Kazarnovskiˇı Pseudovolume, a Clever Use of Support Function

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arXiv:1910.03099v2 [math.CV] 11 Oct 2019 Introduction 1 Contents aanvkı suooueadmxddiscriminants mixed and Kazarnovskˇıi pseudovolume 7 class the on Kazarnovskiˇı pseudovolume 6 in bodies Convex 5 distortion Volume 4 Mixed 3 bodies convex real on Preliminaries 2 2Oe questions Open 12 computations and constructions Useful 11 Examples 10 current The 9 class the on Kazarnovskiˇı pseudovolume 8 22Mntnct ...... inequality . . . . Alexandroﬀ-Fenchel . . . . . 12.3 ...... Monotonicity . . . . 12.2 . . . condition . vanishing . . Non . . . 12.1 ...... groups . Linear . . 11.3 . . . . Hessians . 11.2 . coordinates Polar 11.1 current The 9.4 case The 9.3 case The 9.2 case The 9.1 Keywords: MSC: pseudovolume. mixed h rsn ae rvdsadtie n lotself-conta almost and detailed a provides paper present The ϕ rmr 52A39. Primary -volume aanvkı ie suooue ie oue polytope volume, mixed pseudovolume, Kazarnovskˇıi mixed (dd > k k k 2 = 1 = lvrueo upr function. support of use clever a (dd c h 2 C ...... Γ Kazarnovskiˇı pseudovolume, c n ) h ∧ Γ k ) ∧ k ∧ (dd [email protected] c ae iio PhD Silipo, James h B 2 n ) Abstract ∧ ℓ ...... S K 1 2 ( ( C C n n ) ) nditouto oKazarnovskiˇı to introduction ined ...... 69 66 63 45 42 37 26 24 19 15 72 70 69 68 66 66 62 56 51 46 5 3 List of notation 73

References 76

List of Figures

1 The end-points of the black segment are faces but not exposedones...... 6 2 The three polytopes are combinatorially isomorphic but only the first two are strongly combinatorially isomorphic because the brown face of the third polytope is not parallel tothebrownonesofthefirsttwo...... 6 3 Inducedorientations...... 7 4 A polytope, containing the origin, decomposed into pyramids based onto its facets. . . . 8 5 A cube and parts of the dual cones to the edges passing through the origin...... 8 6 Γ is the light brown polytope with the blue face ∆ and the orange side Λ. The blue, transparent plane is affR ∆, the orange line is affR Λ...... 9 7 A strictly convex body which is not smooth at the black point...... 11 ⊥ 8 On the left the 1-polytope Γ. The half-spaces bounded by the hyperplane E∆ are K0 and Kv. On the right, Supp ς is depicted as the compact set with smooth rounded boundary, (Σ1)ε is the strip with dashed boundary. The blue part is Kv Supp ς (Σ1)ε, the components of its boundary are drawn in orange. The blue/red∩smooth path\ around Kv Supp ς (Σ1)ε is the boundary of Sv,ε. The red segment is the only piece of ∂Sv,ε thatcounts...... ∩ \ . . . 47 9 The 2-polytope Γ isatriangle...... 48

10 Theset Supp ς is depicted as the compact set with smooth rounded boundary, (Σ1)ε is the set with black, dashed boundary. The blue 2-dimensional part of Supp ς is the union of K Supp ς (Σ ) and K Supp ς (Σ ) , the components of ∂[K Supp ς (Σ ) ] v1 ∩ \ 1 ε v2 ∩ \ 1 ε v1 ∩ \ 1 ε and ∂[Kv2 Supp ς (Σ1)ε] are coloured according to the colours of the edges of Γ. The coloured smooth∩ path\ all around K Supp ς (Σ ) is the boundary of S . The red, v1 ∩ \ 1 ε v1,ε brown and green segments are the only pieces of ∂Sv1,ε which count. The same occurs to

Sv2,ε...... 49 11 S = Supp ς is the transparent cylinder with brown contour, Γ is a cube with a vertex in the origin, ∆ is the red face, Λ1 is the orange edge of ∆ with a vertex in the origin o, 3 whereas Λ2 is the green edge parallel to Λ1. The space C is spanned by the orthonormal

basis (u∆,Γ,uΛ1,∆,uo,Λ1 ) or (u∆,Γ,uΛ2,∆,uq,Λ2 ) and three more unit vectors w1, w2, w3 ⊥ 3 spanning EΓ thus getting in both cases positively oriented basis of C . This choice implies that E is oriented by (u ,u ) or (u ,u ). The set K Supp ς (Σ ) is the ∆ Λ1,∆ o,Λ1 Λ2,∆ q,Λ1 Λ1 ∩ \ 2 ε blue slice (on the right) of Supp ς along KΛ1 (Σ2)ε while SΛ1,ε is the set bounded by the \ ε smooth path all around KΛ Supp ς (Σ2)ε. X is the red segment on the boundary 1 ∩ \ Λ1(∆) of SΛ1,ε. Along the black part of the boundary of SΛ1,ε the form ς is zero because this

part is included in ∂ Supp ς. On the blue circular (in fact cylindrical!) pieces of ∂SΛ1,ε, over the singular points of K Supp ς (Σ ) , the integral vanishes in the limit. The Λ1 ∩ \ 2 ε same considerations are valid for Λ2...... 55 12 The rectangle Kλ and the pyramid Γλ for λ =1/4...... 71

List of Tables

1 Pseudovolume of full-dimensional unit balls of Cn...... 63

2 n-pseudovolume of B2n−1 and B2n...... 65

2 1 Introduction

The present article provides a detailed introduction to Kazarnovskiˇımixed pseudovolume. This notion provides a generalisation to Cn of Minkowski mixed volume in Rn; as suggested by the title, it is ultimately the product of a clever use of support functions of convex bodies of Cn. Given a convex body (i.e. a compact convex subset) A Cn, the support function of A is the convex, n⊂ positively 1-homogeneous function hA deﬁned, for z C , by hA(z) = maxu∈A Re z,u , where , is n ∈ h i c h i the standard hermitian product on C . Suppose hA is smooth outside the origin and let d = i(∂¯ ∂), n − let also B2n denote the unit full-dimensional ball about the origin of C and κ2n its Lebesgue measure, then the Kazarnovskiˇıpseudovolume of A is the non-negative real number Pn(A) deﬁned as 1 P (A)= (ddch )∧n . (1) n n!κ A n ZB2n

Accordingly, if A1,...,An are convex bodies whose support functions are smooth outside the origin, their mixed pseudovolume is the non-negative real number Qn(A1,...,An) deﬁned as 1 Q (A ,...,A )= ddch ... ddch . (2) n 1 n n!κ A1 ∧ ∧ An n ZB2n

The mixed version polarizes the unmixed one, i.e. Qn(A,...,A) = Pn(A), so Qn is multilinear with respect to positive scalar multiplication and Minkowski addition. Moreover, for convex bodies included in Rn, Kazarnovskiˇımixed pseudovolume reduces to Minkowski mixed volume, which yields an apparently new integral formula for the latter. The notion of pseudovolume can be defined without the smoothness assumption on support functions, it requires a regularisation argument on positive currents that, for a polytope Γ, yields a remarkable combinatorial formula: Pn(Γ) = ̺(∆) vol n(∆)ψΓ(∆) , (3) ∆∈BXed(Γ,n) where ed(Γ,n) is the set of equidimensional n-faces of Γ, i.e. those faces ∆ of Γ spanning a real B affine subspace affR ∆ and a complex one affC ∆ which satisfy the dimensional equalities dimR affR ∆= dimC affC ∆ = n; ̺(∆) is a weight ascribed to every equidimensional face and equal to the jacobian determinant of a projection (cf. section 4), vol n(∆) is the n-dimensional volume of ∆ and ψΓ(∆) is the outer angle of Γ at ∆ (cf. section 2). Kazarnovskiˇıpsedovolume can be defined for a non convex compact set A Cn with interior points and smooth boundary as ⊂ 1 P (A)= α (dα )∧(n−1) , (4) n n!κ ∂A ∧ ∂A n Z∂A where α is the differential 1-form whose value on the point ζ ∂A equals the projection on the ∂A ∈ characteristic direction of the tangent space Tζ ∂A (cf. section 6), or equivalently as

2n−1(n 1)! P (A)= − K υ , (5) n n!κ ∂A ∂A n Z∂A where K∂A denotes the determinant of the Levi form of ∂A and υ∂A is the standard volume form on the hypersurface ∂A. The resulting object is very interesting: as shown by Alesker [Ale], it yields a valuation on convex bodies, i.e. P (A A )+ P (A A )= P (A )+ P (A ) , n 1 ∪ 2 n 1 ∩ n n 1 n 2 whenever A1, A2 and A1 A2 are convex. This valuation has some remarkable properties, namely: continuity with respect to∪ Hausdorff metric and invariance for translations or unitary transformations. n Moreover Pn cannot be extended to a measure on C , indeed it is not monotonically increasing unless n =1, however, just like ordinary volume, the pseudovolume of a polytope is positive if and only if the polytope spans an n-dimensional complex affine subspace. Although this notion is interesting in integral, tropical and complex analytic geometry, the available literature on the subject is still relatively scarce and this is the reason why we decided to write the present paper. The theory of mixed pseudovolume has been inspired by the so-called BKK theorem ± ± about the number of solutions to a generic system P C[z1 ,...,zn ] of n Laurent polynomials, due to Bernstein [Be], Kushnirenko [Ku] and Khovanski [Kho1⊂ ]. As it is well known, if p(z)= c zλ = λ∈Λp λ P 3 c zλ1 zλn is a Laurent polynomial, the Newton politope ∆ of p is the convex hull in Rn of λ∈Λp λ 1 n p ··· n its (finite) support of summation Λp Z . According to the BKK theorem, the number of solutions in P ∗ n ⊂ the torus (C ) to a generic system P equals the Minkowski mixed volume of the n polytopes ∆p, as p P . ∈ Kazarnovskiˇı[Ka1], by investigating the asymptotic distribution of the solutions to a finite generic h·,λi n system F of exponential sums, (i.e. entire functions of the form f(z)= cλe , where Λf C λ∈Λf ⊂ is a finite subset called the spectrum of f, c C∗ and , is the standard hermitian product of Cn), λ P found that such an asymptotic distribution is controlled∈ byh thei geometry of the convex hulls ∆ Cn of f ⊂ the spectra Λf , f F . In this exponential setting, the mixed pseudovolume plays the role of the mixed volume in the polynomial∈ one and reduces to it when the spectra are included in Rn, in particular the BKK theorem appears as a special instance of Kazarnovskˇıi’s one when the spectra are included in Zn. The link between the asymptotic distribution of zeros and the geometry of the exponents comes from King’s formula for the integration current associated to a complete intersection of analytic hypersurfaces. According to [GK], such a current is nothing but the wedge product of the integration currents associated to the involved hypersurfaces. By Lelong-Poincare formula [Le], the integration current of a single analytic hypersurface f = 0 equals, up to a normalisation constant, the current ddc log f . Loosely speaking, the exponential{ shape} of f and the presence of a logarithm in Lelong-Poincaré formula| | explain why the asymptotic distribution of the zeros of the system is controlled by the geometry of its exponents.

More precisely, let #F = n, F = f1,...,fn and, for 1 ℓ n, let Λℓ =Λfℓ , ∆ℓ = ∆fℓ and hℓ = h∆ℓ . n { } ≤ ≤ If the zero set V (F ) = z C fℓ(z)=0, 1 ℓ n is a complete intersection, then the current n t−1ddc log f (tz) weakly{ ∈ converges| to the current≤ ≤ }n ddch (z), as t + , and it turns out that ℓ=1 | ℓ | ℓ=1 ℓ → ∞ the mixed pseudovolume Qn(∆1,..., ∆n) affects the asymptotic distribution of V (F ). The requirement Vthat V (F ) is a complete intersection is crucial for theV preceding method to work. If one fixes both #F and the number of points in the spectra, this requirement is generically satisfied and can be seen as a sort of non-singularity of V (F ). Of course it depends on both the coefficients and the exponents of the exponential sums. The non-singularity conditions assumed in [Ka1], though generalising those considered in the polynomial case, are merely sufficient conditions which are far from being necessary. Indeed, characterisations of non-singularity are highly unknown and this question is still a source of interest.

Though mixed pseudovolume may be regarded as an accessory tool in the theory of exponential sums, it seems to deserve at least the same attention as the theory it originated from. The notion of pseudovolume first appeared in 1981 in the paper [Ka1] and, as early as its introduction, it was known to the Russian mathematical community as witnessed by Shabat who mentioned pseudovolume in [Sha] p. 199. In 1984 Kazarnovskiˇıgave a more detailed construction of pseudovolume in a second paper [Ka2], however this second article does not provide proofs for all the theorems stated in the first one. The subsequent papers [Ka3; Ka4; Ka5; Ka6; Ka7; Ka8; Ka9] are concerned with interesting analytic and combinatorial investigations in which, however, the pseudovolume does not play a prominent role. Since then Kazarnovskiˇıpseudovolume seemed to have been forgotten until 2003, when the notion got new attention in integral geometry [Ale]. Then the papers [Ka10; Ka11] by Kazarnovskiˇıbriefly provide further details about pseudovolume and its vanishing condition in the polytopal case. In 2001 Alain Yger drew my attention toward exponential sums; in order to investigate their geometry I started my PhD under his supervision by conducting a thorough study of Kazarnovskiˇı’s seminal paper [Ka1] and looking for fully detailed proofs of the main results included therein. The present paper is a substantial expansion of the first chapter of my thesis [Sil], its intentionally elementary style aims at making the reading as accessible as possible, the exposition is sometimes redundant: some results, especially in low dimension, can be deduced by more general ones, nevertheless I inserted them to ease the understanding or to emphasise the use of particular methods which are valid just in special situations. I also tried to give a self-contained presentation with the following two notable exceptions: several complex variables and the theory of currents, at least at an introductory level. The reader should have some acquaintance with this topics and is referred to [Dem], [Ran] or [LeGr]. The paper is organised as follows. Section 2 collects some notions and facts about convexity which will be used in the sequel. Section 3 presents the construction of a class of valuations on convex bodies of Rn generalising ordinary intrinsic volumes and mixed volume. This construction, breafly outlined in [Ka1], admits Kazarnovskiˇıpseudovolume as a special instance and seems, to the best of my knowledge, to have never been systematically investigated before. Section 4 is devoted to the description of the weight function ̺ involved in (3). This section does not go beyond elementary linear algebra, however it provides a useful machinery used throughout the article. Section 5 describes some

4 typical features of convex bodies of Cn thanks to the information from the preceding two sections. Section 6 deals with the notion of pseudovolume on the class of compact subsets of Cn with non empty interior and smooth boundary, it contains the proof of the equality of (4) and (5) as well as that of (1) and (4) for strictly convex bodies with smooth boundary. Section 7 clarifies the relations between mixed pseudovolume and mixed discriminants. Section 8 treats the convex case in full generality, it includes a proof of (3) (following a suggestion by Kazarnovskiˇıhimself) that does not seem to have c ∧k appeared yet. Section 9 is dedicated to a closer study of the current (dd hA) as well as the related c ∧k c ∧ℓ current (dd hA) (dd hB2n ) , with 1 k + ℓ n, in the polytopal case. Both currents are represented as linear∧ combinations of some≤ special current≤ s obtained by “mixing” a volume form with an integration current (cf. (69)). A similar representation is briefly described in [Ka9]. By virtue of such representation (that makes these currents look like those studied in [Ka5]), it is shown that c ∧k (dd hA) enjoys a valuation property (already mentioned in [Ka6] and [Ka7]) implying that revealed by Alesker for Pn. Section 10 provides several examples, including those showing that Pn is neither rational on lattice polytopes nor orthogonally invariant. Section 11 collects some useful notions, facts and computations only occasionally used in the paper. Finally section 12 presents some open problems and partial answers about non vanishing, monotonicity and Alexandrov-Fenchel type inequality for Qn. The article ends with a list of non standard notation employed throughout the paper. I am grateful to Alain Yger, who kindly introduced me to this beautiful subject, and to Boris Yakovlevich Kazarnovskˇıi, who patiently answered the countless questions I have been asking him for the last fifteen years.

2 Preliminaries on real convex bodies

Let n N∗ and give Rn the euclidean structure induced by the standard scalar product ( , ). If A Rn ∈ ⊆ is a non-empty subset, let χA be its characteristic function, i.e. the function defined by χA(x)=1 n if x A and χA(x)=0 if x / A. Let also affR A denote the real affine subspace of R spanned by ∈ ∈ n A, whereas let EA be the R-linear subspace of R parallel to affR A. The set of interior points and the set of boundary points in the relative topology of A, as a subset of affR A, are referred to as the relative interior and the relative boundary of A, they are respectively noted relint A and relbd A. For n n any 0 k n, let k(R ) be the Grassman manifold of the (real) linear k-dimensional subspaces of R endowed≤ with≤ theG usual topology, let also (Rn) denote the whole grassmannian, i.e. the topological n n G ⊥ sum of the k(R )’s. For every E (R ), by E we will denote the orthogonal complement of E. A linear subspaceG E (Rn) is full-dimensional∈ G if its dimension equals n. ∈ G A convex body is a compact convex subset of Rn. The set of convex bodies of Rn is noted (Rn), whereas (Rn) denotes that of nonempty compact subsets of Rn. A polytope is the convex hullK of a finite, possiblyC empty,1 subset of Rn. A lattice polytope is the convex hull of a finite, possibly empty, subset of Zn. Every polytope can be realised as a (bounded) polyhedral set, i.e. a finite intersection of half-spaces. The polytopes of Rn form a set denoted (Rn). For a non-empty subset A Rn, the notions of dimension and full-dimensionality introducedP for R-linear subspaces of Rn, can⊆ be simply extended by just invoking the subspace EA. In particular, this can be done for convex bodies. If n ∗ A (R ), set dA = dimR A = dimR EA. If d N , a d-polytope Γ is a polytope for which dΓ = d. For∈d K= 0, 1, 2, 3, 4, a d-polytope is respectively∈ referred to as a point, segment, polygon, polyhedron, polychoron. If v, w R, the segment with v and w as endponits will be denoted [v, w]. A simplex is the convex hull of affinely∈ independent elements of Rn, a d-simplex is a d-dimensional simplex, it is the convex hull of d +1 affinely independent points. n If A (R ), let us denote hA the support function of A, i.e. hA(z) = supv∈A(z, v) . For example, ∈ K n if Bn is the full-dimensional unit ball in R , then hBn (z) = z . The support function of a convex k k n ⊥ body is a positively 1-homogeneous convex function. Remark that if x = x1 + x2 R = EA EA , ⊥ ∈ ⊕ with x1 EA and x2 EA then hA(x) = hA(x1), so that hA actually depends on dA variables. ∈ n ∈ Given v R 0 , the corresponding supporting hyperplane HA(v) for the convex body A is given by ∈ \{n} HA(v) = z C (z, v) = hA(v) . Observe that, if v = 0, the definition of HA(v) still makes sense, { ∈ | } n nevertheless it is no longer a hyperplane because HA(0) = R . A face ∆ of A, (in symbols ∆ 4 A or A < ∆), is a convex subset such that u, v A and (u+v)/2 ∆ imply u, v ∆. This notion applies also to the case of unbounded polyhedral sets.∈ Every convex body∈ admits two∈ trivial faces, namely ∅ and A; these faces are referred to as the improper faces of A in contrast to the proper ones which are non-empty and different from the convex body A itself. For any proper face ∆ of A we will write ∆ A or A ∆. ≺ ≻ 1By convention, one sets conv (∅) = {0}.

5 n An exposed face ∆ of A is any intersection of the type ∆= A HA(v), for some v R . By taking v =0, it follows that the non-empty improper face is an exposed fa∩ce. The empty face∈ is considered an exposed face too. For polytopes the distinction between faces and exposed faces is not necessary, since the two notions coincide, however Figure 1 shows that this is not the case in general.

Figure 1: The end-points of the black segment are faces but not exposed ones.

n If Γ (R ), any non-empty face of Γ is itself a polytope. If ∆1 4 ∆2 4 Γ, then ∆1 4 Γ and E ∈ PE E . The boundary complex (Γ) of a polytope Γ is the set of its non-empty ∆1 ⊆ ∆2 ⊆ Γ B faces. Two polytopes Γ1 and Γ2 are combinatorially isomorphic if there exists a bijective and inclusion- preserving map from (Γ1) to (Γ2). The polytopes Γ1 and Γ2 are strongly combinatorially isomorphic if there exists a combinatorialB B isomorphism ψ : (Γ ) (Γ ) such that for any ∆ (Γ ) the affine B 1 →B 2 ∈B 1 subspace affR ∆ is parallel to the affine subspace affR ψ(∆). Two triangles are always combinatorially isomorphic but they are strongly combinatorially isomorphic if and only if the sides of the first triangle are respectively parallel to those of the second one. This shows that the preceding notions of isomorphism are different; Figure 2 provides a three-dimensional example.

Figure 2: The three polytopes are combinatorially isomorphic but only the ﬁrst two are strongly combinatorially isomorphic because the brown face of the third polytope is not parallel to the brown ones of the ﬁrst two.

For every 0 k dimR Γ, (Γ, k) will denote the subsets of (Γ) consisting of k-faces (i.e. k- dimensional faces),≤ the≤ cardinal numberB of the set (Γ, k) is denotedBf (Γ). A facet is a face belonging B k to (Γ, dΓ 1), whereas a ridge is a face belonging to (Γ, dΓ 2), a 1-face is called an edge, a 0-face is calledB a vertex− . A polytope is simplicial if all its properB faces− are simplices. A d-polytope is simple if each of its vertices belongs to d 1 facets. The face vector of a polytope Γ (Rn) is the vector − ∈P

f(Γ) = (f0(Γ),...,fdΓ−1(Γ)) .

Setting f (Γ) = f (Γ) = 1 and f (Γ) = 0 for k < 1 or k > d , it is well known that the natural −1 dΓ k − Γ numbers fk(Γ) cannot be arbitrary, indeed they satisfy the Euler’s relation

dΓ ( 1)kf (Γ) = 0 , − k kX=−1 as well as the inequalities d +1 f (Γ) Γ f (Γ) 0 . k +1 ≤ k ≤ k +1 n An oriented d-polytope Γ (R ) is a polytope such that EΓ is an oriented linear subspace. To any complete ﬂag of faces of Γ∈ P

∆ ∆ ... ∆ ∆ ∆ =Γ 0 ≺ 1 ≺ ≺ d−2 ≺ d−1 ≺ d

6 starting with a vertex and ending with Γ, we can associate an orthonormal basis of EΓ made of outer normal unit vectors. In fact, for any vertex ∆0 of Γ, let ∆1 be an edge of Γ admitting ∆0 as a vertex, then we have E = 0 and the trivial orthogonal decomposition ∆0 { } ⊥ n E∆ = E∆ (E E∆ )= 0 (R E∆ )= 0 E∆ . 1 0 ⊕ ∆0 ∩ 1 { }⊕ ∩ 1 { }⊕ 1 By induction suppose we have deﬁned a sequence ∆ ∆ ... ∆ of faces of Γ such that 0 ≺ 1 ≺ ≺ k−1 dimR E∆ℓ = ℓ, for every ℓ 0, 1,...,k 1 . Then, for every k-face ∆k 4 Γ admitting ∆k−1 as a facet, there is an orthogonal decomposition∈{ − }

⊥ E∆ = E∆ − (E E∆ ) , k k 1 ⊕ ∆k−1 ∩ k where the second direct summand is 1-dimensional. The process ends after d steps and yields the orthogonal decomposition d ⊥ EΓ = (E E∆ ) , ∆k−1 ∩ k Mk=1 where each summand is 1-dimensional. For every k 1,...,d , the affine subspace affR ∆ relatively ∈{ } k−1 bounds two relatively open half-spaces of affR ∆k one of which does not intersect ∆k. The outer unit ⊥ normal vector to the face ∆k−1 is the the unit vector u∆ − ,∆ E E∆ pointing towards the k 1 k ∈ ∆k−1 ∩ k relatively open half-space of affR ∆k (relatively bounded by affR ∆k−1) that does not intersect ∆k. For the sake of notation, let us set v = u , for every k 0,...,d 1 . This construction d−k ∆k−1,∆k ∈{ − } provides an orthonormal basis (v1,...,vd) of EΓ depending on the choice of the flag. If Γ is oriented, it is possible to choose a flag such that the obtained basis conforms to the orientation of Γ. Indeed, if the flag ∆ ∆ ... ∆ ∆ ∆ =Γ produces the wrong orientation, it is enough to replace 0 ≺ 1 ≺ ≺ d−2 ≺ d−1 ≺ d the vertex ∆0 with the other vertex of ∆1. Of course when d

v2 v2 aﬀR ∆1 v˜3

v1 we v1

∆1 ∆′ 2 w ∆2 ′ aﬀR ∆2

aﬀR ∆2

Figure 3: Induced orientations.

