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A $t$-$(v,k,\lambda )$ combinatorial design is an incidence structure on $v$ points with blocks containing $k$ points, such that every $t$-element subset lies in exactly $\lambda$ blocks. Repeated blocks may be allowed; a simple design has no repetitions. With $b$ blocks, counting pairs consisting of a block and one of its $t$-subsets givesFor a two-design, each point lies in $r=\lambda (v-1)/(k-1)$ blocks.^[ref]

The seven triples $123,145,167,246,257,347,356$ on $\{1,\dots ,7\}$ form a $2$-$(7,3,1)$ design: every pair occurs once, and every point occurs three times. This is the Fano plane. The equation checks $7\cdot 3=1\cdot 21$. Such exact incidence balance makes combinatorial designs useful for arranging comparisons and constructing highly symmetric finite structures.

If $xyz$ is a hyperedge and the links of $x,y,z$ on four other vertices contain the three perfect matchings of a $K_4$, then those seven vertices contain a Fano plane. There is a second completion: if the link of $v$ contains three disjoint pairs, and the other six vertices contain all transversal triples through those pairs, take four triples of even parity and the three triples containing $v$. This again gives a Fano plane. These finite templates convert forbidden Fano copies into missing cross-part triples.

For all sufficiently large $n$, the extremal number of the Fano plane as a three-uniform hypergraph is the displayed count of all triples crossing a balanced bipartition. The plane has no Property B, so this construction avoids it. A stability argument makes near-extremal Fano-free hypergraphs nearly bipartite; vertex deletion obtains high minimum degree. Optimizing the approximate bipartition and applying Fano completion through three matchings then rules out all internal triples. The large-order qualification matters: for $n=5,6$ the complete hypergraph has no Fano copy but exceeds the displayed count.

There are 21 incident point-line pairs and 28 nonincident pairs in the Fano plane, with $G{L}_3(2)$ transitive on each type. Orbit-stabilizer theorem gives the two orders as $168/21=8$ and $168/28=6$.

A finite linear space consists of a finite set $P$ of points and a set family $\mathcal{L}$ of subsets called lines, each containing at least two points, such that each pair of distinct points lies on exactly one line. Here the nontrivial convention also excludes a line equal to all of $P$. Two distinct lines therefore meet in at most one point, and counting pairs givesNear-pencils and finite projective planes, including the Fano plane, are examples. The De Bruijn--Erdos pair-covering inequality concerns the corresponding point degrees. The original line-count theorem proves $|\mathcal{L}|\ge |P|$, with equality only for a near-pencil or a finite projective plane.^[ref]

An exactly one-intersecting family with empty total intersection gives such a structure by using its members as points. Each original ground element supplies the set of members containing it. Discard the resulting sets of size zero or one; the others are lines, and every pair of member-points belongs to its unique shared-element line. Empty total intersection excludes an all-point line.

For a finite ground set and an indexed set family, this zero-one matrix records membership of points in members. The diagonal entries of $B^{T}B$ are member sizes, and its off-diagonal entries count pairwise intersections. For the Fano plane, either point-line orientation gives the identity $B^{T}B=2I+J$, since every point lies on three lines and each distinct point pair lies on one line.

Put $T=\mathrm{PSL}(3,2)=\mathrm{GL}(3,{\mathbb{F}}_2)$, of order $(8-1)(8-2)(8-4)=168$. It acts on the seven points and seven lines of the Fano plane. Take $U_1$ to be a point stabilizer and $U_2$ to be a line stabilizer; both have index seven and order twenty-four. The fact that point and hyperplane stabilizers are almost conjugate follows by comparing fixed-point counts: for a matrix $g$, the numbers of fixed nonzero vectors and fixed nonzero covectors are bothbecause $g-I$ and its transpose have the same rank, and the dual action $g^{-T}$ has the same fixed space as $g^T$. Over ${\mathbb{F}}_2$, nonzero vectors represent points and nonzero covectors represent lines without any further scalar quotient. The two permutation characters therefore agree. The coset-character formulathen proves Gassmann equivalence. They are not conjugate: a point stabilizer has orbits of sizes one and six on points and hence preserves no three-point line. Its conjugates are again point stabilizers, whereas $U_2$ preserves its line. The point and line stabilizers form the required nonconjugate Gassmann pair.

