On the Shape of a Set In Honour of Lino Di Martino Topics in Groups and their Representations Villa Feltrinelli October 2017 Johannes Siemons, UEA Norwich
→ → ← ← GEOMETRIES ← → GROUPS F − MODULES 1 Mark Kac: The Shape of a Drum 2 Shape in Graphs 3 Some Properties of Shape 4 Incidence Structures 5 Hyperoctahedron
1: Can one hear the Shape of a Drum? Mark Kac posed this question in a 1966 paper. It has over 200 citations and has received several AMS prizes. From the sound of a drum, can we determine its shape? The question remained open for over 25 years. The membrane of a drum D in the ( x, y )-plane is displaced in the z -axis by the amplitude u ( x, y, t ) . The displacement is governed by the wave equation ∂ 2 u ∂t 2 = c 2 � ∂ 2 u ∂x 2 + ∂ 2 u = c 2 ∇ 2 · u = c 2 ∆ · u � ( ∗∗ ) ∂y 2 where ∆ = ∇ 2 is the Laplace operator. Use the boundary condition u ( x, y, t ) = 0 = ˙ u ( x, y, t ) on the rim ∂D of D, and the initial deformation of the membrane by the drum stick as u ( x, y, 0) .
Solving (**) is a standard problem in mechanics: (1) Separate variables, writing u ( x, y, t ) = f ( t ) · w ( x, y ) . (2) Use the standard inner product on the function space, �� � � := f ( x, y ) · g ( x, y ) dx dy. f, g Crucially , the Laplace operator becomes self-adjoint. That is, � � � � ∆ f, g = f, ∆ g for all f and g. (This is due to the boundary condition on ∂D. ) (3) Find the eigenvalues λ k and eigenfunctions w k of ∆ , so ∆ w k = λ k w k for k = 1 , 2 , 3 , .... ∈ N . Finally
(4) Solve (**) for f ( t ) for the eigenvalues λ k , getting f k ( t ) = sin( λ k t ) , and f k ( t ) = cos( λ k t ) . Finally, find the Fourier coeffi- cients a k ∈ R from the initial conditions, writing � u ( x, y, t ) = f ( t ) · w ( x, y ) = a k · f k ( t ) · w k . k ∈ N Kac’s question: Do the λ k determine the contour of D ? The first counter example is due to Gordon, Webb and Wolpert 1992, ’One cannot hear the shape of a drum’
There is a more sophisticated version of Kac’s question: In � u ( x, y, t ) = a k · f k ( t ) · w k , k ∈ N suppose we also know the Fourier coefficients a 1 , a 2 , ..., a k , ... for some suitably chosen functions u ( x, y, t ) . Do the λ k together with the Fourier coefficients a k de- termine the shape of D – for suitably chosen functions u ( x, y, t ) ? This question can be made precise - and is open.
2: Something similar happens in graphs Let Γ be a graph on the vertex set V = { v 1 , ..., v n } . (Assume that Γ is undirected, simple & loop-less.) Write v ∼ u if v is adjacent to u. We imitate what happened in the drum problem. A function f : V → R can be thought of as a formal sum f = � v ∈ V a v v where a v = f ( v ) ∈ R . So let � � � R V := f = with a v ∈ R a v v v ∈ V be the R -vector space of such formal sums, its basis is V. This is the vertex module of Γ . The adjacency in Γ gives rise to the linear adjacency map α : R V → R V defined by � α ( v ) := u . v ∼ u ∈ V (The matrix of α is the adjacency matrix of Γ . )
The vector space R V has the natural inner product, given by ( v, u ) := 1 , if u = v, and = 0 otherwise. Crucially, α is self-adjoint , that is, � � � � α ( v ) , u = v , α ( u ) for all v, u in V. The correspondence now is evident: ↔ ( x, y )-plane of the drum membrane , V ↔ u ( x, y ) amplitude of membrane f ∈ R V α ↔ Laplace Operator ∆ , and so on.
Denote the eigenvalues of α by λ 1 , ..., λ r ∈ R and let the corresponding eigenspaces be E 1 , ..., E r . In particular, any element f ∈ R V has a decomposition f = f 1 + ... + f r � � with spectral components or Fourier components f i ∈ E i . Each f i is an eigenvector of α, unless f i = 0 , and f determines the f i uniquely. Call || f 1 || 2 , ..., || f r || 2 � � sig( f ) := the spectral signature of f.
The function sig: R V → R r can be applied in particular to a single vertex. For instance, if v ∈ V, with v = ( v ) 1 + ( v ) 2 + ... + ( v ) r and ( v ) i ∈ E i , then || ( v ) 1 || 2 , || ( v ) 2 || 2 + ... + || ( v ) r || 2 � � sig( v ) = Shape can be defined for sets of vertices as is the shape of v. well: If S = { u, v, ..., w } ⊆ V consider the characteristic function of S, denoted by f S := u + v + ... + w = f = f 1 + f 2 + ... + f r . Then we call | S | − 1 · sig( f S ) = | S | − 1 · � || f 1 || 2 , ..., || f r || 2 � the shape of S.