The d-volume of a d-polytope is related to the (d 1)-volume of its facets. Indeed, let Γ (Rn) and, for every ∆ (Γ, d 1), let u E⊥ E −be the outer unit normal vector to the∈ facet P ∆. ∈ B − ∆,Γ ∈ ∆ ∩ Γ

7 Then the following equalities are worth-noting:

vol d−1(∆)u∆,Γ =0 , (6) ∆∈BX(Γ,d−1) 1 vol (Γ) = h (u ) vol (∆) . (7) d n Γ ∆,Γ d−1 ∆∈BX(Γ,d−1) Figure 4 provides a geometric interpretation of (6).

u∆,Γ hΓ(u∆,Γ)

Figure 4: A polytope, containing the origin, decomposed into pyramids based onto its facets.

n For any non-empty face ∆ of a polytope Γ (R ), there is a dual cone K∆,Γ (or simply K∆, if ∈ Pn no confusion may arise) deﬁned as K∆,Γ = v R ∆=Γ HΓ(v) . For the improper face Γ, one ⊥ { ∈ | ∩ } has KΓ,Γ = EΓ and this is the only dual cone that is closed. If ∆ is a proper k-face, the subset K∆,Γ ⊥ is a non-empty convex polyhedral cone in E∆ (i.e. an unbounded intersection of ﬁnitely many open half-spaces passing through the origin) and a relatively open (2n k)-submanifold of Rn. Figure 5 shows parts of the dual cones to some edges of a cube. −

Figure 5: A cube and parts of the dual cones to the edges passing through the origin.

On the dual cone K the support function h is linear and, for every z K , h (z)= h (z)= ∆,Γ Γ ∈ ∆,Γ Γ ∆ (z, v), where v is any point of ∆. In fact, the chosen point v ∆ may be represented as v = v1 + v2, ⊥ ∈ with v1 E∆ and v2 E∆, then, on K∆,Γ, one has hΓ(z) = h∆(z) = (z, v) = (z, v2). For any non ∈ ∈ ⊥ empty face ∆ of Γ, let p∆ be the only point in the intersection aﬀR ∆ E∆, then hΓ = ( ,p∆) on the ⊥∩ · whole K∆,Γ. As a consequence, if Λ is a facet of ∆ 4 Γ and uΛ,∆ EΛ E∆ is its outer unit normal vector, there is an orthogonal decomposition: ∈ ∩

pΛ = p∆ + h∆(uΛ,∆)uΛ,∆ . (8)

Figure 6 shows that the points pΛ and p∆ need not belong to Λ and ∆, respectively.

8 ⊥ E E∆ Λ ∩ ⊥ EΛ aﬀR Λ pΛ h∆(uΛ,∆)uΛ,∆

uΛ,∆

E∆ ⊥ E∆ p∆

Λ aﬀR ∆ Γ ∆

EΛ

Figure 6: Γ is the light brown polytope with the blue face ∆ and the orange side Λ. The blue, transparent plane is aﬀR ∆, the orange line is aﬀR Λ.

If ∆ 4 Γ is a k-dimensional face, the topological boundary of the manifold K∆,Γ equals its relative boundary and is given by the disjoint union of the cones which are dual to the faces of Γ admitting ∆ as a proper face, or by the union of the topological closures of the cones which are dual to the faces of Γ admitting ∆ as a facet, i.e.

∂K∆,Γ = relbd K∆,Γ = K∆′,Γ = K∆′,Γ . ′ ′ ∆ ∈B(Γ) ∆ ∈B(Γ,k+1) [ ′ [ ′ ∆≺∆ ∆≺∆

For any integer 0 k dim Γ, the k-star Σ (or simply Σ , if no confusion may arise) is the ≤ ≤ k,Γ k disjoint union of the cones which are dual to the k-dimensional faces of Γ. Notice that Σk is a (2n k)- dimensional (disconnected) manifold and −

n R = Σk = K∆,Γ = Kv,Γ , 0≤[k≤dΓ ∆∈B[(Γ) v∈B[(Γ,0) where, for any vertex v Γ, Kv,Γ is the topological closure of Kv,Γ. It follows that hΓ is a continuous piecewise linear function∈ on the whole Rn with the following representation

hΓ(z)= (p∆,z)χK∆,Γ (z) . ∆∈BX(Γ)

By continuity, on KΛ,Γ, one has hΛ = h∆, for any ∆ 4 Γ such that Λ 4 ∆. If ∆ (Γ, k), the outer angle of ∆ with respect to Γ is the number ∈B ψ (∆) = κ−1 vol (K B ) , (9) Γ n−k n−k ∆,Γ ∩ n ℓ where, for any 0 ℓ n, vol ℓ is the ℓ-dimensional Lebesgue measure in R , whereas κℓ is the ℓ- ≤ ≤ ℓ dimensional Lebesgue measure of the ℓ-dimensional closed unit ball Bℓ about the origin of R . Recall that, if Γ denotes Euler’s gamma function, then

π(ℓ/2) κ = , ℓ Γ (1 + (ℓ/2))

ℓ in particular κ0 = 1, κ1 = 2 and κ2ℓ = π /ℓ!. By Fubini’s theorem together with the properties of Euler’s gamma and beta functions, it is possible to show that, for a d-polytope Γ, outer angles can be −1 computed in the smaller space EΓ, i.e. ψΓ(∆) = κd−k vol d−k(K∆,Γ EΓ Bn). Observe that ψΓ(Γ) = 1 and, for any facet ∆ Γ, ψ (∆) = 1/2. ∩ ∩ ≺ Γ

9 The normal fan Σ(Γ) of Γ is the union of the k-stars Σ , as k runs in the set 0,..., dimR Γ . Two k,Γ { } polytopes Γ1 and Γ2 are strongly combinatorially isomorphic if and only if Σ(Γ1) = Σ(Γ2). For any compact subset A Rn and any λ R, one deﬁnes the multiplication of A by λ as the n ⊂ ∈ subset λA = λz R z A . Of course, if A is a convex body and λ =0, hλA = λhA and when A is a polytope, λA{ is∈ also a| polytope.∈ } For every polytope Γ, any face of the6 polytope λΓ has the form λ∆, for a unique face ∆ of Γ, so that

⊥ ⊥ aﬀR λ∆= λ aﬀR ∆ , Eλ∆ = E∆ , Eλ∆ = E∆ ,Kλ∆,λΓ = K∆,Γ .

n The convex bodies A1 and A2 are (positively) homothetic if A1 = t + λA2, for some t R and λ R , i.e. if one of them is a translate of a multiple of the other one or is reduced to a single∈ point. ∈ ≥0 The Minkowski sum A = ℓ∈I Aℓ of a ﬁnite family of subsets Aℓ ℓ∈I is the zero set 0 if I = ∅, { } { } n whereas is the subset A = ℓ∈I vℓ vℓ Aℓ , ℓ I if I is a non-empty ﬁnite set. The spaces (R ) n { P | ∈ ∈ } C and (R ) are stable with respect to the Minkowski addition. Of course, if I = ∅ and the Aℓ ℓ∈I are K P 6 { } convex bodies, then hA = ℓ∈I hAℓ and when the summands are polytopes, the sum is itself a polytope. In the latter case, for any face ∆ of the Minkowski sum Γ = Γℓ there exists a unique sequence P ℓ∈I of faces ∆ℓ 4 Γℓ, ℓ I for which ∆ = ℓ∈I ∆ℓ, such uniquely determined faces are the (Minkowski) summands of ∆. As∈ a consequence, one gets the equalities P P

affR ∆= affR ∆ℓ , E∆ = E∆ℓ , (10) Xℓ∈I Xℓ∈I ⊥ ⊥ E∆ = E∆ℓ ,K∆,Γ = K∆ℓ,Γℓ . (11) ℓ\∈I ℓ\∈I If #I n, an edge-sum face ∆ of a Minkowski sum of polytopes Γ= Γ is a face whose summands ≤ ℓ∈I ℓ ∆ℓ, respectively, include segments Λℓ, ℓ I, spanning linearly independent directions. ∈ P Notice that the Minkowski sum of finitely many strongly combinatorially isomorphic polytopes yields a polytope that is strongly combinatorially isomorphic to each of its summands. When λ N∗, one has λA = λ A, for any convex body A. If A Rn and ε> 0, the convex body ∈ ℓ=1 ⊂ (A)ε = A + εBn is called the ε-neighborhood of A. It is just the Minkowski sum of A and ε times the P full-dimensional closed unit ball Bn. In particular (Bn)ε is the full-dimensional closed ball around the origin of radius 1+ ε. Remark that the Minkowski sum of a convex body A with a full-dimensional ball has the effect of smoothing the boundary of A. Observe also that, for every ε> 0, the ε-neighborhood of any subset A contains ∂A, in particular, for any polytope Γ, one has

Σ (Σ ) Σ , k,Γ ⊂ k,Γ ε ⊆ 1,Γ ε for any 1 k dim Γ. The class of convex subsets of R n which are ε-neighborhood of polytopes, for some ε> 0≤, will≤ be noted ∞(R). P Example 2.1. The standard n-simplex.

The standard n-simplex is the polytope ∆n deﬁned as

n+1 ∆ = x [0, 1]n+1 x =1 . n ∈ ℓ ( ℓ=1 ) X

For any integer 0 k n 1, the k-th component of the face vector f(∆ ) is given by ≤ ≤ − n n +1 f (∆ )= , k n k +1 and, for any face ∆ (∆ , k), one has vol (∆) = √k +1/k. ∈B n k Example 2.2. The standard n-cube. n n Let In = [ 1, 1] R . For any integer 0 k n 1, the k-th component of the face vector f(In) is given by − ⊂ ≤ ≤ − n f (I )=2n−k , k n k and, for any face ∆ (I , k), one has vol (∆) = 2k and ψ(∆) = 2n−k. ∈B n k

10 Example 2.3. The standard n-crosspolytope. n n Let Θn R be the convex hull of the 2 points whose coordinates are all zero except one of them that runs in⊂ the set 1, 1 . Equivalently Θ = x Rn x 1 . The k-th component of the { − } n { ∈ | 1≤ℓ≤n | ℓ| ≤ } face vector f(Θn) is given by Pn f (Θ )=2k+1 . k n k +1 The unit n-crosspolytope can be constructed recursively as the convex hull of the union Θ1 Θn−1, n−1 n n−1 n ∪ where Θ1 = [ 1, 1] R 0 R and Θn−1 R 0 R are respectively a unit 1-crosspolytope− (i.e.⊂ a segment)× { } and⊂ a unit (n 1)-crosspolytope⊂ × { orthogonal} ⊂ to each other in Rn. It thus follows that vol (Θ )=2n/n!, whereas −vol (∆) = √k +1/k!, for any ∆ (Θ , k), with n n k ∈ B n 1 k hyperplanes to Θn. Each such vector spans the dual cone to a facet of Θn and diﬀerent vectors span P P diﬀerent facets. For example, when n = 4, the outer angle of any 2-face of Θ4 is 1/6. In fact, if ∆ denotes the 2-face obtained by intersecting Θ4 with the supporting hyperplanes corresponding to u1 and v , one realizes that cos u v = (u , v )/ u v =1/2, i.e. u v = π/3, whence 1 1 1 1 1 k 1k·k 1k 1 1 vol (K B ) (1/2)(π/3) 1 dψ(∆) = 2 ∆ ∩ 4 = d= , κ2 π 6 as asserted.

n n For any given A (R ) recall that the subdifferential Subd hA(u) of hA at a point u R is defined as ∈ K ∈ Subd h (u)= v Rn (v,z u) h (z) h (u) , z Rn . A { ∈ | − ≤ A − A ∀ ∈ } It is a non-empty set as long as A is such, moreover it equals the exposed face A H (u). Its elements ∩ A are called the subgradients of hA at u and hA is differentiable at u if and only if # Subd hA(u)=1. For 1 convexity reasons, the condition # Subd hA(u)=1 implies that hA is at u. Anyway, hA will never be differentiable at the origin, unless A is reduced to a single point. C If k N∗ , a connected compact subset of Rn is said to be k-regular if it has non-empty interior and its∈ boundary∪{∞} is a closed k-hypersurface; the class of k-regular subsets will be denoted k(Rn). A connected compact subset of CRn is smooth if it is k-regular for every k N∗, the correspondingS class is ∞ n ∈ k n denoted (R ). Any k-regular subset A admits a (global) defining function ρA (R , R) such that S n ∈C ∗ ρA 0 on ∂A, ρA > 0 on R A, ρA < 0 on relint A and dρA = 0 on ∂A. For k N , we set k(≡Rn)= k(Rn) (Rn). If A\ k(Rn), then any of its defining6 functions has to∈ be convex.∪ {∞} Observe thatK ∞(RSn) ∩K∞(Rn). If a convex∈ K body admits a single supporting hyperplane at each point of its boundary,P then⊂ it K has non-empty interior and, for convexity reasons, this means that it is 1-regular. A convex body is strictly convex if its boundary contains no line segment. In particular such a body has non-empty interior. The support function of a convex body A is differentiable on Rn 0 if 1 n \{ } and only if hA (R 0 , R), or equivalently, if and only if A is strictly convex, (cf. [Sch]). The ∈ C \{ } n n class of strictly convex bodies of R will be denoted 1(R ). Strictly convex bodies and smooth ones are different classes of convex bodies. The convex bodiesK from the class ∞(Rn) are smooth but not strictly convex, whereas the planar convex body depicted in Fig. 7 is strictlyP convex but not 1-regular. So (Rn) = 1(Rn), however taking ε-neighbourhood is a convenient way for smoothing boundaries. K1 6 K

Figure 7: A strictly convex body which is not smooth at the black point.

∗ n n If ℓ N , ℓ(R ) will denote the class of convex bodies of R whose support function belongs to ℓ(R∈n 0∪{∞}). Moreover,K for k,ℓ N∗ we also set k(Rn)= k(Rn) (Rn). C \{ } ∈ ∪ {∞} Kℓ K ∩ Kℓ

11 n n n The Hausdorff metric on (R ) is the mapping h : (R ) (R ) R≥0 defined, for any A1, A2 n C C ×C → n∈ (R ), as h(A1, A2) = inf ε R≥0 A1 (A2)ε and A2 (A1)ε . The Hausdorff metric gives (R ) theC structure of a locally compact{ ∈ metric| space⊆ in which (R⊆n) is a} closed subspace. The subspace C(Rn) ∞ n n K P n (and hence (R )) is dense in (R ). Moreover, any m-tuple of convex bodies A1,...,Am (R ) can be respectivelyP approximatedK by an m-tuple of sequences of polytopes Γ ,..., Γ (R∈n K) such 1,k m,k ∈P that, for every fixed k N, the polytopes Γ1,k,..., Γm,k are strongly combinatorially isomorphic to one-another. ∈ n The convergence of a sequence Am A in (R ) is equivalent to the uniform convergence hAm hA n → K → on compacta of R of the corresponding sequence of support functions. By homogeneity, hAm hA uniformly on compacta of Rn if and only if h h uniformly on the single compactum→∂B . Am → A n Moreover, (for convexity reasons) hAm hA uniformly on ∂Bn if and only if hAm hA converges point- → → n n wise on ∂Bn, (cf. [Sch]). By a regularization argument, one can show that the subset ∞(R ) (R ) n K ⊂ K is dense in (R ) for the Hausdorff metric. Indeed, for any convex body A and every ε R≥0, there exist an ε-regularizationK of A, i.e. a convex body R A (Rn) such that lim R A∈ = A. The ε ∈ K∞ ε→0 ε support function of RεA is given by

hRεA(x)= hA(x + u x )ϕε( u ) υn , Rn k k k k Z where υ = du ... du is the usual standard volume form on Rn and ϕ ∞(R, R) is a non- n 1 ∧ ∧ n ε ∈ C negative function with (compact) support included in the interval [ε/2,ε] such that Rn ϕε( u ) υn =1 . A convenient choice for ϕ is given by the function k k ε R ε2 ϕ (x)= m−1χ (x)exp ε ε [ε/2,ε] (4x 3ε)2 ε2 − − where ε2 mε = χ[ε/2,ε]( u )exp 2 2 υn Rn k k (4 u 3ε) ε Z k k− − and 1 if x [ε/2,ε], χ[ε/2,ε](x)= ∈ 0 if x [ε/2,ε]. ( 6∈

An alternative formula for hRεA(x) is given by

x hRεA(x)= x hA(u)ϕε u υn , k k Rn − x Z k k from which it follows easily that h is differentiable on R n 0 . Observe that, unless A is reduced to RεA \{ } a single point, the regularization hRεA is never differentiable at the origin. In order to get a regularized n version of hA which is differentiable on the whole R one has to drop the requirement of convexity and perform the usual convolution with some regularizing kernel, for example

hA ϕε(x)= hA(u)ϕε( u x )υn , ∗ Rn k − k Z n thus getting, for every ε > 0, a smooth function on the whole R that converges uniformly to hA on each compact set, as ε 0, but such a regularization is generally nor convex neither homogeneous. The equality →

u ϕε( u )υn =0 Rn k k Z has two worth-noting consequences: both regularizations converge monotonically to hA from above; moreover, if h is linear on some open subset Ω Rn, then A ⊂ h (x)= h (x)= h ϕ (x) , A RεA A ∗ ε for any x Ω (∂Ω) . ∈ \ ε ∞ n ∞ n The ε-neighborhood of RεA belongs to 1 (R ) and still converges to A, (as ε 0), so that ∞(R ) n K ∞ n k n →n K ∗ is a dense subspace of (R ). Moreover, the inclusions ∞(R ) ℓ (R ) (R ), for any k,ℓ N , show that k(Rn) (asK well as k(Rn) and (Rn)) is denseK too. ⊂ K ⊂ K ∈ Kℓ K Kℓ We state the following important result.

12 Theorem 2.1. Let r N∗, A ,...,A (Rn) and λ ,...,λ R . Then vol r λ A is a ∈ 1 r ∈ K 1 r ∈ ≥0 n ℓ=1 ℓ ℓ homogeneous polynomial of degre n in λ1,...,λr. P This polynomial can be written as

r vol λ A = V (A ,...,A )λ λ , (12) n ℓ ℓ n ρ(1) ρ(n) ρ(1) ··· ρ(n) ρ Xℓ=1 X where the sum runs on the set of functions ρ : 1,...,n 1,...,r and the coeﬃcients are chosen so as to satisfy, (for any ρ), the condition { }→{ }

Vn(Aρ(1),...,Aρ(n))= Vn(Aς(ρ(1)),...,Aς(ρ(n))) , for every permutation ς of the image of ρ. Notice that, for fixed ρ, the coefficient of the monomial λρ(1) λρ(n) in the expression (12) depends just on Aρ(1),...,Aρ(n) as follows by setting λℓ =0 for every ℓ which··· does not belong to the image of ρ. If n N∗ and A ,...,A (Rn), the n-dimensional ∈ 1 n ∈ K (Minkowski) mixed volume Vn(A1,...,An) of the convex bodies A1,...,An is the coefficient in the expression (12) when r = n and ρ is the identity. For any k N and any convex body A (Rn), for the sake of notation, we set ∈ ∈ K A[k]= A,...,A ,

k times meaning that, for k = 0, the set A is not written| at all.{z For} any integer 0 k n, the k-th intrinsic volume of a convex body A (Rn) is the non-negative real number v (A)≤deﬁned≤ as ∈ K k n v (A)= κ−1 V (A[k],B [n k]) . k n−k k n n −

For k =0 one has v0(A)=1. The number vk(A) does not depend on the ambient space but just on A. v For a polytope Γ, the intrinsic character of k is revealed by the equality

vk(Γ) = vol k(∆)ψΓ(∆) . ∆∈BX(Γ,k) Notice that the k-th intrinsic volume of a k-polytope is just its k-dimensional Lebesgue measure. More- over the k-th intrinsic volume of a (k + 1)-polytope equals half the k-dimensional volume of its relative boundary, in particular the 1-st intrinsic volume of a 2-polytope is the semi-perimeter, whereas the 2-nd intrinsic volume of a 3-polytope is half the surface area of its relative boundary. By continuity of intrinsic volumes for the Hausdorﬀ metric, the preceding remarks apply to any convex body. The notion of valuation on convex bodies will also play an important role in the rest of the paper, so we brieﬂy recall it here. Let G be an abelian group, a valuation on (Rn) with values in G is a n K mapping φ : (R ) ∅ G such that φ(∅)=0 and φ(A1)+ φ(A2)= φ(A1 A2)+ φ(A1 A2), for K ∪{n } → ∪ ∩ n every A1, A2 (R ) such that A1 A2 is convex. By induction, if m 2, let A1,...,Am (R ) be such that A ∈ K... A is convex too.∪ For every non empty subset I ≥ 1,...,m , let ∈ K 1 ∪ ∪ m ⊆{ }

AI = Aj , j\∈I then #I−1 φ(A1 ... Am)= ( 1) φ(AI ) . ∪ ∪ ∅ − 6=I⊆{X1,...,m} A weak valuation on (Rn) is a mapping φ : (Rn) ∅ G such that φ(∅)=0 and K K ∪{ } → φ(A H+)+ φ(A H−)= φ(A H)+ φ(A) , ∩ ∩ ∩ for every A (Rn) and every hyperplane H Rn, where H+ and H− are the closed half-spaces bounded by ∈H. K If G has a topology, (weak) valuations⊂ can be continuous with respect to Hausdorﬀ k n metric. A (weak) valuation φ is k-homogeneous if φ(λA)= λ φ(A), for every λ R>0 and A (R ). A (weak) valuation φ is simple if φ(A)=0 for every A (Rn) which is not full-dimensional.∈ ∈ K These n ∈ K k n n k n n notions can be also deﬁned on proper subsets of (R ), such as (R ), ℓ(R ), ℓ (R ) or (R ). The following results on valuations are of great interest.K K K K P

13 Theorem 2.2 ([Sal]). Every weak valuation on (Rn) with values in a Hausdorﬀ topological R-linear space is a valuation. P

Theorem 2.3 ([Gro]). Every continuous weak valuation on (Rn) with values in a Hausdorﬀ topological R-linear space is a valuation. K

In the following sections we will deﬁne the n-dimensional Kazarnovskiˇıpseudovolume, a notion which generalizes the n-dimensional mixed volume to the complex setting. In order to compare these two notions, it is convenient to list here the most important properties of Vn.

1. Non negativity: Vn is a function which takes on non-negative real values . 2. Continuity: The n-dimensional mixed volume is continuous in the Hausdorﬀ metric .

