Non-commutative computation

with division

Avi Wigderson IAS, Princeton

Pavel Hrubes U. Washington

Arithmetic complexity – why?

- Can’t deal with Boolean complexity- What can be computed with + − ×

÷ ?- Linear algebra, polynomials, codes,

FFT,…- Helps Boolean complexity

(arithmetization)- ………

Arithmetic complexity – basics

X = (X)ij an n×n matrix.

- Detn (X) = Σσ sgn(σ) Πi Xiσ(i) “P”

- Pern (X) = Σσ Πi Xiσ(i) “NP”

- (X)-1 : n2 rational functions “P”

F field

÷×

+− ×

Xi Xj Xic

+S(f) – circuit size“P”: S is poly(n)

L(f) – formula size“NC”: L is poly(n)

n variables,f degree <n

f

X1, X2,… commuting variables: XiXj = XjXi

F[X1, X2,…] polynomial ring: p, q.F(X1, X2,… ) field of rational functions: pq-1

[Strassen’73] Division can be efficiently eliminatedwhen computing polynomials (eg from Gauss elimination for computing Det).

Since then, arithmetic complexity focused on , , We’ll restore division to its former (3rd grade) glory!

Commutative computation

State-of-the-art

F[X1,X2,…] FX1, X2,… F(X1, X2,…) comm, no ÷ non-comm, no ÷ non-comm

Circuit lb Formula lb

NC-hard NP-hard

NC = P? P = NP?

PIT (Word Problem)

S> nlog n [BS] L> n2 [K]

Det [V] Per [V]

P=NC [VSBR] Pern ≤ Detp(n)

BPP [SZ,DL]

X1, X2,… non-commuting vars: XiXj XjXi

FX1, X2,… non-commut. polynomial ring: p, q.

- Order of variables in monomials matter! E.g. Detn (X) = Σσ sgn(σ) X1σ(1) X2σ(2) Xnσ(n)

is just one option (Cayley determinant)

- Weaker model. E.g. X2-Y2 costs 2 multiplications,

but just 1 in the commut. case: X2-Y2 = (X-Y)(X+Y)

Non-commutative computation(groups, matrices, quantum, language theory,…)

State-of-the-art

F[X1,X2,…] F<X1,X2,…> F{X1,X2,…} comm, no ÷ non-comm, no ÷ non-comm

Circuit lb Formula lb

NC-hard NP-hard

NC = P? P = NP?

PIT (Word Problem)

S> nlog n [BS] L> n2 [K]

Det [V] Per [V]

P=NC [VSBR] Pern ≤ Detp(n)?

BPP [SZ,DL]

L(Detn)>2n[N]

Per [HWY]Det [AS]

P NC [N]

BPP [AL,BW]

L(X-

1 )>2n[HW]

X-1 [HW]

P NC [HW]

BPP?

The wonderful wierd world of non-commutative rational

functionsx−1 + y−1 , yx−1y have no expression fg−1 for polys f,g

(x + xy−1x)−1

= x−1 - (x + y)−1 Hua’s identityCan one decide equivalence of 2 expressions?

(x + zy−1w)−1

can’t eliminate this nested inversion!

Reutenauer Thm: Inverting an nxn generic matrix requires n nested inversions.Key to the formula lower bound on X-1

The free skew field (I) [Amitsur]A “circuit complexity” definition!

Field of fractions F(X1, X2,…) of FX1, X2,…

Take all formulae r(X1, X2,…) with , , , ÷ r~s if for all matrices M1, M2,…of all sizes r(M1, M2,…) = s(M1, M2,…) whenever they make sense (no zero division)

Amitsur Thm: F(X1, X2,…) is a skew field –every nonzero element is invertible!

Word problem (RIT): Is r = 0?

The free skew field (II) [Cohn]Matrix inverse definition

R an nxn matrix with entries in FX1, X2,…R is full if R ≠ AB with A nr, B rn, r<n.

Ex: 0 X Y Singular if vars commute -X 0 Z Invertible if vars non-commut. -Y –Z 0Cohn’s Thm: F(X1, X2,…) is the field of entries of inverses of all full matrices over FX1, X2,…Key to formula completeness of X-1

Word problem: Is R invertible (full)?Cohn’s Thm: Decidable (via Grobner basis alg).

Minimal dimension problem

Ex: 0 X Y Singular under M1(F)-substitutions -X 0 Z Invertible with M2(F) substitutions -Y –Z 0

Conjecture: Every full nxn R with entries in {Xi}, F,is invertible under Md(F) substitutions, d=poly(n). - Conjecture true for polynomials [Amitsur-Levizky]- Conjecture implies: 1) RIT BPP2) Efficient elimination of division gates from

non-commutative formulas computing polynomials

3) Degree bounds in Invariant Theory (& GCT )

÷