Example 1: Let Γ = C 4 ∪ K 1 . The spectrum of Γ is 2 1 , 0 3 , − 2 1 . Any C 4 -vertex has shape ( 1 4 , 1 2 , 1 4 ) while the K 1 -vertex has shape (0 , 1 , 0) . Let Γ ∗ be the star on 5 vertices. Its spectrum is Example 2: 2 1 , 0 3 , − 2 1 . The central vertex has shape ( 1 2 , 0 , 1 also 2 ) while the ray-vertices all have shape ( 1 8 , 3 4 , 1 8 ) .
In Example 1 we had Γ = C 4 ∪ K 1 . For instance, if S = { u, v } we need to determine the components of f S = u + v = f 1 + f 2 + f 3 . Accordingly S may consists of • two opposite cycle vertices. Here S has shape ( 1 2 , 0 , 1 2 ); • two adjacent cycle vertices. Here S has shape ( 1 2 , 1 2 , 0); • a cycle vertex and the isolated vertex. Here S has shape ( 1 8 , 3 4 , 1 8 );
Summarising: The eigenspace decomposition of R V relative to the adjacency map α : R V → R V (with t distinct eigenvalues) gives rise to the shape function 2 V − → R t via shape( S ) = | S − 1 | · sig( f S ) . It is not difficult to show that if g is an isomorphism Γ → Γ ′ of two isospectral graphs then shape( S g ) = shape( S ) , for any set S of vertices of Γ . In particular, shape is invariant under graph isomorphism. How good is this invariant? → v (Γ ′ ) is a bijection of Kac’s Question again: If g : V (Γ) − the vertices of two isospectral graph, such that S and S g have the same shape, for all S ⊆ V, is g a graph isomorphism? Shape is very closely related to Delsarte’s inner distribution for association schemes. But we emphasize, shape is defined without any assumption on the graph.
3: Some Properties of Shape Let λ 1 , ..., λ r be the eigenvalues of α : R V → R V, and denote the corresponding eigenspaces by E 1 , ..., E r . For i = 1 , .., r let π i be the orthogonal projection π i : R V → E i . So r π 2 � i = π i , π i ◦ π j = 0 when i � = j, and π i = id . i =1 Any element f ∈ R V has spectral decomposition f = f 1 + ... + f r with f i = π i ( f ) ∈ E i . . For instance, we have the inequality ( f , f i ) = ( f i , f i ) ≥ 0 . This is Delsarte’s linear programming bound when the graph comes from an association scheme.
PROPOSITION 1: Let Γ be a graph with adjacency map α : R V → R V, eigenvalues λ 1 , ..., λ r and eigenspaces E 1 , ..., E r . Then (1) The projection map π i : R V → E i is a polynomial in α of degree r − 1 , with coefficients determined by the λ i . (One can write it down!) (2) Let v = f 1 + ... + f r be a vertex of Γ , where f i ∈ E i . Then || f i || is determined by the number of closed walks at v of length ≤ r − 1 , for all j = 1 , .., r. In particular, if Γ is regular and r ≤ 3 then sig( v ) = sig( w ) for all v, w ∈ V. (3) Let v = f 1 + ... + f r be a vertex of Γ where f i ∈ E i . Suppose that f j = 0 for some j. Then the automorphism group of Γ is not transitive on vertices.
Some Comments: (2) shows that Shape is constant on the vertices of a strongly regular graph. But consider the shape of other sets of vertices! (The smallest pair of non-isomorphic isospectral srgs has pa- rameters (16 , 6 , 2 , 2) . It is the Shrikhande and the line graph of K 4 , 4 . ) Open Question: Is Shape a complete invariant for graphs with all eigenvalues distinct? (Probably not.) Vertex Transitivity: In many computational examples (3) turns out to be a good necessary condition for vertex transitivity. (It is unlikely to be sufficient in general.)
4: Incidence Structures Let X and Y be finite sets and S a subset of X × Y. Then S = ( X, Y ; S ) is an incidence structure on ( X, Y ) . We say that x ∈ X is incident with y ∈ Y if ( x, y ) belongs to S. Incidence structures already appear in Euclid’s Elements. They provide a language that is efficient at handling many problems in combinatorics and geometry. Now S = ( X, Y ; S ) defines a bipartite incidence graph Γ( S ) � � with vertex set X ∪ Y and edge set { x, y } | ( x, y ) ∈ S . Its adjacency matrix is � 0 � S A = . S T 0 Therefore Shape is defined for all subsets of the incidence structure.
Some general comments: (i) Shape is invariant under automorphisms of the incidence structure. There are example of incidence structure where a very strong property holds: A bijection g : X ∪ Y ↔ X ∪ Y is an automorphism of S = ( X, Y ; S ) if and only if and U g U have the same shape, for all U ⊆ X ∪ Y. (ii) Bipartite graphs have many useful spectral properties ! (iii) Shape is very good at revealing regularity conditions in an incidence structure. We mention just two example.
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