ADVANCES ON THE BESSIS- MOUSSA-VILLANI TRACE CONJECTURE

ADVANCES ON THE BESSIS- MOUSSA-VILLANI TRACE CONJECTURE CHRISTOPHER J. HILLAR Abstract. A long-standing conjecture asserts that the polynoial p(t = Tr(A + tb ] has nonnegative coefficients whenever is a positive integer and A and B are any two n n positive seidefinite Heritian atrices. The conjecture arises fro a question raised by Bessis, Moussa, and Villani (1975 in connection with a proble in theoretical physics. Their conjecture, as shown recently by Lieb and Seiringer, is equivalent to the trace positivity stateent above. In this paper, we derive a fundaental set of equations satisfied by A and B that iniize or axiize a coefficient of p(t. Applied to the Bessis- Moussa-Villani (BMV conjecture, these equations provide several reductions. In particular, we prove that it is enough to show that (1 it is true for infinitely any, (2 a nonzero (atrix coefficient of (A + tb always has at least one positive eigenvalue, or (3 the result holds for singular positive seidefinite atrices. Moreover, we prove that if the conjecture is false for soe, then it is false for all larger. Finally, we outline a general progra to settle the BMV conjecture that has had soe recent success. Added in Proof: Exaining the arguents below, it is straightforward to verify that all of the stateents of results hold with positive seidefinite (coplex Heritian replaced with real positive seidefinite. Thus, for instance, if there is a real, negative counterexaple to the BMV conjecture for soe, then there are real counterexaples for all larger powers. 1. Introduction In 1975, while studying partition functions of quantu echanical systes, Bessis, Moussa, and Villania forulated a conjecture regarding a positivity property of traces of atrices 1]. If this property holds, explicit error bounds in a sequence of Padé approxiants follow. Let A and B be n n Heritian atrices with B positive seidefinite, and let φ A,B (t = Trexp (A tb]. The original forulation of the conjecture asserts that the function φ A,B is copletely onotone; in other words, φ A,B is the Laplace transfor of a positive easure µ A,B supported on 0, : Trexp (A tb] = 0 exp( txdµ A,B (x. 1991 Matheatics Subject Classification. 15A24, 15A45, 15A90, 33Cxx, 44A10, 47A50, 47N50, 49J40. Key words and phrases. Bessis-Moussa-Villani (BMV conjecture, positive definite atrices, trace inequality, Euler-Lagrange equations, words in two atrices. Supported under a National Science Foundation Postdoctoral Research Fellowship. 1

2 CHRISTOPHER J. HILLAR Equivalently, the derivatives of the function f(t = φ A,B (t alternate signs: ( 1 f ( (t 0, t > 0, = 0, 1, 2,.... Since its introduction in 1], any partial results and substantial coputational experientation have been given 2, 3, 4, 5, 7, 11, 8, 14, 15], all in favor of the conjecture s validity. However, despite uch work, very little is known about the proble, and it has reained unresolved except in very special cases. Recently, Lieb and Seiringer in 13], and as previously counicated to us 11], have reforulated the conjecture of 1] as a question about the traces of certain sus of words in two positive definite atrices. In what follows, we shall use the standard convention that a positive definite (resp. positive seidefinite atrix is one that is coplex Heritian and has positive eigenvalues (resp. nonnegative eigenvalues. Conjecture 1.1 (Bessis-Moussa-Villani. The polynoial p(t = Tr (A + tb ] has all nonnegative coefficients whenever A and B are n n positive seidefinite atrices. Reark 1.2. Although not iediately obvious, the polynoial p(t has all real coefficients (see Corollary 2.4. The coefficient of t k in p(t is the trace of S,k (A, B, the su of all words of length in A and B, in which k B s appear (it has been called the k-th Hurwitz product of A and B. In 11], aong other things, it was noted that, for < 6, each constituent word in S,k (A, B has nonnegative trace. Thus, the above conjecture is valid for < 6 and arbitrary positive integers n. It was also noted in 11] (see also 1] that the conjecture is valid for arbitrary and