We construct a family using the Lambert quadrilateral construction of a cone torus. Let $\alpha =\pi /14$. We use the elementary hyperbolic tri-rectangle fact that a quadrilateral with three right angles, remaining angle $\alpha$, and side lengths $x,y$ adjacent to the right-angle vertex opposite $\alpha$ exists uniquely forHere $x,y$ are the sides at the opposite right-angle corner, not the sides incident to $\alpha$. Given any $x>0$, choose $y=\mathrm{arsinh}(\cos \alpha /\sinh x)$. Reflect four copies around the opposite right-angle vertex. The result is a symmetric hyperbolic quadrilateral with all four corner angles $\alpha$ and opposite sides of equal lengths. Pair opposite sides by the axis translations. The quotient is an oriented torus with one cone point, whose total angle is $4\alpha =2\pi /7$. Its two central axis loops $a,d$ are embedded closed geodesics, of lengths $2x,2y$, intersecting once at right angles.

Cut this cone torus along $a$. The result is a hyperbolic pair of pants with two geodesic boundaries of length $\ell =2x$ and the cone point. Reglue those equal boundaries with a Fenchel–Nielsen twist $\tau \in \mathbb{R}$. This gives a two-parameter family $B_{\ell ,\tau }$, with the same cone angle, and marked simple classes $a,d$ meeting once. Near any zero-twist member their geodesic representatives remain embedded, transverse and disjoint from the cone point. The length of $a$ remains $\ell$, and the right-angle matrix product givesFor example, the zero-twist holonomies are $\mathrm{diag}(e^x,e^{-x})$ and $\left(\begin{array}{cc}\cosh y& \sinh y\\ \sinh y& \cosh y\end{array}\right)$; multiply the latter by $\mathrm{diag}(e^{\tau /2},e^{-\tau /2})$ to obtain the twist formula. Their commutator has trace $2-4{\sinh }^{2}x{\sinh }^{2}y=-2\cos (\pi /7)$, confirming exact elliptic order seven. The side-pairing construction supplies discreteness; the trace computation alone would not do so.

The orbifold fundamental group isSend $a$ to the supplied generator $A$ and $d$ to $D$. Their commutator $C=[D,A]$ has order seven, so this defines an epimorphism $\rho :\Gamma \to T$. Its kernel avoids all nontrivial cone stabilizers, which are conjugates of powers of $[d,a]$; therefore it is torsion-free. It defines a smooth closed regular hyperbolic cover $X_{\ell ,\tau }\to B_{\ell ,\tau }$ with deck group $T$. The subgroups $U_i$, of order twenty-four, also avoid every order-seven cone stabilizer. Thusare smooth closed seven-sheeted covers of the cone torus. The orbifold Euler characteristic is ${\chi }_{\mathrm{orb}}(B)=-(1-1/7)=-6/7$, givingEquivalently there is one preimage of the cone point, with local degree seven, and its lifted angle is $2\pi$. The Sunada theorem, in the free-subgroup-action form proved above, makes $S_1(\ell ,\tau )$ and $S_2(\ell ,\tau )$ isospectral for every parameter.

Nonisometry requires further work. A cycle of length $r$ in the monodromy permutation of a simple base geodesic gives a primitive simple lifted geodesic of length $r$ times the base length. From the supplied $A$ action, each cover has one lift of $a$ of length $\ell$ and two of length $3\ell$. From the supplied $D$ action, each has one lift of $d$ of each length ${\ell }_d,2{\ell }_d,4{\ell }_d$. Let ${\beta }_i$ be the unique two-cycle lift of $d$, and let ${\alpha }_{i,1},{\alpha }_{i,2}$ be the two three-cycle lifts of $a$. The fact that cycle intersections count intersections of lifted curves gives their intersection numbers: the seven crossings above the one base crossing are labelled by the sheets, so a crossing belongs to both lifted components exactly when its sheet lies in both cycles. ConsequentlyIndeed, in the first cover the two three-cycles have supports $\{1,2,5\},\{3,6,4\}$ and the two-cycle has support $\{0,3\}$; in the second they have supports $\{1,4,3\},\{2,5,6\}$ and $\{2,5\}$. The lifted curves intersect only over the single base crossing, so these are actual intersection counts of the geodesics, not merely abstract sheet-incidence numbers.

To make this a test against every isometry, we must identify these curves intrinsically, rather than require the isometry to respect the covering map. We use generic isolation of lifted simple geodesics, and give the needed genericity argument. Lengths of closed geodesics are real-analytic functions of the cone-torus length and twist parameters. For a primitive hyperbolic class $w$ and an integer $1\le r\le 7$, consider the equalitiesApart from $w=a,r=3$ or $w=d,r=2$, none is an identity on the full two-parameter family. To see this for the first equality, pinch the simple curve $a$ while keeping the cone angle fixed. The collar lemma makes every class crossing $a$ long. A class disjoint from $a$ lies in the cut pair of pants; its limiting holonomy has positive translation length unless it is a power of one of the new cusp loops, which are conjugate to $a$ in the torus group. Cone-peripheral classes are elliptic and do not represent closed geodesics. Thus the only primitive class with length tending to zero in this degeneration is $a$, up to inversion and conjugacy. This excludes an identity with a positive multiple of $\ell$. Pinching $d$ proves the corresponding assertion for ${\ell }_d$. The argument uses the usual hyperbolic collar and cusp degeneration facts, obtainable by the right-angle quadrilateral formulas; the cone angle $2\pi /7<\pi$ remains fixed. Analytic continuation on the connected length-twist parameter space shows that the same identities cannot hold on a small open neighbourhood of a zero-twist member.

