Counting d.o.f.s in periodic frameworks Louis Theran (Aalto - - PowerPoint PPT Presentation

counting d o f s in periodic frameworks
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Counting d.o.f.s in periodic frameworks Louis Theran (Aalto - - PowerPoint PPT Presentation

Counting d.o.f.s in periodic frameworks Louis Theran (Aalto University / AScI, CS) Frameworks Graph G = ( V,E ); edge lengths ( ij ); ambient dimension d Length eqns. ||p i - p j || 2 = ( ij ) 2 The ps are a


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Counting d.o.f.s in periodic frameworks

Louis Theran (Aalto University / AScI, CS)

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Frameworks

  • Graph G = (V,E); edge lengths ℓ(ij); ambient

dimension d

  • Length eqns.
  • The p’s are a “placement” of G / realization of (G, ℓ)

||pi - pj||2 = ℓ(ij)2

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Frameworksʹ

  • Deformation space = local solutions to
  • “mod rigid motions”
  • Degrees of freedom = dim (deformation space)

||pi - pj||2 = ℓ(ij)2

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Rigidity, flexibility

Rigidity question: is the deformation space zero dimensional? Rigid Flexible

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Quiz!

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Quiz!

[Thorpe]

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Quiz!

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Quiz!

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Combinatorial rigidity

Combinatorial rigidity question: which graphs are rigid? Deformation space is a finite-dimensional algebraic set, well-def’d dimension

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  • Each point contributes d variables
  • Each edge contributes 1 equation
  • Always d (d + 1) / 2 rigid motions
  • Don’t waste any

D.o.f. heuristic (“Maxwell counting”)

m’ ≤ d n’ - d(d + 1)/2

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  • Theorem (Laman): Generically, in d = 2, this

implies independence of length equations. (Rigidity if m = 2n – 3.)

Geometry to combinatorics

m’ ≤ d #V(G’) - d(d + 1)/2

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  • Genericity is loosely “no special geometry”
  • Almost all placements are generic
  • Non-generic set is algebraic
  • Non-generically frameworks exhibit

“universality” [Kapovich-Millson]

  • Most general problem very hard

Genericity

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  • Generic frameworks can be general enough
  • Can check Laman “2n – 3” in O(n2) time
  • simple “pebble game” algorithms [Hendrickson-

Jacobs, Berg-Jordán, Lee-Streinu]

  • Useful to know if your problem is non-generic

Why combinatorial rigidity?

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  • Graph is infinite
  • how to compute with it
  • Structure is symmetric
  • any symmetric structure

satisfies lots of extra equations

  • very non-generic looking

Hypothetical zeolite

[Rivin-Treacy-Randall]

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  • A periodic framework (G, ℓ, Γ) is an infinite

framework with

  • Γ < Aut(G)
  • ℓ(γ(ij)) = ℓ(ij)
  • A realization G(p,Λ) is a realization periodic with

respect to a lattice of translations Λ, which realizes Γ

  • Motions preserve the Γ-symmetry

Periodic frameworks

Γ free abelian, rank d finite quotient

[Borcea-Streinu]

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[Whiteley]

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1 vertex orbit 2 edge orbits

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Not allowed

Not one vertex orbit!

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Colored graphs

(0,0) (0,0) (0,0) (1,-1) (0,1) (-1,0)

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  • Each vertex orbit determined by one representative
  • total dn variables from there
  • Lattice representation is a d × d matrix
  • d2 more variables
  • For subgraphs, we will have to distinguish how

much of the symmetry group they “see”

Counting for periodic frameworks

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(0,0) (0,0) (0,0) (1,-1) (0,1) (-1,0)

m ≤ 2n – 3

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(0,0) (0,0) (0,0) (1,-1) (0,1) (-1,0)

m ≤ 2n – 3 + 2

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(0,0) (0,0) (0,0) (1,-1) (0,1) (-1,0)

m ≤ 2n – 3 + 4

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  • Theorem (Malestein-T): For dimension 2

characterizes generic independence of length equations.

  • Minimal rigidity if m = 2n + 1
  • Generic here is choice of vertex orbits
  • Actually an instance of a more general counting heuristic
  • Always necessary
  • Known to be sufficient for more groups in dim 2
  • Similar results for fixed-area unit cell, fixed lattice, etc.

Generic periodic rigidity

m’ ≤ 2(n + k) - 3 - 2(c - 1)

Z2 rank connected comps.

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Allowed!

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  • Let (G, p, L) be a realization of (G,ℓ, Γ)
  • (G, p, L) is (periodically) ultrarigid if
  • it is rigid
  • for any (finite-index) sub-lattice Λ < Γ, (G, p, L) is a

rigid realization of (G, ℓ, Λ)

  • Related concept: “ultra 1-d.o.f.” (in 2d)
  • e.g, 4-regular lattices

Ultrarigidity

[Borcea]

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  • Generic rigidity characterized by the rank of one

matrix (rigidity/compatibility/… matrix)

  • here there is an infinite family of matrices
  • Not completely clear finite ultrarigidity is a generic

property

  • Some evidence towards “no”
  • We don’t know a priori what “extra” bulk modes look

like

Challenges

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Algebraic characterization

  • A realization (G, p, L) is infinitesimally ultrarigid if and
  • nly if:
  • It is infinitesimally periodically rigid
  • The matrix with ijth row, ij ∈ E(G,φ)

! !

has rank dn for all ω ≠ 1

(….. – dij …….. dij ⨂{γij-1,ω} ….) i j

edge direction vector

  • comp. wise

mult

{δ,ω} := (ζ1δ1, …,ζdδd), ζi root of unity

[Connelly-Shen-Smith’14 + Power ’13]

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  • Extra bulk modes fix the lattice
  • [Connelly-Shen-Smith]: Nice geometric argument
  • Direct derivation: Representation theory
  • Can check ultrarigidity in finite time [Malestein-T]
  • new a priori bound on order of ζ’s

Consequences

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  • (G,γ) a colored graph with Γ (≅Zd) colors
  • ψ : Γ ⟶ Δ, epimorphism to a finite cyclic Δ
  • “Ultra Maxwell Count” for (G, ψ (γ))

for all ψ.

  • Finitely many suffice. Sufficient in 2d if (G,γ) is

independent as a periodic framework

Counting

m’ ≤ d n’ – d T(G, ψ (γ))

# c.c.’s w/ Δ rank > 0

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  • For m = 2n + 1, have a combinatorial algorithm

polynomial in m (but not γ) for generic infinitesimal periodic ultra rigidity

  • Useful for “small” colors
  • For m = 2n, have a polynomial time algorithm for fixed-

area periodic ultrarigidity

  • Via some combinatorial equivalences
  • Uses the pebble game, still only O(n4)

Algorithms and combinatorics

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  • Finite vs. infinitesimal ultra-rigidity
  • “Irrational” points on the RUMS
  • Faster algorithms

Questions

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Thanks!