n < 3. Thus, the first case in which prior ethods did not apply and the conjecture was in doubt, is = 6 and n = 3. Even in this case, all coefficients, except TrS 6,3 (A, B], were known to be nonnegative (also as shown in 11]. It was only recently 8], using heavy coputation, that this reaining coefficient was shown to be nonnegative. Much of the subtlety of Conjecture 1.1 can be seen by the fact that S,k (A, B need not have all nonnegative eigenvalues, and in addition that soe words within the S,k (A, B expression can have negative trace (see 11], where it is shown that TrABABBA] can be negative. Our advanceent is the introduction of a fundaental pair of atrix equations satisfied by A and B that iniize or axiize a coefficient of p(t. In what follows, we will be using the natural Euclidean nor on the set of coplex n n atrices: A = TrAA ] 1/2. (Here, C denotes the conjugate transpose of a coplex atrix C. The precise stateent of our ain result is the following. Theore 1.3. Let > k > 0 be positive integers, and let A and B be positive seidefinite atrices of nor 1 that iniize (resp. axiize the quantity TrS,k (A, B] over all positive seidefinite atrices of nor 1. Then, A and B satisfy the following pair of equations: { AS 1,k (A, B = A 2 TrAS 1,k (A, B] (1.1 BS 1,k 1 (A, B = B 2 TrBS 1,k 1 (A, B].

ADVANCES ON THE BESSIS-MOUSSA-VILLANI TRACE CONJECTURE 3 We call (1.1 the Euler-Lagrange equations for Conjecture 1.1. The nae coes fro the reseblance of our techniques to those of coputing the first variation in the calculus of variations. We should reark that there have been other variational approaches to this proble 2, 3]; a review can be found in 15]. Although we are otivated by Conjecture 1.1, we discovered that these equations are also satisfied by a iniization (resp. axiization over Heritian atrices A and B of nor 1 (see Corollary 3.7, and it is natural to consider this ore general situation. In this regard, we present the following application of the Euler-Lagrange equations. Theore 1.4. If A and B are Heritian atrices of nor 1 and > 1, then ( TrS,k (A, B]. k Moreover, if > k > 0, then equality holds only when A = ±B, and if in addition > 2, then A has precisely one nonzero eigenvalue. Reark 1.5. When = 1, this theore fails to hold. For exaple, let A be the n n diagonal atrix A = diag(n 1/2,..., n 1/2. Then A = 1, but TrS 1,0 (A, B] = TrA] = n 1/2 > 1 for n > 1. It is easy to see that this axiu is at least ( k, and using eleentary considerations involving the Cauchy-Schwartz inequality, one can show that TrS,k (A, B] S,k (A, B n 1/2 ( k A k B k n 1/2 = ( k n 1/2. However, we do not know if a dependency on the size of the atrices involved can be reoved without appealing to equations (1.1. As a strategy to prove Conjecture 1.1, we offer the following. Conjecture 1.6. Let > k > 0 be positive integers. Positive seidefinite (resp. Heritian atrices A and B of nor 1 that satisfy the Euler-Lagrange equations coute. Fro this result, Conjecture 1.1 would be iediate. Of course, Theore 1.4 iplies that Conjecture 1.6 holds for the case of Heritian axiizers and iniizers. We next list soe of the ajor consequences of the equations found in Theore 1.3. The first one iplies that counterexaples to Conjecture 1.1 are closed upwards. The precise stateent is given by the following. Theore 1.7. Suppose that there exist integers M, K and n n positive definite atrices A and B such that TrS M,K (A, B] < 0. Then, for any M and k K such that k M K, there exist n n positive definite A and B aking TrS,k (A, B] negative. Corollary 1.8. If the Bessis-Moussa-Villani conjecture is true for soe 0, then it is also true for all < 0. Corollary 1.8 also reduces the BMV conjecture to its asyptotic forulation. Corollary 1.9. If the Bessis-Moussa-Villani conjecture is true for infinitely any, then it is true for all. Corollary 1.9 otivates a general progra to solve the BMV conjecture, and there is evidence that this approach is ore than a theoretical possibility. For

