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 [3]). 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 [3]. 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 [3] 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 [5] 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 ([7]). 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 ([6]). 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 [1]). 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 [2]. Theorem 3.1 ([2]). 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 [2]. 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 [2]. A graph is called claw free if it does contain an induced graph isomorphic to K1,3 , a claw. Theorem 4.1 ([2]). 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 [2]. 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 [4]: 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 [1] Alon, Noga, and Michael Tarsi. "Colorings and orientations of graphs." Combinatorica 12.2 (1992): 125–134. [2] 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. [3] 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

[4] Heilmann, Ole J., and Elliott H. Lieb. "Theory of monomer-dimer systems." Statistical Mechanics. Springer Berlin Heidelberg, 1972. 45–87. [5] Lang, Serge. "Algebra", volume 211 of Graduate Texts in Mathematics. (2002). [6] 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. [7] Lovász, László. "Stable sets and polynomials." Discrete mathematics 124.1-3 (1994): 137153.

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