis the same map has eigenvalues $e^{i\theta },e^{-i\theta }$, and its trace is independent of basis. ThusThese possibilities give rotation angles, modulo a full turn, $\pi$, $\pm 2\pi /3$, $\pm \pi /2$, $\pm \pi /3$, and $0$, respectively. Consequently the crystallographic restriction theorem isThe restriction comes from integrality of the Euclidean lattice action, not simply from finiteness of the point group.