4 CHRISTOPHER J. HILLAR instance, Hägele 6] has used this approach and Corollary 1.8 to prove the conjecture for all 7 (and all n. Inspired by Hägele s ideas, Klep and Schweighofer 12] used seidefinite prograing techniques to prove the conjecture for all 9. It should be noted that these techniques provably fail for the difficult = 6 case, aking the appeal to Corollary 1.8 fundaental. A next result characterizes the BMV conjecture in ters of the eigenvalues of the atrix S,k (A, B. Theore 1.10. Fix positive integers k and n. Then, TrS,k (A, B] 0 for all n n positive seidefinite A and B if and only if whenever S,k (A, B 0, it has at least one positive eigenvalue. Reark 1.11. This theore can be viewed as a transfer principle for the BMV conjecture: instead of proving positivity for the su of all the eigenvalues, we need only show it for at least one of the. Thus, our original conjecture can be ade to reseble a variant of Perron s Theore for nonnegative atrices. Conjecture 1.12. Fix positive integers k and n, and positive seidefinite n n atrices A and B. Then S,k (A, B either has a positive eigenvalue or is the zero atrix. Our final result generalizes a fact first discovered in 8] (there only the real case was considered, and it iplies that it is enough to prove the Bessis-Moussa-Villani conjecture for singular A and B. Theore 1.13. Let, n be positive integers, and suppose that Tr (A + tb 1] has nonnegative coefficients for each pair of n n positive seidefinite atrices A and B. If p(t = Tr (A + tb ] has nonnegative coefficients whenever A, B are singular n n positive seidefinite atrices, then p(t has nonnegative coefficients whenever A and B are arbitrary n n positive seidefinite atrices. The organization of this paper is as follows. In Section 2, we recall soe facts about Hurwitz products, and in Section 3 we derive the two equations found in Theore 1.3. Finally, in Section 4, we use these equations to prove our ain Theores 1.4, 1.7, 1.10, and 1.13. 2. Preliinaries We begin with a review of soe basic facts involving Hurwitz products; soe of this aterial can be found in 8]. The coefficients S,k (A, B ay be generated via the recurrence: (2.1 S,k (A, B = AS 1,k (A, B + BS 1,k 1 (A, B. The following leas will be useful for coputing the traces of the S,k. Lea 2.1. Fix integers > k 0. For any two n n atrices A and B, we have Tr S,k (A, B] = k Tr AS 1,k(A, B].

ADVANCES ON THE BESSIS-MOUSSA-VILLANI TRACE CONJECTURE 5 Proof. Consider the following chain of equalities: ] 0 = Tr (A + tb i 1 (A A (A + tb i = Tr A (A + tb 1] ] Tr (A + tb i 1 A (A + tb i = Tr A (A + tb 1] d Tr dy y=1 (Ay + tb]. y=1 = Tr A (A + tb 1] d dy Tr (Ay + tb ] Since S,k (Ay, B = y k S,k (A, B, it follows that the coefficient of t k in the last expression above is just TrAS 1,k (A, B] ( ktrs,k (A, B]. Lea 2.2. Fix integers k > 0. For any two n n atrices A and B, we have Tr S,k (A, B] = k Tr BS 1,k 1(A, B]. Proof. Follows fro Lea 2.1 by taking the trace of both sides of equation (2.1. Let A and B be n n Heritian atrices. Since S,k (A, B is the su of all words of length in A and B with k B s, it follows that the conjugate transpose of S,k (A, B siply perutes its constituent suands. This verifies the following fact. Lea 2.3. If A and B are n n Heritian atrices, then the atrix S,k (A, B is Heritian. Corollary 2.4. The polynoial p(t = Tr (A + tb ] has all real coefficients whenever A and B are n n Heritian atrices. Although S,k (A, B is Heritian for Heritian A and B, it need not be positive definite even when A and B are n n positive definite atrices, n > 2. Exaples are easily generated, and coputational experients suggest that it is usually not positive definite. Finally, we record a useful fact about positive definite congruence. Lea 2.5. Let C be any coplex n n atrix and let A be an n n positive seidefinite atrix. Then CAC is positive seidefinite. Proof. See 9, p. 399]. 3. Derivation of the Euler-Lagrange Equations The arguents for our ain theores are based on a variational observation. It says that an expression TrS,k (A, B] is iniized or axiized when A and B satisfy the Euler-Lagrange equations (see Corollary 3.6. Before presenting a proof of this fact, we give a series of technical preliinaries.

