Supplementary lecture notes AMC 2017∗ Guus Regts May 14, 2017

1

The cap set problem

A cap set is a set A ⊆ F3n such that for any triple ( a, b, c) ∈ A3 with not all three terms equal, one has a + b + c 6= 0. As in F3 we have 2 = −1, a set is a cap set if and only it does contain a nontrivial arithmetic progressions of length 3. The main goal of this section is to show the following breakthrough result of Ellenberg and Gijswijt: Theorem 1.1 (Ellenberg and Gijswijt ). There exists a constant c < 3 such that for any n ∈ N and any cap set A ⊂ F3n , | A| ≤ cn . The proof we present here is based on the polynomial method and follows . Lemma 1.2. Let q be a prime power. Let Mn be the set of monomials in x1 , . . . , xn whose degree in each variable is at most q − 1. Let Vn be the vector space they span. Then Vn is linearly isomorphic to the space of functions f : Fnq → Fq . Proof. Define the evaluation map Fn

e : Vn → Fq q

e( p) := ( p( a)) a∈Fnq ,

by

for p ∈ Vn . We claim that e is a linear isomorphism. Indeed, first note that e is surjective as for each a ∈ Fnq , the polynomial ∏in=1 (1 − ( xi − ai )q−1 ) is mapped to the indicator function of the point a. (Where we use that in Fq one has x q−1 = 1 for any x ∈ Fq \ {0}.) As the Fn

dimension of Vn is qn and the dimension of Fq q is also clearly qn , the lemma is proven. For any d ∈ [0, (q − 1)n] we let Mnd be the set of monomials in Mn of degree at most d and Vnd ⊆ Vn the subspace they span. Proposition 1.3. Let α, β, γ ∈ Fq such that α + β + γ = 0. Let A ⊂ Fnq . Suppose p ∈ Vnd satisfies p(αa + βb) = 0 for each pair of distinct elements a, b ∈ A. Then the number of elements a ∈ A such that p(−γa) 6= 0 is at most 2| Mnd/2 |. ∗ These

are supplementary lecture notes for the spring 2017 Mastermath course Algebraic Methods in Combinatorics. Please send corrections, errors, typos, etc to [email protected]

1

Proof. As p ∈ Vnd , it is a linear combination of monomials of degree at most d. Hence we can write p(αx + βy) =

cm m(αx + βy)

m∈ Mnd

cm,m0 m( x )m0 (y).

(1)

m,m0 ∈ Mnd deg(mm0 )≤d

Now in each summand of (1), one of m and m0 has degree at most d/2. This implies that we can write p(αx + βy) = ∑ m( x ) f m (y) + ∑ m(y) gm ( x ), m∈ Mnd/2

m∈ Mnd/2

for certain families of polynomials f m and gm both indexed by m ∈ Mnd/2 . Let now B be the A × A matrix defined by Ba,b := p(αa + βb). Then Ba,b =

m( a) f m (b) +

m ( b ) gm ( a ),

m∈ Mnd/2

m∈ Mnd/2

a decomposition of B as a sum of 2| Mnd/2 | rank one matrices. Hence the rank of B is at most 2| Mnd/2 |. On the other hand by assumption, B is a diagonal matrix and hence at most 2| Mnd/2 | of its diagonal entries are nonzero. In other words p(α + β) a) = p(−γa) 6= 0 for at 2| Mnd/2 | elements a ∈ A. Theorem 1.4. Let α, β, γ ∈ Fq such that α + β + γ = 0. Let A ⊂ Fnq be such that the equation αa + βb + γc = 0, (q−1)n/3

has no solutions ( a, b, c) ∈ A3 unless a = b = c. Then | A| ≤ 3| Mn

|.

Proof. We may assume that γ 6= 0. Let d ∈ [0, (q − 1)n]. Let V denote the space of polynomials in Vnd vanishing on the complement of −γA. Then dim(V ≥ | Mnd | − qn + | A|.

(2)

To see this let us write X = Fnq \ (−γA). Consider the space Vnd /V , which we may identify with a subspace of the space of functions X → Fq . Hence its dimension is at most | X |. As we have Vnd = Vnd /V ⊕ V , it follows that dim(V ) = dim(Vn d) − dim(Vnd /V ) ≥ | Mnd | − | X | = | Mnd | − qn + | A|, as desired. We view elements of V as functions on Fnq . Let p ∈ V be an element of maximal support (i.e. as a function on Fnq ). Let Σ := { a ∈ Fnq | p( a) 6= 0} be the support of p. Then

|Σ| ≥ dim(V ).