There are countably many class equations, and each nontrivial real-analytic equation has a closed zero set with empty interior. The Baire category theorem therefore gives a parameter $({\ell }_0,{\tau }_0)$ in that small neighbourhood at which both target lengths are attained only by the prescribed lift components. This isolation persists on a smaller open neighbourhood. Indeed, near a fixed compact hyperbolic metric there are uniformly only finitely many primitive geodesic classes below a fixed length bound, by discreteness of the covering group and a compact fundamental region; nearby metrics are uniformly bilipschitz. Only finitely many classes can therefore approach the two fixed finite target lengths. Their strict inequalities persist under a sufficiently small parameter change.

Now keep $\tau ={\tau }_0$ and let $\ell$ vary in a sufficiently small interval $I$ around ${\ell }_0$. For each $\ell \in I$, the only primitive geodesics of length $3\ell$ in either cover are the two $\alpha$ components, and the only primitive geodesic of length $2{\ell }_d$ is $\beta$. An isometry would have to preserve these length classes and their intersection-count multiset, contradicting $\{0,1\}\ne \{0,2\}$. Henceis an interval-parametrized family of isospectral nonisometric hyperbolic Riemann surfaces of genus four. Shrink $I$ again so a small length window around $3{\ell }_0$ contains only those two components for every parameter. Their common length $3\ell$ varies strictly, so members at distinct parameters cannot all be the same surface in different markings. The spectra agree within each pair; this construction does not assert a nontrivial continuously isospectral deformation of one hyperbolic surface.

The unrestricted identity needs qualification. A Fano plane has seven vertices of a hypergraph, so the complete three-uniform hypergraph on six vertices of a hypergraph avoids it and has twenty edges, whereas the displayed balanced-part count is eighteen. On five vertices of a hypergraph the corresponding numbers are ten and nine. Thus the intended extremal theorem is the sufficiently-large-$n$ Fano-plane Turán theorem. We describe its stability proof, including the exact counting step that upgrades an asymptotic result to equality.

Identify the Fano points with the nonzero vectors of ${\mathbb{F}}_2^3$, and its hyperedges with triples summing to zero. It has no Property B. Indeed, a line-free set has at most four points: if a set $A$ of nonzero vectors contains no zero-sum triple and $a\in A$, the sets $A\setminus \{a\}$ and $a+(A\setminus \{a\})$ are disjoint subsets of the six nonzero vectors other than $a$, so $|A|\le 4$. Moreover, if $|A|=4$, those two three-element sets partition those six vectors. Write $A=\{a,b,c,d\}$. The sum $b+c$ lies outside $A$, so belongs to $a+(A\setminus \{a\})$. It is neither $a+b$ nor $a+c$, and must therefore equal $a+d$. Thus $a+b+c+d=0$. The other three points have sum zero because the sum of all seven nonzero vectors is zero. They therefore form a line. In any two-colouring one colour has at least four points: if it contains a line we are done, and otherwise the other colour contains a line. This proves non-two-colourability.

Partition the host vertices of a hypergraph into sizes $a,b$ and take all triples meeting both parts. This bipartite three-uniform hypergraph has Property B and is consequently Fano-free. Its edge count isThe product $ab$ is maximized at balanced integer sizes. Writingwe have the exact lower bound $\mathrm{ex}(n,F)\ge b(n)$, and $b(n)=(3/4+o(1))\left(\genfrac{}{}{0ex}{}{n}{3}\right)$.

The structural step is Fano stability: a Fano-free hypergraph with density sufficiently close to $3/4$ has a bipartition with at most $\rho n^3$ internal triples, for any prescribed $\rho >0$. Its proof studies four hypergraph link graphs around a complete four-vertex three-uniform configuration. Colour a pair by the link vertices of a hypergraph containing it. Three perfect matchings in a graph on four other vertices of a hypergraph, with distinct colours corresponding to a triple of the configuration, would complete a Fano copy. Thus these rainbow patterns are forbidden. The near-extremal total link degree forces, after removing a small exceptional set, an almost balanced pattern with pair multiplicity two within the two classes and four between them. Internal triple density bounded away from zero produces many complete tripartite six-vertex templates; the nearly complete links then supply three paired neighbours and complete a Fano plane. Consequently the internal triple density tends to zero. These are the main ingredients of the stability argument: link degree counting supplies the approximate partition, and the forbidden completion templates eliminate positive internal density.

