Amoebas, Monge-Amp`ere measures and triangulations of the ...

Electronic versions of this document are available athttp://www.matematik.su.se/reports/2000/10Date of publication: June 21, 20001991 Mathematics Subject Classification: Primary 32A07 Secondary 32A05, 26A51,52B20.Postal address:Department of MathematicsStockholm UniversityS-106 91 StockholmSwedenElectronic addresses:http://www.matematik.su.seinfo@matematik.su.se

Proof. The linear function c α +〈α, x〉 coincides with N f on an open set, hencec α + 〈α, x〉 ≤ N f (x) for all x. Taking the supremum over all α in A proves part(i). If E α denotes the component of Ω \ A f of order α, thenc α + 〈α, x〉 ≤ S(x) = N f (x) = c α + 〈α, x〉, for x ∈ E α ,which shows that E α ⊂ {α} ∗ . Since grad N f = β in E β we have a strictinequality c α + 〈α, x〉 < N f (x) = S(x) for x ∈ E β and all β ≠ α, whichimplies that E α = {α} ∗ \ A f . If x ∈ S f then there exist α, β ∈ A such thatx ∈ {α} ∗ ∩ {β} ∗ ⊂ Ω \ ⋃ γ≠α E γ \ ⋃ γ≠β E γ = A f . Finally we note that theconnected components of Ω \ S f are precisely the sets {α} ∗ . Let us now turn to the question of how to compute the coefficients c α correspondingto a given Laurent polynomial f. If Γ is a face of the Newton polytopeof f(z) = ∑ α∈C w αz α , where C denotes a finite subset of Z n , let f| Γ denotethe truncation of f to Γ, that is f| Γ (z) = ∑ α∈Γ w αz α . It is well known thatwhen α ∈ Γ ∩ Z n , the complement of the amoeba of f has a component of orderα precisely if the complement of the amoeba of f| Γ does (see [3] Prop. 2.6 andalso [4]).Proposition 2. Let f be a Laurent polynomial and let Γ be a face of the Newtonpolytope of f. If α ∈ Γ and the complement of A f has a component of order α,then c α (f) = c α (f| Γ ). In particular, if α is a vertex of the Newton polytope off, then c α = log |w α |.Proof. Take an outward normal v to the Newton polytope at Γ. If α ∈ Γ andx is in the component of R n \ A f of order α, it is known that x + tv is also inthat component for all t > 0. Thereforec α (f) − c α (f| Γ ) = 1(2πi)∫Log n logf(z)dz 1 . . . dz n∣−1 (x+tv) f| Γ (z) ∣ .z 1 . . . z nand here the integrand clearly tends to 0 when t → ∞. To further describe the dependence of the coefficients c α on the polynomialf(z) = ∑ α∈C w αz α it is useful to introduce the functionsΦ α (w) = 1log(f(z)/z(2πi)∫Log α ) dz 1 . . . dz nn , (5)−1 (x) z 1 . . . z nwhere x is in the component of order α. This means that c α = Re Φ α . Noticethat Φ α is a holomorphic function in the coefficients w with values in C/2πiZ,defined whenever the complement of the amoeba of f has a component of orderα.Example. Suppose f(z) = (z + a 1 ) . . . (z + a N ) = w 0 + . . . + w N−1 z N−1 + z Nis a polynomial in one variable. Let 0 ≤ k ≤ N and assume that 0 < |a 1 |