(3)

Indeed, for otherwise, by the Vanishing Lemma there exists a nonzero polynomial q ∈ V such that q( x ) = 0 for all x ∈ Σ. Then the support of p + q would strictly contain Σ, contradicting our choice of p. 2

Write S( A) for the set of all elements of Fnq of the form αa + βb with a and b distinct elements of A. Then S( A) is disjoint from −γA, as by assumption A does not contain solutions to αa + βb + γc = 0 unless a = b = c. This implies that p vanishes on S( A). By Proposition 1.3 we know that p(−γa) 6= 0 for at most 2| Mnd/2 | elements a ∈ A. This implies that |Σ| ≤ 2| Mnd/2 | (4) Combining (2), (3) and (4) we obtain

| Mnd | − qn + | A| ≤ dim(V ) ≤ |Σ| ≤ 2| Mnd/2 |, and hence

| A| ≤ 2| Mnd/2 | + (qn − | Mnd |). qn

(5)

| Mnd |

is the number of monomials where no variable has degree more Now note that − than q − 1, whose degree is greater than d. These are in natural bijection with those mono( q −1) n − d mials whose degree is less than (q − 1)n − d, of which there are at most | Mn |. Plugging this in with d = 2(q − 1)n/3 into (5) we obtain (q−1)n/3

| A | ≤ 2 | Mn

(q−1)n−2(q−1)n/3

| + | Mn

(q−1)n/3

| = 3 | Mn

|,

and finishes the proof. Taking α = β = γ = 1 in F3 , we now have a bound on the maximum size of a cap set in F3n we just need to realize that | Mn2n/3 |3−n is exponentially small as n grows. To see this note that | Mn2n/3 |3−n is equal to the probability that when selecting exponents of the variables x1 , . . . , xn uniformly at random and independently, that the monomial formed this way has degree bounded by 2/3n, or equivalently, by symmetry, that monomial formed this way has degree at least 4/3n. Let Xi be independent random variables, each taking values 0, 1, 2 with probability 1/3. Let X = ∑in=1 Xi . Then

| Mn4n/3 |3−n = Pr[ X ≥ 4/3n]. Note that E[ X ] = n(1/3 + 2/3) = n. So we can rewrite Pr[ X ≥ 4n/3] = Pr[ X − E[ X ] ≥ n/3]. Now Yi = Xi − E[ Xi ] has mean zero and takes values in {−1, 0, 1}. Therefore we can apply the following concentration inequality (which is a very useful tool in several areas of combinatorics). Theorem 1.5 (Hoeffding bound). Let for i = 1, . . . , n Xi be a mean zero random variable taking values in [ ai , bi ] ⊂ R. Then for any λ > 0, n

Pr[ ∑ Xi > λ] ≤ e−λ

2 /2

∑in=1 (bi − ai )2 .

i =1

We give the proof below, but let us first see how this helps us. From the Hoeffding bound we obtain that with λ = 1/3n, and ai = −1, bi = 1, Pr[ X − E[ X ] ≥ n/3] < e−n/72 , From which we deduce that indeed | Mn4n/3 |3−n is exponentially small and form which we can derive the existence of a constant c < 3 such that Theorem 1.1 holds. We refer to  for a more precise estimate. We now give a proof of the Hoeffding bound. 3

Proof of Theorem 1.5. Let λ > 0. Then for any i, E[eλXi ] ≤

− ai λbi bi eλai + e . bi − a i bi − a i

(6)

Indeed, for any x we have by convexity of the exponential function, bi − x

eλx = e bi −ai

x−a

λai + b −ai λbi i

i

bi − x λai x − ai λbi e + e , bi − a i bi − a i

Taking the expectation on both sides and using that E[ Xi ] = 0, (6) now follows. We next claim that for or any p ∈ [0, 1] and t > 0, pe−t(1− p) + (1 − p)etp ≤ 1/2(et + e−t ) ≤ et

2 /2

.

