On the Shape of a Set In Honour of Lino Di Martino Topics in Groups - - PowerPoint PPT Presentation

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 (∗∗) ∂2u ∂t2 = c2 ∂2u ∂x2 + ∂2u ∂y2

• = c2 ∇2 · u = c2 ∆ · u

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, g
• :=
• f(x, y) · g(x, y) dx dy.

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 λk and eigenfunctions wk of ∆, so ∆wk = λkwk for k = 1, 2, 3, .... ∈ N. Finally

(4) Solve (**) for f(t) for the eigenvalues λk, getting fk(t) = sin(λkt), and fk(t) = cos(λkt). Finally, find the Fourier coeffi- cients ak ∈ R from the initial conditions, writing u(x, y, t) = f(t) · w(x, y) =

• k∈N

ak · fk(t) · wk . Kac’s question: Do the λk 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) =

• k∈N

ak · fk(t) · wk , suppose we also know the Fourier coefficients a1, a2, ..., ak, ... for some suitably chosen functions u(x, y, t). Do the λk together with the Fourier coefficients ak 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 = {v1, ..., vn}. (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 av v where av = f(v) ∈ R. So let

RV :=

• f =
• v∈V

av v with av ∈ R

• 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 : RV → RV defined by α(v) :=

• v∼u∈V

u . (The matrix of α is the adjacency matrix of Γ.)

The vector space RV 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: V ↔ (x, y)-plane of the drum membrane, f ∈ RV ↔ u(x, y) amplitude of membrane α ↔ Laplace Operator ∆, and so on.

Denote the eigenvalues of α by λ1, ..., λr ∈ R and let the corresponding eigenspaces be E1, ..., Er. In particular, any element f ∈ RV has a decomposition f = f1 + ... + fr with spectral components

• r Fourier components
• fi ∈ Ei.

Each fi is an eigenvector of α, unless fi = 0, and f determines the fi uniquely. Call sig(f) :=

• ||f1||2, ..., ||fr||2

the spectral signature of f.

The function sig: RV → Rr 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 ∈ Ei, then sig(v) =

• ||(v)1||2, ||(v)2||2 + ... + ||(v)r||2

is the shape of v. Shape can be defined for sets of vertices as well: If S = {u, v, ..., w} ⊆ V consider the characteristic function

• f S, denoted by

fS := u + v + ... + w = f = f1 + f2 + ... + fr . Then we call |S|−1 · sig(fS) = |S|−1 ·

• ||f1||2, ..., ||fr||2

the shape of S.

Example 1: Let Γ = C4∪K1. The spectrum of Γ is 21, 03, −21. Any C4-vertex has shape (1

4, 1 2, 1 4) while the K1-vertex has shape

(0, 1, 0). Example 2: Let Γ∗ be the star on 5 vertices. Its spectrum is also 21, 03, −21. The central vertex has shape (1

2, 0, 1 2) while

the ray-vertices all have shape (1

8, 3 4, 1 8).

In Example 1 we had Γ = C4 ∪ K1. For instance, if S = {u, v} we need to determine the components of fS = u + v = f1 + f2 + f3. 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 RV relative to the adjacency map α : RV → RV (with t distinct eigenvalues) gives rise to the shape function 2V − → Rt via shape(S) = |S−1| · sig(fS). It is not difficult to show that if g is an isomorphism Γ → Γ′ of two isospectral graphs then shape(Sg) = shape(S), for any set S of vertices of Γ. In particular, shape is invariant under graph

• isomorphism. How good is this invariant?

Kac’s Question again: If g : V (Γ) − → v(Γ′) is a bijection of the vertices of two isospectral graph, such that S and Sg 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 λ1, ..., λr be the eigenvalues of α: RV → RV, and denote the corresponding eigenspaces by E1, ..., Er. For i = 1, .., r let πi be the orthogonal projection πi: RV → Ei. So π2

i = πi ,

πi ◦ πj = 0 when i = j, and

r

• i=1

πi = id. Any element f ∈ RV has spectral decomposition f = f1 + ... + fr with fi = πi(f) ∈ Ei. . For instance, we have the inequality (f , fi) = (fi, fi) ≥ 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 : RV → RV, eigenvalues λ1, ..., λr and eigenspaces E1, ..., Er. Then (1) The projection map πi : RV → Ei is a polynomial in α of degree r − 1, with coefficients determined by the λi. (One can write it down!) (2) Let v = f1 + ... + fr be a vertex of Γ, where fi ∈ Ei. Then ||fi|| 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 = f1 + ... + fr be a vertex of Γ where fi ∈ Ei. Suppose that fj = 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

• ther 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 K4,4.) Open Question: Is Shape a complete invariant for graphs with all eigenvalues distinct? (Probably not.) Vertex Transitivity: In many computational examples (3) turns

• ut 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

A = S ST

• .