To see concretely why stability gives the exact bound, suppose $e(H)\ge b(n)$, and first obtain high minimum hypergraph vertex degree. Delete any vertex of degree at most $3m^2/8-2m$ when the current order is $m$. Directly from the formula for $b$, for large $m$,After deletions down to order $m$, the remaining hypergraph therefore has at least $b(m)+n-m$ edges. It cannot reach $m=\lceil n^{1/3}\rceil$, since this exceeds $\left(\genfrac{}{}{0ex}{}{m}{3}\right)$ for large $n$. The process stops at an order still tending to infinity, with minimum degree of a graph greater than $3m^2/8-2m$. If the high-minimum-degree case has at most $b(m)$ edges, any deletion gives a contradiction. Thus it suffices to treat that case; rename its order $n$.

Apply stability with a very small constant, for example $\rho =10^{-12}$, and choose a partition $A\bigsqcup B$ minimizing the number of internal hyperedges. Its part sizes are close to $n/2$, since its cross-part capacity is $|A||B|(n-2)/2$, and $e(H)\ge b(n)$ leaves at most $\rho n^3$ available compensation for any imbalance. In particular $||A|-n/2|,||B|-n/2|\le 2\sqrt{\rho }n$ for large $n$. Optimality also giveswith the analogous inequality for $x\in B$: otherwise moving $x$ would reduce the internal count.

Suppose some $x\in A$ has more than $n^2/100$ edges in its same-side link. Its opposite-side link then has at least as many. Each link has a matching in a graph of size at least $\lfloor n/300\rfloor$: the endpoints of a maximal matching in a graph cover every edge, so a matching in a graph of size $s$ bounds the link edge count by $2sn$. For each choice of one matching in a graph edge in $A$ and two matching in a graph edges in $B$, the six vertices of a hypergraph plus $x$ would form a Fano plane if all transversal triples through the three pairs were present. Specifically take the three triples through $x$ and the four even-parity transversal triples. Hence at least one of these cross-part triples is missing. The matchings in a graph are disjoint, so different choices give different missing triples. There are at leastFor large $n$ this exceeds $\rho n^3$, contradicting the edge count: the missing cross triples already outweigh every internal triple. Thus every same-side link has at most $n^2/100$ edges.

If an internal triple $xyz\subseteq A$ remains, high minimum degree of a graph gives, for each of its vertices of a hypergraph,The intersection of the three opposite-side links therefore has at least their total edge count minus $2\left(\genfrac{}{}{0ex}{}{|B|}{2}\right)$. Using $|B|\le (1/2+2\sqrt{\rho })n$, this exceeds $|B{|}^2/3$ for large $n$. By the quadratic Turan theorem bound, their intersection contains $K_4$. Its three perfect matchings in a graph, assigned to $x,y,z$, together with $xyz$ are the seven edges of a Fano plane. This is precisely Fano completion through three matchings, and gives a contradiction. The same argument excludes internal triples in $B$.

Thus the high-minimum-degree hypergraph is genuinely bipartite, not merely approximately so, and has at most $b(n)$ edges. The deletion argument proves the same bound for the original hypergraph. Combining with the balanced construction yieldsThis proof description keeps the essential roles distinct: non-two-colourability gives the lower construction, link-graph stability gives an approximate partition, and the minimum-degree/matching in a graph argument removes the last internal edges. The ceiling in the final term is confirmed by the PDF; the converted TeX incorrectly repeats the floor.

We construct the two domains explicitly from seven congruent triangles. Let a scalene reference triangle have sides $a,b,c$ opposite its three labelled vertices. Choose its three side lengths in a small open neighbourhood of $(1,1,1)$, with a fixed strict ordering. Thus the three angles are distinct and lie between $\pi /4$ and $\pi /2$. Each copy keeps these side labels.

For the first assembly glue the following tile pairs, reflecting across the indicated side:Each coloured adjacency graph is a tree with central tile $7$ and three two-tile arms. Place tile $7$ in the plane and obtain every other tile by the prescribed reflections. At the equilateral reference shape the copies have disjoint interiors and form a polygonal disk; the same holds throughout a sufficiently small neighbourhood, since the only contacts between tiles are the prescribed edges and their incident vertices. In particular each assembly is a bounded connected planar domain.

