Decision Problems in Borel Combinatoricsims.nus.edu.sg/events/2019/recur/files/su.pdfBernoulli Shift and Cayley Graph Three Decision Problems The Twelve Tiles Theorem Undecidability

Bernoulli Shift and Cayley GraphThree Decision Problems

The Twelve Tiles TheoremUndecidability of the Graph Homomorphism Problem

Decision Problems in Borel Combinatorics

Su Gao

College of ScienceUniversity of North Texas

NUSJune 3, 2019

Su Gao Decision Problems in Borel Combinatorics



All results in this talk are joint work with Steve Jackson, EdKrohne, and Brandon Seward. My research was supported by NSFgrants DMS-1201290 and DMS-1800323. Some results appeared inEd Krohne’s Ph.D. dissertation, for which Steve Jackson and I wereco-advisors. All results are to appear in a manuscript ContinuousCombinatorics of Abelian Group Actions by all four authors.




Bernoulli Shift and Cayley Graph

For a countable group G , the Bernoulli shift on G is the dynamicalsystem · : G × 2G → 2G defined by

(g · x)(h) = x(g−1h).

When G = 〈S〉 for a finite S = S−1, one can define a Cayley graphon 2G by

(x , y) ∈ E (2G ) ⇐⇒ ∃s ∈ S s · x = y .

Examples 2Zn, F (2Z

2), 2Fn , F (2Fn)




Bernoulli Shift and Cayley Graph

The Cayley graph on F (2Z2) consists of continuum many

components, with each component a grid resembling Z2.

r r r r r r r r rr r r r r r r r rr r r r r r r r rr r r r r r r r rr r r r r r r r r




Subshift of Finite Type

A Zn-subshift of finite type is a Y ⊆ bZn

for which there is a finiteset {p1, . . . , pn} of patterns such that

Y = {x ∈ bZn

: none of the patterns p1, . . . , pk occur in x}.

We denote such a subshift as Yb,~p.







Y = {x ∈ bZn









Y = {x ∈ bZn






Three Decision Problems

Question (Subshift Problem). Given a subshift of finite type Y , isthere a continuous (or Borel) equivariant map π : F (2Z

n)→ Y ?

Question (Graph Homomorphism Problem). Given a finite graph Γ,is there a continuous (or Borel) graph homomorphism from F (2Z

n)

into Γ?

Question (Tiling Problem). Given a finite set of finite tiles, is therea clopen (or Borel) tiling of all of F (2Z

n)?






n)→ Y ?


n)

into Γ?


n)?






n)→ Y ?


n)

into Γ?


n)?






n)→ Y ?


n)

into Γ?


n)?




Examples

Question (Chromatic Number Problem). What is the continuous(or Borel) chromatic number of F (2Z

n) for n ≥ 2?

This problem is equivalent to the instances of the GraphHomomorphism Problem:

Question Is there a continuous (or Borel) graph homomorphismfrom F (2Z

n) into K3 or K4?

Answer The continuous chromatic number of F (2Zn) is 4, and the

Borel chromatic number of F (2Z2) is 3.




Examples


n) for n ≥ 2?



n) into K3 or K4?






Examples


n) for n ≥ 2?



n) into K3 or K4?






Examples


n) for n ≥ 2?



n) into K3 or K4?






Examples

There is no continuous graph homomorphism from F (2Z2) into the

Petersen graph.

0

1

20

11

0

02

2

Figure: The Petersen graph with a three-coloring.




Examples

There is a continuous graph homomorphism from F (2Z2) into the

Grotzsch graph.

0

1

2

34

5

6

7

89

10

Figure: The Grotzsch Graph. The odd cycle γ = (0, 1, 2, 3, 9, 0) has order2 in the homotopy group modded out by four-cycles.




Examples

Fact There is a continuous tiling of F (2Z2) by tiles of dimensions:

2× 2, 2× 3, 3× 2, 3× 3.

Fact There is no continuous tiling of F (2Z2) by tiles of dimenstions:

2× 3, 3× 2, 3× 3;

or2× 2, 2× 3, 3× 2.

Question Is there a continuous tiling of F (2Z2) by tiles of

dimensions:2× 2, 3× 3?




Examples


2× 2, 2× 3, 3× 2, 3× 3.


2× 3, 3× 2, 3× 3;

or2× 2, 2× 3, 3× 2.






Examples


2× 2, 2× 3, 3× 2, 3× 3.


2× 3, 3× 2, 3× 3;

or2× 2, 2× 3, 3× 2.






Examples


2× 2, 2× 3, 3× 2, 3× 3.