3. Symmetry: Vn is a symmetric function of its arguments .

4. Multilinearity: Vn is multilinear with respect to Minkowski sum and non-negative real multiplication . 5. Diagonal property: V (A[n]) = vol (A) , for every convex body A (Rn) . n n ∈ K 6. Translation invariance:

Vn(A1,...,An)= Vn(x1 + A1,...,xn + An) , for any A ,...,A (Rn) and x ,...,x Rn. 1 n ∈ K 1 n ∈ 7. Orthogonal invariance: Vn(A1,...,An)= Vn(F (A1),...,F (An)) , for any A ,...,A belonging to (Rn) and any orthogonal transformation F : Rn Rn. 1 n K → 8. Symmetric formula:

n 1 ∂n Vn(A1,...,An)= vol n λℓAℓ , n! ∂λ1 ...∂λn ! Xℓ=1 for any A ,...,A (Rn). 1 n ∈ K 9. Polarization formula:

1 n−#I Vn(A1,...,An)= ( 1) vol n Aℓ , n! ∅ − 6=I⊆{X1,...,n} Xℓ∈I for any A ,...,A (Rn). 1 n ∈ K n 10. Non-degeneracy: Let A1,...,An (R ), then Vn(A1,...,An) > 0 if and only if there exist segments Λ A ,..., Λ A spanning∈ K lines with linearly independent directions. 1 ⊆ 1 n ⊆ n 11. Monotonicity: The n-dimensional mixed volume is non-decreasing, with respect to inclusion, in each of its arguments. 12. Recursive formula: If Γ ,..., Γ (Rn) are full-dimensional and Γ(2) = n Γ , then 1 n ∈P ℓ=2 ℓ 1 V (Γ ,..., Γ )= h (u )V (∆ ,..., ∆P) , (13) n 1 n n Γ1 ∆ n−1 2 n (2) ∆∈B(ΓX,n−1) where, for any ∆ (Γ(2),n 1), ∆ ,..., ∆ is the unique sequence of faces summing up to ∆, ∈B − 2 n Vn−1(∆2,... ∆n) is the (n 1)-dimensional mixed volume of the projections of ∆2,..., ∆n onto n−1 ⊥ − (2) E R and u E E (2) is the outer unit normal vector to the facet ∆ of Γ . ∆ ≃ ∆ ∈ ∆ ∩ Γ 13. Mixed discriminant formula: If A ,...,A 2(Rn) then 1 n ∈ K1 1 V (A ,...,A )= h D (HessR h ,..., HessR h )υ , (14) n 1 n n A1 n−1 A2 An ∂Bn Z∂Bn where, for every u ∂B and every 2 ℓ n, HessR h is the restriction to T ∂B of the (real) ∈ n ≤ ≤ Aℓ u n Hessian matrix of hAℓ at u, υ∂Bn is the standard volume form on ∂Bn and Dn−1(HessR hA2 ,..., HessR hAn ) is the mixed discriminant of HessR h ,..., HessR h (cf. section 7), i.e. 1/(n 1)! times the A2 An − coeﬃcient of λ1λ2 ...λn in the polynomial det(λ1In + λ2 HessR hA2 + ... + λn HessR hAn ).

14 14. Intrinsic formula: If Γ ,..., Γ (Rn) and Γ= k Γ , then 1 k ∈P ℓ=1 ℓ n P κ−1 V (Γ ,..., Γ ,B [n k]) = V (∆ ,..., ∆ )ψ (∆), (15) n−k k n 1 k n − k 1 k Γ ∆∈BX(Γ,k) where, for any ∆ (Γ, k), ∆ ,..., ∆ is the unique sequence of faces summing up to ∆, ∈ B 1 k and Vk(∆1,... ∆k) is the k-dimensional mixed volume of the projections of ∆1,..., ∆k onto E Rk. ∆ ≃ 15. Rationality on lattice polytopes: If Γ ,..., Γ (Rn) are lattice polytopes, then 1 n ∈P n!V (Γ ,..., Γ ) Z . n 1 n ∈

n 16. Valuation property: Let n N 0, 1 , m 1,...,n 1 and let Am+1,...,An in (R ) be ﬁxed. Then the mapping g ∈: (\{Rn) } R, given∈ { for any−A } (Rn) by K m K → ∈ K

gm(A)= Vn(A[m], Am+1,...,An) ,

is a valuation.

17. Alexandrov-Fenchel inequality: If n 2, for any A , A , A ,...,A (Rn), ≥ 1 2 3 n ∈ K V (A , A , A ,...,A )2 V (A [2], A ,...,A )V (A [2], A ,...,A ) . (16) n 1 2 3 n ≥ n 1 3 n n 2 3 n The interested reader is referred to [Sch], [Ewa], [Sal] and [Gro] for further details on the general theory of convex bodies as well as for the proofs of the results recalled so far.

3 Mixed ϕ-volume

Let ϕ be an arbitrary, but fixed, real function defined on the Grassmannian (Rn). On the basis of [Ka1] we propose the following definition. G Definition 3.1. For any integer 0 k n, the k-th intrinsic ϕ-volume of a polytope Γ (Rn) is the vϕ ≤ ≤ ∈P real number k (A) defined as

vϕ k (Γ) = ϕ(E∆) vol k(∆)ψΓ(∆) . (17) ∆∈BX(Γ,k) vϕ vϕ Of course k (Γ) = 0 as soon as dim Γ < k. For k = 0 (resp. k = n) one has 0 (Γ) = ϕ( 0 ) vϕ vϕ { } (resp. n(Γ) = ϕ(EΓ) vol n(Γ)). Setting k (∅)=0, the k-th intrinsic ϕ-volume is k-homogeneous, i.e. vϕ kvϕ k (λΓ) = λ k (Γ), for every λ R≥0. The k-th intrinsic ϕ-volume is just a weighted version of the usual k-th intrinsic volume, if ϕ ∈ 1, vϕ(Γ) = v (Γ). ≡ k k Lemma 3.1. Let r N∗, Γ ,..., Γ (Rn) and λ ,...,λ R . Then vϕ r λ Γ is a ∈ 1 r ∈ P 1 r ∈ ≥0 k ℓ=1 ℓ ℓ homogeneous polynomial of degree k in λ1,...,λr. P r ′ r Proof. Let Γ = ℓ=1 Γℓ and Γ = ℓ=1 λℓΓℓ. If λ1,...,λr are all positive, then there is a bijection ′ r ′ r ′ from (Γ, k) to (Γ , k) mapping the face ∆ = ∆ℓ of Γ to the face ∆ = λℓ∆ℓ of Γ . Asa B BP P ℓ=1 ℓ=1 consequence E∆ = E∆′ and K∆,Γ = K∆′,Γ′ , so that P P ϕ v ′ ′ ′ ′ ′ k (Γ )= ϕ(E∆ ) vol k(∆ )ψΓ (∆ ) ′ ′ ∆ ∈BX(Γ ,k) r = ϕ(E∆) vol k λℓ∆ℓ ψΓ(∆) . ! ∆∈BX(Γ,k) Xℓ=1 ′ ′ ′ For each ∆ (Γ , k), up to a translation carrying ∆ in E∆′ = E∆, we can use Theorem 2.1 in E Rk thus∈ getting B the equality ∆ ≃ r vol λ ∆ = V ∆ ,..., ∆ λ λ k ℓ ℓ n ρ(1) ρ(k) ρ(1) ··· ρ(k) ℓ=1 ! ρ X X 15 with ρ running in the set of all the functions 1,...,k 1,...,r and the coeﬃcients chosen so as to satisfy, (for any ρ), the condition { }→{ }

Vk(Γρ(1),..., Γρ(n))= Vk(Γς(ρ(1)),..., Γς(ρ(n))) , for every permutation ς of the image of ρ. It follows that

vϕ(Γ′)= ϕ(E )V ∆ ,..., ∆ ψ (∆) λ λ . (18) k  ∆ k ρ(1) ρ(k) Γ  ρ(1) ··· ρ(k) ρ ∆∈B(Γ,k) X X   If there is a j 1,...,r such that λ =0, then it is enough to drop Γ in the sum Γ. ∈{ } j j n Deﬁnition 3.2. For any integer 1 k n, and polytopes Γ1,..., Γk (R ), the k-th mixed ϕ-volume ϕ ≤ ≤ ∈P of Γ1,..., Γk is the coeﬃcient Vk (Γ1,..., Γk) of λ1 λk in the polynomial expression (18) when r = k and ρ is the identity, i.e. ···

ϕ Vk (Γ1,..., Γk)= ϕ(E∆)Vk (∆1,..., ∆k) ψΓ(∆) , (19) ∆∈BX(Γ,k) k with Γ= ℓ=1 Γℓ.

RemarkP 3.1. By the non-degeneracy property of Vk, the sum in (19) can be performed just on those edge-sum k-faces ∆ 4 k Γ such that ϕ(E ) =0. ℓ=1 k ∆ 6 Lemma 3.2. The k-thP mixed ϕ-volume has the following properties:

ϕ ϕ n (a) symmetry, i.e. Vk (Γ1,..., Γk)= Vk (Γς(1),..., Γς(k)) , for every sequence Γ1,..., Γk (R ) and every permutation ς of 1,...,n ; ∈P { } (a) diagonal property, i.e. Vϕ(Γ[k]) = vϕ(Γ) , for every Γ (Rn); k k ∈P (a) translation invariance.

Proof. The claim (a) follows at once from the second of (10), the second of (11) and the deﬁnition of k-dimensional mixed ϕ-volume. The statement (b) is a consequence of the diagonal property of Vk and ϕ vϕ the deﬁnition of Vk and k . The claim (c) follows from the translation invariance of ϕ, Vk and the outer angle.

Lemma 3.3. The k-th mixed ϕ-volume satisﬁes the following formula:

k k ϕ 1 ∂ vϕ Vk (Γ1,..., Γk)= k λℓΓℓ , (20) k! ∂λ1 ...∂λk ! Xℓ=1 for every Γ ,..., Γ (Rn). 1 k ∈P Proof. The equality is straightforward if one realizes that the right member of (20) is the coeﬃcient of λ λ in the polynomial (18). 1 ··· k The idea of the proof of the following theorem comes from [Tho]. Theorem 3.1. The k-th mixed ϕ-volume satisﬁes the following polarization formula:

1 V ϕ(Γ ,..., Γ )= ( 1)k−#I vϕ Γ , (21) k 1 k k! − k ℓ I⊆{1,...,k} ℓ∈I ! IX6=∅ X for every Γ ,..., Γ (Rn). 1 k ∈P n Proof. For every Γ (R ), we introduce a shift operator σΓ on the space of real functions deﬁned n ∈ P n n on (R ). For each w : (R ) R set σΓw(A) = w(A + Γ), for every A (R ). Then σ{0} is the identityP operator and σ Pσ = σ→ , indeed, for every w : (Rn) R, one∈ P has Γ1 Γ2 Γ1+Γ2 P → σ w( )= w( + 0 )= w( ) {0} · · { } ·

16 and σ σ w( )= σ w( +Γ )= w( +Γ +Γ )= σ w( ) . Γ1 Γ2 · Γ1 · 2 · 2 1 Γ1+Γ2 · If D = σ σ , an integration of the equality (20) on [0, 1]k yields Γ Γ − {0} V ϕ(Γ ,..., Γ )= D D vϕ ( 0 ) . k 1 k Γk ··· Γ1 k { } Indeed

ϕ Vk (Γ1,..., Γk) k k 1 ∂ vϕ = k λℓΓℓ dλ1 ... dλk [0,1]k k! ∂λ1 ...∂λk ! Z Xℓ=1 k−1 k k 1 ∂ vϕ vϕ = k Γ1 + λℓΓℓ k λℓΓℓ dλ2 ... dλk [0,1]k−1 k! ∂λ2 ...∂λk " ! − !# Z Xℓ=2 Xℓ=2 k−1 k 1 ∂ vϕ = (DΓ1 k ) λℓΓℓ dλ2 ... dλk [0,1]k−1 k! ∂λ2 ...∂λk ! Z Xℓ=2 k−2 k 1 ∂ vϕ = (DΓ2 DΓ1 k ) λℓΓℓ dλ3 ... dλk [0,1]k−2 k! ∂λ3 ...∂λk ! Z Xℓ=3 = ...... 1 ∂ = D D vϕ (λ Γ ) dλ k! ∂λ Γk−1 ··· Γ1 k k k k Z[0,1] k 1 = D D vϕ (Γ ) D D vϕ ( 0 ) k! Γk−1 ··· Γ1 k k − Γk−1 ··· Γ1 k { } 1 ϕ ϕ = D D v ( 0 +Γ ) D D v ( 0 ) k! Γk−1 ··· Γ1 k { } k − Γk−1 ··· Γ1 k { } 1 ϕ = D D v ( 0 ) . k! Γk ··· Γ1 k { } vϕ It remains to show that DΓk DΓ1 k ( 0 ) equals k! times the right hand side of (21). For every I 1,...,k , set Γ = Γ··· and observe{ } that, as σ ,...,σ commute with one-another, one gets ⊆{ } I ℓ∈I ℓ Γ1 Γk P k D D = D Γk ··· Γ1 Γℓ ℓY=1 k = (σ σ ) Γℓ − {0} ℓY=1 = ( 1)k−#I σ − Γℓ I⊆{1,...,k} ℓ∈I IX6=∅ Y = ( 1)k−#I σ . − ΓI I⊆{1,...,k} IX6=∅

Now the conclusion follows easily, because σ vϕ( 0 )= vϕ(Γ ). ΓI k { } k I Theorem 3.2. The k-th mixed ϕ-volume is k-linear.

ϕ ϕ Proof. As Vk is symmetric, it is enough to prove the linearity of Vk in its ﬁrst argument, i.e.

ϕ ′ ′ ϕ ′ ϕ ′ Vk (µ1∆1 + µ1∆1, Γ2,..., Γk)= µ1Vk (∆1, Γ2,..., Γk)+ µ1Vk (∆1, Γ2,..., Γk),

′ n ′ ′ ′ for every ∆1, ∆1, Γ2,..., Γk (R ) and any µ1,µ1 R≥0. Set Γ1 = µ1∆1 + µ1∆1 and observe that vϕ k vϕ ∈ P ′ ′ k ∈ k ( ℓ=1 λℓΓℓ)= k (λ1µ1∆1 +λ1µ1∆1 + ℓ=2 λℓΓℓ), for every λ1, λ2,...,λk > 0. Then the coeﬃcients of λ1λ2 λk in both sides of this equality must agree. P ··· P ϕ Corollary 3.1. The k-th mixed ϕ-volume Vk is the only k-linear symmetric function which agrees with vϕ k on the diagonal.

17 n vϕ Proof. Let L be a k-linear symmetric function on (R ) agreeing with k on the diagonal. Then, on one hand, P k k k k k ∂ ∂ vϕ L λℓΓℓ,..., λℓΓℓ = k λℓΓℓ , ∂λ1 ...∂λk ! ∂λ1 ...∂λk ! Xℓ=1 Xℓ=1 Xℓ=1 for every Γ ,..., Γ (Rn). On the other hand, as L is multilinear and symmetric, one has 1 k ∈P ∂k k k L λℓΓℓ,..., λℓΓℓ ∂λ1 ...∂λk ! Xℓ=1 Xℓ=1 k k ∂k = L λℓ1 Γℓ1 ,..., λℓk Γℓk , ∂λ1 ...∂λk ! ℓX1=1 ℓXk=1 k k ∂k = ... L(Γℓ1 ,..., Γℓk )λℓ1 λℓk ∂λ1 ...∂λk ··· ℓX1=1 ℓXk=1 = k! L(Γ1,..., Γk) , so that the claim follows by Lemma 3.3.

Remark 3.2. On the monotonicity of the k-th mixed ϕ-volume. Let Γ , Γ′ ,..., Γ (Rn) be such that Γ Γ +Γ′ . Then, by linearity, 1 1 k ∈P 1 ⊂ 1 1 V ϕ(Γ +Γ′ , Γ ,..., Γ )= V ϕ(Γ ,..., Γ )+ V ϕ(Γ′ ,..., Γ ) V ϕ(Γ ,..., Γ ) . k 1 1 2 k k 1 k k 1 k ≥ k 1 k ϕ Without any further hypothesis on the shape of ϕ, this is the only monotonicity property Vk can satisfy. vϕ Theorem 3.3. For every 1 k n, the k-th intrinsic ϕ-volume k is a k-homogeneous, translation n ≤ ≤ vϕ invariant valuation on (R ). If ϕ is continuous, k provides a k-homogeneous, translation invariant, continuous valuation onP (Rn). K vϕ Proof. The translation invariance follows at once from the deﬁnition of k . The k-homogeneity is easy, indeed, for every Γ (Rn) and λ> 0 ∈P vϕ 1 k (λΓ) = ϕ(E∆) vol k(∆)ψΓ(∆) κn−k ∆∈BX(λΓ,k) 1 = ϕ(Eλ∆) vol k(λ∆)ψΓ(λ∆) κn−k ∆∈BX(Γ,k) 1 k = ϕ(E∆)λ vol k(∆)ψΓ(∆) κn−k ∆∈BX(Γ,k) kvϕ = λ k (Γ) .

vϕ Since R is a Hausdorﬀ space, by virtue of Theorem 2.2, it is enough to show that k is a weak k-homogeneous valuation. Let H Rn be a hyperplane intersecting Γ, then Γ H, Γ H+ and Γ H− are polytopes too. All the k-dimensional⊂ faces of Γ H are also k-faces of Γ∩ H+ ∩and Γ H−∩. For ∩ ∩ ∩ each such common k-face ∆ one has a disjoint union K∆,Γ∩H+ K∆,Γ∩H− = K∆,Γ∩H . Moreover, Γ H+ and Γ H− may respectively have k-faces ∆ and ∆ such∪ that ∆ ∆ is a k-face of Γ not ∩ ∩ 1 2 1 ∪ 2 included in H. In this case, ϕ(E∆1 ) = ϕ(E∆2 ) = ϕ(E∆1∪∆2 ), vol k(∆1) + vol k(∆2) = vol k(∆1 ∆2) + − ∪ + − and K∆1,Γ∩H = K∆2,Γ∩H = K∆1∪∆2,Γ. Any other k-face of Γ H and Γ H which does not intersect H is also a k-face of Γ. The preceding analysis implies that∩ ∩

vϕ(Γ H+)+ vϕ(Γ H−)= vϕ(Γ H)+ vϕ(Γ) . k ∩ k ∩ k ∩ k vϕ The equality k (∅)=0 completes the proof of the ﬁrst statement. The continuity hypothesis on ϕ vϕ n implies the continuity of the weak valuation k on (R ). By Theorem 2.3, this produces a continuous valuation on the whole (Rn) which is still k-homogeneousP and translation invariant. K ϕ We have shown that the Vk and Vk share some important properties, nevertheless the validity of other properties (like monotonicity, continuity, non-degeneracy, etc) heavily depends on the shape of the function ϕ. In the following section we will present a geometrically interesting instance of such a function, playing a fundamental role in the construction of Kazarnovskiˇıpseudovolume.

18 4 Volume distortion

∗ n Let n N and let E1, E2 R be a couple linear subspaces of positive dimensions d1 = dim E1 and ∈ n ⊂ d2 = dim E2. If R is endowed with the usual scalar product ( , ), consider the R-linear mapping

p ̺ : E ι E E⊥ = Rn id Rn = E E⊥ E , E1,E2 1 −→ 1 ⊕ 1 −→ 2 ⊕ 2 −→ 2 where ι is the inclusion, id is the identity mapping and p is the projection. To the linear mapping ̺E1,E2 one can associate the R-linear operator

t −1 ̺E ,E ψ (̺E ,E ) ψ ̺˜ : E 1 2 E 2 E∗ 1 2 E∗ 1 E , E1,E2 1 −−−−−−→ 2 −−−−→ 2 −−−−−−→ 1 −−−−−→ 1 where ψ1 (resp. ψ2) is the natural isomorphism mapping any element v of E1 (resp. E2) to the linear form (v, ) on E (resp. E ). The coeﬃcient of volume distortion under the projection of E on E is · 1 2 1 2 the non-negative real number ̺(E1, E2) deﬁned as

̺(E1, E2)= det̺ ˜E1,E2 . p The number det̺ ˜E1,E2 is sometimes called the Gramian of the linear mapping ̺E1,E2 . Of course the rank of ̺˜E1,E2 cannot exceed min(d1, d2), and, in fact, ̺(E1, E2)=0 as soon as d2 < d1. If d1 = d2 = d and a1,...,ad (resp. b1,...,bd) is an orthonormal basis of E1 (resp. E2), then

̺(E , E )= det ̺ = det(a ,b ) ; 1 2 | E1,E2 | | ℓ j | this value equals the d-dimensional non-oriented volume of the parallelotope generated in E2 by the row- vectors of the matrix ((aℓ,bj ))ℓ,j and, of course, it does not depend on the choice of the orthonormal basis in E1 and E2. As the norm of each row-vector in this matrix cannot exceed 1, the same happens to ̺(E1, E2). By giving Cn the euclidean structure induced by the standard scalar product of R2n, one can define the coefficient of volume distortion for R-linear subspaces of Cn. Such a scalar product on Cn can be realized as the real part of the standard hermitian product of Cn, i.e. Re , . h· ·i If A Cn is a non-empty subset, let affR A (resp. affC A) denote the real (resp. complex) affine ⊂ n n subspace of C spanned by A, whereas EA (still) denotes the R-linear subspace of C parallel to affR A. Let (Cn, R) be the Grassmanian of the real linear subspaces of Cn. For every E (Cn, R), the G ∈ G C-linear subspace spanned by E will be noted linC E, of course it is equal to E + iE. The dimension n dimR E of a subspace E (C , R) is just the real dimension of E, whereas the complex dimension ∈ G dimC E is meant to be the complex dimension of linC E. In general dimR E dimR(linC E) 2 dimR E C ≤ ≤ and, setting E = E iE, the equality dimR(linC E) = 2 dimR E occurs if and only if and only if ∩C linC E = E iE, i.e. E = 0 . ⊕ { } By E⊥ (resp. E⊥C ) we will denote the orthogonal complement of E respect to the scalar product Re , (resp. the hermitian product , ). In general one has the relations (linC E)⊥C = E⊥C and ⊥hC i ⊥ h i ′ ⊥ E E , the latter inclusion becoming an equality as soon as E = linC E. By setting E = E linC E, ⊆ ∩ the orthogonal decomposition linC E = E E′ yields dimR E′ dimR E, because ⊕ ≤

′ 1 C dimR E = dimR(linC E) dimR E dimR E. 2 − ≤ C In particular, dimR E′ = dimR E if and only if E = 0 , although in general iE = E′. Anyway { } 6 dimR E′ dimR E mod 2, since otherwise the complex space linC E would have an odd real dimension. ≡ The Cauchy-Riemann dimension of E is just the complex dimension of EC . A linear subspace E (Cn, R) is full-dimensional if its real dimension equals 2n, it is purely real if it is included in Rn. A∈ subspace G E (Cn, R) is equidimensional if its real and complex dimensions coincide. Of course ∈ G n 1-dimensional subspaces are such, as well as purely real ones. If E1, E2 (C , R) and E1 E2, C C ∈ G ⊂ then E1 E2 , whence the equidimensionality of E2 is inherited by each of its subspaces. A subspace ⊆ C E (Cn, R) with dimR E>n cannot be equidimensional, for in this case E = 0 . ∈ G 6 { } The orthogonal decompositions E E ′ = linC E = iE iE ′ imply that iE = (iE′)⊥ linC E, so that ⊕ ′⊕ ∩ ′ one can consider the R-linear projection ̺E,iE′ : E iE . If v ker̺E,iE′ , then Re v,iE =0 so that → C ∈ Ch i v (iE′)⊥ E (iE′)⊥ linC E = iE, hence v E . On the other hand, v E implies iv E, ∈ ∩ ′ ⊆ ∩ ′ ∈ C ∈ − ∈ then Re v,iE = Re iv,E = 0, i.e. ker̺E,iE′ = E . It follows that ̺E,iE′ is an isomorphism if and onlyh if iE is equidimensional.h− i The coeﬃcient of volume distortion ̺(E) is by deﬁnition the

19 ′ ′ coeﬃcient of volume distortion under the projection ̺E,iE′ of E onto iE , i.e. ̺(E) = ̺(E,iE ). A subspace E (Cn, R) is real similar if ̺(E)=1. The trivial subspace 0 is conventionally assumed to be real similar,∈ G so that we set ̺( 0 )=1. Purely real subspaces{ are} real similar as well as 1- dimensional ones, nevertheless there are{ } equidimensional subspaces which are not real similar. Remark that if E is equidimensional, then E′ cannot be a complex subspace, since otherwise E′ = iE′ which ′ ′ yields the contradiction ̺(E,iE ) = ̺(E, E )=0. If E is equidimensional and d = dimR E > 0, let v1,...,vd, w1,...,wd be a basis of linC E over R such that v1,...,vd is an orthonormal basis of E and ′ w1,...,wd an orthonormal basis of E , then ̺(E) = det(Re vℓ, iwj )ℓ,j . In this case, the equality |′ h i | Re vℓ, iwj = Re wj , ivℓ implies that ̺(E) = ̺(E ). In order to achieve a standard formula for ̺(Eh), it isi useful− to ﬁndh a standardi basis for E′.

n C Lemma 4.1. Let E (C , R) with dimR E =2c < d = dimR E. If the vectors v1,...,v2c, v2c+1,...,vd ∈ G C form an orthonormal basis of E such that v1,...,v2c span E , then the vectors

d t = iv Re iv , v v , ℓ ℓ − h ℓ si s s=1 X ′ ′ for ℓ = 1,...,d span E and t2c+1,...,td provide a basis of E . In particular E is equidimensional if and only if t1,...,td are R-linearly independent.