(7)

Indeed, since (0, pe(1− p)t + (1 − p)e− pt ) = (1 − p)(− p, e− pt ) + p(1 − p, e(1− p)t ) it is below the line connecting (−1, e−t ) and (1, et ) in R2 by the convexity of the exponential function. This shows the first inequality. The second inequality follows by looking at the Taylor series. bi Now by taking p = bi − ai , t = λ ( bi − ai ) and combining (6) and (7) we obtain, that for any i, 2 2 (8) E[eλXi ] ≤ eλ (bi −ai ) /2 . Now we turn to the proof of the theorem. Let µ := ∑in=1 (bi − ai )2 . eλ

2 /µ

n

Pr[ ∑ Xi > λ] = eλ

2 /µ

n

Pr[eλ ∑i=1 Xi /µ > eλ

2 /µ

n

] ≤ E[eλ ∑i=1 Xi /µ ]

i =1

n

n

= ∏ E[eλXi /µ ] ≤ ∏ e i =1

λ2 (bi − ai )2 /2 µ2

=e

∑in=1

λ2 (bi − ai )2 /2 µ2

= eλ

2 /2µ

,

(9)

i =1

where the first inequality is due to Markov’s inequality, the second equality by indepen2 dence of the Xi and the last inequality by (8). Now multiplying (9) with e−λ /µ gives the theorem.

2

Hilbert’s Nullstellensatz

In this section we will very briefly discuss Hilbert’s Nullstellensatz. Let F be an algebraically closed field. You may think of F = C. A set I ⊂ F[ x1 , . . . , xn ] is called an ideal if for all p, q ∈ I, p + q ∈ I and if for each p ∈ I and q ∈ F[ x1 , . . . , xn ], pq ∈ I. A typical example of an ideal is a vanishing ideal defined for a subset X ⊂ Fn by

I( X ) := { p ∈ F[ x1 , . . . , xn ] | p( x ) = 0 for all x ∈ X }. Another common way of obtaining ideals is by having collection of polynomials { p j } ⊂ F[ x1 , . . . , xn ] and letting I be the smallest ideal containing all the p j . Hilbert’s Nullstellensatz says something about common zeros of ideals. Theorem 2.1 (Weak Nullstellensatz). Let F be an algebraically closed field. Let I ⊆ F[ x1 , . . . , xn ] be an ideal. Then 1 ∈ / I if and only if there exists a point a ∈ Fn such that for each p ∈ I p( a) = 0. We refer to Lang  for a proof of this result. You will give a prove of this theorem for F = C in one of the exercises. We can use the weak Nullstellensatz for combinatorial applications. 4

Proposition 2.2 (). Let F be an algebraically closed field. A graph G = (V, E) has stability number at least k if and only if 1 is not contained in the ideal generated by the following system of equations: xi2 − xi = 0 xi x j = 0

for all i ∈ V for all {i, j} ∈ E

∑ xi = k.

i ∈V

Proof. First note that G has stability number at least k if and only if there exists a solution to the above system of equations. Indeed, if G has a stable set of size k, assigning 1 to the variables corresponding to the members of this stable set and 0 to the remaining variables we obtain a solution. Conversely, if there exists a solution, then the variables taking the value 1 correspond to an independent set of size k. The Weak Nullstellensatz now implies that 1 is contained in the ideal if and only if G has no independent set of size k. For the next application we will assume that F is of characteristic zero. Proposition 2.3 (). Let F be an algebraically closed field of characteristic zero. A graph G = (V, E) with vertex set {1, . . . , n} is k-colorable if and only if 1 is not contained in the ideal generated by following system of equations: xik − 1 = 0 xik−1

+

xik−2 x j

+···+

xi x kj −2

+

x kj −1

=0

for all

i = 1, . . . , n

(10)

for all

ij ∈ E.

(11)

Proof. Let I be the ideal generated by the polynomials describing the equations. First of all note that the system of equations has a solution if and only if G is k-colorable. Indeed, suppose G is k-colorable. Fix a coloring f : V → [k ], let ζ be a primitive kth root of unity and let xi = ζ f (i) . Then x = ( x1 , . . . , xn ) is a common zero of I. It clearly satisfies (10). To see that it also satisfies (11) note first that x k − 1 = ( x − 1)( x k−1 + x k−2 + · · · + x + 1).

