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Algebra in Flatland

Connie Dennis*, Cassie Stamper*

Department of MathematicsKansas State UniversityManhattan, KS 66506

Mentor: Dr. David Yetter*July 24, 2012

* Supported by NSF grant # DMS1004336

Motivation

The starting point is to think of computations as physical things thattake place in space and time, rather than abstractions.

Secondly, we wanted to compute things by representing mathematicalquantatities by quantum mechanical states.

This is the idea of quantum computing.

Physics in two spatial dimensions is different than physics in threeand higher dimensions.

In three and higher dimensions there are only two types of particles:bosons and fermions.

C. Dennis, C.Stamper (KSU) SUMaR 2012 July 24, 2012 2 / 35

Motivation

Motivation Continued

If two bosons in identical states are swapped, the state remains thesame.

If two fermions in identical states are swapped, the state negates.

In a plane, quasi particles can exhibit the fractional quantum halleffect.

When two quasi particles are swapped, the state is multiplied by acomplex number with absolute value 1.

The motivation for this project is to represent math in quantummechanical states, where when two things are swapped, the state ismultiplied by a phase ζ which is a primitive Nth root of unity.

Need to Know

Definition

Let Z be the set of all integers. A Z graded vector space is a vector space,V which decomposes into a direct sum of the form:

V =⊕n∈Z

Where each Vn is a vector space. For a given n the elements of Vn arethen called homogeneous elements of degree n.

Need to Know Continued

We will use || to represent degree.

The anyonic braiding associated to ζ (ζN = 1 primitive) is given by:∀ a,b homogeneous σ(a⊗ b) = ζ |a||b|b ⊗ a

Need to Know Continued

We will use || to represent degree.

The anyonic braiding associated to ζ (ζN = 1 primitive) is given by:∀ a,b homogeneous σ(a⊗ b) = ζ |a||b|b ⊗ a

Questions

Question

What happens to ordinary commutativity and associativity in the anyonicsetting?

Answer 1:Commutative became a · b = ζ |a||b| b · aAnswer 2:Associative stays the same unless the multiplication changes degrees.

Answer 3:If |m| 6= 0 we will answer later.

Question

Do the resulting axioms give reasonable algebraic systems?

We used rewrite systems as a tool to answer this question.

Questions

Question

Answer 1:Commutative became a · b = ζ |a||b| b · a

Answer 2:Associative stays the same unless the multiplication changes degrees.

Question

Questions

Question

Questions

Question

Questions

Question

Questions

Question

Definitions

Definition

A rewrite system is a collection of directed rules for replacing parts ofsymbol strings with other symbol strings.

Definition

Descending Chain Condition- given a word there are no infinite strings ofsuccessive rule applications which can be made starting at the word.

Definition

Local Confluence- If two instances apply to a word, then there aresequences of rule applications to each of the results which give equalresults.

Definitions

Definition

Definitions

Definition

Definitions

Definition

Knuth - Bendix

Theorem

Knuth - Bendix Theorem:If a rewrite system satisfies two properties, the Descending ChainCondition (DCC) and Local Confluence, then:(a) Given any word, there is a unique reduced word. The reduced word isthe canonical representative of the equivalence class.(b) If two words are equivalent, the reduced word reachable from each isthe same.

Free Anyonic Commutative Algebra

C[x(d1)1 ....x

(dn)n ]ζ

freely generated by x1......xn

of degrees d1......dn

ζN = 1

∀a, b homogenous :

a · b = ζ |a||b|b · ab · a = ζ |b||a|a · bSo by solving:

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ |a||b|b · a

b · a = ζ |b||a|a · bSo by solving:

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ |a||b|b · ab · a = ζ |b||a|a · b

So by solving:

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · a

a · b 6= 0⇒ N | 2|a||b|

C[x(d1)1 ....x

(dn)n ]ζ

ζN = 1

a · b = ζ−|a||b|b · aa · b 6= 0⇒ N | 2|a||b|

Conditions

N - |a|2 ⇒ a2 = 0

N - 2|a||b| ⇒ ab = 0

N|2|a||b|and N - |a||b| ⇒ ab = −ba

N||a||b| ⇒ ab = ba

Conditions

N - |a|2 ⇒ a2 = 0

N - 2|a||b| ⇒ ab = 0

N|2|a||b|and N - |a||b| ⇒ ab = −ba

N||a||b| ⇒ ab = ba

Conditions

N - |a|2 ⇒ a2 = 0

N - 2|a||b| ⇒ ab = 0

N|2|a||b|and N - |a||b| ⇒ ab = −ba

N||a||b| ⇒ ab = ba

Conditions

N - |a|2 ⇒ a2 = 0

N - 2|a||b| ⇒ ab = 0

N|2|a||b|and N - |a||b| ⇒ ab = −ba

N||a||b| ⇒ ab = ba

n 0 1 2

0 C C C1 C 0 02 C 0 0

All phase orders that are odd primes behave this way

n 0 1 2

0 C C C1 C 0 02 C 0 0

All phase orders that are odd primes behave this way

n Must square to 0 ?