C Proof. The linear span of v2c+1,...,vd is equidimensional, otherwise E would not be maximal among the complex subspaces contained in E. This means that dimR(linC E)=2d 2c, so that dimR E′ = d 2c. − d − Observe also that c> 0 implies t1 = ... = t2c =0. Suppose, by contradiction, that ℓ=2c+1 aℓtℓ =0 is a non trivial linear combination, then P d d d i a v = a Re iv , v v . (22) ℓ ℓ ℓ h ℓ si s s=1 ! ℓ=2Xc+1 X ℓ=2Xc+1

The vector on the right of (22) cannot be 0, (otherwise v1,...,vd would be linearly dependent), so equality (22) implies that d C a v E 0 . (23) ℓ ℓ ∈ \{ } ℓ=2Xc+1 If c =0 (i.e. EC = 0 ) the relation (23) is impossible, if c> 0 the vector d a v must be a non { } ℓ=2c+1 ℓ ℓ trivial linear combination of v1,...,v2c, which is also impossible since v1,...,vd are linearly independent. P ′ ′ In particular, when E is equidimensional, t1,...,td E provide a basis of E . Conversely, the linear ∈′ independence of t1,...,td means that dimR E = dimR E , i.e. E is equidimensional.

Remark 4.1. On the natural orientation of complex spaces. Every complex subspace of Cn has a natural orientation coming from the complex structure. This orientation will be referred to as the positive orientation. If e1,...,ec is a sequence of C-linearly independent vectors of Cn, then the corresponding C-linear span is positively oriented, as a real 2c-dimensional subspace, by the basis e1,ie1,...,ec,iec.

n C Corollary 4.1. Let E (C , R) with dimR E = 2c < d = dimR E. If v1,...,v2c, v2c+1,...,vd ∈ G C is an orthonormal basis of E such that v1,...,v2c is an orienting basis of E , then the sequence v ,...,v , v ,t ,...,v ,t is an orienting basis of E E′ = linC E. In particular, the sequence 1 2c 2c+1 2c+1 d d ⊕ v1,...,vd,t2c+1,...,td yield the positive orientation of linC E if and only if

(d 2c) 0 or (d 2c) 1 . (24) − ≡4 − ≡4 C Proof. Up to a unitary transformation, we may suppose linC E = Cd−c 0 n−d+c and E = Cc 0 d−2c 0 n−d+c. As shown by Lemma 11.4, from the real point of× view, { } a unitary transformation×{ } is an×{ orthogonal} transformation of R2n with positive determinant, so it preserves the (real) orthonormality of the basis v1,...,v2c, v2c+1,...,vd. By virtue of the preceding argument we can carry out the computation in the smaller space Cd−c. As t iv E, for every 2c +1 ℓ d, it follows that ℓ − ℓ ∈ ≤ ≤

det(v1,...,v2c, v2c+1,t2c+1,...,vd,td) (25)

= det(v1,...,v2c, v2c+1, iv2c+1,...,vd, ivd) , (26)

20 and of course (25) cannot be zero because v1,...,v2c, v2c+1,t2c+1,...,vd,td is a basis. In fact (26) equals the determinant of the matrix

(v1)1 ... (v2c)1 (v2c+1)1 (iv2c+1)1 ... (vd)1 (ivd)1 ......  ......  (v ) ... (v ) (v ) (iv ) ... (v ) (iv )  1 2c 2c 2c 2c+1 2c 2c+1 2c d 2c d 2c   0 ... 0 (v ) (iv ) ... (v ) (iv )   2c+1 2c+1 2c+1 2c+1 d 2c+1 d 2c+1   ......   ......     0 ... 0 (v ) (iv ) ... (v ) (iv )   2c+1 2d 2c+1 2d 2c+1 2d 2c+1 2d    where both the north-west block and the south-east one cannot have zero determinant. Indeed, the C north-west block has positive determinant because v1,...,v2c is an orienting basis of E , while that of the south-east one equals

Re v Im v ... Re v Im v 2c+1 2c+1 − 2c+1 2c+1 d 2c+1 − d 2c+1 Im v2c+1 2c+1 Re v2c+1 2c+1 ... Im vd1 Re vd1  . . . . .  det ...... ,    Re v Im v ... Re v Im v   2c+1 d − 2c+1 d dd − dd   Im v Re t ... Im v Re v   2c+1 d 2c+1 d dd dd    which, in the notation of Lemma 11.4, is just

2 v2c+1 2c+1 ... vd 2c+1 . . . det ψd−2c  . .. .  .

v ... v  2c+1 d dd   

The last statement follows easily by the relation

det(v1,...,v2c, v2c+1,t2c+1,...,vd,td) (d−2c−1)(d−2c) = ( 1) 2 det(v ,...,v ,t ,...,t ) . − 1 d 2c+1 d

Indeed the number (d 2c 1)(d 2c)/2 is even if and only if (d 2c) 4 0 or (d 2c) 4 1. The proof is thus complete. − − − − ≡ − ≡

Theorem 4.1. Let E (Cn, R). If v ,...,v is an orthonormal basis of E, then ∈ G 1 d ̺(E)= det(Re t ,t ) . (27) h ℓ j i q Proof. Observe, ﬁrst of all, that det(Re t ,t ) 0 because it is the square of the volume of the h ℓ j i ≥ parallelotope spanned by t1,...,td. Let us ﬁrst suppose E is not equidimensional. In this case the vectors ′ t1,...,td E are R-linearly dependent and then det(Re tℓ,tj ) vanishes. If E is equidimensional, the ∈ ′ h i vectors t1,...,td provide a basis of E . This basis is generally not orthogonal (unless d = 2) nor orthonormal, so setting u = t and, for 1 < j d, 1 1 ≤ j−1 Re t ,u u = t h j si u , j j − Re u ,u s s=1 s s X h i ′ yields an orthogonal basis of E , and the vectors wj = uj / uj , 1 j d, provide an orthonormal basis of E′. As a consequence, ̺(E)= det(Re v , iw ) . Observek k that,≤ for≤ any 1 j, ℓ d, | h ℓ j i | ≤ ≤ d Re t ,t = δ Re iv , v Re iv , v h ℓ j i ℓ,j − h j si h ℓ si s=1 X d = δ Im v , v Im v , v ℓ,j − h j si h ℓ si s=1 X = δ + v , v v , v , (28) ℓ,j h ℓ sih j si 1≤s≤d j6=Xs6=ℓ

21 Re v ,it = Re t ,t and, moreover (by induction on j) h ℓ j i − h ℓ j i Re v , iw = Re t , w h ℓ j i − h ℓ j i j−1 = u −1 Re t ,t Re v , iw Re v , iw . −k jk h ℓ j i− h j si h ℓ si " s=1 # X −1 The j-th column of the matrix (Re vℓ, iwj ) is the sum of uj Re tℓ,tj and a linear combination of the preceding columns, so its determinanth i equals that of−k the matrixk h( ui −1 Re t ,t ) whence −k jk h ℓ j i d −1 det(Re v , iw ) = det( u −1 Re t ,t )= u det(Re t ,t ) . | h ℓ j i | k jk h ℓ j i k sk h ℓ j i s=1 ! Y d On the other hand, the real number s=1 us equals the d-dimensional (non-oriented) volume of the ′ k k parallelotope of E generated by the vectors u1,...,ud and such a volume can be written as Q d u = det(Re t , w )= det(Re t ,t ) , k sk h ℓ j i h ℓ j i s=1 Y q so that the preceding computation implies that ̺(E)= det(Re t ,t ). h ℓ j i Remark 4.2. For every orientation or(E) 1, 1 pon E it is possible to choose an orientation ∈ {− } or(E′) 1, 1 such that E E′ = linC E gets the positive orientation. This can be achieved, ∈ {− } ⊕ C for example, by choosing any orthonormal basis v1,...,vd of E such that v1,...,v2c spans E over R, followed by the basis tσ(2c+1),...,tσ(d), where t2c+1,...,td are the vectors constructed in Lemma 4.1 and σ is a permutation of 2c +1,...,d with sign(σ) = ( 1)(d−2c−1)(d−2c)/2. In particular, if (d 2c) 0 { } − − ≡4 or (d 2c) 4 1, σ can simply be the identity, whereas it can be a transposition if (d 2c) 4 2 or − ≡ (d−2c−1)(d−2c)/2 − ≡ (d 2c) 4 3. Even more simply, the basis v1,...,v2c, ( 1) t2c+1,t2c+2,...,td will always do.− ≡ −

Remark 4.3. Observe that if d =2 and E is equidimensional, the equality (28) shows that the vectors t ,t are orthogonal to each-other and that t = t = ̺(E). By(24), the basis v , v ,t ,t yields 1 2 k 1k k 2k 1 2 2 1 the positive orientation of linC E. p Corollary 4.2. Let E (Cn, R), d = dimR E and m = dimC(linC E). Let also v ,...,v be an ∈ G 1 d orthonormal basis of E with respect to Re , , e1,...,em an orthonormal basis of linC E with respect to , . If A = ( v , v ) and B = ( v ,eh i ) , then h i h ℓ j i 1≤ℓ,j≤d h ℓ ki 1≤ℓ≤d,1≤k≤m ̺(E) = det A = det(B tB¯). (29)

In particular, if d = m, ̺(E) equals the real jacobian of the C-linear operator on linC E mapping eℓ to v , for every 1 ℓ d. ℓ ≤ ≤ Proof. Let M be the d d matrix whose (ℓ, j)-entry is × v , v v , v = δ + v , v v , v , h ℓ sih j si ℓ,j h ℓ sih j si 1≤s≤d 1≤s≤d X j6=Xs6=ℓ then, by (27) and (28), it follows that ̺(E)= √det M. Observing that det A is real (since A a hermitian matrix) and M equals the product of A with tA, it follows that ̺(E)= √det M = (det A)2 = det A . Nevertheless det A 0. Indeed A = B tB¯ and rank A = rank B = m, so det A = det(B tB¯)=0| if E| p is not equidimensional≥ (i.e. d>m) and det A = det(B tB¯) = det B 2 = 0 as soon as d = m. In the | | 6 latter case, mapping eℓ to vℓ yields a C-linear automorphism, say L, on linC E whose matrix in the basis d d e1,...,ed is just ( vℓ,ej ). As happens to every other holomorphic mapping C C , the square of the modulus of theh complexi jacobian of L (i.e. det L 2) equals its real jacobian, i.e.→ the jacobian of L regarded as a mapping R2d R2d. | | → Remark 4.4. Corollary 4.2 shows that E (Cn, R) ̺(E)=0 is a Zariski closed subset, thus equidimensionality is a generic property{ of R∈-linear G subspaces| of C}n. Moreover it implies that the coeﬃcient of volume distortion is continuous and unitarily invariant.

22 Remark 4.5. ̺(E) is unitarily invariant but not orthogonally invariant. Corollary 4.2 shows that the coeﬃcient of volume distortion is unitarily invariant. Indeed, if F : Cn Cn is a unitary transformation, then → ̺(F (E)) = det F (v ), F (v ) = det F det A det tF = det A = ̺(E) . h ℓ j i This is generally false if F is an orthogonal transformation of R2n. For instance, consider the real subspace E = Rn Cn and the orthogonal R-linear mapping F : R2n R2n permuting the elements ⊂ 2n → ie1 and en of the canonical basis of R . As F (E) contains the complex line spanned by e1, it follows that ̺(F (E))=0, whereas ̺(E)=1.

Remark 4.6. Two further formulas for ̺(E)

Up to a renumbering of v1,...,vd, one can suppose that v1,...,vm are C-linearly independent, so the basis e1,...,em can be chosen as an orthonormalization of v1,...,vm with respect to , . Setting e =0, for m < k d, a further formula for ̺(E) would read as follows h i k ≤ 2 v ,e v ,e 0 0 h 1 1i ··· h 1 mi ··· 2 ...... ̺(E)= det(B 0) = det  ......  , | | | v ,e v ,e 0 0  h d 1i ··· h d mi ···    where the last zero d (d m) block is absent if d = m. Using the Gram-Schimdt method one sets × − e = v , ω = v ℓ−1 v ,e e and e = ω / ω , for 1 <ℓ m. This choice has the advantage of 1 1 ℓ ℓ − k=1h ℓ ki k ℓ ℓ k ℓk ≤ producing some zero entries in the matrix B, namely vℓ,ek = 0, for every 2 ℓ < k m, so that, when d = m the matrixP B is lower triangular and h i ≤ ≤

d 2 d d 2 2 2 ̺(E)= det B = vℓ,eℓ = vℓ,eℓ = ωℓ . | | h i |h i| k k ℓ=1 ℓ=2 ℓ=2 Y Y Y Another formula for ̺(E) can be found in [Ale]. Let B = B E and iB = iz iE z B . The E 2n ∩ E { ∈ | ∈ E} subset BE + iBE has the structure of a (possibly degenerate) cylinder on the base BE whose dimension equals 2d if and only if E is equidimensional. Since 0 vol (B + iB )= ̺(E)κ2 κ2 , ≤ 2d E E d ≤ d −2 one can set ̺(E)= κd vol 2d(BE + iBE). Example 4.1. The coeﬃcient of length, area and volume distortion.

When d =1 we already know that ̺(E)=1, so there is no length distortion. In fact ̺(E)= v1, v1 =1. If d =2, ̺(E)=1+ v , v 2 =1 (Im v , v )2. Since Im v , v = Re iv , v equals the cosineh i of the h 1 2i − h 1 2i h 1 2i h 1 2i (convex) angle of the vectors iv1 and v2, det A is the square of the sine of that angle. As a consequence, E is real similar if and only if v2 and iv1 are orthogonal to each other. For d =3, one gets the similar formula: ̺(E)=1+ v , v 2 + v , v 2 + v , v 2 h 1 2i h 1 3i h 2 3i =1 (Im v , v )2 (Im v , v )2 (Im v , v )2 . − h 1 2i − h 1 3i − h 2 3i So, for 1 d 3, one can write the following simple formula: ≤ ≤ ̺(E)=1+ v , v 2 =1 (Im v , v )2 , h ℓ j i − h ℓ j i 1≤Xℓ

Example 4.2. The coeﬃcient of 4-dimensional volume distortion. Unlike the preceding two cases, if d =4 the formula for ̺(E) involves also some mixed products: ̺(E)=1+ v , v 2 + v , v 2 + v , v 2 + v , v 2 + v , v 2 + v , v 2+ h 1 2i h 1 3i h 1 4i h 2 3i h 2 4i h 3 4i + v , v 2 v , v 2 + v , v 2 v , v 2 + v , v 2 v , v 2 h 1 2i h 3 4i h 1 3i h 2 4i h 1 4i h 2 3i 2 v , v v , v v , v v , v − h 1 2ih 1 3ih 2 4ih 3 4i +2 v , v v , v v , v v , v h 1 2ih 1 4ih 2 3ih 3 4i 2 v , v v , v v , v v , v . − h 1 3ih 1 4ih 2 3ih 2 4i

23 Example 4.3. Real similar n-dimensional coordinate subspaces of Cn. n 2n The number of real n-dimensional coordinate subspaces of C equals n . If n = 1 the coordinate axes are both real similar, whereas for n > 1 only 2n are equidimensional. In fact, if e ,...,e is the 1 n canonical basis of Cn, a real n-dimensional coordinate subspace of Cn is obtained by choosing n distinct elements from the set e1,ie1,...,en,ien avoiding to choose two elements sharing the same index. This yields 2n diﬀerent choices,{ each of which} produces a real similar subspace because the chosen vectors are orthogonal to one another with respect to the hermitian product of Cn.

5 Convex bodies in Cn

The sets of convex bodies, polytopes and k-regular compact subsets of Cn, k N∗ , are respectively denoted (Cn), (Cn) and k(Cn). For a non-empty subset A Cn,∈ the∪{∞} notions of dimension, complexK dimension,P full-dimensionality,S pure reality, equidimensionality⊆ and real similarity as well as the coeﬃcient of volume distortion introduced for R-linear subspaces of Cn, can be simply extended by just invoking the real subspace EA. In particular, this can be done for convex bodies. For any n A (C ), set dimR A = dimR EA, dimC A = dimC EA and ̺(A) = ̺(EA). In the complex setting (R∈n K)= A (Cn) A Rn , i.e. the set of purely real convex bodies; in the same way we will use K { ∈k K n | ⊂n } k n ∞ n n the notation (R ), ℓ(R ), ℓ (R ), (R ) and (R ). In the complex setting a lattice polytope Γ (Cn) isK a polytopeK whoseK vertices haveP coordinatesP in the ring Z + iZ of Gauss’ integers. ∈P n If A (C ), the support function hA of A is computed with respect to the scalar product Re , , ∈ K n h i i.e. hA(z) = supv∈A Re z, v , so that, given v C 0 , the corresponding supporting hyperplane H (v) for the convex bodyh Ai is H (v)= z C∈n Re\{z, v} = h (v) . A A { ∈ | h i A } n If Γ (C ), and ∆1 4 ∆2 4 Γ, the inclusions E∆1 E∆2 EΓ imply linC E∆1 linC E∆1 ∈ P C C C ⊆ ⊆ ⊆ ⊆ linC EΓ and E E E . ∆1 ⊆ ∆2 ⊆ Γ For every 0 k dimR Γ, (Γ, k) will denote the subset of (Γ) consisting of equidimensional ≤ ≤ Bed B k-faces. Notice that, for any (non-empty) polytope, ed(Γ, 0) = (Γ, 0) = ∅ and, if Γ is not reduced to a single point, (Γ, 1) = (Γ, 1) = ∅. B B 6 Bed B 6 For any proper face ∆ Γ, one has dimC ∆ dimC Γ, but if Γ is purely real the equality cannot occur. The following theorem suggested≺ by Kazarnovskiˇıin≤ a private communication clariﬁes the situation in the general case.

n Theorem 5.1. Let Γ (C ) be a d-polytope and let 2 k dimC Γ an integer. If ed(Γ, k) is empty, then the following statements∈P hold true. ≤ ≤ B

1. For any ℓ k +1,...,d , the set (Γ,ℓ) is empty; in particular Γ is not equidimensional. ∈{ } Bed 2. There exists a ∆ (Γ, k) such that linC E = linC E ; in particular dimC Γ < k. ∈B Γ ∆ 3. dimC Γ = max 0 j n (Γ, j) = ∅ . { ≤ ≤ |Bed 6 } ′ ′ C C ′ Proof. If ∆ is an ℓ-face of Γ and ∆ is a k-face of ∆ , then 0 = E∆ E∆′ , so ∆ cannot be equidimensional. In particular this occurs to the improper face Γ,{ which} 6 proves⊆ the ﬁrst statement.

In order to prove the second statement, observe that by first statement one has d > dimC Γ and C C C EΓ = 0 . Now it may happen that EΓ = EΓ or EΓ ) EΓ . In both cases we construct a strictly decreasing6 { } sequence ∆ ... ∆ of faces of Γ such that, for every ℓ 1,...,d k , one has d−1 ≻ ≻ k ∈ { − } dimR ∆d−ℓ = d ℓ and linC E∆ℓ = linC EΓ. The last term ∆k in the sequence will be the desired k-face. C− C If E = E , then E = linC E = E and d is an even number. For any facet ∆ Γ one has Γ Γ Γ Γ Γ d−1 ≺ E ( linC E linC E = E . The first inclusion is strict because E has odd real dimension ∆d−1 ∆d−1 ⊆ Γ Γ ∆d−1 whereas linC E∆d−1 (as every complex subspace) has even real dimension. Computing real dimensions yields (d 1) < dimR linC E dimR linC E = d, i.e. dimR linC E = d, i.e. linC E = linC E . − ∆d−1 ≤ Γ ∆d−1 ∆d−1 Γ In the second case, let v be a vertex of Γ and consider all the facets of Γ admitting v as a vertex. The intersection of the real affine subspaces spanned by such facets is reduced to the single vertex v (otherwise v would not be a vertex), so the intersection of the corresponding real linear subspaces is C reduced to 0 . As EΓ = 0 , there will be a facet ∆d−1 of Γ admitting v as a vertex such that E∆d−1 { } C 6 { } C does not contain EΓ . This implies that the real linear subspace E∆d−1 EΓ has real codimension 1 C C ∩ C in EΓ and, though it is not a complex subspace, it spans the whole EΓ over C. As a consequence EΓ is a complex subspace of linC E∆d−1 and so E∆d−1 must be a proper real subspace of EΓ linC E∆d−1 , C ∩ since otherwise it would contain EΓ . Let v1,...,vd−1 be a basis of E∆d−1 and let vd be an element of

24 E linC E not belonging to E . The real subspace spanned by v ,...,v equals E so that Γ ∩ ∆d−1 ∆d−1 1 d Γ E E linC E , i.e. E linC E , whence linC E = linC E . Γ ⊆ Γ ∩ ∆d−1 Γ ⊆ ∆d−1 Γ ∆d−1 Repeating the whole argument on the facet ∆d−1, one ﬁnds a ridge ∆d−2 of Γ, such that linC EΓ = linC E = linC E . By induction, in (d k) steps one gets a k-face ∆ of Γ such that linC E = ∆d−1 ∆d−2 − k Γ linC E∆k . Like any other k-face of Γ, ∆k is not equidimensional, so dimC Γ = dimC ∆k < k. As for the last statement, let j be the minimal integer such that (Γ, j + 1) is empty. Then Bed (Γ, j) is non-empty and, by the preceding statement, j = dimC Γ. Bed Dual cones to the faces of a polytope as well as the corresponding stars are deﬁned like in the real case. For any λ R 0 and any polytope Γ, every face of the polytope λΓ has the form λ∆, for a unique face ∆ of∈Γ, so\{ that}

affR λ∆= λ affR ∆ , affC λ∆= λ affC ∆ , Eλ∆ = E∆ ,

⊥ ⊥ ⊥C ⊥C linC Eλ∆ = linC E∆ , Eλ∆ = E∆ , Eλ∆ = E∆ ,Kλ∆,λΓ = K∆,Γ .

n s If Γ1,..., Γs (C ) and Γ= ℓ=1 Γs there exists a unique sequence of faces ∆ℓ 4 Γℓ, ℓ =1,...,s, ∈Ps such that ∆= ∆ℓ. As a consequence, one gets the equalites ℓ=1 P P s s s

affR ∆= affR ∆ℓ , affC ∆= affC ∆ℓ , E∆ = E∆ℓ , Xℓ=1 Xℓ=1 Xℓ=1 s s s ⊥ ⊥ ⊥C ⊥C linC E = linC E , E = E , E = E , ∆ ∆ℓ ∆ ∆ℓ ∆ ∆ℓ Xℓ=1 ℓ\=1 ℓ\=1 and s

K∆,Γ = K∆ℓ,Γℓ . ℓ\=1 n v̺ The mapping ̺ on the real grassmanian (C , R) yields a corresponding k-th intrinsic ̺-volume k, v̺ G for 0 k 2n, nevertheless k 0 if n < k 2n, since there is no equidimensional real subspace in Cn of≤ (real)≤ dimension greater than≡ n. Moreover,≤ as ̺ 1, by Theorem 3.3 one has ≤ V̺(A ,...,A ) V (A ,...,A ,B [2n k]) , k 1 k ≤ 2n 1 k 2n − n for every A1,...,Ak (C ). The following result provides a non vanishing condition for the k-th mixed ̺-volume. ∈ K

n k Corollary 5.1. Let Γ ,..., Γ (C ) and let Γ = Γ . Then dimC Γ < k if and only if 1 k ∈ P j=1 j V̺(Γ ,..., Γ )=0. k 1 k P

Proof. Suppose dimC Γ < k and, for every non empty subset I 1,...,k , set ΓI = j∈I Γj . Then, ⊆{ } v̺ for every I, dimC ΓI dimC Γ < k and by Theorem 5.1 ed(ΓI , k) = ∅ whence k(ΓI )=0 and, by ̺ ̺ P (21), V (Γ ,..., Γ )=0≤ . On the other hand, if V (Γ ,...,B Γ )=0, then (Γ, k) = ∅ and again by k 1 k k 1 k Bed Theorem 5.1 we obtain dimC Γ < k.