6 CHRISTOPHER J. HILLAR Proposition 3.1. Let > k > 0 be positive integers. Let B be any given Heritian n n atrix, and suppose that A is a positive seidefinite atrix of nor 1 that iniizes (resp. axiizes TrS,k (A, B] over all positive seidefinite atrices of nor 1. Let ε > 0, and let C := C(x = (c rs (x be an n n atrix with entries c rs (x = u rs (x + iv rs (x in which u rs and v rs are differentiable functions u rs, v rs : ε, ε] R. Moreover, suppose that C(0 = I and CAC 0 for all x ε, ε]. Then Tr d dx ( CAC CAC S 1,k (A, B] x=0 = 0. Proof. Let A, B, and C be as in the stateent of the theore. Keeping in ind Corollary 2.4, we ay consider the differentiable function f : ε, ε] R given by ( ] CAC f(x = Tr S,k CAC, B. By hypothesis, the iniu (resp. axiu of f is achieved at x = 0. Consequently, it follows that df(x (3.1 dx = 0. x=0 Next, notice that, ( d CAC ] dx Tr CAC + tb ( CAC i 1 = Tr CAC + tb d dx When x = 0, the above expression evaluates to ( Tr (A + tb i 1 d CAC dx CAC = Tr d ( CAC dx CAC It follows, therefore, fro (3.1 that ( d CAC Tr dx CAC ( d CAC = Tr dx ] CAC + tb ( CAC ( CAC CAC + tb CAC + tb (A + tb i ] x=0 (A + tb 1 ] x=0 S 1,k (A, B] x=0 = 0. i ] A corresponding stateent can be ade by fixing A and iniizing (resp. axiizing over B. Proposition 3.2. Let > k > 0 be positive integers. Let A be any given Heritian n n atrix, and let B be a positive seidefinite atrix of nor 1 that iniizes (resp. axiizes TrS,k (A, B] over all positive seidefinite atrices of nor 1. Let ε > 0, and let C := C(x = (c rs (x be an n n atrix with entries c rs (x = u rs (x + iv rs (x in which u rs.

ADVANCES ON THE BESSIS-MOUSSA-VILLANI TRACE CONJECTURE 7 and v rs are differentiable functions u rs, v rs : ε, ε] R. Moreover, suppose that C(0 = I and CBC 0 for all x ε, ε]. Then ( d CBC Tr dx CBC S 1,k 1 (A, B] = 0. Proof. The proof is siilar to that of Proposition 3.1, so we oit it. In our next lea, we copute the derivative found in Propositions 3.1 and 3.2. For notational siplicity, the entry-wise derivative of the atrix C evaluated at the point x = 0 will be denoted by C. Lea 3.3. With the hypotheses as in Proposition 3.1, we have ( d CAC (3.2 dx CAC x=0 = C A + AC TrC A 2 ]A TrC A 2 ]A. Proof. A straightforward application of the product rule 10, p. 490] for (atrix differentiation gives ( d CAC dx CAC = d ( ( 1 dx CAC CAC 1 dc + CAC dx AC + CA dc. dx Next, we copute d dx CAC 1 = CAC 2 = CAC 2 d dx x=0 d dx CAC (TrCAC CAC ] 1/2 = (1/2 CAC 2 TrCAC CAC ] 1/2 d dx TrCAC CAC ]. d The product expansion of dx TrCAC CAC ] occurring in this last line is: dc Tr dx AC CAC + CA dc dx CAC + CAC dc ] dx AC + CAC CA dc dx ] ] dc dc = 2Tr dx AC CAC + 2Tr dx AC CAC. Finally, setting x = 0 and using the assuptions that A = 1 and C(0 = I, equation (3.2 follows. We now have enough to prove the