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 U and Ug 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.

4.1 Designs Let Γ be a bipartite graph with part X and Y. Let B ⊆ Y. Then (X, B) is a design with block set B if the induced subgraph Γ[X ∪ B] ⊆ Γ is bi-regular. This is the usual definition of t − (n, k, λ) designs when X and Y are the t- and k-subsets of an n-set, respectively. It covers many other types of designs. As before let fB =

y∈B y.

PROPOSITION 2: Let Γ be a bipartite and biregular graph with vertex set X ∪ Y. Then there is an index set J such that the following is true: (X, B) is a design if and only if (fB)i = 0 for all i ∈ J. This generalizes the Graver-Jurkat Theorem of 1971, the proof now is very short. The proposition does make the point: Shape is exactly right to capture the definition of a design.

4.2: Orbits Let Γ again be bipartite on X ∪ Y, the parts of Γ. Let G be a group of automorphisms which maps X to X and Y to Y. PROPOSITION 3: (Orbit Numbers) Suppose that the map α

• RX: RX → RY is injective. Then

(∗) #

• G-orbits on X
• ≤

#

• G-orbits on Y
• .

Furthermore, there is an index i = i0 so that (∗) is an equality if and only if the following is true: For every G-orbit B on Y we have (fB)i = 0. We emphasize once more: (fB)i = 0 is an explicit equation for the orbit B in terms of certain powers of the adjacency map α: R(X∪Y ) → R(X∪Y ) and its eigenvalues. The computation

• f eigenvalues is essential.

Recall:

GEOMETRY

← →

GROUP

← →

← →

F − MODULES General problem: Compute eigenvalues and eigenspace decomposition!

5: The Hyperoctahedron

and its incidence graph.

The definition: For the integer n ≥ 1 let V = {α1, ¯ α1, ..., αn, ¯ αn} be a set of 2n distinct elements, called vertices. For 0 ≤ k ≤ n let Xk be the collection of all k-element subsets x of V with |x ∩ {αi, ¯ αi}| ≤ 1 for all 1 ≤ i ≤ n. Such sets are called faces. The set n

k=0 Xk of all faces is the

face complex of the n-dimensional hyperoctahedron. It is a ranked partially ordered set under subset inclusion, denoted Hn. Get many Incidence Structures. Determine their Spectra!

The automorphism group of Hn is the hyperoctahedral group Bn, it plays a distinguished role in geometry. The interesting part of the spectrum occurs between adjacent layers (Xk, Xk+1). So consider the adjacency map α = αk,k+1: R(Xk ∪ Xk+1) → R(Xk ∪ Xk+1) , find its eigenvalues and eigenspaces. It turns out be very efficient to work with α2 =: ν+

k : RXk → RXk

and α2 = ν−

k+1: RXk+1 → RXk+1 .

[The eigenvalues of α then are ± √ λ∗ where λ∗ ≥ 0 is an eigen- value of ν.]

The key to the problem is the map σk : RXk → RXk defined by σk(x) :=

• α∈x

(x \ {α}) ∪ {¯ α} for x ∈ Xk. Its eigenvalues and eigenspaces are easy to compute. It satisfies B-TYPE LEMMA: Let 0 ≤ k ≤ n. Then ν+

k − ν− k = (2n − 3k)I − σk

where I denotes the identity map on RXk. This lemma gives the main result on the hyperoctahedral complex.

PROPOSITION 4: Spectral Decomposition of Hn. Let k < n. The eigenvalues of ν+

k : CXk → CXk are of the form

λ∗

k,j,i = 2(k − j + 1 − i)(n − k − i)

where 0 ≤ j ≤ n and 0 ≤ i ≤ min{k − j, n − k}, with multiplicity mult(λ∗

k,j,i) =

• n

j + i

j + i

j

• −
• n

j + i − 1

j + i − 1

j

• .

PROPOSITION 5: For fixed k and j, i as above the eigenspaces E∗

k,j,i are pairwise non-isomorphic and irreducible as Bn-modules.

Furthermore, E∗

k,j,i ≈ E∗ k′,j′,i′ iff j = j′ and i = i′.

Conclusion: GEOMETRY ← →

GROUP

← →

← →

F − MODULES Irreducibles can be computed without the group.

Thank You