. . . < |a k | < r < |a k+1 | < . . . < |a N |. Then one finds thatΦ k (w) = 1 ∫2πik∑∫==j=1N∑j=k+1|z|=r|z|=rlog(f(z)/z k )dzzlog((z + a j )/z)dzz+log a j = log(a k+1 . . . a N ).N∑j=k+1∫|z|=rlog(z + a j )dzzIn this case, we observe that Φ k (w) may be continued as a finitely branchedholomorphic function, whose branches correspond to various permutations ofthe roots of f. Incidentally, the sum of all branches of exp Φ k (w) is equal to w k .We now return to the general case and show that the functions Φ α verynearly satisfy a system of GKZ hypergeometric equations. We remark that thefunctions Φ α , or rather ∂Φ α /∂w α were used in [2] in the study of constant termsin powers of Laurent polynomials. The main result was obtained by showing,essentially, that the second term in the representation (6) is non-constant alongevery complex line parallel to the w α axis, provided that α is not a vertex ofthe Newton polytope of f.Theorem 2. Let f(z) = ∑ γ∈C w γz γ be a Laurent polynomial with C being afixed finite subset of Z n . Then the holomorphic functions Φ α have the powerseries expansionwhereΦ α (w) = log w α + ∑ (−k α − 1)!k∈K α∏β≠α k β! (−1)kα−1 w k , (6)K α = {k ∈ Z C ; k α < 0, k β ≥ 0 if β ≠ α, ∑ γk γ = 0, ∑ γγk γ = 0}. (7)The series converges for example when |w α | > ∑ β≠α |w β|. Moreover, Φ α satisfiesthe differential equations∑∑(∂ u − ∂ v )Φ α = 0 if (u γ − v γ ) = 0 and γ(u γ − v γ ) = 0 (8)γγand∑w γ ∂ γ Φ α = 1 (9)γ∑γw γ ∂ γ Φ α = α. (10)γHere we have used the notation ∂ γ = ∂/∂w γ and ∂ u for u ∈ N C is the obviousmultiindex notation for a higher partial derivative.6

Proof.∑Let M be an invertible n×n matrix with integer entries and let f(z) =∑α a αz α be any Laurent polynomial. Define a new Laurent polynomial Mf =α a αz Mα where a multiindex α is regarded as a column vector. It is notdifficult to show that the linear mapping x ↦→ M T x (where T denotes transpose)takes the amoeba of Mf onto the amoeba of f.Now let f be the special polynomial in the proposition and consider M whichmap the Newton polytope of f onto a translate of itself. The set of such Mcan be identified with the symmetric group S n+1 via its action on the vertices.Then Mf coincides with f up to an invertible factor, hence they have the sameamoeba. In particular, M T maps the component of order (1, . . . , 1) in thecomplement of A f (if it exists) onto itself. If x is any point in that component,then the convex hull of the points M T x, where M ranges over S n+1 , containsthe origin. Since every component in the complement of the amoeba is convex,it follows that the component of order (1, . . . , 1) contains the origin. Conversely,if some component E contains the origin, then M T E also contains the originand hence coincides with E. This implies that E has order (1, . . . , 1).Now, the amoeba of f contains the origin precisely ifa = −(1 + z1 n+1 + . . . + znn+1 )/z 1 . . . z nwith |z 1 | = . . . = |z n | = 1, or equivalently,−a ∈ {t 0 + . . . + t n ; |t 0 | = . . . = |t n | = 1, t 0 . . . t n = 1}. We depict the sets K n for a few small values of n. Note that the cusps on theboundary correspond to polynomials with singular hypersurfaces.Figure 2: The sets K n for n = 2 and 3.Let us in particular consider the case n = 2. The coefficients c α are equal to0 when α is a vertex of the Newton polytope whilec (1,1) = log |a| + Re ∑ k>0(3k − 1)!(k!) 3 (−1) k−1 a −3kwhen |a| > 3, by Theorem 2. In fact it can easily be checked that the seriesconverges even when |a| = 3. What happens with the spine when a approachesthe boundary of K 2 from the outside? For example, when a → −3 a numerical8