0 No1 Yes2 Yes

n 0 1 2 3

0 C C C C1 C 0 A 02 C A C A3 C 0 A 0

0 No1 Yes2 No3 Yes

n 0 1 2 3 4 5 6 7

0 C C C C C C C C1 C 0 0 0 A 0 0 02 C 0 A 0 C 0 A 03 C 0 0 0 A 0 0 04 C A C A C A C A5 C 0 0 0 A 0 0 06 C 0 A 0 C 0 A 07 C 0 0 0 A 0 0 0

All phase orders that are a power of 2 behave this way

n 0 1 2 3 4 5 6 7

0 C C C C C C C C1 C 0 0 0 A 0 0 02 C 0 A 0 C 0 A 03 C 0 0 0 A 0 0 04 C A C A C A C A5 C 0 0 0 A 0 0 06 C 0 A 0 C 0 A 07 C 0 0 0 A 0 0 0

All phase orders that are a power of 2 behave this way

0 No1 Yes2 Yes3 Yes4 No5 Yes6 Yes7 Yes

N = 2k

k > 2µ2(|a|)⇒ a2 = 0

k − 1 > µ2(|a|) + µ2(|b|)⇒ ab = 0

k − 1 = µ2(|a|) + µ2(|b|)⇒ ab = −ba

k 5 µ2(|a|) + µ2(|b|)⇒ ab = ba

N = 2k

k > 2µ2(|a|)⇒ a2 = 0

k − 1 > µ2(|a|) + µ2(|b|)⇒ ab = 0

k − 1 = µ2(|a|) + µ2(|b|)⇒ ab = −ba

k 5 µ2(|a|) + µ2(|b|)⇒ ab = ba

N = 2k

k > 2µ2(|a|)⇒ a2 = 0

k − 1 > µ2(|a|) + µ2(|b|)⇒ ab = 0

k − 1 = µ2(|a|) + µ2(|b|)⇒ ab = −ba

k 5 µ2(|a|) + µ2(|b|)⇒ ab = ba

N = 2k

k > 2µ2(|a|)⇒ a2 = 0

k − 1 > µ2(|a|) + µ2(|b|)⇒ ab = 0

k − 1 = µ2(|a|) + µ2(|b|)⇒ ab = −ba

k 5 µ2(|a|) + µ2(|b|)⇒ ab = ba

n 0 1 2 3 4 5

0 C C C C C C1 C 0 0 A 0 02 C 0 0 C 0 03 C A C A C A4 C 0 0 C 0 05 C 0 0 A 0 0

All phase orders that are a power of 2 multiplied with an odd prime behavethis way

n 0 1 2 3 4 5

0 C C C C C C1 C 0 0 A 0 02 C 0 0 C 0 03 C A C A C A4 C 0 0 C 0 05 C 0 0 A 0 0

All phase orders that are a power of 2 multiplied with an odd prime behavethis way

0 No1 Yes2 Yes3 Yes4 Yes5 Yes

n 0 1 2 3 4 5 6 7 8

0 C C C C C C C C C1 C 0 0 0 0 0 0 0 02 C 0 0 0 0 0 0 0 03 C 0 0 C 0 0 C 0 04 C 0 0 0 0 0 0 0 05 C 0 0 0 0 0 0 0 06 C 0 0 C 0 0 C 0 07 C 0 0 0 0 0 0 0 08 C 0 0 0 0 0 0 0 0

All phase orders that are a product of two odd primes behave this way

n 0 1 2 3 4 5 6 7 8

0 C C C C C C C C C1 C 0 0 0 0 0 0 0 02 C 0 0 0 0 0 0 0 03 C 0 0 C 0 0 C 0 04 C 0 0 0 0 0 0 0 05 C 0 0 0 0 0 0 0 06 C 0 0 C 0 0 C 0 07 C 0 0 0 0 0 0 0 08 C 0 0 0 0 0 0 0 0