ain results of this section. Theore 3.4. Let > k > 0 be positive integers. Let B be any given Heritian n n atrix, and let A be a positive seidefinite atrix of nor 1 that iniizes (resp. axiizes TrS,k (A, B] over all positive seidefinite atrices of nor 1. Then (3.3 AS 1,k (A, B = A 2 TrAS 1,k (A, B]. Proof. Let A and B be as in the hypotheses of the theore. By using different atrices C in the stateent of Proposition 3.1, we will produce a set of equations satisfied by the entries of AS 1,k (A, B that cobine to ake the single atrix equation (3.3. For ease of presentation, we introduce the following notation. For

8 CHRISTOPHER J. HILLAR integers r, s, let E rs denote the n n atrix with all zero entries except for a 1 in the (r, s entry. Fix integers 1 r, s n and take C = I + xe rs. Since C is invertible for all x 1/2, 1/2], it follows that CAC 0 for all such x. Therefore, the hypotheses of Lea 3.3 are satisfied. The forula there is ( d CAC dx CAC x=0 = C A + AC TrC A 2 ]A TrC A 2 ]A, in which C is the entry-wise derivative of the atrix C evaluated at the point x = 0. Additionally, Proposition 3.1, along with a trace anipulation, tells us that ( TrC A 2 ] + TrC A 2 ] Tr AS 1,k (A, B] (3.4 = Tr C AS 1,k (A, B + AC S 1,k (A, B] = Tr C AS 1,k (A, B] + Tr S 1,k (A, BAC ] = Tr C AS 1,k (A, B] + Tr C AS 1,k (A, B]. Since C = E rs, a coputation shows that for any atrix N, the trace of C N is just the (s, r entry of N. In particular, it follows fro (3.4 that the (s, r entries of AS 1,k (A, B + AS 1,k (A, B and (A 2 + A 2 TrAS 1,k (A, B] coincide. We have therefore proved that (3.5 AS 1,k (A, B + AS 1,k (A, B = (A 2 + A 2 TrAS 1,k (A, B]. We next perfor a siilar exaination using the atrices C = I + ixe rs to arrive at a second atrix identity. Cobining equation (3.2 and Proposition 3.1 as before, we find that (3.6 AS 1,k (A, B AS 1,k (A, B = (A 2 A 2 TrAS 1,k (A, B]. The theore now follows by adding these two equations and dividing both sides of the result by 2. Siilar arguents using Proposition 3.2 in place of Proposition 3.1 produce the following results. Theore 3.5. Let > k > 0 be positive integers. Let A be any given Heritian n n atrix, and let B be a positive seidefinite atrix of nor 1 that iniizes (resp. axiizes TrS,k (A, B] over all positive seidefinite atrices of nor 1. Then BS 1,k 1 (A, B = B 2 TrAS 1,k 1 (A, B]. Cobining the stateents of this section, we have finally derived the Euler- Lagrange equations (1.1 for Conjecture 1.1. Corollary 3.6. Let > k > 0 be positive integers, and let A and B be positive seidefinite atrices of nor 1 that iniize (resp. axiize the quantity TrS,k (A, B] over all positive seidefinite atrices of nor 1. Then A and B satisfy the following pair of equations: { AS 1,k (A, B = A 2 TrAS 1,k (A, B] BS 1,k 1 (A, B = B 2 TrBS 1,k 1 (A, B].