˜M(u 1 , . . . , u n ) is a positive measure which depends multilinearly on the argumentsu j . We call ˜M the mixed real Monge-Ampère operator.The real Monge-Ampère operator is related to its complex counterpart asfollows. Suppose u 1 , . . . , u n are smooth convex functions on the domain Ω.Let U 1 , . . . , U n be plurisubharmonic functions on Log −1 (Ω) defined by U j (z) =u j (Log z). Then∫∫n! ˜M(u 1 , . . . , u n ) = dd c U 1 ∧ . . . ∧ dd c U n (12)ELog −1 (E)where d c = (∂ − ¯∂)/2πi. This remains true if one of the functions U j , say U 1 , isallowed to be an arbitrary smooth plurisubharmonic function in Log −1 (Ω), u 1being defined byu 1 (x) = 1U 1 (z)dz 1 . . . dz n(2πi)∫Log n .−1 (x) z 1 . . . z nMore generally, if U 1 , . . . , U n are arbitrary smooth plurisubharmonic functionson Log −1 (Ω), andthen∫n!E˜M(u 1 , . . . , u n ) =u j (x) = 1U j (z)dz 1 . . . dz n(2πi)∫Log n (13)−1 (x) z 1 . . . z n∫T n2 ∫dd c U 1 (t (1) z) ∧ . . . ∧ dd c U n (t (n) z)dλ(t).Log −1 (E)(14)Here T n2 denotes the real n 2 -dimensional torus {t = (t (k)j ); |t (k)j | = 1, j, k =1, . . . n} equipped with the usual normalized Haar measure λ, and each t (k) =(t (k)1 , . . . , t(k) n ) acts on C n by componentwise multiplication. These formulascan be checked by direct computation. A proof of (12) when u 1 = . . . =u n can be found in [6]. The general case follows by polarization since bothsides are multilinear. Formula (14) follows by reversing the order of integrationon the right. Since the inner integral is constant along certain n dimensionalsubmanifolds of T n2 it is actually possible to omit some of the variables in theouter integration. This also proves the generalised version of (12) with U 1 anarbitrary smooth plurisubharmonic function.Theorem 4. Let E be any Borel set in Ω, and let ∆ denote the Laplace operator.Then∫ ∫(n − 1)! ∆N f =ω n−1ELog −1 (E)∩f −1 (0)where ω = (|z 1 | −2 d¯z 1 ∧ dz 1 + . . . + |z n | −2 d¯z n ∧ dz n )/2πi.Proof. If u is any convex function, then ∆u = n ˜M(u, |x| 2 , . . . , |x| 2 ) and ω =dd c | Log z| 2 . Since dd c log |f| is equal to the current of integration along f −1 (0)it follows from (12) that∫∫n! ∆N f = n dd c log |f| ∧ ω n−1E∫= nLog −1 (E)Log −1 (E)∩f −1 (0)ω n−1 .10

The following theorem can be thought of as a local analog of Bernstein’stheorem [1] relating the number of solutions to a system of polynomial equationsto the mixed volume of their Newton polytopes.Theorem 5. Let f 1 , . . . , f n be holomorphic functions in Log −1 (Ω) and let E ⊂Ω be a Borel set. Then n! ˜M(N f1 , . . . , N fn )(E) is equal to the average numberof solutions in Log −1 (E) to the system of equationsas t = (t (j)kf j (t (j)1 z 1, . . . , t (j)n z n ) = 0 j = 1, . . . , n (15)) ranges over the torus {t; |t(j) | = 1, j, k = 1, . . . , n}.kProof. If U j are smooth plurisubharmonic functions which converge to log |f j |,then u j defined by (13) converge to N fj . By the general properties of the realMonge-Ampère operator this implies that ˜M(u1 , . . . , u n ) converges weakly to˜M(N f1 , . . . , N fn ). Also dd c U 1 (t (1) z) ∧ . . . ∧ dd c U n (t (n) z) converges weakly tothe sum of point masses at the solutions of f 1 (t (1) z) = . . . = f n (t (n) z) = 0.Hence the theorem follows by passing to the limit in (14) if we only show that∫dd c U 1 (t (1) z) ∧ . . . ∧ dd c U n (t (n) z)Log −1 (E)remains uniformly bounded as U j → log |f j |. Here we may assume that E iscompact and that U j is of the form U j = ψ(|f j |) where ψ will converge to log.Let f t (z) = (f 1 (t (1) z), . . . , f n (t (n) z)). Then f t (z) is a holomorphic functionin z and t defined for z in a neighbourhood of Log −1 (E) and t in a complexneighbourhood of T n2 . Using compactness arguments it is not difficult to showthat there exists a constant C such that the number of solutions in Log −1 (E)to the equation f t (z) = w is bounded above by C for almost all t ∈ T n2 andw ∈ C n . Since η = dd c ψ(|w 1 |) ∧ . . . ∧ ψ(|w n |) induces a positive measure on C nwith total mass 1, it follows that∫0 ≤ ft ∗ η ≤ Cand this completes the proof. EIf E is a compact component of A f1 ∩ . . . ∩ A fn , then for topological reasons,the number of solutions to (15) in Log −1 (E) does not depend on t. Hence weobtain the following corollary.Corollary 1. Suppose K is a compact component of A f1 ∩ . . . ∩ A fn . Thenn! ˜M(N f1 , . . . , N fn )(K) is a positive integer, which is equal to the number ofsolutions of the system f 1 (z) = . . . = f n (z) = 0 in Log −1 (K).Let us now see what the measure µ f looks like in some specific cases. First, ifn = 1, the amoeba is a discrete point set and µ f is a sum of point masses. Moreprecisely, µ f ({x}) is equal to the number of zeros of f on the circle log |z| = x. Inthe case of two variables there is an interesting estimate on the Monge-Ampèremeasure.Theorem 6. Let f be a holomorphic function in two variables defined on acircular domain Log −1 (Ω). Then µ f is greater than or equal to π −2 timesLebesgue measure on the amoeba of f.11