To prove equality of spectra, index tiles by the seven nonzero vectors $x\in {\mathbb{F}}_2^3$, using $x=(x_1,x_2,x_3)$ with label $x_1+2x_2+4x_3$. LetThese involutions act on points by $x\mapsto s_{s}x$ and on nonzero covectors by $y\mapsto s_s^{-T}y=s_s^{T}y$. Their swaps are exactly the two columns of the table. Write $P_s,Q_s$ for the corresponding permutation matrices. The Fano plane incidence matrixsatisfies $B{P}_s=Q_{s}B$, since incidence is preserved by a linear transformation and its inverse transpose. Each point lies on three lines, while two different points have one common line. Thereforewhere $J$ is the all-ones matrix, proving that $B$ is invertible.

For the Dirichlet boundary condition, a glued side has a positive swap and an exposed side has diagonal entry $-1$. Both trees have the same bipartition, with sign matrixTheir signed reflection matrices are $M_s=-D{P}_{s}D$ and $N_s=-D{Q}_{s}D$. On a glued edge its endpoint signs are opposite, so the off-diagonal entries are $+1$; on an exposed edge the entry is $-1$. Consequently the invertible matrix $C=DBD$ satisfies

We spell out why this gives isospectrality. Pull each function back from the seven tiles to the reference triangle. Its seven boundary values along side $s$ form a vector $u_s$ satisfying $M_s{u}_s=u_s$: this imposes equal values at glued edges and zero values at exposed edges. Outward normal derivatives of a smooth eigenfunction satisfy $M_s{v}_s=-v_s$, giving opposite normal derivatives on glued edges. The identities above preserve both conditions under $u\mapsto Cu$. Since $C$ is constant, it also commutes with the scalar Laplacian on each triangle.

One can include all weak eigenfunctions without any corner regularity issue by orthogonalization of a transplantation matrix. The matrix $C^{T}C$ commutes with the symmetric $M_s$, so $O=C(C^{T}C{)}^{-1/2}$ is orthogonal and still intertwines $M_s,N_s$. On the disjoint tiles, $O$ preserves the $L^2$ norm and the sum of the Dirichlet energies; it maps the trace-matching form domain $H_0^1({\Omega }_P)$ bijectively to $H_0^1({\Omega }_L)$. Thus it is a unitary equivalence of the Dirichlet Laplacians, proving equality of every eigenvalue with its multiplicity.

Finally, the only reentrant vertices of either polygon are the three vertices of tile $7$, with interior angles four times the corresponding reference-triangle angles. All other boundary angles are less than $\pi$. These three distinct angles force any hypothetical isometry to map each labelled central vertex to its counterpart. An isometry between connected open planar domains is a restriction of a rigid motion: its derivative is orthogonal, and preservation of the flat Levi-Civita connection makes that derivative constant. After aligning the two central triangles, this rigid motion fixes three noncollinear points and is the identity. But the two domains do not coincide: the tile beyond side $a$ of the central triangle has its next attachment along $b$ in the point assembly and along $c$ in the line assembly. These occupy different open regions. This contradiction proves nonisometry.

Therefore the three freely varying side lengths give a three-parameter family of nonisometric Dirichlet-isospectral bounded planar domains. The parameters include scale; no congruence normalization removes it.

We construct a family by seven-triangle Dirichlet transplantation. Start with a scalene triangle $\tau$ sufficiently close to an equilateral triangle. Its three side lengths are the three independent parameters; restrict to an open set where its angles $\alpha ,\beta ,\gamma$ are distinct and lie between $\pi /4$ and $\pi /2$. Label the sides $a,b,c$ opposite the corresponding vertices $A,B,C$.

For each of two planar domains, begin with tile $7$ and attach the remaining congruent copies by Euclidean reflection across the indicated sides. The attachment tables areFor example, $(6,7)$ in the first $a$ row means to reflect tile $7$ across its $a$ side to obtain tile $6$. Both attachment graphs are trees: there is a central tile with three arms, each consisting of two tiles. Every side not used in the table is an exterior boundary side.

At the equilateral triangle, these seven reflected tile interiors are disjoint; at each central vertex four sectors of angle $\pi /3$ meet, leaving an exterior gap $2\pi /3$. Other boundary vertices have angles $\pi /3$ or $2\pi /3$, and distinct nonincident pieces are separated. The positive angular gaps and separations persist under sufficiently small perturbations of the three side lengths. Thus this construction really gives connected polygonal planar domains on an open three-dimensional parameter set. Their only reentrant vertices are the central vertices $A,B,C$, with interior angles $4\alpha ,4\beta ,4\gamma$; all other boundary angles are one or two triangle angles and are less than $\pi$.