ADVANCES ON THE BESSIS-MOUSSA-VILLANI TRACE CONJECTURE 9 Our proof generalizes to show that the sae equations hold for Heritian iniizers (resp. axiizers, or ore generally, for classes of unit nor atrices with the sae inertia. This result is the ain ingredient in our proof of Theore 1.4 concerning the axiu of TrS,k (A, B]. Corollary 3.7. Let > k > 0 be positive integers, and let A and B be Heritian atrices of nor 1 that iniize (resp. axiize TrS,k (A, B] over all Heritian atrices of nor 1. Then A and B ust satisfy the following pair of equations: { AS 1,k (A, B = A 2 TrAS 1,k (A, B] BS 1,k 1 (A, B = B 2 TrBS 1,k 1 (A, B]. In general, we conjecture that trace iniizers coute (Conjecture 1.6, a clai that would iply Conjecture 1.1. We close this section with one ore application of the Euler-Lagrange equations. Corollary 3.8. Suppose that the iniu of TrS,k (A, B] over the set of positive seidefinite atrices is zero, and let A and B be positive seidefinite atrices that achieve this iniu. Then, S,k (A, B = 0. Proof. When k = or k = 0, the clai is clear. Therefore, suppose that > k > 0. Let A and B be positive seidefinite atrices with TrS,k (A, B] = 0. If either A or B is zero then the corollary is trivial. Otherwise, consider 0 = TrS,k(A, B] A k B k = TrS,k(Ã, B], in which à = A/ A and B = B/ B. Cobining equations (1.1 with the assuptions, it follows that ÃS 1,k(Ã, B = 0 and BS 1,k 1 (Ã, B = 0. Moreover, equation (2.1 iplies that S,k (Ã, B = ÃS 1,k(Ã, B + BS 1,k 1 (Ã, B = 0. Multiplying both sides of this identity by A k B k copletes the proof. 4. Proofs of the Main Theores We first use the Euler-Lagrange equations to prove Theore 1.4. Proof of Theore 1.4. Let > 1 and n be positive integers. Since our arguents are the sae in both cases, we consider deterining the axiu of TrS,k (A, B]. Let M be the copact set of Heritian atrices with nor 1 and choose (A, B M M that axiizes TrS,k (A, B]. If k = 0, then the desired inequality is of the for n n TrA ] λ i λ 2 i = A = 1, in which λ 1,..., λ n are the eigenvalues of A. A siilar arguent holds for = k. Therefore, we assue below that > k > 0. The Euler-Lagrange equations fro Corollary 3.7 iply that (4.1 AS 1,k (A, B = A 2 TrAS 1 (A, B]. Perforing a unifor, unitary siilarity, we ay assue that A is diagonal of the for A = diag(λ 1,..., λ r, 0,..., 0, in which λ 1,..., λ r are nonzero. Let

10 CHRISTOPHER J. HILLAR Ã = diag(λ 1 1,..., λ 1 r, 0,..., 0 be the pseudo-inverse of A, and set D = ÃA. Multiplying both sides of (4.1 by Ã, it follows that DS 1,k (A, B = ATrAS 1,k (A, B]. Taking the nor of both sides of this expression and applying Lea 2.1, we have ( k 1 (4.2 TrS,k(A, B] = DS 1,k (A, B S 1,k (A, B. k It follows that ( ( k TrS,k (A, B] 1 ( k k = k as desired. We next verify the final assertions in the stateent of the theore. Fro above, every inequality in the chain (4.2 is an equality. Thus, each ter occurring in S 1,k (A, B = W (A, B, W a su over length 1 words W with k B s, takes the value 1. In particular, we have that 1 = A k 1 B k. When 1 > k > 1, an application of Lea 4.1 below copletes the proof of the theore. The reaining cases k = 1 or = k + 1 are dealt with as follows. Without loss of generality, we ay suppose that k = 1 (interchange the roles of the atrices A and B. Applying the Cauchy-Schwartz inequality, we obtain the following chain of inequalities: ( n 2 n n (4.3 1 = TrA 1 B] 2 = λ 1 i b ii (λ 2 i 1 b 2 ii A B = 1. It follows that each inequality in (4.3 is an equality. In particular, the second-tolast identity says that B is diagonal. Moreover, equality in Cauchy-Schwartz iplies that λ 1 i = δb ii for soe real nuber δ and all i. Since 1 = n λ 1 i b ii = δ 2, it follows that A = ±B. If, in addition, > 2 and A has ore than 1 nonzero eigenvalue, then n n n 1 = λ 2 i > (λ 2 i 1 = b 2 ii = 1, a contradiction. Therefore, the conclusions of the theore hold for k = 1. Lea 4.1. Suppose that A and B are Heritian atrices of nor 1 and r > 0 and s > 1 are integers such that A r B s = 1. Then, A = ±B has only 1 nonzero eigenvalue. Proof. Perforing a unifor, unitary siilarity, we ay suppose that B is a diagonal atrix with entries less than or equal to 1 in absolute value. Fro the hypotheses, we have 1 = A r B s A r B s A r B s = 1. Therefore, B s = 1 = B, and since s > 1, this iplies that B has a single nonzero eigenvalue. It follows that A r B s = A r B = 1 is equal to the absolute value of the (1, 1 entry of A r. Finally, since A r = 1, the atrix A r has only one nonzero entry, and therefore, A has only one nonzero eigenvalue. Thus, A r = ±A and since A r = ±B, it follows that A = ±B. The arguent for our next result uses the following well-known fact; we provide a proof for copleteness (see also Theore 7.6.3 and Proble 9, p. 468 in 9].