Proof. We prove the inequality in a neighbourhood of a point x ∈ A f whereLog −1 (x) intersects f −1 (0) transversely in a finite number of points. Since thisis true for almost all x it will establish the inequality.Write log z j = x j + iy j for j = 1, 2 and assume that the hypersurface f −1 (0)is given locally as the union of graphs y = φ k (x). We shall express the Hessianof N f in terms of the functions φ k .Differentiating the integral (1) defining N f with respect to x 1 we obtain∂N f∂x 11∂f/∂z 1 dz 1 dz 2= Re(2πi)∫Log 2 −1 (x) f(z)z 2= 1 ∫n(f(·, z 2 ), x 1 ) dz 22πi log |z 2|=x 2z 2= 12π∫ 2π0n(f(·, e x2+iy2 ), x 1 )dy 2 .Here n(f(·, z 2 ), x 1 ) is the number of zeros minus the number of poles of thefunction z 1 ↦→ f(z 1 , z 2 ) inside the disc {log |z 1 | < x 1 } provided that it is meromorphicin that domain. In general, n(f(·, z 2 ), x 1 ) − n(f(·, z 2 , x ′ 1) is equal tothe number of zeros in the annulus {x ′ 1 < log |z 1| < x 1 } when x ′ 1 < x 1. Theintegrand in the last integral is a piecewise constant function with a jump ofmagnitude 1 at y 2 = φ k,2 (x). It follows that the gradient of ∂N f /∂x 1 is given bya sum of terms ±(2π) −1 grad φ k,2 . The correct sign of each term can be foundby observing that n(f(·, e x2+iy2 ), x 1 ) is increasing as a function of x 1 , hence allthe terms contributing to ∂ 2 N f /∂x 2 1 should be positive. A similar computationapplies to ∂N f /∂x 2 .Assume now that f −1 (0) is given locally by an equationa log z 1 + b log z 2 + higher terms = constant.Solving for y in this equation yields that( )∂φ k∂x = 1 Re(a¯b) |b|2Im(a¯b) −|a| 2.− Re(a¯b)The crucial observation here is that( )∂φk,2 /∂x±1 ∂φ k,2 /∂x 2−∂φ k,1 /∂x 1 −∂φ k,1 /∂x 2(16)is a positive definite matrix with determinant 1, and that 2π Hess(N f ) is a sumof such matrices.The inequality now follows from the following lemma since Log −1 (x) intersectsf −1 (0) in at least two points for generic x in A f . Lemma. If A 1 , A 2 are 2 × 2 positive definite matrices, then det(A 1 + A 2 ) ≥det A 1 + det A 2 + 2 √ det A 1 det A 2 . Equality holds if and only if A 1 and A 2 arereal multiples of one another.Proof. To see this, write A j = ( a j b j)b j c jand apply the Cauchy-Schwarz inequalityto the vectors (b j , a j c j − b 2 j ) to obtain b 1b 2 + √ det A 1 det A 2 ≤ √ a 1 a 2 c 1 c 2√.12