(12)

For an edge ij we have f (i ) 6= f ( j) and so we may assume that xi = ζ a and x j = ζ b with a < b. Then by (12) we have 1 + ζ b−a + ζ 2b−2a + ζ 3b−3a + · · · + ζ (k−1)(b−a) = 0. Multiplying this by (ζ a )k−1 we see that (11) is satisfied. Conversely, by the same argument, any solution to the system directly gives a k-coloring of G. Now utilizing the Weak Nullstellensatz we see that if 1 ∈ I, then the system has no solution and hence G is not k-colorable. While if 1 ∈ / I, this implies there exists a solution, which implies G is k-colorable. From the Weak Nullstellensatz one can deduce the original Nullstellensatz by Rabinowitch’ trick. We first need a definition. For an ideal I we define its vanishing set

V ( I ) := { a ∈ Fn | p( a) = 0 for all p ∈ I }. We clearly have that I ⊆ I(V ( I )). 5

Theorem 2.4 (Nullstellensatz). Let F be an algebraically closed field. Let I ⊆ F[ x1 , . . . , xn ] be an ideal. Then √ I(V ( I )) = I := { p | there exists k ∈ N such that pk ∈ I }. Proof. Let f ∈ I(V ( I )). Consider the ideal I 0 ⊆ F[ x0 , x1 , . . . , xn ] generated by I and 1 − x0 f . Then there exists no a = ( a0 , a1 , . . . , an ) ∈ Fn+1 such that for each p ∈ I 0 p( a) = 0 as by definition we have that if p ∈ I and p( a) = 0, then f ( a) = 0, but then 1 − a0 f ( a) 6= 0. So by the Weak Nullstellensatz we have that 1 is contained in I 0 . In other words there exists p1 , . . . pt ∈ I and q0 , q1 , . . . , qt ∈ F[ x0 , . . . , xn ] such that t

1=

∑ q i p i + q0 (1 − x0 f ).

i =1

Plugging in x0 = 1/ f in this equation we obtain the equation, t

1=

∑ qi ( f (x1 , . . . , xn )−1 , x1 , . . . , xn )) pi (x1 , . . . , xn ).

i =1

Multiplying both sides with large enough powers of f ( x1 , . . . , xn ) we obtain that f k ∈ I for some k. For a graph G = (V, E) on vertex set {1, . . . , n} define the following polynomial in n variables: f G : = ∏ ( x i − x j ). ij∈ E,i < j

Utilizing the Nullstellensatz we obtain the following: Proposition 2.5. Let F be an algebraically closed field of characteristic zero. A graph G on the n vertices {1, 2, ..., n} is not k-colorable if and only if there exists ` ∈ N such that the polynomial f G` lies in the ideal Ik generated by the polynomials xik − 1 (i = 1, . . . , n). Proof. Suppose we can find a proper coloring f : V → {1, . . . , k }. Let ζ be a primitive k-th root of unity. Then evaluating f G at the point a ∈ Fn with ai = ζ f (i) we have that since f is a proper coloring that f G ( a) 6= 0. But then f G` cannot be contained in Ik for any `, as each element of Ik vanishes at all points whose coordinates are roots of unity. Suppose conversely that is not k-colorable. Then f G ( a) = 0 for each a ∈ Fn for which each ai a kth root of unity. This implies that f G ∈ I(V ( Ik )), which by the Nullstellensatz means that some power of f is contained in Ik . This result can be strengthened by using a combinatorial form of the Nullstellensatz: Proposition 2.6 (Alon and Tarsi ). Let F be an algebraically closed field of characteristic zero. A graph G on the n vertices {1, 2, ..., n} is not k-colorable if and only if the polynomial f G lies in the ideal Ik generated by the polynomials xik − 1 (i = 1, . . . , n).

3

Real rooted polynomials and interlacing

It is clear that if one has two polynomials f , g with only real roots, this does not automatically imply that f + g is real rooted (we call a polynomial f ∈ R[ x ] real rooted if all 6