Because these three reentrant angles are distinct, an isometry of the two planar domains would send each labelled central vertex to its counterpart. A planar isometry is the restriction of a Euclidean rigid motion: its differential preserves the Euclidean connection, hence is constant on the connected domain. After aligning the central triangles, such a motion fixes three noncollinear points and is the identity. But the domains are not identical. On the arm attached to side $a$, the point construction attaches the last tile across side $b$, whereas the line construction attaches it across side $c$. The former side is therefore interior in one domain and exterior in the other. This proves the two domains are nonisometric throughout this three-parameter family.

We now prove equality of the Dirichlet Laplacian spectra, including multiplicities. Index $1,\dots ,7$ by the nonzero binary vectors $x=(x_1,x_2,x_3)$, with integer label $x_1+2x_2+4x_3$. The transformationsare involutions over ${\mathbb{F}}_2$. Their point permutations $P_s$ are precisely the transpositions in the left attachment table, with the remaining points fixed. On covectors the inverse-transpose transformations give permutations $Q_s$ with exactly the right table. Let $B$ be the point-line incidence matrix of a set system of the Fano plane:The incidence relation is invariant under transforming $x$ and its covector $y$ contragrediently, so $B{P}_s=Q_{s}B$ for all three sides. Each point lies on three lines, and two distinct points lie on exactly one common line. Consequentlywhere $J$ is the all-ones matrix. Its eigenvalues are $9$ once and $2$ six times, so $B$ is invertible.

The Dirichlet boundary condition requires a sign change at an unglued side. Give each tile the sign $(-1{)}^{\text{distance from tile }7}$ and putThe signs alternate across each glued side in both domains. The signed side matrices are thereforeThey have entry $+1$ across a glued pair and entry $-1$ on a fixed exterior tile. The invertible seven-triangle Dirichlet transplantation matrix $C=DBD$ satisfiesThis is an actual Dirichlet boundary condition intertwiner, not merely the unsigned incidence relation.

To justify the spectral conclusion even at the reentrant corners, use the closed Dirichlet energy form. Pull a function on either domain back to its seven reference triangles and write its components as a vector $F$. The form domain consists of $H^1$ components whose side traces agree on glued pairs and vanish on exterior sides; equivalently $F=S_{s}F$ on each side for the point domain, and $G=T_{s}G$ for the line domain. These trace conditions exactly characterize $H_0^1$ of the glued polygon. Both its $L^2$ norm and its Dirichlet energy are the sums of their seven reference-tile integrals.

Since the signed side matrices are symmetric, the intertwining identities also give $S_s{C}^{T}C=C^{T}C{S}_s$. Thus $(C^{T}C{)}^{-1/2}$ commutes with every $S_s$, and the polar factoris an orthogonal matrix with $U{S}_s=T_{s}U$. Multiplication of the component vector by this constant matrix preserves the $L^2$ inner product and the integrated squared gradients, and maps the two form domains bijectively. The self-adjoint operators associated with these forms are consequently unitarily equivalent, so their eigenvalues and multiplicities agree. The eigenfunctions are smooth in the interiors; no unwarranted smoothness at polygonal corners is required. We have proved

The Fano plane has as points the seven nonzero vectors of ${\mathbb{F}}_2^3$, because every one-dimensional vector subspace has just one nonzero vector. Its lines are the seven two-dimensional vector subspaces. Every line has three points, every point lies on three lines, two points determine the line $\{u,v,u+v\}$, and two lines meet in one point.

The two order-24 classes are therefore stabilizer subgroups of points and of lines. Each is isomorphic to $S_4$: fixing a vector gives the block group ${\mathbb{F}}_2^2⋊G{L}_2(2)$, the full affine group on four points; inverse transpose gives the dual line stabilizer subgroup. Their distinct fixed-point behavior in part (i) distinguishes the two conjugacy classes.

For a point $p$ and line $L$, $G_p\cap G_L$ stabilizes the ordered pair $(p,L)$. There are $7\cdot 3=21$ incident pairs and $7\cdot 4=28$ nonincident pairs. A change of basis takes any pair to a fixed model of its type, so each type is a single orbit. Orbit-stabilizer theorem gives the Fano point-line stabilizer intersection:Both occur: take $L=⟨e_1,e_2⟩$ and choose $p=⟨e_1⟩$ or $p=⟨e_3⟩$.

For the extension, identify vectors and covectors by the standard pairing. The inverse-transpose automorphism exchanges a point with its orthogonal plane and exchanges the two order-24 classes. Thus it is outer. The Fano plane duality group $\overline{G}=G⋊⟨\tau ⟩$ has order 336 and acts on points and lines, preserving incidence while allowing their types to be swapped.