2× 3, 3× 2, 3× 3;

or2× 2, 2× 3, 3× 2.








n)→ Y ?


n)

into Γ?


n)?




In the rest of this talk we concentrate on the continuous version ofthe Graph Homomorphism Problem for n = 2.

Question For which finite graph Γ is there a continuous graphhomomorphism from F (2Z

2) to Γ?

Main Theorem This is a Σ01-complete problem.






2) to Γ?







2) to Γ?





The Twelve Tiles Theorem

We give a complete (but theoretical) answer to the ContinuousGraph Homomorphism Problem for F (2Z

2).

Theorem There are finite graphs Γn,p,q, for each triple (n, p, q) ofpositive integers with n < p, q, such that for all finite graphs Γ thefollowing are equivalent:

1. there is a continuous graph homomorphism from F (2Z2) to Γ;

2. there is a graph homomorphism from Γn,p,q to Γ for somen < p, q with (p, q) = 1;

3. for all n and sufficiently large p, q with (p, q) = 1, there is agraph homomorphism from Γn,p,q to Γ.




The Twelve Tiles Theorem

We give a complete (but theoretical) answer to the ContinuousGraph Homomorphism Problem for F (2Z

2).

Theorem There are finite graphs Γn,p,q, for each triple (n, p, q) ofpositive integers with n < p, q, such that for all finite graphs Γ thefollowing are equivalent:

1. there is a continuous graph homomorphism from F (2Z2) to Γ;

2. there is a graph homomorphism from Γn,p,q to Γ for somen < p, q with (p, q) = 1;

3. for all n and sufficiently large p, q with (p, q) = 1, there is agraph homomorphism from Γn,p,q to Γ.




The Twelve Tiles Graph

Fix n < p, q, we define Γn,p,q.

The definition involves 12 tiles (finite grid graphs):

I 4 torus tiles

I 4 commutativity tiles

I 2 long horizontal tiles

I 2 long vertical tiles







I 4 torus tiles










I 4 torus tiles










I 4 torus tiles










I 4 torus tiles










I 4 torus tiles







Torus Tiles

R×

R×

R×

R×

Ra Ra

Rc

Rc

Gca=ac

n

p

n p

R×

R×

R×

R×

Rb Rb

Rc

Rc

Gcb=bc

n

q

n p

R× : n × n, Ra : n × (p − n), Rb : n × (q − n)Rc : (p − n)× n, Rd : (q − n)× n




Torus Tiles (continued)

R×

R×

R×

R×

Ra Ra

Rd

Rd

Gda=ad

n

p

n q

R×

R×

R×

R×

Rb Rb

Rd

Rd

Gdb=bd

n

q

n q




Commutativity Tiles

R× R× R×

R× R× R×

Ra Ra

Rd

RdRc

Rc

Gdca=acd

R×

R×

R×

R×

R×

R×

Rc

Rc

Ra

Ra

Rb

Rb

Gcba=abc




Commutativity Tiles (continued)

R× R× R×

R× R× R×

Ra Ra

Rc

RcRd

Rd

Gcda=adc

R×

R×

R×

R×

R×

R×

Rc

Rc

Rb

Rb

Ra

Ra

Gcab=bac




Long Horizontal Tiles

R× R× R× R× R× R×

R× R× R× R× R×

Ra Ra

Rc Rc Rc Rc Rc

Rd Rd Rd Rd Rd

· · ·

q copies of Rc , q + 1 copies of R×

p copies of Rd , p + 1 copies of R×

Gcqa=adp




Long Horizontal Tiles (continued)

R× R× R× R× R× R×

R× R× R× R× R×

Ra Ra

Rc Rc Rc Rc Rc

Rd Rd Rd Rd Rd

· · ·

p copies of Rd , p + 1 copies of R×

q copies of Rc , q + 1 copies of R×

Gdpa=acq




R×

R×

R×

R×

R×

R×

R×

R×

R×

R×

R×

Rc

Rc

Ra

Ra

Ra

Ra

Ra

Rb

Rb

Rb

Rb

Rb

...

qcop

iesof

Ra ,

q+

1cop

iesof

R×

pcop

iesof

Rb ,

p+

1cop

iesof

R×

Gcbp=aqc

Figure: The long vertical tiles in Γn,p,q.




R×

R×

R×

R×

R×

R×

R×

R×

R×

R×

R×

Rc

Rc

Ra

Ra

Ra

Ra

Ra

Rb

Rb

Rb

Rb

Rb

...

qcop

iesof

Ra ,

q+

1cop

iesof

R×

Gcaq=bpc

Figure: The long vertical tiles in Γn,p,q.