Then it follows that det(A 1 + A 2 ) − det A 1 − det A 2 = a 1 c 2 + c 1 a 2 − 2b 1 b 2 ≥2 √ a 1 a 2 c 1 c 2 − 2b 1 b 2 ≥ 2 √ det A 1 det A 2 . The conditions for equality are that(b 2 1, a 1 c 1 − b 2 1) is proportional to (b 2 2, a 2 c 2 − b 2 2) and that (a 1 , c 1 ) is proportionalto (a 2 , c 2 ) which clearly is equaivalent to A 1 being proportional to A 2 . As an immediate consequence of Theorem 3 and Theorem 6 we have thefollowing estimate.Corollary 2. Let f be a Laurent polynomial in two variables. Then the area ofthe amoeba of f is not greater than π 2 times the area of the Newton polytope off.On the contrary, when n ≥ 3 the volume of the amoeba of a polynomial isalmost always infinite.Example. As an illustration of the last theorem we consider the polynomialsf(z 1 , z 2 ) = f a (z 1 , z 2 ) = a + z 1 + z 2 + z 1 z 2 where a is assumed to be a realnumber.We want to compute the number of points in Log −1 (x) ∩ f −1 (0) for a givenpoint x ∈ R n . For points z in this set it must hold that |a + z 1 | = |1 + z 1 |e x2 .Conversely, if this equation holds, then (z 1 , (a + z 1 )/(1 + z 1 )) is in this set. Ifθ = arg z 1 , the equation can be rewritten 2e x 1 (e2x2 −a) cos θ = a 2 +e 2x1 −e 2x2 −e 2x1+2x2 . From this it follows immediately that the amoeba of f is defined bythe inequality 4e 2x1 (e 2x2 − a) 2 ≥ (a 2 + e 2x1 − e 2x2 − e 2x1+2x2 ) 2 . Moreover, thefollowing can be deduced when we assume that a ≠ 1. If x is in the interiorof the amoeba, then Log −1 (x) ∩ f −1 (0) has precisely two points. If x is in theboundary of the amoeba, and not equal to (log a, log a)/2 when a is positive,then Log −1 (x) ∩ f −1 (0) has exactly one point. If a > 0 and x = (log a, log a)/2,then Log −1 (x) ∩ f −1 (0) contains a real curve.By the preceding theorem, µ f is greater than or equal to π −2 times theLebesgue measure on the amoeba of f. Assume now that a < 0. For x in theinterior of A f , 2π Hess(N f ) is a sum of two matrices of the form (16). Since fhas real coefficients it follows that φ 1 = −φ 2 , hence the two matrices are equal.This means that all inequalities in the proof of Theorem 6 actually becomeequalities. It follows also from Theorem 5 that µ f has no mass on the boundaryof the amoeba. Hence the area of the amoeba is equal to π 2 .When a is positive and not equal to 1, the same conciderations hold awayfrom the special point (log a, log a)/2. On the other hand, the amoeba of f ais strictly smaller than the amoeba of f −a . The remaining mass of µ fa , whichmust have the same total mass as µ f−a resides as a point mass at (log a, log a)/2.In the particular case a = 1, there is a factorization f(z) = (z 1 + 1)(z 2 + 1)and the amoeba consists of the two lines {x 1 = 0} and {x 2 = 0}. The Monge-Ampère measure µ f then degenerates into a single point mass at the intersectionpoint of the two lines.References[1] David Bernstein: The number of roots of a system of equations, FunctionalAnal. Appl. 9 (1975), 183 - 185.13

Figure 3: The amoebas of a + z 1 + z 2 + z 1 z 2 for a = −5 and a = 5.[2] Johannes Duistermaat, Wilberd van der Kallen: Constant terms in powersof a Laurent polynomial, Indag. Math. 9 (1998), 221 - 231.[3] Mikael Forsberg, Mikael Passare, August Tsikh: Laurent determinants andarrangements of hyperplane amoebas, Adv. in Math. 151 (2000), 45 - 70.[4] Israel Gelfand, Mikhail Kapranov, Andrei Zelevinsky: Discriminants, resultantsand multidimensional determinants, Birkhäuser, Boston, 1994.[5] Grigory Mikhalkin: Real algebraic curves, moment map and amoebas,Manuscript, Harvard University, 1998. (To appear in Ann. of Math.)[6] Alexander Rashkovskii: Indicators for plurisubharmonic functions of logarithmicgrowth. Available at http://xxx.lanl.gov/abs/math/9911240[7] Jeffrey Rauch, Alan Taylor: The Dirichlet problem for the multidimensionalMonge-Ampère equation, Rocky Mountain J. Math. 7 (1977), 345 -364.[8] Lev Ronkin: On zeros of almost periodic functions generated by holomorphicfunctions in a multicircular domain, To appear in “Complex Analysisin Modern Mathematics”, Fazis, Moscow, 2000, pp. 243-256.[9] Lev Ronkin: Introduction to the theory of entire functions of several variables,Translations of mathematical monographs, AMS, Providence, 1974.14