The stabilizer subgroup of an unordered point-line pair has twice the corresponding ordered-pair stabilizer subgroup: some element outside $G$ swaps the pair, since duality produces a pair of the same incidence type and $G$ is transitive on such pairs. Hence these stabilizer subgroups have orders 16 and 12. Also every outside conjugate of a Sylow 7-subgroup can be moved back by an element of $G$, giving

We prove these exhaust the maximal subgroups other than $G$. Let $K$ be such a maximal subgroup and $H=K\cap G$. Since $K$ is not contained in $G$, $H$ is normal in $K$ and $|K:H|=2$. If $7\mid |H|$, part (i) gives a unique Sylow 7-subgroup $P$ of $H$. It is characteristic in $H$, hence normalized by $K$. Therefore maximality forces $K=N_{\overline{G}}(P)$.

Suppose $7\nmid |H|$ and $H\ne 1$. The classification argument in part (i) says $H$ fixes a point or a line. Because $H$ is normal in $K$, an element outside $G$ swaps its fixed-point and fixed-line sets, so it fixes at least one of each. WriteFixed lines are kernels of fixed nonzero covectors, since the only nonzero scalar is one. Duality gives $2^d-1=2^{d^{\ast }}-1$, hence $d=d^{\ast }\in \{1,2\}$.

If $d=1$, there is a unique fixed point and unique fixed line; $K$ preserves their unordered pair. If $d=2$, put $W=V^H$ and $A=((V^{\ast }{)}^H{)}^{\perp }$. For every $h\in H$, $\ker (h-I)$ contains the plane $W$ and $\mathrm{im}(h-I)$ lies in the line $A$. Over ${\mathbb{F}}_2$ there is only one nonzero linear map $V/W\to A$, so $H$ has order two. Invertibility forces $A\subseteq W$: for $h=I+vf$ one needs $f(v)=0$. Thus $A$ and $W$ are its unique image point and kernel line. These are canonically swapped by duality, so $K$ again preserves their unordered incident pair.

If $H=1$, then $K$ has order two, generated by an involutory duality $u$. For any point $p$, the pair $\{p,u(p)\}$ is preserved by $u$. Hence $K$ lies in an order-12 or order-16 pair stabilizer subgroup and is not maximal. We have exhausted all cases.

The pair stabilizer subgroups are themselves maximal: a proper maximal overgroup is one of the just-listed types, and divisibility excludes orders 12 inside 16 or 42, and 16 inside 12 or 42. The order-42 normalizer is maximal by the unique-Sylow argument. Incidence types give one conjugacy class each; Sylow conjugacy gives one normalizer class. Thus, apart from the normal index-two subgroup $G$,

A useful precise form of the transplantation theorem is as follows. Take $N$ congruent copies of a Euclidean tile with labelled flat faces. For each face colour $s$, encode the gluings by a symmetric signed permutation matrix $M_s$: a glued pair gives off-diagonal entries $+1$ in the two corresponding positions; an unglued face gives a diagonal entry $-1$ for a Dirichlet boundary condition, or $+1$ for a Neumann boundary condition. Corresponding face identifications must use the same maps in the tile coordinates. Suppose two assemblies have matrices $M_s$ and $M_s^{\prime }$ and there is an invertible matrix $C$ withThen, for every eigenvalue $\lambda$, $f=(f_1,\dots ,f_N)\mapsto Cf$ is an isomorphism between their Laplacian eigenfunction spaces. The two assemblies are isospectral, including multiplicities.

To see why, pull every tile restriction back to the same Euclidean tile. A boundary value vector $u$ and an outward normal derivative vector $v$ satisfyAt an internal face these are continuity of the function and equality of oppositely directed derivatives. At a diagonal $-1$ they impose $u_i=0$; at a diagonal $+1$ they impose $v_i=0$. The intertwining identities preserve both equations, and the Euclidean Laplacian commutes with constant linear combinations of tile functions. The resulting piecewise function is a weak eigenfunction, with the prescribed boundary conditions, and interior elliptic regularity makes it smooth wherever the assembled manifold is smooth. At polygonal corners one uses the usual weak domain of the Dirichlet Laplacian or Neumann Laplacian, rather than assuming smoothness through a corner. The inverse matrix gives the inverse transplantation.

Here is a completely specified propeller pair. Label the seven triangles $0,\dots ,6$, with central triangle $0$, and colour the three sides $a,b,c$. In the left and right assemblies the internal-edge pairings areEvery pair means reflection of one triangle across the indicated side; the other faces are external boundary. Each assembly has a central triangle and three two-triangle arms. These are the propeller domains. The drawing uses one acute scalene triangle, and each colour always denotes the side opposite the same marked vertex of the starting triangle.