The smallest 12-tiles graph is Γ1,2,3: 60 nodes and 180 edges.




Undecidability of the Continuous Graph HomomorphismProblem

The Twelve-Tiles Theorem implies that the set of all Γ for whichthere is a continuous graph homomorphism from F (2Z

2) is Σ0

1.

Theorem The set of all finite Γ for which is there a continuousgraph homomorphism from F (2Z

2) to Γ is not computable.

There is not a computable bound of how large p and q will be forthe first Γn,p,q to admit a graph homomorphism to a given finite Γ.






2) is Σ0

1.









2) is Σ0

1.








We use

Theorem The word problem for finitely presented torsion-freegroups is undecidable.

We define a computable reduction of this word problem to theContinuous Graph Homomorphism Problem for F (2Z

2).





We use

Theorem The word problem for finitely presented torsion-freegroups is undecidable.

We define a computable reduction of this word problem to theContinuous Graph Homomorphism Problem for F (2Z

2).





Start with a finite presentation

G = 〈a1, . . . , ak | r1, . . . , rl〉

of a torsion-free group Gn, and

a distinguished word w = w(a1, . . . , ak).

(*) There is (a lower bound) α > 0 such that, if the distinguishedword w 6= e in G , then for all integer m ≥ 1, wm is not equal inGn to any word of length ≤ αm.





Start with a finite presentation

G = 〈a1, . . . , ak | r1, . . . , rl〉

of a torsion-free group Gn, and

a distinguished word w = w(a1, . . . , ak).

(*) There is (a lower bound) α > 0 such that, if the distinguishedword w 6= e in G , then for all integer m ≥ 1, wm is not equal inGn to any word of length ≤ αm.





Consider

G ′ = 〈a1, . . . , ak , z |r1, . . . , rl , z2w−1〉 = 〈a1, . . . , ak+1 | r1, . . . , rl , rl+1〉.

Construct a graph Γ′. Γ′ will have a distinguished vertex v0. Foreach of the generators of G ′, we add a sufficiently long cycle βi oflength `i > 4 that starts and ends at the vertex v0. We make theedge sets of these cycles pairwise disjoint. This gives a naturalnotion of length `(ai ) = `i which extends in the obvious manner toreduced words in the free group generated by the ai . For eachword rj , we wish to add to Γ′ a rectangular grid-graph Rj whoselength and width are both > 4 and whose perimeter is equal to`(rj). In order for this to be possible, we will need to make certainthat each `(rj) is a large enough even number.




The edges used in the various Rj are pairwise disjoint, and aredisjoint from the edges used in the cycles corrresponding to thegenerators ai . We then label the edges (say going clockwise,starting with the upper-left vertex) of the boundary of Rj with theedges occurring in the concatenation of the paths corrrespondingto the generators in the word rj .Finally, Γ is obtained from Γ′ by forming the quotient graph wherevertices on the perimeters of the Rj are identified with thecorresponding vertex in one of the ai .




Instead of using the Twelve Tiles Theorem directly, the proof usessome corollaries of the Twelve Tiles Theorem that give positive andnegative conditions in terms of the homotopy group of the graph Γ.

Theorem If there is an odd-length cycle γ which has finite order inπ∗1(Γ), then there is a continuous graph homomorphism from

F (2Z2) to Γ.

Theorem Suppose for every n there are p, q > n with (p, q) = 1such that, for any p-cycle γ in Γ, γq is not a p-th power in π∗1(Γ).

Then there is no continuous graph homomorphism from F (2Z2) to

Γ.




What about F (2Z)?

Theorem The Continuous Graph Homomorphism Problem forF (2Z) is decidable.




What about F (2Z)?

Theorem The Continuous Graph Homomorphism Problem forF (2Z) is decidable.




Summary

For n = 1, both the continuous and the Borel versions of theSubshift Problem are decidable.

For n = 2, the continuous Subshift Problem and the continuousGraph Homomorphism Problem are Σ0

1-complete.

All other cases are open.




Summary

For n = 1, both the continuous and the Borel versions of theSubshift Problem are decidable.

For n = 2, the continuous Subshift Problem and the continuousGraph Homomorphism Problem are Σ0

1-complete.

All other cases are open.




Thank You!


Documents

Decision Problems in Borel Combinatoricsims.nus.edu.sg/events/2019/recur/files/su.pdfBernoulli Shift and Cayley Graph Three Decision Problems The Twelve Tiles Theorem Undecidability