ADVANCES ON THE BESSIS-MOUSSA-VILLANI TRACE CONJECTURE 11 Lea 4.2. If P and Q are positive seidefinite atrices, then P Q has all nonnegative eigenvalues. Proof. Suppose first that P is positive definite. Then P Q is siilar to P 1/2 P QP 1/2 = P 1/2 QP 1/2. In particular, P Q is siilar to a positive seidefinite atrix by Lea 2.5. Therefore, in this case P Q has all nonnegative eigenvalues. The general version of the clai now follows fro continuity. We are now prepared to present a proof that counterexaples to Conjecture 1.1 are closed upward. Theore 1.10 closely follows. Proof of Theore 1.7. Suppose that Conjecture 1.1 is false for soe and k and let A and B be real positive seidefinite atrices of unit nor that iniize TrS,k (A, B] = k TrAS 1,k(A, B] = k TrBS 1,k 1(A, B]. We show that for these sae atrices A and B, we have TrS +1,k (A, B] < 0 and TrS +1,k+1 (A, B] < 0. Cobining equation (2.1 and the identities (1.1 fro Corollary 3.6, it follows that (4.4 S,k (A, B = A 2 TrAS 1,k (A, B] + B 2 TrBS 1,k 1 (A, B]. This atrix is negative seidefinite since it is the su of two such atrices. Hence, the product AS,k (A, B has all non-positive eigenvalues by Lea 4.2. Thus, Lea 2.1 iplies that (4.5 TrS +1,k (A, B] = + 1 k TrAS,k(A, B] 0. In the case of equality, ultiplying equation (4.4 on the left by A and taking the trace of both sides, it follows that 0 = TrA 3 ]TrAS 1,k (A, B] + TrAB 2 ]TrBS 1,k 1 (A, B]. However, TrAB 2 ] 0 by Lea 4.2 and since A is nonzero, we ust have TrA 3 ] > 0. This gives a contradiction to equality in (4.5. It follows that TrS +1,k (A, B] < 0 as desired. In the sae anner, we can also prove that TrS +1,k+1 (A, B] = + 1 k + 1 TrBS,k(A, B] is negative. The conclusions of the theore now follow iediately. Proof of Theore 1.10. We first prove the direction ( using the contrapositive. Suppose that TrS,k (A, B] can be ade negative. The proof of Theore 1.7 shows that there exist positive seidefinite atrices A and B such that S,k (A, B is negative seidefinite and TrS,k (A, B] < 0 (so that S,k (A, B is nonzero. It follows that the second iplication in the stateent of the theore is false. The converse is clear. Finally, we work out the proof of Theore 1.13; the arguent is siilar in spirit to the proof of Theore 1.4.