roots of f are real). Also it is possible for three polynomials to sum to a polynomial that is not real rooted, while every two of the three polynomials sum op to a real rooted polynomial. For example, if f 1 = x2 + 2x, f 2 = x2 − 2x, and f 3 = − x2 + 1 then the roots of f 1 + f 2 + f 3 = x2 + 1 are i and −i and hence not real. There are however situations where the sum of several real-rooted polynomials is real rooted. This is the topic of the current section. Let f 1 , . . . , f k be polynomials in R[ x ]. They are called compatible if ∑ik=1 ci f i is real rooted for all c1 , . . . , ck ≥ 0. Let ( a1 , . . . , am ) and (b1 , . . . , bn ) be two sequences that are monotonically non-increasing (i.e. a1 ≥ a2 ≥ . . . ≥ am ). The first sequence is said to interlace the second sequence if n ≤ m ≤ n + 1 and if the sequence ( a1 , b1 , a2 , b2 , . . .) is another monotonically non-increasing sequence. Let f ∈ R[ x ] be real rooted of degree d. Let r1 ≥ r2 ≥ . . . ≥ rd be the roots of f . Then (r1 , . . . , rd ) is called the root sequence of f . Let f 1 , . . . , f k ∈ R[ x ] be real-rooted polynomials with positive leading coefficients. A common interlacer for f 1 , . . . , f k is a sequence (c1 , c2 , . . .) that interlaces the root sequence of each of the f i . The aim of this section is to prove the following theorem due to Chudnovsky and Seymour . Theorem 3.1 (). Let f 1 , . . . , f k ∈ R[ x ] be real-rooted polynomials with positive leading coefficients. The the following are equivalent: (1) for each 1 ≤ s < t ≤ k, f s , f t are compatible, (2) for all 1 ≤ s < t ≤ k, the polynomials f s , f t have a common interlacer, (3) f 1 , . . . , f k have a common interlacer, (4) f 1 , . . . , f k are compatible. To prove this theorem, we shall first need various other results from . Lemma 3.2. If f and g are compatible polynomials with positive leading coefficients, then | deg( f ) − deg( g)| ≤ 1. Proof. The proof is by induction on max{deg( f ), deg( g)}. Assume first that one of f , g is the constant function, say f ( x ) = c. Then c > 0, as f has positive leading coefficient. Looking at h = g + λ f for λ large enough, it follows that h has at most one real root. Hence by compatibility of f and g, it follows that the degree of h is at most one. So therefore the degree of g is at most one, as desired. So we may now assume that both the degree of f and g are at least one. We next claim that the derivatives f 0 and g0 are compatible. This follows from the fact that for a, b ≥ 0, a f 0 + bg0 = ( a f + bg)0 has only real roots, as between any pair of roots of a f + bg there is a root of its derivative (counting multiplicities). As both f 0 and g0 have positive leading coefficient, we obtain by induction that

| deg( f ) − deg( g)| = | deg( f 0 ) − deg( g0 )| ≤ 1, as desired. For a polynomial f and a ∈ R we denote by n f ( a) the number of roots of f in [ a, ∞), counting multiplicities. Lemma 3.3. If f and g are compatible polynomials with positive leading coefficients, then for any c ∈ R, |n f (c) − n g (c)| ≤ 1. 7

Proof. The proof is by induction on max{deg( f ), deg( g)}. We may assume that f and g do not have any common roots, as these contribute the same number to both n f (c) and n g (c). (Factoring out the greatest common divisor does not affect the compatibility, nor the positivity of the leading coefficient.) Suppose for contradiction that n f (c) − n g (c) ≥ 2 for some c ∈ R. We may assume that c is a root of f and in fact we may assume that c is the largest root of f for which n f (c) − n g (c) ≥ 2. As c is a root of f , it is not a root of g. We next claim that n f (c) − n g (c) = 2.

(13)

Indeed suppose that n f (c) − n g (c) ≥ 3. We have n f 0 (c) = n f (c) − 1 and n g0 (c) ≤ n g (c), as between every pair of real roots there is a root of the derivative. This implies that n f 0 (c) − n g0 ≥ n f (c) − n g (c) − 1 ≥ 2, contradicting our inductive hypothesis. This shows (13). Now choose b strictly larger than all roots of f and all roots of g. Since both g and f have positive leading coefficient, it follows that both f (b) and g(b) are positive. Since n f (c) − n g (c) = 2 (which is even), we can choose a < c such that both f and g have no roots in the interval [ a, c) and such that f ( a) and g( a) have the same sign. Now we have n f ( a ) − n f ( b ) = n f ( c ) 6 = n g ( c ) = n g ( a ) − n g ( b ).