Corollary 5.2. Let Γ ,..., Γ (Cn). The following assertions are equivalent. 1 k ∈P ̺ (a) Vk (Γ1,..., Γk) > 0; (b) there are line segments Λ Γ ,..., Λ Γ with linearly independent complex directions; 1 ⊂ 1 k ⊂ k

(c) For every non empty subset I 1,...,k , dimC Γ #I. ⊆{ } j∈I j ≥ P Proof. (a) (b). By Corollary 5.1, the Minkowski sum Γ=Γ +...+Γ admits at least an equidimen- ⇒ 1 k sional k-face ∆ = ∆1 +...+∆k such that Vk(∆1,..., ∆k) > 0. By the non-degeneracy property of mixed volume, this implies that ∆ is an edge-sum k-face, i.e. there exist line segments Λ1 ∆1,..., Λk ∆k spanning over R lines with R-linearly independent directions. Among all such sequence⊆ s, there must⊆ be at least one sequence of segments spanning over C lines with C-linearly independent directions, otherwise ∆ would not be equidimensional. ̺ (b) (a). As k = dimC(Λ + ... +Λ ) dimC(Γ), then V (Γ ,..., Γ ) > 0 by Corollary 5.1. ⇒ 1 k ≤ k 1 k

25 (b) (c). For every I 1,...,k , dimC Γ dimC Λ #I. ⇒ ⊆{ } j∈I j ≥ j∈I j ≥ (c) (b) is a consequence of Lemma 5.1.8P in [Sch]. P ⇒ n For any ε R>0, the ε-regularization of a convex body A (C ) is deﬁned as in the real case, its support function∈ is ∈ K

hRεA(z)= hA(z + u z )ϕε( u ) υ2n , Cn k k k k Z where n n υ = (i/2)du d¯u = d(Re u ) d(Im u ) (30) 2n ℓ ∧ ℓ ℓ ∧ ℓ ℓ^=1 ℓ^=1 n 2n ∞ is the usual standard volume form on C = R and ϕε (R, R) is a non-negative function with (compact) support included in the interval [ε/2,ε] such that∈ C ϕ ( u ) υ = 1 . The regularization Cn ε k k 2n hRεA is convex but not diﬀerentiable at the origin, whereas the usual one, hA ϕε, provides a plurisub- harmonic smooth function on the whole Cn. R ∗

6 Kazarnovskiˇıpseudovolume on the class S2(Cn)

n n 2n We will use the identiﬁcation C = (R + iR) R , so that zℓ = xℓ + iyℓ = ξ2ℓ−1 + iξ2ℓ and i : R2n R2n acts on the vector ξ in the usual way,≃ i.e. iξ = ( ξ , ξ ,..., ξ , ξ ). Moreover → − 2 1 − 2n 2n−1 ∂ 1 ∂ ∂ = i , dz = dx + idy , ∂z 2 ∂x − ∂y ℓ ℓ ℓ ℓ ℓ ℓ ∂ 1 ∂ ∂ = + i , d¯z = dx idy , ∂z¯ 2 ∂x ∂y ℓ ℓ − ℓ ℓ ℓ ℓ for every ℓ =1,...,n. In consequence, the operators ∂, ∂¯, d= ∂ + ∂¯ and dc = i(∂¯ ∂) are given by − n n 1 ∂ ∂ ∂ ∂ ∂ ∂ = dxℓ + dyℓ + i dyℓ dxℓ = dzℓ , 2 ∂xℓ ∂yℓ ∂xℓ − ∂yℓ ∂zℓ Xℓ=1 Xℓ=1 n n 1 ∂ ∂ ∂ ∂ ∂ ∂¯ = dxℓ + dyℓ i dyℓ dxℓ = d¯zℓ . 2 ∂xℓ ∂yℓ − ∂xℓ − ∂yℓ ∂z¯ℓ Xℓ=1 Xℓ=1 n n ∂ ∂ ∂ ∂ d= dxℓ + dyℓ = dzℓ + d¯zℓ , ∂xℓ ∂yℓ ∂zℓ ∂z¯ℓ Xℓ=1 Xℓ=1 n n c ∂ ∂ ∂ ∂ d = dyℓ dxℓ = i d¯zℓ dzℓ . ∂xℓ − ∂yℓ ∂z¯ℓ − ∂zℓ Xℓ=1 Xℓ=1 Since dc equals its conjugate operator, it follows that dc and ddc (often referred to as the Bott-Chern operator) are real operators, moreover ddc =2i∂∂¯ is given by

n 2 2 c ∂ ∂ dd = dxℓ dyj + dyℓ dyj ∂xℓ∂xj ∧ ∂yℓ∂xj ∧ ℓ,jX=1 (31) ∂2 ∂2 dxℓ dxj dyℓ dxj − ∂xℓ∂yj ∧ − ∂yℓ∂yj ∧ or, in complex coordinates, by n 2 c ∂ dd =2i dzℓ d¯zj . ∂zℓ∂z¯j ∧ ℓ,jX=1 It should be mentioned that very often in the literature the operator dc is deﬁned in diﬀerent ways, namely (i/2)(∂¯ ∂) or even (i/2π)(∂¯ ∂), nevertheless we will conform to the choice of [Ka1], where dc = i(∂¯ ∂). − − − Remark that for every h 2(Cn, C), ddch is a 2-form so that the 2n-form (ddch)∧n is identically zero as soon as h depends on∈ less C than n-complex variables.

26 The integration with respect to Lebesque measure in Cn will be usually performed by integrating the standard volume form

in n n υCn = υ = dz d¯z = dx dy , 2n 2n ℓ ∧ ℓ ℓ ∧ ℓ ℓ^=1 ℓ^=1 then, for h 2(Cn, C), ∈C ∂2h (ddch)∧n = (2i∂∂h¯ )∧n =4nn! det υ . (32) ∂z ∂z¯ 2n ℓ k If f : Cn Cn is a holomorphic mapping, then → (ddc(h f))∧n = det J 2n(ddch)∧n , ◦ | f | where Jf is the (complex) jacobian determinant of f. n For every differential form ω on C , by ω we will denote the unique form such that ω ω = υ2n. The operator on forms is well defined and∗ known as Hodge -operator 2. It can be proved∧∗ that it defines a linear∗ operator on (p, q)-forms with values on (n p,n∗ q)-forms. − − For any j =1,...,n, we will also use the (2n 1)-forms − n υ [x ]= dx = dy dx dy , (33) 2n j ∗ j j ∧ ℓ ∧ ℓ ℓ=1 ℓ^6=j n υ [y ]= dy = dx dx dy , (34) 2n j −∗ j j ∧ ℓ ∧ ℓ ℓ=1 ℓ^6=j n in υ [z ]= dz = d¯z dz d¯z , 2n j ∗ j 2n j ∧ ℓ ∧ ℓ ℓ=1 ℓ^6=j in n υ [¯z ]= d¯z = dz dz d¯z , 2n j −∗ j 2n j ∧ ℓ ∧ ℓ ℓ=1 ℓ^6=j with the obvious relations

υ [z ] υ [¯z ]= υ [x ] and υ [z ]+ υ [¯z ]= iυ [y ] . 2n j − 2n j 2n j 2n j 2n j 2n j

n For every R-linear oriented subspace E C let υE denote the volume form on E. Observe that, for every v Cn the dc of the linear form Re⊂ v,z is given by ∈ h i dc Re v,z = i(∂¯ ∂) Re v,z h i − h i i = (∂¯ ∂)( v,z + z, v ) 2 − h i h i i n = vℓd¯zℓ v¯ℓdzℓ 2 − ! Xℓ=1 so that, for any w Cn, ∈ n i (dc Re v,z )(w)= v w¯ v¯ w h i 2 ℓ ℓ − ℓ ℓ Xℓ=1 i = ( v, w w, v ) 2 h i − h i i = v, w v, w 2 h i− h i = Im v, w − h i = Re iv,w . h i 2The symbol ∗ has already been employed to denote the convolution, however this will not cause confusion.

27 c This shows that, if v =1 and E is the line spanned by v, the 1-form d Re v,z is a volume form υE′ on the line E′ = iEkspannedk by iv. h i Unlike the one dimensional case, if E Cn is an equidimensional real subspace with d > 1 and ⊂ ⊥ v1,...,vd is an orthonormal basis of E one gets the following equalities valid just on E :

d(d−1)/2 d(d−1)/2 c c ̺(E)υ ′ = ( 1) υ = ( 1) d Re v ,z ... d Re v ,z . (35) E − iE − h 1 i ∧ ∧ h d i Indeed, by using the notation of Theorem 4.1,

d(d−1)/2 υ ′ = ( 1) Re w , ... Re w , E − h 1 −i ∧ ∧ h d −i ( 1)d(d−1)/2 = − Re u , ... Re u , ̺(E) h 1 −i ∧ ∧ h d −i ( 1)d(d−1)/2 = − Re t , ... Re t , . ̺(E) h 1 −i ∧ ∧ h d −i

When a ,...,a E⊥, for any 1 j, ℓ d, 1 d ∈ ≤ ≤ d Re t ,a = Re iv Re iv , v v ,a h j ℓi h j − h ℓ si s ℓi s=1 X d = Re iv ,a Re iv , v Re v ,a h j ℓi− h ℓ si h s ℓi s=1 X = Re iv ,a , h j ℓi because Re v ,a =0, then h s ℓi ( 1)d(d−1)/2 υ ′ (a ,...,a )= − (Re iv , ... Re iv , ) (a ,...,a ) E 1 d ̺(E) h 1 −i ∧ ∧ h d −i 1 d ( 1)d(d−1)/2 = − υ (a ,...,a ) ̺(E) iE 1 d ( 1)d(d−1)/2 = − (dc Re v ,z ... dc Re v ,z ) (a ,...,a ) . ̺(E) h 1 i ∧ ∧ h d i 1 d

Moreover, as E⊥ = E′ E⊥C and iE E⊥C , it follows that iE E⊥ = iE (E′ E⊥C ) = iE E′, ⊕ ⊥ ⊥ ∩ ′ ∩ ⊕ ∩ whence the restriction of υiE to E coincides with the restriction to E . Let A 2(Cn), then ∂A admits a Gauss map ν : ∂A ∂B , given by ν = ρ / ρ , ∈ S ∂A → 2n ∂A ∇ A k∇ Ak for every fixed defining function ρA. The Gauss map does not depend on the choice of the defining function used to represent it. Infact, if ρÃ is another defining function for A, then there exists a positive function g 1(Cn, R) such that ρ˜ = gρ on Cn, g = 0 on ∂A, and ρ˜ = g ρ on ∂A. Asa ∈ C A A 6 ∇ A ∇ A consequence, we may choose a defining function ρA such that ρA 1 on ∂A, but we will not assume this normalization. k∇ k≡ It follows that ∂ρ ∂ρ ∂ρ ∂ρ ∂ρ ∂ρ ρ (z)= A , A ,..., A , A =2 A ,..., A , ∇ A ∂x ∂y ∂x ∂y ∂z¯ ∂z¯ 1 1 n n 1 n 1 ∂ρ ∂ρ ∂ρ ∂ρ 2 ∂ρ ∂ρ ν (z)= A , A ,..., A , A = A ,..., A . ∂A ρ ∂x ∂y ∂x ∂y ρ ∂z¯ ∂z¯ k∇ Ak 1 1 n n k∇ Ak 1 n so that , if ι : ∂A Cn is the inclusion mapping, the volume form υ on ∂A is given by ∂A → ∂A (dρ ) υ = ι∗ ∗ A ∂A ∂A ρ k∇ Ak n ∗ 1 ∂ρA ∂ρA = ι∂A υ2n[xℓ] υ2n[yℓ] ρA ∂xℓ − ∂yℓ ! k∇ k Xℓ=1 n ∗ 2 ∂ρA ∂ρA = ι∂A υ2n[zℓ] υ2n[¯zℓ] . ρA ∂z¯ℓ − ∂zℓ ! k∇ k Xℓ=1 The unit vector ν∂A(z) is, by its definition, orthogonal to the tangent hyperplane Tz∂A and the latter C subspace admits a maximal complex subspace Tz∂A , which has real codimension 1 in Tz∂A. It follows

28 C that there is a unit vector in Tz∂A spanning the orthocomplement of Tz ∂A in Tz∂A, i.e. spanning the so called characteristic direction of Tz∂A. The real plane spanned by ν∂A(z) and the characteristic C ⊥C direction of Tz∂A is just the complex line (Tz∂A ) , so that the characteristic direction of Tz∂A is indeed spanned by iν∂A(z). The central object in the definition of Kazarnovskiˇıpseudovolume on the class 2(Cn) is the differential form α defined, for every z ∂A and every ζ T ∂A, as S ∂A ∈ ∈ z i (α ) (ζ)=Re ζ,iν (z) = ( ν (z), ζ ζ,ν (z) ) . ∂A z h ∂A i 2 h ∂A i − h ∂A i

The real number (α∂A)z(ζ) measures the projection of iζ along the outward unit normal vector ν∂A(z) and, as such, it vanishes if and only if ζ T C∂A. Equivalently,− (α ) (ζ) measures the projection of ζ ∈ z ∂A z on the characteristic direction of Tz∂A. In global coordinates, this form is given by

n ∗ 1 ∂ρA ∂ρA α∂A = ι∂A dyℓ dxℓ ρA ∂xℓ − ∂yℓ ! k∇ k Xℓ=1 dcρ = ι∗ A ∂A ρ k∇ Ak i(∂ρ¯ ∂ρ ) = ι∗ A − A ∂A ρ k∇ Ak 2i∂ρ¯ = ι∗ A , ∂A ρ k∇ Ak ∗ ∗ where the last equality follows from the relation ι∂AdρA = ι∂A(∂ρA +∂ρA)=0, on Tz∂A. The preceding expressions of α∂A have some remarkable consequences, namely 1 ddcρ dα = ι∗ ρ d α + A , (36) ∂A ∂A k∇ Ak ρ ∧ ∂A ρ k∇ Ak k∇ Ak

ddcρ ∧(n−1) α (dα )∧(n−1) = ι∗ α A ∂A ∧ ∂A ∂A ∂A ∧ ρ " k∇ Ak # 2i n−1 = ι∗ α (∂∂ρ¯ )∧(n−1) (37) ∂A ∂A ∧ ρ A " k∇ Ak # and n ∧(n−1) 1 ∗ i υ∂A = ι∂A α∂A dzℓ d¯zℓ (38) (n 1)!  ∧ 2 ∧ !  − Xℓ=1 C   Remark 6.1. As (α∂A)z vanishes on Tz ∂A, the only part of (dα∂A)z giving a non zero contribution to ∧(n−1) C the product (α∂A (dα∂A) )z is its restriction to Tz ∂A. It follows that the value of the (2n 1)- ∧∧(n−1) − form α∂A (dα∂A) at the point z ∂A is proportional to the value of the volume form υ∂A at ∧ ∈ n−1 the same point and the proportionality constant equals (n 1)!(2i/ ρA(z) ) times the product k (z) k (z) of the eighenvalues of the complex hessian matrix− ∂∂ρ¯k∇ restrictedk to T C∂A. 1 ··· n−1 A z Deﬁnition 6.1. Let A 2(Cn). The n-dimensional Kazarnovskiˇıpseudovolume (or simply the ∈ S n-pseudovolume) of A is the real number Pn(A) deﬁned as 1 P (A)= α (dα )∧(n−1) . (39) n n!κ ∂A ∧ ∂A n Z∂A Lemma 6.1. Let A 2(C). Then the 1-dimensional pseudovolume is the semi-perimeter of A. ∈ S Proof. In this case we have i ∂ρ ∂ρ α = ι∗ A d¯z A dz = υ . ∂A ∂A ρ ∂z¯ − ∂z ∂A k∇ Ak

As κ1 =2, P1(A) equals the half of the integral of υ∂A over ∂A, i.e the semi-perimeter.

29 Remark 6.2. The coeﬃcient in the deﬁnition of Pn.

In the english version of the original article [Ka1], the coeﬃcient in the formula (39) is denoted cn and about this coeﬃcient one reads: “..... cn =1/(σnn!), where σn is the volume of the unit n-sphere."

In this point there might be a problem with the translation from Russian into English. Infact σn must be the volume of the unit n-ball (i.e. our κn), otherwise Lemma 6.1, (whose statement is present in the ﬁrst page of [Ka1]), should have claimed that P1(A) equals (1/π) times the semi-perimeter of A.

Lemma 6.2. The 1-form α∂A is translation invariant. In particular, the pseudovolume is such.

n n Proof. If F : C C is a translation, for every z ∂A and ζ Tz∂A, one has ν∂F (A)(F (z)) = ν∂A(z) and → ∈ ∈

∗ (F α∂F (A))z(ζ) = (α∂F (A))F (z)(dFz (ζ))

= (α∂F (A))F (z)(ζ) = Re ζ,iν (F (z)) h ∂F (A) i = Re ζ,iν (z) h ∂A i = (α∂A)z(ζ) , so that 1 P (T (A)) = α (dα )∧(n−1) n n!κ ∂F (A) ∧ ∂F (A) n Z∂F (A) 1 ∗ ∧(n−1) = F α∂F (A) (dα∂F (A)) n!κn ∂A ∧ Z = Pn(A) .

The proof is complete.

Lemma 6.3. The 1-form α∂A is unitarily invariant. In particular, the pseudovolume is such. n n −1 t Proof. If F : C C is a unitary transformation, one has ρF (A) = ρA F = ρA F . If w1,...,wn are the coordinates→ in the range of F , by the chain rule, one gets ◦ ◦

∂(F −1) ρ (F (z)) = ρ (z) j (F (z)) ∇ F (A) ∇ A · ∂w¯ ℓ 1≤j,ℓ≤n −1 ∂(F )j + ρA(z) (F (z)) . ∇ · ∂w¯ℓ !1≤j,ℓ≤n

As F −1 is C-linear, (in particular holomorphic), by confusing F with its jacobian matrix one gets

∂(F −1) ∂ F −1 ∂(F −1) j = 0 and j = j = (F −1) = (tF ) , ∂w¯ ∂w¯ ∂w j,ℓ j,ℓ ℓ ℓ ℓ for every 1 j, ℓ n, then ≤ ≤ ρ (F (z)) = ρ (z) tF = F t ρ (z)= F ( ρ (z)) , ∇ F (A) ∇ A · · ∇ A ∇ A and so ρ (F (z)) ν (F (z)) = ∇ F (A) ∂F (A) ρ (F (z)) k∇ F (A) k F ( ρ (z)) = ∇ A F ( ρ (z)) k ∇ A k F ( ρ (z)) = ∇ A ρ (z) k∇ A k ρ (z) = F ∇ A ρ (z) k∇ A k = F (ν∂A(z)) , (40)

30 for every z ∂A. It follows that, for any ζ T ∂A, ∈ ∈ z ∗ (F α∂F (A))z(ζ) = (α∂F (A))F (z)(dFz(ζ))

= (α∂F (A))F (z)(F (ζ)) = Re F (ζ),iν (F (z)) h ∂F (A) i = Re F (ζ), iF (ν (z)) h ∂A i = Re F (ζ), F (iν (z)) h ∂A i = Re ζ,iν (z) h ∂A i = (α∂A)z(ζ) , so that 1 P (F (A)) = α (dα )∧(n−1) n n!κ ∂F (A) ∧ ∂F (A) n Z∂F (A) 1 ∗ ∧(n−1) = F α∂F (A) (dα∂F (A)) n!κn ∂A ∧ Z = Pn(A) . The proof is complete.

Lemma 6.4. The 1-form α∂A is (positively) 1-homogeneous and so the pseudovolume is (positively) n-homogeneous.

n n Proof. Let λ R>0 and let F : C C be the homothety corresponding to the multiplication by λ. For every z ∂A∈ and ζ T ∂A, one→ has ν (F (z)) = ν (λz)= ν (z) and ∈ ∈ z ∂F (A) ∂λA ∂A ∗ (F α∂F (A))z(ζ) = (α∂F (A))F (z)(dFz (ζ))

= (α∂λA)λz(λζ) = Re λζ,iν (λz) h ∂λA i = Re λζ,iν (z) h ∂A i = λ Re ζ,iν (z) h ∂A i = λ(α∂A)z(ζ) , so that 1 P (λA)= α (dα )∧(n−1) n n!κ ∂F (A) ∧ ∂F (A) n Z∂λ(A) 1 ∗ ∧(n−1) = F α∂F (A) (dα∂F (A)) n!κn ∂A ∧ n Z = λ Pn(A) . The proof is complete. When n> 1, an alternative formula for the pseudovolume involves the second quadratic form of ∂A. Recall that, the second quadratic form of ∂A is the twice covariant tensor II∂A on T ∂A whose value at a point z ∂A is given, for any ζ, η T ∂A, by ∈ ∈ z (II ) (ζ, η)=Re (dν ) (ζ), η , ∂A z h ∂A z i with ρ (dν ) (ζ) = d ∇ A (ζ) ∂A z ρ k∇ Ak 2 ∂ρ 2 ∂ρ = d A (ζ),..., d A (ζ) . ρ ∂z¯ ρ ∂z¯ k∇ Ak 1 k∇ Ak n Notice that the second quadratic form of ∂A behaves well with respect to unitary transformations. Infact, if F : Cn Cn is such a mapping, then, by virtue of (40), → ∗ F (II∂F (A))z(ζ, η) = (II∂F (A))F (z)(F (ζ), F (η)) = Re (dν ) (F (ζ)), F (η) h ∂F (A) F (z) i = Re F (dν ) (ζ), F (η) h ◦ ∂A z i = Re (dν ) (ζ), η h ∂A z i = (II∂A)z(ζ, η) .