All phase orders that are a product of two odd primes behave this way

0 No1 Yes2 Yes3 No4 Yes5 Yes6 No7 Yes8 Yes

n 0 1 2 3 4 5 6 7 8 9 10 11 12 13 140 C C C C C C C C C C C C C C C1 C 0 0 0 0 0 0 0 0 0 0 0 0 0 02 C 0 0 0 0 0 0 0 0 0 0 0 0 0 03 C 0 0 0 0 C 0 0 0 0 C 0 0 0 04 C 0 0 0 0 0 0 0 0 0 0 0 0 0 05 C 0 0 C 0 0 C 0 0 C 0 0 C 0 06 C 0 0 0 0 C 0 0 0 0 C 0 0 0 07 C 0 0 0 0 0 0 0 0 0 0 0 0 0 08 C 0 0 0 0 0 0 0 0 0 0 0 0 0 09 C 0 0 0 0 C 0 0 0 0 C 0 0 0 010 C 0 0 C 0 0 C 0 0 C 0 0 C 0 011 C 0 0 0 0 0 0 0 0 0 0 0 0 0 012 C 0 0 0 0 C 0 0 0 0 C 0 0 0 013 C 0 0 0 0 0 0 0 0 0 0 0 0 0 014 C 0 0 0 0 0 0 0 0 0 0 0 0 0 0C. Dennis, C.Stamper (KSU) SUMaR 2012 July 24, 2012 22 / 35

0 No1 Yes2 Yes3 Yes4 Yes5 Yes6 Yes7 Yes8 Yes9 Yes10 Yes11 Yes12 Yes13 Yes14 Yes

Associativity

C{x (d1)1 ....x

(dk )k }ζ,dm

freely generated by x1......xk

of degrees d1......dk with multiplication of degree dm

Convention: All braidings are positive.

The opposite convention would be the same except with respect toζ−1

Weakly associative with respect to the anyonic braiding for ζ

Associativity

C{x (d1)1 ....x

(dk )k }ζ,dm

Associativity

C{x (d1)1 ....x

(dk )k }ζ,dm

Associativity

C{x (d1)1 ....x

(dk )k }ζ,dm

Associativity

C{x (d1)1 ....x

(dk )k }ζ,dm

Associativity

C{x (d1)1 ....x

(dk )k }ζ,dm

Elements

Elements are complex linear combinations of binary trees.

The final nodes are labeled from x1....xk

A final node is a node whose subtrees are both empty.

There is a height function on the nodes with range {0, 1, ...,M} or ∅

Elements

Elements Continued

All final nodes are at the same height.

No two non-final nodes are at the same height.

The children of a node are higher than the parent.

The empty tree names the identity.

Elements Continued

Rewrite Rules

a b c a b c

ζ|m||a|- - - - - - - - - - - - -- - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - -

a b c d a b c d

ζ - (|m||m|)

- - - - - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - -

Multiplication

T1 T2 = T1

Unital Rewrite Rules

rewrites.pdf

Local Confluence Check 1

Proof 1.pdf

a b c d e

((ab)1(cd)2 e)3

a b c d e

((ab)1(c(de)2)3

ζ (|m||c|)

a b c d e

((ab)2 (c(de)1)3

ζ (|m||c|) – (|m||m|)

a b c d e

ζ (|m||c|) – 2(|m||m|)

(ab)3(c(de)1)2

a b c d e

(ab)3((cd)1e)2

ζ -2(|m||m|)

a b c d e

((ab)2(cd))1e)3

ζ -(|m||m|)

Proof 2.pdf

a b c d e

ζ(|m||a|)

a b c d e

ζ(|m||a|) - |m||m|

a b c d e

ζ(|m||a|) – 2(|m||m|)

a b c d e

ζ – 2(|m||m|)

a b c d e

ζ - (|m||m|)

Proof 3.pdf

a b c d e f

ζ - (|m||m|)

a b c d e f

ζ -2(|m||m|)

a b c d e f

ζ -3(|m||m|)

a b c d e f

ζ -2(|m||m|)

a b c d e f

ζ - (|m||m|)

Proof 4.pdf

a b c d

a b c d a b c d

a b c d

a b c da b c d

ζ |m|(|m|+|a|+|b|)

ζ |m|(|a|+|b|)

ζ 2(|m|+|a|)+|b|

ζ |m|(|m|+|a|)ζ (|m|+|a|)

Conclusion

Every element in a free-weakly associative algebra can be uniquelyrepresented as a linear combination of reduced trees.

Reduced means all left subtrees are empty or have exactly one node,and all non-final nodes have two children.

Conclusion

Every element in a free-weakly associative algebra can be uniquelyrepresented as a linear combination of reduced trees.

Reduced means all left subtrees are empty or have exactly one node,and all non-final nodes have two children.

Thank You!

connie dennis*, cassie stamper* › research › reu › results2012_files › final... · complex...

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connie dennis, cassie stamper › research › reu › results2012_files › final... · complex...