(14)

Now define for t ∈ [0, 1] the polynomial pt := t f + (1 − t) g. Then for each t ∈ [0, 1], pt is real rooted, as f and g are compatible. Also pt ( a) and pt (b) are nonzero, as f and g have the same sign at a and b. Since the roots of pt depend continuously on its coefficients and hence on t, it follows that the number of roots of pt in the interval ( a, b) is independent of t. However p1 has n f ( a) − n f (b) roots and p0 has n g ( a) − n g (b) roots in ( a, b). This contradicts (14) and finishes the proof. Lemma 3.4. Let f , g be two real-rooted polynomials. Then f and g have a common interlacer if and only if |n f (c) − n g (c)| ≤ 1 for all c ∈ R. Proof. We may assume that d = deg( f ) ≥ deg( g). Let ( a1 , . . . , ad ) be the root sequence of f and let (b1 , b2 . . . , bdeg( g) ) be the root sequence of g. Suppose (c1 , c2 , . . . , cd ) is an interlacer for f and g. Let c ∈ R. We may assume c1 ≥ c ≥ cd . Choose i, largest such that ci ≥ c ≥ ci+1 . Then n f (c) ∈ {i − 1, i } and n g (c) ∈ {i − 1, i }, which implies |n f (c) − n g (c)| ≤ 1, as desired. Conversely, suppose |n f (c) − n g (c)| ≤ 1 for all c ∈ R. Then we can build a common interlacer by choosing c1 = max{ a1 , b1 } and c2 = max{ a2 , b2 } etc. If deg( f ) > deg( g), we set cd = ad . It is easy to see that ak ≥ bk+1 and bk ≥ ak+1 for any k. This implies that (c1 , c2 , . . . , cd ) is a common interlacer for f and g. We need one more observation before we prove Theorem 4.1. Given a real-rooted polynomial f with root sequence (r1 , . . . , rd ) we define the root intervals of f , I1 , . . . , Id+1 as follows: I1 := [r1 , ∞), for d > i > 1 we set Ii := [ri , ri−1 ] and Id+1 := (−∞, rd ]. (In case d = 0 we set I1 = R). Then a sequence (c1 , . . . , cm ) is an interlacer for f if and only if m ∈ {d, d + 1} and ci ∈ Ii for each i = 1, . . . , m.

8

Proof of Theorem 4.1. It is clear that (4) implies (1). We will show that (1) implies (2), (2) j j implies (3) and (3) implies (4). Let di = deg( f i ) and let d := maxik=1 di . Let (r1 , . . . , rd j ) be j

j

the root sequence of f j for j = 1, . . . , k. Let for j = 1, . . . , k, I1 , . . . , Id j +1 be the root intervals of f j . Proof of (1) implies (2). Let 1 ≤ s < t ≤ k. Let d∗ = min{ds , dt }. Then d∗ ≥ max{ds , dt } − 1 by Lemma 3.2. By the observation above it suffices to show that for each 1 ≤ j ≤ d∗ + 1, Ijs ∩ Ijt 6= ∅. Suppose for contradiction that the intersection Ijs ∩ Ijt is empty for some j. Let j be the smallest j such that Ijs ∩ Ijt = ∅. Then j ≥ 2. By symmetry we may assume that r sj−1 ≤ r tj−1 . In particular, r tj exists and r sj−1 < r tj (otherwise Ijs ∩ Ijt would not be empty.) But then n f t (r tj ) = j and n f s (r sj ) = j − 2, contradicting Lemma 3.3. Proof of (2) implies (3). From (2) it follows that for each 1 ≤ s < t ≤ k and each j = T T 1, . . . , d, Ijs ∩ Ijt 6= ∅. This implies that ks=1 Ijs 6= ∅ for each j. So choosing p j ∈ ks=1 Ijs for each j we have that ( p1 , . . . , pd ) is a common interlacer for f 1 , . . . , f k . Proof of (3) implies (4). The proof is by induction on d. We may assume that no x0 ∈ R is a root of all the f i . Otherwise we set gi = f i /( x − x0 ), and g1 , . . . , gk still has a common interlacer by (2) and Lemma 3.4. Consider for certain ci ≥ 0, f := ∑ik=1 ci f i . We need to show that all roots of f are real. We may assume that all ci are positive. As the f i have a common interlacer it follows that for each i = 1, . . . , k, d − 1 ≤ di ≤ d. So we may assume that there exists a common interlacer with d terms, say ( p1 , . . . , pd ). Fix i ∈ {1, . . . , k }. Since the leading term of f i is positive, and since ( p1 , . . . , pd ) interlaces f i , we have f i ( p j ) ≥ 0 if j is odd and f i ( p j ) ≤ 0 if j is even. As this holds for all i and as the f i do not have a common root, it follows that f ( p j ) > 0 if j is odd and f ( p j ) < 0 is j is even. So for j = 1, . . . , d − 1 there exists p j > r j > p j+1 such that f (r j ) = 0. That is, we found d − 1 real roots of f . Since f is a real polynomial of degree d it has an even number of non-real roots. So this number must be zero and it follows that all roots of f are real. This completes the proof.