31 II C By restricting ( ∂A)z to complex tangent vectors ζ, η Tz ∂A, one can deduce an R-bilinear symmetric IIC ∈ form ∂A by setting C 1 (II ) (ζ, η)= (II ) (ζ, η) + (II ) (iζ,iη) . ∂A z 2 ∂A z ∂A z IIC IIC Since ( ∂A)z(ζ, η) = ( ∂A)z(iζ,iη), it follows that C C (ζ, η) = (II ) (ζ, η)+ i(II ) (ζ,iη) , L∂A,z ∂A z ∂A z C deﬁnes a hermitian form on Tz ∂A known as the Levi form of ∂A at the point z. Notice that, for n =1, T C∂A = 0 , so 0 in this case. z { } L∂A,z ≡ In order to ﬁnd a simpler expression of , let us carry on some computation: L∂A,z 2 (ζ, η)=Re (dν ) (ζ), η + Re (dν ) (iζ), iη L∂A,z h ∂A z i h ∂A z i + i Re (dν ) (ζ), iη + i Re (dν ) (iζ),i2η h ∂A z i h ∂A z i = Re (dν ) (ζ), η + Re (dν ) (iζ), iη h ∂A z i h ∂A z i + i Im (dν ) (ζ), η + i Im (dν ) (iζ), iη h ∂A z i h ∂A z i = (dν ) (ζ), η + (dν ) (iζ), iη h ∂A z i h ∂A z i = (dν ) (ζ) i(dν ) (iζ), η h ∂A z − ∂A z i = (∂ν ) (ζ) + (∂ν¯ ) (ζ) i(∂ν ) (iζ) i(∂ν¯ ) (iζ), η h ∂A z ∂A z − ∂A z − ∂A z i = (∂ν ) (ζ) + (∂ν¯ ) (ζ) i2(∂ν ) (ζ)+ i2(∂ν¯ ) (ζ), η h ∂A z ∂A z − ∂A z ∂A z i =2 (∂ν ) (ζ), η . h ∂A z i

For any k =1,...,n, the k-th component of ∂ν∂A equals

n −1 2 2 ∂ρA ∂ ρA ∂ρA 1 ∂ ρA ∂ =2 k∇ k + dzℓ , ρA ∂z¯k ∂zℓ ∂z¯k ρA ∂zℓ∂z¯k k∇ k Xℓ=1 k∇ k so that n n −1 2 ∂ ρA ∂ρA 1 ∂ ρA ∂A,z(ζ, η)=2 k∇ k + ζℓη¯k L ∂zℓ ∂z¯k ρA ∂zℓ∂z¯k Xk=1 Xℓ=1 k∇ k n −1 n 2 ∂ ρA 2 ∂ ρA = ρA, η k∇ k ζℓ + ζℓη¯k h∇ i ∂zℓ ρA ∂zℓ∂z¯k Xℓ=1 k∇ k ℓ,kX=1 n 2 2 ∂ ρA = ζℓη¯k , ρA ∂zℓ∂z¯k k∇ k ℓ,kX=1 as ρ , η =0 for η T C∂A. It follows that the Levi form of ∂A has the following expression: h∇ A i ∈ z n 2 2 ∂ ρA ∂A = dzℓ d¯zk . L ρA ∂zℓ∂z¯k ⊗ k∇ k ℓ,kX=1 It is important to remark that, though involving global coordinates, the preceding expression is always C ¯ restricted to Tz ∂A, on which ∂ρA = ∂ρA =0. It follows that ∂A actually depends on at most (n 1) complex coordinates. L − ∗ Thanks to (40) and the unitary invariance of II∂A, it follows that F ∂F (A) = ∂A , for any unitary transformation F : Cn Cn. L L → For any z ∂A, let K∂A(z) denote the product of the eigenvalues of ∂A,z. Remark that K∂A(z) ∈ n−1 L equals (2/ ρA ) times the product of the eigenvalues of the complex Hessian of ρA restricted to C k∇ k 2 n Tz ∂A. We recall that a subset A (C ) is (Levi) pseudoconvex if, for every z ∂A, the Levi form ∈ S ∈ C ∂A,z is semi-positive deﬁnite, i.e. ∂A,z(ζ, η) 0, for every z ∂A and any ζ, η Tz ∂A. It is also usefulL to recall that a subset A L2(Cn) is strictly≥ (Levi) pseudoconvex∈ if, for every∈ z ∂A, the Levi ∈ S ∈ C form ∂A,z is positive deﬁnite, i.e. ∂A,z(ζ, η) > 0, for every z ∂A and any ζ, η Tz ∂A 0 . Any (strictly)L convex subset A 2(CLn) is (strictly) pseudoconvex,∈ but the converse is∈ generally\{ false} already for n = 1, in which case∈ S the class of k-regular pseudoconvex subsets coincide with the space k 2 n (C). Of course, if A (C ) is pseudoconvex, then K∂A(z) 0 for every z ∂A, with a strict Sinequality when A is strictly∈ S pseudoconvex. ≥ ∈

32 Theorem 6.1. Let n N∗ and A 2(Cn). Then ∈ ∈ S 2n−1(n 1)! P (A)= − K υ . (41) n n!κ ∂A ∂A n Z∂A In particular, if A is pseudoconvex then P (A) 0. n ≥ ∧(n−1) Proof. Observe that both the forms α∂A (dα∂A) and K∂A(z)υ∂A have maximal degree, so they have to be proportional. By (38), Remark∧6.1 and (37), one has 2n−1(n 1)!K υ − ∂A ∂A n ∧(n−1) n−1 ∗ = i K∂A ι∂A α∂A dzℓ d¯zℓ  ∧ ∧ !  Xℓ=1  n−1  n ∧(n−1) ∗ 2i = ι∂A α∂A k1(z) kn−1(z) dzℓ d¯zℓ  ∧ ρA ··· ∧ !  k∇ k Xℓ=1  n−1  2i = ι∗ α (∂∂ρ¯ )∧(n−1) ∂A ∂A ∧ ρ A " k∇ Ak # = α (dα )∧(n−1) . ∂A ∧ ∂A When A is pseudoconvex, the integrand in (41) is non negative, so that P (A) 0. n ≥ Remark 6.3. The dimensional coefficient (n 1)! in formula (41). − Kazarnovskiˇıin his paper [Ka1] states formula (41) without the dimensional coefficient (n 1)!. The − reason should be in the definition of the volume form υ∂A or in that of the Levi form ∂A, on which the reference [Ka1] lacks a fully detailed description. Any way, as shown by Example 10.1L, our presentation is correct. Theorem 6.2. The Kazarnovskiˇı n-pseudovolume is continuous on ∞(Cn) for the Hausdorff metric. K Proof. Let A ∞(Cn) and let A N be a sequence of convex bodies from ∞(Cn) converging ∈ K { m}m∈ K to A in the Hausdorff metric. We want to show that limm→+∞ Pn(Am) = Pn(A). Let zo an interior point of A and Am, for sufficiently big m N. For any such m, let γm : ∂A ∂Am map the point z ∂A to the point γ (z) ∂A at which∈ the ray issuing from z and pointing→ to z intersects ∂A . ∈ m ∈ m o m The smoothness of A and Am imply that γm is an orientation preserving diffeomorphism, so that

∧(n−1) ∗ ∧(n−1) α∂Am (dα∂Am ) = γm α∂Am (dα∂Am ) ∂A ∧ ∂A ∧ Z m Z = γ∗ α (dγ∗ α )∧(n−1) . m ∂Am ∧ m ∂Am Z∂A For every z ∂A and any ζ T ∂A, ∈ ∈ z (γ∗ α ) (ζ)= α ((dγ ) (ζ)) = Re (dγ ) (ζ),iν (γ (z)) m ∂Am z ∂Am,γm(z) m z h m z ∂Am m i ∗ As m + , the mapping γm approaches the identity mapping on ∂A, hence γmα∂Am approaches α∂A and the→ theorem∞ follows. Remark 6.4. The Kazarnovskiˇı n-pseudovolume is continuous on ∞(Cn) for the Hausdorff metric. S Theorem 6.2 can be extended to the whole ∞(Cn) and even to the larger space 2(Cn). The proof of this more general result would differ from theS one presented above just in the choiceS of the diffeomorphism γm.

The following theorem provides an easy formula for computing the n-pseudovolume on the smaller space ∞(Cn) 2(Cn). P ⊂ S Theorem 6.3. Let n N∗, Γ (Cn) and ε> 0. For each 0 k< 2n and any ∆ (Γ, k), ∈ ∈P ≤ ∈B 1 α (dα )∧(n−1) = εn−k̺(∆) vol (∆)ψ (∆) , (42) n!2n−kκ ∂(Γ)ε ∧ ∂(Γ)ε k Γ 2n−k Z∆+(K∆∩∂εB2n) then, by neglecting non equidimensional faces,

n n−k 2 κ2n−k v̺ n−k Pn((Γ)ε)= k(Γ)ε . (43) κn Xk=0

33 Proof. The hypersurface ∆ + (K ∂εB ) has the structure of a cylinder on the (2n k 1)- ∆ ∩ 2n − − dimensional base K∆ ∂εB2n. As the outward unit normal vector to ∆ + (K∆ ∂εB2n) belongs to ⊥ ∩ ∩ E∆, it follows that α∂(Γ)ε depends on (2n k) variables. If k>n, one has (2n k)= n + (n k) n, then the statement holds true for n k > 0. Up to non singular R-linear transformation, (aﬀecting our computation by a non zero mul≥ tiplicative constant), we may suppose that

⊥ n ′ ′ ′ E∆ = z C z1 = ... = zk = Re zk +1 = ... = Re zk−k =0 { ∈ n | } = z C x = ... = x ′ = y = ... = y ′ =0 . { ∈ | 1 k−k 1 k } ⊥ For the sake of notation, set ν = ν∂(Γ)ε . As ν(z) E∆, for every z ∆ + (K∆ + ∂εB2n), we deduce that ∈ ∈ n n

α∂(Γ)ε = (Re νℓ) dyℓ (Im νℓ) dxℓ , ′ − ′ ℓ=kX−k +1 ℓ=Xk +1 on ∆ + (K ∂εB ). Moreover ν(z) = ν(z + u), for every u E , so the partial derivatives of the ∆ ∩ 2n ∈ ∆ components of ν with respect to x ,...x ′ and y ,...,y ′ are all zero on ∆ + (K ∂εB ). This 1 k−k 1 k ∆ ∩ 2n implies that dα∂(Γ)ε has the following expression

n n n ∂(Re νℓ) ∂(Re νℓ) dxj dyℓ + dyj dyℓ ′  ′ ∂xj ∧ ′ ∂yj ∧  ℓ=kX−k +1 j=kX−k +1 j=Xk +1 n  n n  ∂(Im νℓ) ∂(Im νℓ) dxj dxℓ + dyj dxℓ , − ′  ′ ∂xj ∧ ′ ∂yj ∧  ℓ=Xk +1 j=kX−k +1 j=Xk +1 ′ ′   ∧(n−1) with (k k ) > 0 and k > 0. As a consequence the (2n 1)-form α∂(Γ)ε (dα∂(Γ)ε ) vanishes − − ∧ ′ identically on ∆ + (K∆ ∂εB2n), because it depends on no more than (2n 2k ) variables and, of course, 2k′ > 1. We deduce∩ that the statement of the theorem holds true for− every face which is not equidimensional. Finally, if ∆ (Γ, k), up to a non singular R-linear transformation, say L , we may suppose ∈Bed ∆ E⊥ = z Cn Re z = ... = Re z =0 ∆ { ∈ | 1 k } = z Cn x = ... = x =0 . { ∈ | 1 k } Indeed, by the unitary invariance of α , we can make linC E = z Cn z = ... = z = 0 , ∂(Γ)ε ∆ { ∈ | k+1 n } so we need to deﬁne L∆ just on the smaller subspace linC E∆. Let e1,...,ek be the canonical basis of linC E = Ck 0 n−k, let also v ,...,v be an orthonormal basis of E (with respect to the ∆ ×{ } 1 k ∆ scalar product Re , ) and consider the operator L : linC E linC E mapping e to v . As ∆ is h i ∆ ∆ → ∆ ℓ ℓ equidimensional, L∆ is an isomorphism of complex linear spaces, with det L∆ = det( vℓ,ej ). If we 2k 2 h i look at L∆ as a mapping on R , its jacobian equals det( vℓ,ej ) which, by virtue of Corollary 4.2, is nothing but ̺(∆). | h i |

From now on we should systematically consider the pullback of α∂(Γ)ε via L∆ thus changing the names of variables, however, for the sake of notation, we will leave those names unchanged remembering to append the jacobian ̺(∆) to the computation of the integral on the left of (42). Arguing as before, in the simpliﬁed coordinates we now get n n α = (Re ν ) dy (Im ν ) dx ∂(Γ)ε ℓ ℓ − ℓ ℓ ℓ=Xk+1 Xℓ=1 so that, on ∆ + (K ∂εB ), the form dα is given by ∆ ∩ 2n ∂(Γ)ε n n ∂(Re ν ) n ∂(Re ν ) ℓ dx dy + ℓ dy dy  ∂x j ∧ ℓ ∂y j ∧ ℓ j j=1 j ℓ=Xk+1 j=Xk+1 X   n n ∂(Im ν ) n ∂(Im ν ) ℓ dx dx + ℓ dy dx . −  ∂x j ∧ ℓ ∂y j ∧ ℓ j j=1 j Xℓ=1 j=Xk+1 X  

34 Notice that in the present situation the (2n 1)-form α (dα )∧(n−1) depends on the full array − ∂(Γ)ε ∧ ∂(Γ)ε of 2n variables. In order to complete the computation we need to specify a deﬁning function of (Γ)ε or at least its restriction to the open subset ∆+ K∆. Such a restriction, noted ρ∆, can surely be chosen as the function n n ρ (z)= ε2 + x2 + y2 , ∆ − ℓ ℓ ℓ=Xk+1 Xℓ=1 then ρ (z) = 2ε and ν(z) = ε−1(0,y ,..., 0,y , x ,y ,...,x ,y ), for every z ∆ + (K k∇ ∆ k 1 k k+1 k+1 n n ∈ ∆ ∩ ∂εB2n). This yields n n −1 α∂(Γ)ε = ε xℓdyℓ yℓdxℓ − ! ℓ=Xk+1 Xℓ=1 and k n dα = ε−1 dx dy +2 dx dy , ∂(Γ)ε  ℓ ∧ ℓ ℓ ∧ ℓ Xℓ=1 j=Xk+1 for every z ∆ + (K ∂εB ). Observe that  ∈ ∆ ∩ 2n n (dα )∧n = ε−n2n−kn! dx dy = ε−n2n−kn!υ , ∂(Γ)ε ℓ ∧ ℓ 2n ℓ^=1 so, by Stokes’ theorem, 1 α (dα )∧(n−1) n!2n−kκ ∂(Γ)ε ∧ ∂(Γ)ε 2n−k Z∆+(K∆∩∂εB2n) 1 = (dα )∧n n!2n−kκ ∂(Γ)ε 2n−k Z∆+(K∆∩εB2n) ε−n = ̺(∆) vol 2n(∆ + (K∆ εB2n)) κ2n−k ∩ ε−n = ̺(∆) vol k(∆) vol 2n−k(K∆ εB2n) κ2n−k ∩ −n ε 2n−k = ̺(∆) vol k(∆)ε vol 2n−k(K∆ B2n) κ2n−k ∩ n−k = ε ̺(∆) vol k(∆)ψΓ(∆) , so that (42) is proved. Notice that (42) is a non trivial equality if and only if ∆ is equidimensional. Moreover 2n−1 ∂(Γ) = ∆ + (K ∂εB ) , ε ∆ ∩ 2n k[=0 ∆∈B[(Γ,k) so, by summing, as k runs in the set 0,..., 2n 1 , the relations (42) and neglecting non equidimensional faces, one gets (43). The proof is thus{ complete.− }

∞ n n The notion of pseudovolume applies in particular to convex bodies belonging to + (C ) (C ). The following results of [Ka1] imply a further formula for P on ∞(Cn). K ⊂ S n K+ ∞ n Lemma 6.5. Let A 1 (C ). Then the Gauss map ν∂A and the restriction to ∂B2n of the gradient mapping h are inverse∈ K diﬀeomorphisms. ∇ A ∞ Proof. Let us start with ν∂A : ∂A ∂B2n. We already know that the Gauss map is . For any −1 → C u ∂B2n, the ﬁber ν∂A(u) is non-empty since it equals the exposed face A HA(u). By the strict convexity∈ of A, A H (u) is reduced to a single point, so ν is bijective. The∩ exposed face A H (u) ∩ A ∂A ∩ A is nothing but the sub-gradient of hA at the point u, so that

h (u) Subd h (u)= A H (u)= ν−1(u) , ∇ A ∈ A ∩ A { ∂A } i.e. ν−1 = h . It follows that ν−1 is also diﬀerentiable. ∂A ∇ A ∂A Lemma 6.6. For every A ∞(Cn), the forms dch and (ν−1)∗α are cohomologous. ∈ K1 A ∂A ∂A

35 Proof. Let u ∂B and ζ T ∂B be ﬁxed. Then ∈ 2n ∈ u 2n n c ∂hA ¯ ∂hA (d hA)u (ζ)= i (u)ζℓ (u)ζℓ ∂u¯ℓ − ∂uℓ Xℓ=1 = (i/2) ( h (u), ζ ζ, h (u) ) , h∇ A i − h ∇ A i whereas

−1 ∗ −1 (ν ) α∂A = (α∂A) −1 (dν )u(ζ) ∂A u ν∂A(u) ∂A

= Re d( hA)u(ζ),iu h ∇ i = Im d( h ) (ζ),u h ∇ A u i = (i/2) ( u, d( h ) (ζ) d( h ) (ζ),u ) . h ∇ A u i − h ∇ A u i It follows that dch (ν−1)∗α (ζ) A − ∂A ∂A u = (i/2) [ h (u), ζ ζ, h (u) u, d( h ) (ζ) + d( h ) (ζ),u ] h∇ A i − h ∇ A i − h ∇ A u i h ∇ A u i = (i/2) [ h (u), ζ + d( h ) (ζ),u u, d( h ) (ζ) ζ, h (u) ] h∇ A i h ∇ A u i − h ∇ A u i − h ∇ A i = (i/2) [d ( h (u),u ) d ( u, h (u) )] (ζ) h∇ A i − h ∇ A i = d [Im u, h (u) ] (ζ) . h ∇ A i The lemma is thus proved.

Remark 6.5. Lemma 6.6 can be found in [Sha] pag. 199 but, although I cannot provide a precise reference, Kazarnovskˇıtold me it’s originally due to Koushnirenko. Lemma 6.7. For every A ∞(Cn), the form dch is unitarily invariant. ∈ K1 A n n −1 Proof. Let F : C C a unitary transformation. Then hF (A) = hA F , F (∂B2n) = ∂B2n and → −1 ◦ (arguing as in the proof of Lemma 6.3) (hA F )(F (z)) = F ( hA(z)), so that, for every z ∂B2n and any ζ T ∂B , ∇ ◦ ∇ ∈ ∈ z 2n ∗ c c F (d hF (A))z(ζ) = (d hF (A))F (z)(F (ζ)) i = h (F (z)), F (ζ) F (ζ), h (F (z)) 2 h∇ F (A) i − h ∇ F (A) i i = (h F −1)(F (z)), F (ζ) F (ζ), (h F −1)(F (z)) 2 h∇ A ◦ i − h ∇ A ◦ i i = [ F ( h (z)), F (ζ) F (ζ), F ( h (z)) ] 2 h ∇ A i − h ∇ A i i = [ h (z), ζ ζ, h (z) ] 2 h∇ A i − h ∇ A i c = (d hA)z(ζ) . The proof is complete.

Corollary 6.1. For any A ∞(Cn), one has ∈ K1 1 1 P (A)= dch (ddch )∧(n−1) = (ddch )∧n . (44) n n!κ A ∧ A n!κ A n Z∂B2n n ZB2n In particular Pn(A) > 0.

Proof. By Lemma 6.6,

1 1 ∗ dch (ddch )∧(n−1) = ν−1 α (dα )∧(n−1) n!κ A ∧ A n!κ ∂A ∂A ∧ ∂A n Z∂B2n n Z∂B2n = Pn(A) .

Moreover, the only singularity of hA is at the origin, where the second derivatives of hA are O(1/ z ), c ∧n k k (because hA is positively homogeneous of degree 1). It follows that the form (dd hA) has a locally integrable density and so, by Stokes’ theorem, one obtains the right-most term in the equality (44). Since a convex body is pseudoconvex, Theorem 6.1 implies that P (A) 0. However, the strict convexity of n ≥ A implies its strict pseudoconvexity, hence Pn(A) is strictly positive.

36 7 Kazarnovskˇıi pseudovolume and mixed discriminants

In this section we investigate the relations between mixed pseudovolume and mixed discriminants in order to get an analogue of (14) for the mixed pseudovolume. We ﬁrst recall some information about mixed discriminants. Let S be the symmetric group over the set 1,...,n . For ℓ = 1,...,n, let M = (m(ℓ) ) be an n { } ℓ j,k n-by-n complex matrix and, for every σ Sn, consider the n-by-n matrices mixσ(M1,...,Mn) and σ ∈ mix (M1,...,Mn) respectively deﬁned by

m(1) m(n) 1,σ(1) ··· 1,σ(n) . . . mixσ(M1,...,Mn)=  . .. .  , (1) (n)  m m   n,σ(1) ··· n,σ(n)    m(σ(1)) m(σ(n)) 1,1 ··· 1,n σ . . . mix (M1,...,Mn)=  . .. .  . (σ(1)) (σ(n))  m mn,n   n,1 ···  σ   The matrices mixσ(M1,...,Mn) and mix (M1,...,Mn) are mixed versions of M1,...,Mn, indeed the σ ℓ-th column of mixσ(M1,...,Mn) is the σ(ℓ)-th column of Mℓ and the ℓ-th column of mix (M1,...,Mn) σ is ℓ-th column of Mσ(ℓ). In general mixσ(M1,...,Mn) = mix (M1,...,Mn) unless σ is the identity. σ 6 When M1 = ... = Mn = M one has mix (M,...,M)= M too. If sgn(σ) denotes the sign of σ Sn, then it is easy to see that ∈

− σ 1 sgn(σ) det (mixσ(M1,...,Mn)) = det mix (M1,...,Mn) . The mixed discriminant of M1,...,Mn is the complex number Dn(M1,...,Mn) given by

1 σ Dn(M1,...,Mn)= det (mix (M1,...,Mn)) (45) n! S σX∈ n 1 = sgn(σ) det (mixσ(M1,...,Mn)) . n! S σX∈ n

It follows that, for every n-by-n complex matrices M,N,M1,M2,...,Mn,

n 1 (ℓ) 1. Dn(M1,...,Mn)= sgn(σ τ) mσ(ℓ),τ(ℓ); n! S · σ,τX∈ n ℓY=1 2. n!Dn(M1,...,Mn) is the coeﬃcient of t1 ...tn in the polynomial det(t1M1 + ... + tnMn); ∂ 3. n!Dn(M1,...,Mn)= det(t1M1 + ... + tnMn); ∂t1 ...∂tn

n−#I 4. n!Dn(M1,...,Mn)= ( 1) det Mℓ ; ∅ − 6=I⊆{X1,...,n} Xℓ∈I

5. Dn(M,...,M) = det M;

6. Dn(In,...,In)=1; 7. D (M ,...,M )= D (M ,...,M ), for every σ S ; n σ(1) σ(n) n 1 n ∈ n 8. D (aM + bN,M ,...,M )= aD (M,M ,...,M )+ bD (N,M ,...,M ) for every a,b C; n 2 n n 2 n n 2 n ∈ 9. If M ,...,M are positive semideﬁnite, then D (M ,...,M ) 0; 1 n n 1 n ≥ 10. D (NM ,...,NM ) = (det N) D (M ,...,M )= D (M N,...,M N). n 1 n · n 1 n n 1 n

37 Il M is an n-by-n matrix and 1 j, k n, let M [j,k] denote the submatrix obtained by deleting the j-th row and the k-th column of ≤M. For≤ every 1 ℓ n, by Laplace expansion on the ℓ-th column of ≤ ≤ the matrices in (45) it follows that Dn(M1,...,Mn) is a linear form in the entries of Mℓ:

Dn(M1,...,Mn) 1 n = m(ℓ) ( 1)k+ℓD (M [j,k],...,M [j,k],M [j,k],...,M [j,k]) , (46) n j,k − n−1 1 ℓ−1 ℓ+1 n j,kX=1 [j,k] [j,k] [j,k] [j,k] (ℓ) where the coeﬃcient (1/n)Dn−1(M1 ,...,Mℓ−1 ,Mℓ+1 ,...,Mn ) is the mixed minor of mj,k or the (ℓ) principal mixed minor of mj,k, in case j = k. The mixed discriminant has several other properties, the interested reader is referred to [A2], [Bap] or [FMS] for further information.

If A ,...,A 2(Cn) and HessC h ,..., HessC h are the respective complex hessian matrices 1 n ∈ K1 A2 An of the support functions hA1 ,...,hAn , then c c n dd h ... dd h =4 n!D (HessC h ,..., HessC h ) υ . A1 ∧ ∧ An n A1 An 2n Indeed, both sides depend multilinearly on A1,...,An and, for A1 = ... = An = A, by (32) and the n equality Dn(HessC hA,..., HessC hA) = det(HessC hA), one gets 4 n! det(HessC hA)υ2n on both sides. It follows that 4n Q (A ,...,A )= D (HessC h ,..., HessC h ) υ , n 1 n κ n A1 An 2n n ZB2n which provides a further formula for Q on 2(Cn). n K1 Let us recall the following result.

