A toy example Function of Qi’ s in basis! Algebraic rank -> linear rank, {Q1, Q2, …, Qk} is in linear span of {Q1, Q2,…, Qd} • Q1.Q2.Q3……Qk = f(Q1.Q2……Qd) T T ∑ reduces to ∑ C = ⋅ Q i 2 ! Q ik C = ⋅ Q i 2 ! Q id ) Q i 1 F i ( Q i 1 i = 1 i = 1
A toy example Function of Qi’ s in basis! Algebraic rank -> linear rank, {Q1, Q2, …, Qk} is in linear span of {Q1, Q2,…, Qd} • Q1.Q2.Q3……Qk = f(Q1.Q2……Qd) T T ∑ reduces to ∑ C = ⋅ Q i 2 ! Q ik C = ⋅ Q i 2 ! Q id ) Q i 1 F i ( Q i 1 i = 1 i = 1 Prove lower bounds for the new model.
A toy example Function of Qi’ s in basis! Algebraic rank -> linear rank, {Q1, Q2, …, Qk} is in linear span of {Q1, Q2,…, Qd} • Q1.Q2.Q3……Qk = f(Q1.Q2……Qd) T T ∑ reduces to ∑ C = ⋅ Q i 2 ! Q ik C = ⋅ Q i 2 ! Q id ) Q i 1 F i ( Q i 1 i = 1 i = 1 Prove lower bounds for the new model. Partial derivative based methods prove useful!
Key idea
Key idea • Product of polynomials of linear rank d, is a function of the polynomials in the basis.
Key idea • Product of polynomials of linear rank d, is a function of the polynomials in the basis. • Use this to reduce express each product gate as a polynomial in few variables (polynomials in the basis).
Key idea • Product of polynomials of linear rank d, is a function of the polynomials in the basis. • Use this to reduce express each product gate as a polynomial in few variables (polynomials in the basis). • Such functions are ‘simple’ even though they could be of high degree.
Key question
Key question • Q = {Q1, Q2, …, Qr} - algebraically independent
Key question • Q = {Q1, Q2, …, Qr} - algebraically independent • F such that {F , Q1, Q2, …, Qr} - algebraically dependent
Key question • Q = {Q1, Q2, …, Qr} - algebraically independent • F such that {F , Q1, Q2, …, Qr} - algebraically dependent • Can we infer something about the structure of F , from Q?
Key question • Q = {Q1, Q2, …, Qr} - algebraically independent • F such that {F , Q1, Q2, …, Qr} - algebraically dependent • Can we infer something about the structure of F , from Q? If this is linear dependence, then yes!
Key question • Q = {Q1, Q2, …, Qr} - algebraically independent • F such that {F , Q1, Q2, …, Qr} - algebraically dependent • Can F be expressed a polynomial function of Q?
Key question • Q = {Q1, Q2, …, Qr} - algebraically independent • F such that {F , Q1, Q2, …, Qr} - algebraically dependent • Can F be expressed a polynomial function of Q? No! Q = {x^2}, F = x
Key lemma • For almost all , there is a polynomial b ∈ F n H(y1, y2, …, yr), such that F(x+b) equals the low degree component of H(Q1(x+b), Q2(x+b), …, Qr(x+b)).
Key lemma • For almost all , there is a polynomial b ∈ F n H(y1, y2, …, yr), such that F(x+b) equals the low degree component of H(Q1(x+b), Q2(x+b), …, Qr(x+b)). Algebraic dependence implies functional dependence!
Toy example
Toy example • Q(x) = x^2, F(x) = x
Toy example • Q(x) = x^2, F(x) = x • Pick a non zero b.
Toy example • Q(x) = x^2, F(x) = x • Pick a non zero b. • Take H(y) = 1/2b (y^2 + b^2)
Toy example • Q(x) = x^2, F(x) = x • Pick a non zero b. • Take H(y) = 1/2b (y^2 + b^2) • H(Q) = (x + b) + degree 2 terms!
Subsequently, Pandey, Saxena, Sinhababu showed that the converse of the lemma is also true. In summary, functional dependence is (almost) captured by algebraic dependence and vice versa.
Proof of key lemma
Proof sketch
Proof sketch A(x), B(x) are algebraically dependent. Goal is to express B • as a polynomial in A; upto a translation, taking homog. components.
Proof sketch A(x), B(x) are algebraically dependent. Goal is to express B • as a polynomial in A; upto a translation, taking homog. components. Let G(y, z) be a non-zero polynomial such that G(A, B) = 0 •
Proof sketch A(x), B(x) are algebraically dependent. Goal is to express B • as a polynomial in A; upto a translation, taking homog. components. Let G(y, z) be a non-zero polynomial such that G(A, B) = 0 • Let G be the minimum degree annihilator of {A, B} •
Proof sketch A(x), B(x) are algebraically dependent. Goal is to express B • as a polynomial in A; upto a translation, taking homog. components. Let G(y, z) be a non-zero polynomial such that G(A, B) = 0 • Let G be the minimum degree annihilator of {A, B} • Consider the univariate polynomial in z, G(A(x), z) •
Proof sketch A(x), B(x) are algebraically dependent. Goal is to express B • as a polynomial in A; upto a translation, taking homog. components. Let G(y, z) be a non-zero polynomial such that G(A, B) = 0 • Let G be the minimum degree annihilator of {A, B} • Consider the univariate polynomial in z, G(A(x), z) • z-B(x) is a root of G(A(x), z) •
Proof sketch
Proof sketch Can we say something interesting about the structure of • roots of a polynomial?
Proof sketch Can we say something interesting about the structure of • roots of a polynomial? Yes! •
Proof sketch Can we say something interesting about the structure of • roots of a polynomial? Yes! • [Dvir et al.] Under mild constraints* the roots of G(A(x), z) • can be written as the low deg. component of a polynomial in A(x)
Proof sketch Can we say something interesting about the structure of • roots of a polynomial? Yes! • [Dvir et al.] Under mild constraints* the roots of G(A(x), z) • can be written as the low deg. component of a polynomial in A(x) (*) is where a translation is needed •
Over fields of characteristic zero, the converse is also easy • to show. Relies on Jacobian criterion for algebraic independence
Over fields of characteristic zero, the converse is also easy • to show. Relies on Jacobian criterion for algebraic independence Proofs over positive characteristic are much more technical •
Questions
Questions • Depth 4 circuits over small finite fields:
Questions • Depth 4 circuits over small finite fields: • What can we say about product of low degree polynomials of large algebraic rank over small fields ?
Questions • Depth 4 circuits over small finite fields: • What can we say about product of low degree polynomials of large algebraic rank over small fields ? • Can they be ‘approximated’ by simple functions?
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