4

Roots of independence polynomials

Let G = (V, E) be a graph. A set U ⊂ V is called independent is no edge of G has both endpoints in U. A set F ⊂ E is called a matching if no two edges in F share a vertex. Alternatively, if F is an independent set in the linegraph of G. The independence polynomial of G is defined as IG ( x ) = ∑ x| I | . I ⊆V independent

The matching polynomial of G is defined as

MG ( x ) =

x| F| .

F⊆E matching

In this section we will prove some results on the location of roots of independence and matching polynomials, starting with the following beautiful result of Chudnovsky and 9

Seymour . A graph is called claw free if it does contain an induced graph isomorphic to K1,3 , a claw. Theorem 4.1 (). Let G be a claw free graph. Then all roots of IG ( x ) are real. To prove Theorem 4.1 we need some auxiliary results from . We start with the fundamental identity for the independence polynomial. Let G = (V, E) be a graph and let v be a verte of G. Then IG ( x ) = IG−v ( x ) + xIG\ N [v] ( x ),

(15)

where N [v] = {v} ∪ {u ∈ V | {u, v} ∈ E}, closed neighbourhood of v, where for a subset S ⊆ V, G \ S denotes the graph induced by V \ S and where G − v := G \ {v}. To see this split the collection of independent sets of G into two parts: one consisting of independent sets containing v and the other not containing v. We can split the sum for IG ( x ) accordingly, which gives exactly the right-hand side of (15). Generalizing (15), we have for a clique K of G (a complete subgraph), IG ( x ) = IG \ K ( x ) +

∑ xIG\ N[k] (x).

(16)

k∈K

A clique K is called simplicial of for each k ∈ K, the set N [k ] \ K is a clique. Lemma 4.2. Let G be a claw-free graph and let K be a simplicial clique in G. Then N [k ] \ K is a simplicial clique in G \ K for each k ∈ K. Proof. Let k ∈ K. By definition, N [k ] \ K is a clique in G. So it suffices to show it is simplicial. To this end let n ∈ N [k ] \ K and suppose that N [n] \ K ∪ N [k ] is not a clique. Choose two non-adjacent vertices x, y ∈ N [n] \ K ∪ N [k]. Then { x, y, n, k } induces a claw in G. Indeed, x, y, k are all connected to n but are pairwise not connected, as by definition x, y are elements of N [n] \ K ∪ N [k ] and hence they are no neighbours of k. However we assumed that G is claw-free so this is a contradiction and we must conclude that N [n] \ K ∪ N [k ] is a clique in G and hence N [n] \ Kis a clique in G \ K. This finishes the proof. The following result about the matching polynomial was proved by Heilmann and Lieb : Theorem 4.3. Let G be a graph of maximum degree at most ∆ with ∆ ≥ 2. Then every root x0 of MG ( x ) is real and satisfies x0 ≤ 4∆−−1 1 . You are asked to prove this in one of the exercises.

References  Alon, Noga, and Michael Tarsi. "Colorings and orientations of graphs." Combinatorica 12.2 (1992): 125–134.  Chudnovsky, Maria, and Paul Seymour. "The roots of the independence polynomial of a clawfree graph." Journal of Combinatorial Theory, Series B 97.3 (2007): 350–357.  Ellenberg, Jordan S., and Dion Gijswijt. "On large subsets of Fnq with no three-term arithmetic progression." Annals of Mathematics 185.1 (2017): 339–343. 10

 Heilmann, Ole J., and Elliott H. Lieb. "Theory of monomer-dimer systems." Statistical Mechanics. Springer Berlin Heidelberg, 1972. 45–87.  Lang, Serge. "Algebra", volume 211 of Graduate Texts in Mathematics. (2002).  Loera, J. A., et al. "Expressing combinatorial problems by systems of polynomial equations and Hilbert’s nullstellensatz." Combinatorics, Probability and Computing 18.04 (2009): 551-582.  Lovász, László. "Stable sets and polynomials." Discrete mathematics 124.1-3 (1994): 137153.

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## Supplementary lecture notes AMC 2017

May 14, 2017 - âThese are supplementary lecture notes for the spring 2017 Mastermath course Algebraic Methods in Com- binatorics. ..... It is clear that if one has two polynomials f, g with only real roots, this does not auto- ..... Mechanics.

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