Theorem 7.1 (Alexandroﬀ). For any complex n-by-n positive hermitian matrices M,M3,...,Mn and every complex n-by-n hermitian matrix N,

2 Dn(M,N,M3 ...,Mn) Dn(M,M,M3,...,Mn)Dn(N,N,M3,...,Mn) , (47) ≥ with equality if and only if N = λM for some λ R. ∈ The following lemma gives an equivalent formulation of Theorem 7.1.

Lemma 7.1. Given complex n-by-n hermitian matrices M,N,M3,...,Mn with M,M3,...,Mn positive deﬁnite and N arbitrary, then(47) holds true if and only if

D (M,N,M ...,M )=0 implies D (N,N,M ...,M ) 0 , (48) n 3 n n 3 n ≤ with equality if and only if N =0.

Li [Ping] obeserved that, by continuity, the inequality (47) remains true if M,M3,...,Mn are only positive semi-definite. Theorem 7.1 for real symmetric matrices has been proved by Alexandroff in the article [A2] in which he gave his second proof of (16). In the second page, lines 20-23, of the original paper [A2], Alexandroff states that Theorem 7.1 can be proved by adapting the proof given for real symmetric matrices, anyway Li [Ping] as well as Shenfeld and Van Handel [SvH] found alternative proofs. Lemma 7.2. Let n 2 and A , A ,...,A 2(Cn), then ≥ 1 2 n ∈ K1 ι∗ (dch ddch ... ddch ) ∂B2n A1 ∧ A2 ∧ ∧ An n−1 ∗ t C C =4 n!ι∂B2n Dn(M + M, Hess hA2 ,..., Hess hAn )υ∂B2n , where ∂hA1 ∂hAn z1 ... z1 z1 ∂z1 ∂zn ∂hA1 ∂hA1   M = . ... = . .. . . (49)  .  ∂z ∂z . . . 1 n  ∂h ∂h  zn  A1 A1    zn ... zn     ∂z1 ∂zn  In particular  

n−1 4 ∗ t Q (A ,...,A )= ι D (M + M, HessC h ,..., HessC h )υ . (50) n 1 n κ ∂B2n n A2 An ∂B2n n Z∂B2n

38 n Proof. In the global coordinates of C , for every u ∂B2n, the standard volume form on ∂B2n at the point u reads ∈

n υ = ι∗ (Re u )υ [x ] (Im u )υ [y ] ∂B2n ∂B2n ℓ 2n ℓ − ℓ 2n ℓ ℓ=1 Xn ∗ = ι uℓυ2n[zℓ] u¯ℓυ2n[¯zℓ] . ∂B2n − Xℓ=1

On each open subset of the family z ∂B2n Re zk = 0 , (resp. z ∂B2n Im zk = 0 ), as ∗ c { ∈ c | 6 c } { ∈ | 6 } 1 k n, both υ∂B and ι (d hA dd hA ... dd hA ) get simpler expressions because they ≤ ≤ 2n ∂B2n 1 ∧ 2 ∧ ∧ n both become multiples of υ2n[xk], (resp. υ2n[yk]). Indeed, for u ∂B2n, the tangent space Tu∂B2n is the zero-set of the equation ∈ n (Re um)xm + (Im um)ym =0 , m=1 X so if 1 r n is such that Re u =0 then, on T ∂B , one has ≤ ≤ r 6 u 2n n n Re u Im u dx = m dx m dy , r − Re u m − Re u m m=1 r m=1 r mX6=r X

n n n Re u Im u υ [x ] = dy m dx m dy dy dx dy 2n ℓ ℓ ∧ − Re u m − Re u m ∧ k ∧ j ∧ j m=1 r m=1 r j=1  mX6=r X  r6=^j6=ℓ  n  Re u = dy ℓ dx dy dx dy ℓ ∧ −Re u ℓ ∧ r ∧ j ∧ j r j=1 r6=^j6=ℓ n Re u = ℓ dy dx dy Re u r ∧ j ∧ j r j=1 j^6=r Re u = ℓ υ [x ] , Re u 2n r r for ℓ = r, and 6

n Re u n Im u n υ [y ] = dx m dx m dy dy dx dy 2n ℓ ℓ ∧ − Re u m − Re u m ∧ r ∧ j ∧ j m=1 r m=1 r j=1  mX6=r X  r6=^j6=ℓ   Im u n = dx ℓ dy dy dx dy ℓ ∧ −Re u ℓ ∧ r ∧ j ∧ j r j=1 r6=^j6=ℓ Im u n = ℓ dy dx dy −Re u r ∧ j ∧ j r j=1 j^6=r Im u = ℓ υ [x ] , −Re u 2n r r for every ℓ, whence

n (Im u )2 (Re u )2 (Im u )2 υ = Re u + r + ℓ + ℓ υ [x ] ∂B2n  r  2n r Re ur Re ur Re ur ℓ=1  Xℓ6=r   1  = υ2n[xr] . Re ur

39 A similar computation on the open subset Im z =0 yields r 6 n n Re u Im u dy = m dx m dy , r − Im u m − Im u m m=1 r m=1 r X mX6=r

n n n Re u Im u υ [x ] = dy dx m dx m dy dx dy 2n ℓ ℓ ∧ r ∧ − Im u m − Im u m ∧ j ∧ j m=1 r m=1 r j=1  X mX6=r  r6=^j6=ℓ  n  Re u = dy dx ℓ dx dx dy ℓ ∧ r ∧ −Im u ℓ ∧ j ∧ j r j=1 r6=^j6=ℓ n Re u = ℓ dx dx dy −Im u r ∧ j ∧ j r j=1 j^6=r Re u = ℓ υ [y ] , −Im u 2n r r for every ℓ, and

n n n Re u Im u υ [y ] = dx dx m dx m dy dx dy 2n ℓ ℓ ∧ r ∧ − Im u m − Im u m ∧ j ∧ j m=1 r m=1 r j=1  X mX6=r  r6=^j6=ℓ  n  Im u = dx dx ℓ dy dx dy ℓ ∧ r ∧ −Im u ℓ ∧ j ∧ j r j=1 r6=^j6=ℓ n Im u = ℓ dx dx dy Im u r ∧ j ∧ j r j=1 j^6=r Im u = ℓ υ [y ] , Im u 2n r r for ℓ = r, whence 6 2 n 2 2 (Re ur) (Re uℓ) (Im uℓ) υ∂B2n = Im ur + υ2n[yr] − Im ur − − Im ur Im ur  ℓ=1  Xℓ6=r   1  = − υ2n[yr] . Im ur On the other hand, as any (2n 1)-form on Cn, the form − dch ddch ... ddch A1 ∧ A2 ∧ ∧ An can be expressed in the basis (33), (34). In fact, dch ddch ... ddch equals A1 ∧ A2 ∧ ∧ An n n n n ∂h ∂h ∂2h i A1 d¯z A1 dz 2i Ak dz d¯z   ∂z¯ j1  − ∂z ℓ1  ∧ ∂z ∂z¯ ℓk ∧ jk j =1 j1 ℓ1 ! ℓk jk X1 ℓX1=1 k^=2 ℓkX,jk =1   n  n n ∂h ∂2h =2n−1in i A1 d¯z Ak dz d¯z  ∂z¯ j1  ∧ ∂z ∂z¯ σ(k) ∧ τ(k) j =1 j1 S σ(k) τ(k) ! X1 σ,τX∈ n kY=2 k^=2  n  n 2 n n−1 n ∂hA1 ∂ hAk 2 i i dzℓ1 dzσ(k) d¯zτ(k) − ∂zℓ1 ! ∧ S ∂zσ(k)∂z¯τ(k) ! ∧ ℓX1=1 σ,τX∈ n kY=2 k^=2 n 2 n n−1 n ∂hA1 ∂ hAk =2 i d¯zτ(1) dzσ(k) d¯zτ(k) S ∂z¯τ(1) ∂zσ(k)∂z¯τ(k) ! ∧ ∧ σ,τX∈ n kY=2 k^=2 n 2 n n−1 n ∂hA1 ∂ hAk 2 i dzσ(1) dzσ(k) d¯zτ(k) . − S ∂zσ(1) ∂zσ(k)∂z¯τ(k) ! ∧ ∧ σ,τX∈ n kY=2 k^=2

40 Now

n d¯z dz d¯z τ(1) ∧ σ(k) ∧ τ(k) k=2 ^ n n (n−1)(n−2) = ( 1) 2 d¯z dz d¯z − τ(1) ∧ σ(ℓ) ∧ τ(j) j=2 ℓ^=2 ^ n n n(n−1) = ( 1) 2 sgn(τ) dz d¯z − σ(ℓ) ∧ j j=1 ℓ^=2 ^ n n − n(n 1) σ(1)−1 = ( 1) 2 ( 1) sgn(στ) dz d¯z − − ℓ ∧ j ℓ=1 j=1 ℓ6=^σ(1) ^ n n (n−1)(n+2) = ( 1) 2 sgn(στ)d¯z dz d¯z − σ(1) ∧ ℓ ∧ j ℓ=1 j=1 ℓ6=^σ(1) j6=^σ(1) n = sgn(στ)d¯z dz d¯z σ(1) ∧ ℓ ∧ ℓ ℓ=1 ℓ6=^σ(1) = ( 2i)nsgn(στ)υ [z ] − 2n σ(1) υ2n[x ]+ iυ2n[y ] = ( 2i)nsgn(στ) σ(1) σ(1) − 2 and

n dz dz d¯z σ(1) ∧ σ(k) ∧ τ(k) k=2 ^ n n (n−1)(n−2) = ( 1) 2 dz d¯z − σ(ℓ) ∧ τ(j) j=2 ℓ^=1 ^ n n (n−1)(n−2) = ( 1) 2 sgn(σ) dz d¯z − ℓ ∧ τ(j) j=2 ℓ^=1 ^ n n − − (n 1)(n 2) τ(1)−1 = ( 1) 2 ( 1) sgn(σ)dz dz d¯z − − τ(1) ℓ ∧ τ(j) ℓ=1 j=2 ℓ6=^τ(1) ^ n n (n−1)(n−2) = ( 1) 2 sgn(στ)dz dz d¯z − τ(1) ∧ ℓ ∧ j ℓ=1 j=1 ℓ6=^τ(1) j6=^τ(1) n = sgn(στ)dz dz d¯z τ(1) ℓ ∧ ℓ ℓ=1 ℓ6=^τ(1) = ( 2i)nsgn(σ τ)υ [¯z ] − · 2n τ(1) υ2n[x ]+ iυ2n[y ] = ( 2i)nsgn(σ τ) − τ(1) τ(1) . − · 2 If u ∂B and Re u =0, on T ∂B one gets ∈ 2n r 6 u 2n Re u Im u υ [x ]+ iυ [y ]= σ(1) i σ(1) υ [x ] 2n σ(1) 2n σ(1) Re u − Re u 2n r r r u¯ = σ(1) υ [x ] Re u 2n r r

=u ¯σ(1)υ∂B2n

41 and Re u Im u υ [x ]+ iυ [y ]= τ(1) i τ(1) υ [x ] − 2n τ(1) 2n τ(1) − Re u − Re u 2n r r r u = − τ(1) υ [x ] Re u 2n r r = u υ . − τ(1) ∂B2n On the open subset Im u =0 one gets r 6 Re u Im u υ [x ]+ iυ [y ]= σ(1) + i σ(1) υ [y ] 2n σ(1) 2n σ(1) − Im u Im u 2n r r r u¯ = − σ(1) υ [y ] Im u 2n r r

=u ¯σ(1)υ∂B2n and Re u Im u υ [x ]+ iυ [y ]= τ(1) + i τ(1) υ [y ] − 2n τ(1) 2n τ(1) Im u Im u 2n r r r u = τ(1) υ [y ] Im u 2n r r = u υ . − τ(1) ∂B2n In both cases, it follows that

ι∗ (dch ddch ... ddch ) ∂B2n A1 ∧ A2 ∧ ∧ An n 2 n−1 ∂hA1 ∂hA1 ∂ hAk =4 sgn(σ τ) u¯σ(1) + uτ(1) υ∂B2n S · ∂z¯τ(1) ∂zσ(1) ∂zσ(k)∂z¯τ(k) ! σ,τX∈ n kY=2 n−1 t =4 n!Dn(M + M, HessC hA2 ,..., HessC hAn )υ∂B2n , where in the last equality we have used property 1 of Dn. The lemma follows by the arbitrary choice of u ∂B ; in particular (50) is a consequence of the deﬁnition of Q . ∈ 2n n

8 Kazarnovskiˇıpseudovolume on the class K(Cn)

Corollary 6.1 suggests a way to extend the deﬁnition of Pn to all convex bodies, even those with the worst ∞ n n boundary structure. The passage from 1 (C ) to (C ) requires the regularization of the complex Monge-Ampère operator as described inK [BT] or[DemK ]. Deﬁnition 8.1. Let A ,...,A (Cn). The Kazarnovskiˇı n-dimensional mixed pseudovolume 1 n ∈ K of A1,...,An is the (non-negative) real number, noted Qn(A1,...,An), given by 1 Q (A ,...,A )= ddch ... ddch . (51) n 1 n n!κ A1 ∧ ∧ An n ZB2n

If A = A1,...,An, set 1 P (A)= Q (A[n]) = (ddch )∧n , (52) n n n!κ A n ZB2n and call it the Kazarnovskiˇı n-dimensional pseudovolume of A. Remark 8.1. Kazarnovskiˇı n-dimensional mixed pseudovolume is well defined. All the currents involved in Definition 8.1 are positive, so that their wedge product is legitimate. Indeed, it is defined inductively by setting

ddch ddch ... ddch = ddc(h ddch ... ddch ) . (53) A1 ∧ A2 ∧ ∧ An A1 A2 ∧ ∧ An

Although this definition is not symmetric in A1,...,An, it can be shown that the resulting current is such. Indeed, let ε > 0 be fixed and, for any 1 ℓ n, let us replace in the definition (53) ≤ ≤

42 the function hAℓ with its regularization hAℓ ϕε. As shown in [BT] or [Dem], the resulting relation yields an equality of smooth diﬀerential forms∗ the left side of which is clearly symmetric. Since the form c c c c dd (hA1 ϕε) ... dd (hAn ϕε) is symmetric and weakly converges to the current dd hA1 ... dd hAn , as ε 0∗, it follows∧ ∧ that the∗ limiting current is symmetric too. ∧ ∧ → Positive currents have measure coeﬃcients, so the current κ = ddch ... ddch is a positive A1 ∧ ∧ An measure and the integral on the right side of (51) stands for the κ-measure of the unit ball B2n. In the sequel, the pairing between a current T and a test form ς will be denoted T, ς . The mixed hh ii pseudovolume of A1,...,An, can be computed as

1 c c lim dd hA1 ... dd hAn , ςm , n!κn m→∞hh ∧ ∧ ii

where (ςm) is a regularization of χB2n . It follows that Qn is non-negative, symmetric, multilinear and translation invariant.

Remark 8.2. Kazarnovskiˇı n-dimensional mixed pseudovolume is continuous for the Hausdorﬀ metric.

If, for every 1 ℓ n, (Am,ℓ)m∈N is a sequence of convex bodies belonging to some dense subset of n ≤ ≤ (C ) and converging to Aℓ in the Hausdorﬀ metric, then the corresponding sequence hAm,ℓ of support K n c functions converges uniformly to hAℓ on each compact set of C and so the current dd hAm,ℓ converges weakly to the current ddch , as m + . This fact implies the continuity of Q for the Hausdorﬀ Aℓ → ∞ n metric, so that Qn(A1,...,An) can be computed as the limit

Qn(A1,...,An) = lim Qn(Am,1,...,Am,n) . m→∞

n As a consequence, any property of Qn on some dense subspace of (C ) can be extended to the whole (Cn). K K Remark 8.3. Kazarnovskiˇı n-dimensional pseudovolume on ∞(Cn). K ∞ n ∞ n On the dense subspace 1 (C ) (C ) both Definition 8.1 and Definition 6.1 may apply and, in fact, by virtue of CorollaryK 6.1, the⊂K two definitions agree, in particular

(ddch )∧n = α (dα )∧(n−1) , (54) (RεA)ε ∂(RεA)ε ∧ ∂(RεA)ε ZB2n Z∂(RεA)ε for every A ∞(Cn) and any ε > 0. Theorem 6.2 and Remark 8.2 imply that the two deﬁnitions agree on the∈ whole K ∞(Cn), indeed passing to the limit in (54), as ε 0, yields K →

(ddch )∧n = α (dα )∧(n−1) . (55) A ∂A ∧ ∂A ZB2n Z∂A In particular, (55) holds true on ∞(Cn), so for every Γ (Cn) and any ε> 0, P ∈P

(ddch )∧n = α (dα )∧(n−1) . (56) (Γ)ε ∂(Γ)ε ∧ ∂(Γ)ε ZB2n Z∂(Γ)ε Then, as ε 0, Remark 8.2, (56) and Theorem 6.3 yield the equality → 1 (ddch )∧n = v̺ (Γ) , (57) n!κ Γ n n ZB2n which reveals a remarkable combinatorial feature of Kazarnovskiˇıpseudovolume.

Inspired by a privately communicated idea of Kazarnovskiˇı’s, the following Theorem 8.1 and Corol- lary 8.1 generalise (57).

Theorem 8.1. Let n N∗, Γ (Cn) and ε> 0. Then ∈ ∈P κ n n Q (Γ[k],B [n k]) = v̺(Γ) . (58) 2n−kκ k n 2n − k 2n−k

43 Proof. By virtue of 55 we can compute Pn((Γ)ε) in two ways. On one hand 1 P ((Γ) )= (ddch )∧n n ε n!κ (Γ)ε n ZB2n 1 = (ddc(h + εh ))∧n n!κ Γ B2n n ZB2n 1 = (ddch + εddch )∧n n!κ Γ B2n n ZB2n n n n−k 1 c k c ∧(n−k) = ε (dd hΓ) (dd hB2n ) k n!κn ∧ k=0 ZB2n Xn n = εn−kQ (Γ[k],B [n k]) , (59) k n 2n − Xk=0 on the other hand, (thanks to Theorem 6.3), 1 P ((Γ) )= α (dα )∧(n−1) n ε n!κ ∂(Γ)ε ∧ ∂(Γ)ε n Z∂(Γ)ε n n−k 2 κ2n−k v̺ n−k = k(Γ)ε . (60) κn kX=0 The theorem follows by comparing coeﬃcients in (59) and (60).

Corollary 8.1. For any 1 k n and A ,...,A (Cn), ≤ ≤ 1 k ∈ K κ n n Q (A ,...,A ,B [n k]) = V̺(A ,...,A ) . (61) 2n−kκ k n 1 k 2n − k 1 k 2n−k

Proof. Both sides of (61) are continuous in each of the k arguments A1,...,Ak, so it’s enough to prove the equality in the case of polytopes. In this case we know that both sides of (61) are symmetric and k-linear, so, by Corollary 3.1, they agree as soon as they do it on the diagonal, i.e. for A1 = ... = Ak = A (Cn). The claim follows from Theorem 8.1. ∈P Corollary 8.2. For any A ,...,A (Rn), 1 n ∈ K

Qn(A1,...,An)= Vn(A1,...,Ak) .

Proof. By continuity, it’s enough to prove the equality for polytopes. In this case the statement is an easy consequence of Corollary 8.1 with k = n and purely real polytopes.

Remark 8.4. An integral formula for the Minkowski mixed volume. If A ,...,A 2(Rn), the corresponding support functions are constant with respect to the variables 1 n ∈ K1 y1,...,yn so 1 HessC h = HessR h Ak 4 Ak and, by the expression (31) of ddc in real coordinates, it follows that

n 2 c ∂ hAk dd hAk (z)= dxℓ dyj ∂xℓ∂xj ∧ ℓ,jX=1 whence 1 V (A ,...,A )= ddch ... ddch (62) n 1 k n!κ A1 ∧ ∧ An n ZB2n 4n = D (HessC h ,..., HessC h )υ κ n A1 An 2n n ZB2n 1 = D (HessR h ,..., HessR h )υ . κ n A1 An 2n n ZB2n By continuity, (62) remains valid for arbitrary real convex bodies A ,...,A (Rn). 1 n ∈ K

44 Remark 8.5. Qn is unitarily invariant and satisﬁes both the polarization formula both the symmetric one.

By the continuity of the mixed n-pseudovolume, it follows that Qn is unitarily invariant and satisﬁes the usual polarization and symmetric formulas,

1 n−#I Qn(A1,...,An)= ( 1) Pn Aℓ , (63) n! ∅ − ! 6=I⊆{X1,...,n} Xℓ∈I 1 ∂n n Qn(A1,...,An)= Pn λℓAℓ . (64) n! ∂λ1 ...∂λn ! Xℓ=1 ∞ n Indeed, by Lemma 6.7, Qn is unitarily invariant on + (C ), whereas by Theorem 3.1, Lemma 3.3 and Corollary 8.1, it satisﬁes (63) and (64) on (Cn). K P

c ∧k 9 The current (dd hΓ)

c ∧k n The present section is devoted to a closer study of the current (dd hΓ) , for any Γ (C ). The case in which the polytope Γ is replaced by a general convex body is much harder, but it∈P can be viewed as c a weak limit of the polytopal case. Let us first consider the case k =1 and observe that d hΓ is simply c a 1-form the coefficients of which are locally constant functions, so that the current dd hΓ is supported c on the corner locus of hΓ. Such a corner locus equals the set of points where the coefficients of d hΓ are not continuous, i.e. the closure of the 1-star of Γ, hence

c Supp dd hΓ = Σ1,Γ . (65)

Moreover, if A , A (Cn) are such that A A is convex, the well-known equality 1 2 ∈ K 1 ∪ 2 A + A = (A A ) + (A A ) 1 2 1 ∪ 2 1 ∩ 2 implies that hA1 + hA2 = hA1∪A2 + hA1∩A2 , whence

c c c c dd hA1 + dd hA2 = dd hA1∪A2 + dd hA1∩A2 . (66)

c Setting h∅ 0, the relation (66) shows that taking the dd of support functions yields a 1-homogeneous valuation on≡ (Cn) with values in the space of (1, 1)-currents. In this section we will prove the following generalizations,K for k> 1,of (65) and (66):

c ∧k Supp (dd hΓ) = Σk,Γ , (67)

c ∧k c ∧k c ∧k c ∧k (dd hA1 ) + (dd hA2 ) = (dd hA1∪A2 ) + (dd hA1∩A2 ) . (68)

n If Γ (C ) is ﬁxed, for every 1 k n and any ∆ ed(Γ, k), let λ∆ be the (k, k)-current of measure∈ type P on Cn acting on any continuous≤ ≤ and compactly∈ B supported (n k,n k)-form ς as − −

∗ λ∆, ς = ι (υE′ ς) , (69) hh ii ∆ ∆ ∧ ZK∆ n ′ where ι∆ : K∆ C is the inclusion, K∆ is oriented by some orthonormal basis w1,...,wk of E∆ and ∗ → υE′ = ι (Re w1, ... Re wk, ). Let us point out some remarks about the current λ∆. ∆ ∆ h −i ∧ ∧ h −i The current λ is a kind of integration current on the manifold K obtained as a restriction to • ∆ ∆ K of the current induced by the volume form υ ′ . ∆ E∆

The deﬁnition of λ∆ makes sense if and only if ∆ is equidimensional. Indeed, the degree of the • ∗ form ι (υE′ ς) equals the dimension of K∆ if and only if ∆ is equidimensional. ∆ ∆ ∧ ′ ⊥C n ⊥ ′ ⊥C For any equidimensional face ∆ of Γ, one has E∆ E∆ E∆ = C and K∆ E∆ = E∆ E∆ , • ′ ⊕ ⊥C⊕ ⊂ ⊕ so the orientation of K∆ comes from that of E∆ E∆ . Since the mutually orthogonal C-linear ′ ⊥C ⊕ ⊥ subspaces E∆ E∆ and E∆ are naturally positively oriented, it follows that the orientation of E∆ ⊕ ′ ultimately depends on the orientation of E∆. As shown in Remark 4.2, choosing an orientation ′ ′ on E∆ (resp. E∆) implies a corresponding choice of orientation on E∆ (resp. E∆) yielding the

45 natural positive orientation on E E′ . The volume form υ ′ involved in the definition of the ∆ ∆ E∆ ⊕ ′ current λ∆ conforms to the chosen orientation on E∆. This way the integral in (69) is not affected by the choice of such an orientation and the current λ∆ is thus well defined. According to de Rham [dR], the form υ ′ is an example of odd form, i.e. a form changing its sign as E′ (an thus E∆ ∆ E∆) changes its orientation. A convenient way to make a coherent choice is to fix an orthonormal basis v1,...,vk of E∆ and set

k(k−1)/2 υE′ = ( 1) Re w1, ... Re wk, , (70) ∆ − h −i ∧ ∧ h −i ′ ⊥ where w1,...,wk is the basis of E∆ deﬁned in Theorem 4.1. Accordingly, the space E∆ (and hence K∆) will get the orientation associated to the basis

( 1)k(k−1)/2w ,...,w ,b ,...,b , − 1 k 2k+1 2n ⊥C where b2k+1,...,b2n is a positive basis of E∆ .