12 CHRISTOPHER J. HILLAR Proof of Theore 1.13. Suppose we know that Conjecture 1.1 is true for the power 1 and also suppose that for soe k there exist n n positive definite atrices A and B such that TrS,k (A, B] is negative. Clearly, we ust have > k > 0. By hoogeneity, there are positive definite A and B with nor 1 such that TrS,k (A, B] is negative. Let M be the set of positive seidefinite atrices with nor 1 and choose (A, B M M that iniizes TrS,k (A, B]; our goal is to show that A and B ust both be singular. Suppose by way of contradiction that A is invertible. The Euler-Lagrange equations say that AS 1,k (A, B = A 2 TrAS 1,k (A, B]. Multiplying both sides of this equation by A 1 and taking the trace, it follows that TrS 1,k (A, B] = TrA]TrAS 1,k (A, B]. By hypothesis, TrS 1,k (A, B] is nonnegative. Therefore, using Lea 2.1, we have k Tr S,k(A, B] = Tr AS 1,k (A, B] = TrS 1,k(A, B] TrA] 0, a contradiction (TrA] is nonzero since A is nonzero. It follows that A ust be singular. A siilar exaination with B also shows that it ust be singular. Thus, if Conjecture 1.1 is true for singular A and B, it ust be true for invertible A and B as well. This copletes the proof of the theore. 5. Acknowledgents We would like to thank Scott Arstrong for several interesting and useful rearks concerning a preliinary version of this anuscript. References 1] D. Bessis, P. Moussa and M. Villani, Monotonic converging variational approxiations to the functional integrals in quantu statistical echanics, J. Math. Phys. 16 (1975, 2318 2325. 2] K. J. Le Couteur, Representation of the function Tr(exp(A λb as a Laplace transfor with positive weight and soe atrix inequalities, J. Phys. A: Math. Gen. 13 (1980, 3147 3159. 3] K. J. Le Couteur, Soe probles of statistical echanics and exponential operators, pp 209 235 in Proceedings of the International Conference and Winter School of Frontiers of Theoretical Physics, eds. F. C. Auluck, L.S. Kothari, V.S. Nanda, Indian National Acadey, New Dehli, 1977, Published by the Mac Millan Copany of India, 1978. 4] M. Drota, W. Schacherayer and J. Teichann, A hyper-geoetric approach to the BMVconjecture, Monatshefte fur Matheatik 146 (2005, 179 201. 5] M. Fannes and D. Petz, Perturbation of Wigner atrices and a conjecture, Proc. Aer. Math. Soc. 131 (2003, 1981 1988. 6] D. Hägele, Proof of the cases p 7 of the Lieb-Seiringer forulation of the Bessis-Moussa- Villani conjecture, J. Stat. Phys., to appear. (DOI: 10.1007/s10955-007-9327-8. 7] F. Hansen, Trace functions as Laplace transfors, J. Math. Phys., 47 043504 (2006. 8] C. Hillar and C. R. Johnson, On the positivity of the coefficients of a certain polynoial defined by two positive definite atrices, J. Stat. Phys., 118 (2005, 781 789. 9] R. Horn and C. R. Johnson, Matrix analysis, Cabridge University Press, New York, 1985. 10] R. Horn and C. R. Johnson, Topics in atrix analysis, Cabridge University Press, New York, 1991.

ADVANCES ON THE BESSIS-MOUSSA-VILLANI TRACE CONJECTURE 13 11] C. R. Johnson and C. Hillar, Eigenvalues of words in two positive definite letters, SIAM J. Matrix Anal. Appl., 23 (2002, 916 928. 12] I. Klep and M. Schweighofer, private counication, 2007. 13] E. H. Lieb and R. Seiringer, Equivalent fors of the Bessis-Moussa-Villani conjecture, J. Stat. Phys., 115 (2004, 185-190. 14] Nathan Miller, 3 3 cases of the Bessis-Moussa-Villani conjecture, Princeton University Senior Thesis, 2004. 15] P. Moussa, On the representation of Tr `e A λb as a Laplace transfor, Rev. Math. Phys. 12, 621 655 (2000. Departent of Matheatics, Texas A&M University, College Station, TX 77843 E-ail address: chillar@ath.tau.edu