Let $P_s,Q_s$ be the ordinary permutation matrices obtained by adding fixed points to the pairings. The following matrix supplies the Neumann boundary condition transplantation:There is a short way to construct and check it. Start with the support $\{1,2,4\}$ in row $0$, and require the support in row $Q_s(i)$ to be $P_s$ applied to the support in row $i$. Traversing the six edges of the right-hand tree gives exactly the displayed rows; checking the remaining fixed-point relations verifies $N{P}_s=Q_{s}N$ for all three colours. Every row has three ones and every two different rows meet in exactly one column, soThe right-hand side has eigenvalues $9$ once and $2$ six times, proving $N$ is invertible, with $|\det N|=24$. Its incidence pattern is that of the Fano plane, which explains the uniform intersections.

For the Dirichlet boundary condition, put $D=\mathrm{diag}(1,-1,-1,1,-1,1,1)$. In both assemblies this gives opposite signs to adjacent triangles. Thus their signed face matrices are $M_s=-D{P}_{s}D$ and $M_s^{\prime }=-D{Q}_{s}D$: the off-diagonal gluing entries are positive and the unpaired diagonal entries are negative. The matrixthen satisfies $C{M}_s=M_s^{\prime }C$ and is invertible. The transplantation theorem proves equality of the Dirichlet Laplacian spectra; $N$ proves equality of the Neumann Laplacian spectra as well.

Finally vary the three labelled side lengths $(a,b,c)$ of the initial triangle in an open region of strict triangle inequalities, with an acute scalene shape sufficiently close to an equilateral triangle. Reflection constructs both domains continuously from these three independent real parameters. At the equilateral configuration the tiles have disjoint interiors. Choose a sufficiently small neighbourhood of that configuration with every triangle angle ${\alpha }_i$ in $(\pi /4,\pi /2)$. At each central vertex exactly four tile angles meet, forming a sector of angle $4{\alpha }_i<2\pi$, while nonincident pieces stay separated. Thus there is an open three-dimensional parameter region in which both assemblies are embedded planar domains.

For a scalene triangle in this region, the three central vertices are exactly the three reflex boundary vertices: their interior angles are $4{\alpha }_i$, while the other boundary vertices have angles ${\alpha }_i$ or $2{\alpha }_i<\pi$. Hence a planar isometry of the domains must map the central triangle onto the central triangle. Put those two central triangles in the same position. Their unequal side lengths force any such isometry to fix the central triangle pointwise. Yet the first arm across, for example, side $a$ attaches its outer triangle across colour $c$ in $L$, and across colour $b$ in $R$. These occupy different positions, so the identity does not identify the domains. An intrinsic Riemannian isometry between connected planar domains also restricts locally to a rigid Euclidean motion and hence is one globally. The two propellers are nonisometric throughout this open three-parameter family. Scaling is included as one of the parameters; modulo scaling the family has two parameters.

Take seven copies of a scalene triangle, with sides labelled $a,b,c$ opposite its three vertices, and reflect adjacent copies across their glued side. For the first planar domain, the glued pairs are $a:(2,3),(6,7)$, $b:(4,6),(5,7)$, $c:(1,5),(3,7)$; for the second they are $a:(1,3),(5,7)$, $b:(2,6),(3,7)$, $c:(4,5),(6,7)$. Both coloured trees have central tile $7$. For triangles sufficiently close to equilateral the interiors do not overlap. If its three angles are distinct and between $\pi /4$ and $\pi /2$, the only reentrant vertices are the original central vertices, with angles four times the corresponding triangle angles. They force any proposed isometry to preserve the labelled central triangle, which is incompatible with the different attachments on its arms.

For the spectral proof index the nonzero vectors and covectors of ${\mathbb{F}}_2^3$ by $1,\dots ,7$, using binary coordinates. The Fano plane incidence matrix is $B_{y,x}=1_{y\cdot x=0}$. It satisfies $B^{T}B=2I+J$, so is invertible, and intertwines the point and line permutations of each of the three elementary transvections. Put $D=\mathrm{diag}(1,1,-1,1,-1,-1,1)$. The signed reflection matrices for the Dirichlet boundary condition are $-D{P}_{s}D$ and $-D{Q}_{s}D$, hence $C=DBD$ intertwines them. Its polar orthogonal factor gives a unitary equivalence of the Dirichlet Laplacians, preserving both the $L^2$ inner product and the Dirichlet energy of the piecewise functions. This produces a three-parameter family of noncongruent isospectral planar domains.