The only terms of ς giving a non zero contribution to ι∗ (υ ′ ς)= ι∗ υ ′ ι∗ ς are those involving ∆ E∆ ∆ E∆ ∆ • ⊥C ∧ ∧ the diﬀerentials of the coordinate functions of E∆ , nevertheless the coeﬃcients of such terms do depend on the full array of coordinates of K E⊥. ∆ ⊂ ∆ The support of the current λ is the closure of K but λ is concentrated just on K . • ∆ ∆ ∆ ∆

9.1 The case k = 1.

Lemma 9.1. Let Γ (C), then ddch = vol (Λ)λ . In particular Supp ddch = Σ . ∈P Γ Λ∈B(Γ,1) 1 Λ Γ 1 P c Proof. If Γ is a point, the lemma is trivial. If Γ is not a point, hΓ is piece-wise linear, the current dd hΓ is zero on C Σ1 and it acts on test functions, so let ς be a smooth function with compact support. If \ c d = 1, Γ is a line segment and, up to a translation (which do not aﬀect d hΓ), we may suppose that ∗ ⊥ Γ = [0, v] with v C . Then EΓ is spanned by v and, according to our convention, iv is a basis of EΓ ∈ ⊥ such that v,iv gives EΓ EΓ = C the natural orientation. The closure of the 1-star of Γ is the closure ⊕ ′ ⊥ i Arg v of the dual cone to Γ and, in this case, it is nothing but EΓ = EΓ = iEΓ = ie ρ C ρ R . Then { ∈ | ∈ }

ddch , ς = dch , dς hh Γ ii hh Γ ii = dch dς Γ ∧ ZSupp ς c = lim d hΓ dς ε→0 ∧ ZSupp ς\(Σ1)ε c = lim d hΓ dς ε→0 ∧ ZK0∩Supp ς\(Σ1)ε c + lim d hΓ dς . ε→0 ∧ ZKv ∩Supp ς\(Σ1)ε

Since hΓ vanishes on K0,

c dd hΓ, ς = lim (i/2)(vd¯z v¯dz) dς hh ii ε→0 − ∧ ZKv∩Supp ς\(Σ1)ε = lim (i/2)d[(¯vdz vd¯z)ς] ε→0 − ZKv∩Supp ς\(Σ1)ε = lim (i/2)d[(¯vdz vd¯z)ς] ε→0 − ZSv,ε = lim (i/2)(¯vdz vd¯z)ς . ε→0 − Z∂Sv,ε where Sv,ε is the (ε/2)-neighbourhood of Kv Supp ς (Σ1)ε. Of course Kv Supp ς (Σ1)ε ( Sv,ε ( Kv and both K Supp ς (Σ ) and S approach∩ the\ set K Supp ς, as∩ ε 0+\ . However, unlike v ∩ \ 1 ε v,ε v ∩ → Kv Supp ς (Σ1)ε, the set Sv,ε has smooth boundary. This boundary can be decomposed into three parts:∩ the pieces\ which belong to ∂ Supp ς (on which ς is zero), the spherical pieces over the singular points of ∂[K Supp ς (Σ ) ], (which shrink to the corresponding singular points as ε 0) and the v ∩ \ 1 ε →

46 ﬂat pieces parallel and near to KΓ (in Figure 8 the set ∂Sv,ε admits just one of such parts). Each of the latter components is a relatively open line segment inheriting from ∂Sv,ε an orientation which is opposite ⊥ to that of EΓ . As ε 0, each of these segments approaches the subset KΓ Supp ς = KΓ Supp ς, ∗ i Arg→v ∗ −i Arg v ∗ ∩ ∩ on which ι dz = ie dρ, ι d¯z = ie dρ and ι vE′ = dρ, thus Γ Γ − Γ Γ

ddch , ς = (i/2)(¯viei Arg v + vie−i Arg v)ςdρ hh Γ ii − ZKΓ∩Supp ς = v ςdρ | | ZKΓ∩Supp ς = v ςdρ | | ZKΓ∩Supp ς = vol (Γ) λ , ς , 1 hh Γ ii c i.e. dd hΓ = vol 1(Γ)λΓ.

Observe that giving EΓ the opposite orientation means choosing v as a basis of EΓ and, conse- ⊥ − quently, v, iv as a basis of EΓ EΓ = C. In both cases the two basis of the latter space are positively oriented− and− the proof proceeds⊕ in the same way.

Kv E⊥ Γ E⊥ Γ Kv

Kv ∩ Supp ς \ (Σ1)ε v Γ

0 ε K K 0 0

⊥ Figure 8: On the left the 1-polytope Γ. The half-spaces bounded by the hyperplane E∆ are K0 and Kv . On the right, Supp ς is depicted as the compact set with smooth rounded boundary, (Σ1)ε is the strip with dashed boundary. The blue part is Kv ∩ Supp ς \ (Σ1)ε, the components of its boundary are drawn in orange. The blue/red smooth path around Kv ∩ Supp ς \ (Σ1)ε is the boundary of Sv,ε. The red segment is the only piece of ∂Sv,ε that counts.

Let us now suppose d = dim Γ = 2, (cf. Figure 9). If Γ is oriented and v0, v1,...,vs is a numbering of the vertices of Γ which agrees with the orientation of ∂Γ induced by unit outer normal vectors, the sides of Γ are the oriented line segments Λ = [v , v ], 1 ℓ s, with v = v and v = v . ℓ ℓ ℓ+1 − ≤ ≤ s+1 0 −1 s

47 v2 + Kv2

Λ1 Λ2 Γ

v v0 Λ0 1

v0 + Kv0 v1 + Kv1

Figure 9: The 2-polytope Γ is a triangle.

i Arg(vℓ+1−vℓ) The dual cone to the side Λℓ is spanned by the unit outer normal vector uΛℓ,Γ = ie i Arg(vℓ −vℓ) ∗ i Arg(vℓ −vℓ) ∗ −−i Arg(vℓ −vℓ) hence KΛ = ie +1 ρ C ρ > 0 , ι dz = ie +1 dρ, ι d¯z = ie +1 dρ ℓ {− ∈ | } Λℓ − Λℓ and vE′ = dρ. Λℓ

ddch , ς = dch , dς hh Γ ii hh Γ ii = dch dς Γ ∧ ZSupp ς c = lim d hΓ dς ε→0 ∧ ZSupp ς\(Σ1)ε s c = lim d hΓ dς . ε→0 Kv ∩Supp ς\(Σ1)ε ∧ Xℓ=0 Z ℓ

The domains of linearity of hΓ are the dual cones to the vertices of Γ, so

s c dd hΓ, ς = lim (i/2)(vℓd¯z v¯ℓdz) dς hh ii ε→0 K ∩Supp ς\(Σ ) − ∧ ℓ=0 Z vℓ 1 ε Xs = lim (i/2)d[(vℓd¯z v¯ℓdz)ς] − ε→0 K ∩Supp ς\(Σ ) − ℓ=0 Z vℓ 1 ε Xs = lim (i/2)d[(vℓd¯z v¯ℓdz)ς] − ε→0 S − ℓ=0 Z vℓ,ε sX = lim (i/2)(¯vℓdz vℓd¯z)ς , (71) ε→0 ∂Sv ,ε − Xℓ=0 Z ℓ where S is the (ε/2)-neighbourhood of K Supp ς (Σ ) . As before, the only parts of the ∂S vℓ,ε vℓ ∩ \ 1 ε vℓ,ε which really count are those which are parallel and near to KΛℓ and KΛℓ−1 respectively, with the ﬁrst one inheriting from ∂Svℓ,ε an orientation which is opposite to that of KΛℓ and the second one inheriting from ∂Svℓ,ε an orientation which is equal to that of KΛℓ−1 , (cf. Figure 10). Observe that giving Γ the opposite orientation simply switches the orientations of the latter boundary pieces.

48 KΛ1

K (Σ ) v2 \ 1 ε

K (Σ ) v1 \ 1 ε

KΛ2 K (Σ ) v0 \ 1 ε

ε ε 2

KΛ0

Figure 10: The set Supp ς is depicted as the compact set with smooth rounded boundary, (Σ1)ε is the set with

black, dashed boundary. The blue 2-dimensional part of Supp ς is the union of Kv1 ∩ Supp ς \ (Σ1)ε

and Kv2 ∩ Supp ς \ (Σ1)ε, the components of ∂[Kv1 ∩ Supp ς \ (Σ1)ε] and ∂[Kv2 ∩ Supp ς \ (Σ1)ε] are coloured according to the colours of the edges of Γ. The coloured smooth path all around Kv1 ∩ Supp ς \ (Σ1)ε is the boundary of Sv1,ε. The red, brown and green segments are the only pieces of ∂Sv1,ε which count. The same occurs to Sv2,ε.

As ε 0, such boundary pieces are line segments approaching respectively KΛℓ Supp ς and K Supp→ ς, thus (71) becomes ∩ Λℓ−1 ∩ s i Arg(vℓ+1−vℓ) −i Arg(vℓ+1−vℓ) (i/2)(¯vℓie + vℓie )ςdρ K ∩Supp ς ℓ=0 Z Λℓ X s i Arg(vℓ−vℓ−1) −i Arg(vℓ−vℓ−1) + (i/2)( v¯ℓie vℓie )ςdρ K ∩Supp ς − − ℓ=0 Z Λℓ−1 X s i Arg(vℓ+1−vℓ) −i Arg(vℓ+1−vℓ) = (1/2) ( v¯ℓe vℓe )ςdρ K ∩Supp ς − − ℓ=0 Z Λℓ Xs i Arg(vℓ−vℓ−1) −i Arg(vℓ−vℓ−1) + (1/2) (¯vℓe + vℓe )ςdρ K ∩Supp ς ℓ=0 Z Λℓ−1 Xs i Arg(vℓ+1−vℓ) −i Arg(vℓ+1−vℓ) = (1/2) ( v¯ℓe vℓe )ςdρ KΛ ∩Supp ς − − Xℓ=0 Z ℓ s−1 i Arg(vj+1−vj ) −i Arg(vj+1 −vj ) + (1/2) (¯vj+1e + vj+1e )ςdρ . j=−1 KΛj ∩Supp ς X Z

Since Λ−1 =Λs, we simply get

s i Arg(vℓ+1−vℓ) −i Arg(vℓ+1−vℓ) (1/2) [(¯vℓ+1 v¯ℓ)e + (vℓ+1 vℓ)e ]ςdρ , KΛ ∩Supp ς − − Xℓ=0 Z ℓ

49 where the term in brackets is nothing but 2 v v = 2 vol (Λ ). It then follows that | ℓ+1 − ℓ| 1 ℓ s c dd hΓ, ς = vℓ+1 vℓ ςdρ = vol 1(Λ) λΛ, ς , (72) hh ii | − | KΛ hh ii Xℓ=0 Z ℓ Λ∈BX(Γ,1) c c i.e. dd hΓ = Λ∈B(Γ,1) vol 1(Λ)λΛ. The support of dd hΓ is the union of the supports of the currents λ , as Λ (Γ, 1), i.e. Σ . Λ ∈B P 1 Theorem 9.1. Let Γ (Cn), then ∈P c dd hΓ = vol 1(Λ)λΛ . (73) Λ∈BX(Γ,1) c In particular Supp dd hΓ = Σ1.

Proof. If Γ is reduced to a single point, hΓ is linear and both sides of (73) are zero. If Γ is not reduced c n to a single point, the current d hΓ is just a smooth form almost everywhere, i.e. on C Σ1, so, for any test (n 1,n 1)-form ς, \ − − ddch , ς = dch , dς hh Γ ii hh Γ ii c = lim d hΓ dς ε→0 ∧ ZSupp ς\(Σ1)ε c = lim d hΓ dς ε→0 Kv∩Supp ς\(Σ1)ε ∧ v∈BX(Γ,0) Z c = lim d hΓ dς , ε→0 Sv,ε ∧ v∈BX(Γ,0) Z where Sv,ε denotes the (ε/2)-neighbourhood of Kv Supp ς (Σ1)ε. Of course Kv Supp ς (Σ1)ε ( S ( K and both K Supp ς (Σ ) and S approach∩ the\ set K Supp ς, as∩ε 0+.\ However, v,ε v v ∩ \ 1 ε v,ε v ∩ → unlike Kv Supp ς (Σ1)ε, the set Sv,ε has smooth boundary. This boundary can be decomposed into three∩ parts: the\ pieces which belong to ∂ Supp ς (on which ς is zero), the spherical pieces over the singular points of ∂[Kv Supp ς (Σ1)ε], (which shrink to the corresponding singular points of ∂[K Supp ς (Σ ) ] as ε ∩ 0) and the\ ﬂat pieces parallel and near to ∂K v ∩ \ 1 ε → v For every v (Γ, 0), we have ∈B n dch = (i/2) v d¯z v¯ dz Γ ℓ ℓ − ℓ ℓ Xℓ=1 c c c on Kv, so that d hΓ dς = d(d hΓ ς) on such a cone, then by Stokes’ theorem dd hΓ, ς equals the limit, as ε 0, of∧ − ∧ hh ii → n (i/2) vℓd¯zℓ v¯ℓdzℓ ς . (74) − ∂Sv,ε − ! ∧ v∈BX(Γ,0) Z Xℓ=1 For any ﬁxed v (Γ, 0) the only parts of the ∂S which really count are those which are parallel ∈ B vℓ,ε and near to the 1-codimensional parts of ∂Kv, if any. Since, for any vertex v,

∂Kv = KΛ , Λ∈B(Γ,1) v[≺Λ it follows that, for any ﬁxed side Λ of Γ, the set

∂Sv,ε (75) v∈B[(Γ,0) may admit components parallel to KΛ and belonging to diﬀerent half-spaces. If Λ is the oriented line −1 ε segment [wΛ, vΛ], then aΛ = (vΛ wΛ) vol 1(Λ) is an orienting orthonormal basis of EΛ. Let Xv(Λ) ε − and Xw(Λ) be, respectively, the subsets of (75) parallel and near to KΛ and belonging to diﬀerent half- ε ⊥ ε spaces, then Xv(Λ) inherits from (75) an orientation which is opposite to that of EΛ , whereas Xw(Λ)

50 ⊥ inherits from (75) an orientation which equals that of EΛ , so that (74) becomes

n lim (i/2) vΛ,ℓd¯zℓ v¯Λ,ℓdzℓ ς ε→0 Xε − ! ∧ Λ∈BX(Γ,1) Z v(Λ) Xℓ=1 n lim (i/2) wΛ,ℓd¯zℓ w¯Λ,ℓdzℓ ς . − ε→0 Xε − ! ∧ Λ∈BX(Γ,1) Z w(Λ) Xℓ=1 ε ε As ε 0, both Xv(Λ) and Xw(Λ) approach KΛ Supp ς, then the continuity of the integrands and the compactness→ of their supports imply that the sum∩ of the preceding terms becomes

n vol 1(Λ) (i/2) aΛ,ℓd¯zℓ a¯Λ,ℓdzℓ ς , KΛ∩Supp ς − ! ∧ Λ∈BX(Γ,1) Z Xℓ=1 whence c ∗ dd h , ς = vol (Λ) ι (υ ′ ς)= vol (Λ) λ , ς . Γ 1 Λ EΛ 1 Λ hh ii KΛ ∧ hh ii Λ∈BX(Γ,1) Z Λ∈BX(Γ,1) Observe that giving Λ the opposite orientation does not aﬀect the computation because both vol 1(Λ) and λΛ do not depend on the chosen orientation. c The support of dd hΓ is the union of the supports of the currents λΛ, as Λ (Γ, 1), i.e. Σ1. The proof is thus complete. ∈B

Remark 9.1. Lemma 9.1 and Theorem 73 can be viewed as a generalisation of the following formula known as la formule des sautes in the french literature. Let S R be a discrete subset and g : R R be a ⊂ ′ → ′ continuous piecewise smooth function admitting a corner point at each s S, (i.e. g−(s) = limx→s− g (x) ′ ′ ∈ and g+(s) = limx→s+ g (x) are both ﬁnite but distinct). Then, it is well-known that the second derivative of g is a measure µ with the following decomposition

′′ ′ ′ µ = g χR + g (s) g (s) δ , (76) \S + − − s s∈S X where χR is the characteristic function of the set R S, (i.e. χR (x)=1 if x R S and χR (x)=0 \S \ \S ∈ \ \S if x S) and δs is the Dirac delta on the point s S. The ﬁrst term in (76) is the ordinary second derivative∈ of g on the points where f is smooth, whereas∈ the second one gives the contribution of the singular points, each singular point producing a Dirac delta multiplied by the corresponding jump. If g is convex then µ is a positive measure, indeed both the second derivative of g (on the points where g′ is smooth) and the jumps of g′ (at singular points) are non-negative. When g is a convex piecewise linear function, the absolutely continuous part of µ vanishes, and µ becomes a positive Borel measure which is singular with respect to the Lebesgue measure.

9.2 The case k = 2.

Lemma 9.2. Let ∆ (Cn) be an oriented polygon, then ∈P

vol (Λ)υ ′ =0 . 1 EΛ Λ∈BX(∆,1) Proof.

Let us ﬁrst suppose that ∆ is a triangle with vertices a,b,c. If we denote Λ1, Λ2 and Λ3 the oriented sides [a,b], [b,c] and [c,a] respectively, we get

n i ′ ¯ vol 1(Λ1)υE = (bℓ aℓ)d¯zℓ (bℓ a¯ℓ)dzℓ , Λ1 2 − − − ℓ=1 Xn i ′ ¯ vol 1(Λ2)υE = (cℓ bℓ)d¯zℓ (¯cℓ bℓ)dzℓ , Λ2 2 − − − ℓ=1 Xn i vol 1(Λ3)υE′ = (aℓ cℓ)d¯zℓ (¯aℓ c¯ℓ)dzℓ , Λ3 2 − − − Xℓ=1

51 and consequently

vol 1(Λ1)υE′ + vol 1(Λ2)υE′ + vol 1(Λ3)υE′ Λ1 Λ2 Λ3 i n = (b a + c b + a c )d¯z (¯b a¯ +¯c ¯b +¯a c¯ )dz =0 . 2 ℓ − ℓ ℓ − ℓ ℓ − ℓ ℓ − ℓ − ℓ ℓ − ℓ ℓ − ℓ ℓ Xℓ=1 If ∆ is given the opposite orientation, a minus sign will affect each of the three terms involved in the preceding sum without changing the result. If ∆ is not a triangle, it can be decomposed into two or more triangles and by applying the preceding argument to each triangle one realises that the contributions of common sides cancel out for orientation reasons. Lemma 9.3. Let ∆ (Cn) an equidimensional oriented polygon then ∈P ∗ c vol 1(Λ)υE′ ι d hΓ = vol 1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, , Λ ∧ Λ Λ ∧ h −i Λ∈BX(∆,1) Λ∈BX(∆,1) where, for every Λ (∆, 1), uΛ,∆ denotes the unit outer normal vector to the side Λ of ∆ and ι : K Cn is the∈ inclusion. B Λ Λ → Proof. Let Λ (∆, 1) be fixed. Recall that E⊥ affR Λ= p and E⊥ affR ∆= p . Then, by ∈B Λ ∩ { Λ} ∆ ∩ { ∆} virtue of (8), pΛ = p∆ + h∆(uΛ,∆)uΛ,∆ and ∗ c υE′ ι d hΓ = υE′ Re ipΛ, Λ ∧ Λ Λ ∧ h −i = υE′ Re ip∆, + h∆(uΛ,∆)υE′ Re iuΛ,∆, , Λ ∧ h −i Λ ∧ h −i By Lemma 9.2,

vol 1(Λ)υE′ Re ip∆, = Re ip∆, vol 1(Λ)υE′ =0 Λ ∧ h −i − h −i ∧ Λ Λ∈BX(∆,1) Λ∈BX(∆,1) whence ∗ c vol 1(Λ)υE′ ι d hΓ = vol 1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, Λ ∧ Λ Λ ∧ h −i Λ∈BX(∆,1) Λ∈BX(∆,1) which proves the lemma. Lemma 9.4. Let ∆ (Cn) an equidimensional oriented polygon then, for every Λ (∆, 1), ∈P ∈B vol 1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, = ̺(∆) vol 1(Λ)h∆(uΛ,∆)υE′ . (77) Λ ∧ h −i ∆ In particular

vol 1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, =2̺(∆) vol 2(∆)υE′ . (78) Λ ∧ h −i ∆ Λ∈BX(∆,1) Proof. As ∆ is oriented, each of its facets gets the orientation induced by the outer unit normal vector. If aΛ is an orienting basis of EΛ, then (uΛ,∆,aΛ) is an orienting orthonormal basis of E∆, υE′ = Re iaΛ, and vol 1(Λ) Re iaΛ, Re iuΛ,∆, is a non-zero 2-form. Consider the vectors Λ h −i h −i ∧ h −i t = iu Re iu ,a a , Λ,1 Λ,∆ − h Λ,∆ Λi Λ t = ia Re ia ,u u Λ,2 Λ − h Λ Λ,∆i Λ,∆ wΛ,1 = tΛ,1/ ̺(∆) and wΛ,2 = tΛ,2/ ̺(∆). According to Theorem 4.1, since ∆ is equidimensional, the sequence uΛ,∆,aΛ, wΛ,1, wΛ,2 is an orthonormal basis of linC E∆ over R such that wΛ,1, wΛ,2 span ′ p p E∆. For dimensional reasons, (k =2), this basis does not belong to the positive orientation of linC E∆ so, by (70), we have υ ′ = Re w , Re w , . Let us complete the preceding basis by adding E∆ Λ,1 Λ,2 − h −i∧ h⊥C −i an orienting orthonormal basis b5,...,b2n of E∆ over R. Since iaΛ and iuΛ,∆ belong to linC E∆, both the linear forms Re ia , and Re iu , vanish on E⊥C , i.e. they both vanish on each of the h Λ −i h Λ,∆ −i ∆ vectors b5,...,b2n. A direct computation shows that Re iaΛ, wΛ,2 = Re iuΛ,∆, wΛ,1 = ̺(∆) and Re ia , w =0. Indeed, h i h i h Λ Λ,1i p Re iaΛ,iaΛ Re iaΛ,uΛ,∆ uΛ,∆ Re iaΛ, wΛ,2 = h − h i i h i ̺(∆) 1 (Re ia ,u )2 = − h Λ pΛ,∆i ̺(∆) = ̺(∆)p, p 52 Re iuΛ,∆,iuΛ,∆ Re iuΛ,∆,aΛ aΛ Re iuΛ,∆, wΛ,1 = h − h i i h i ̺(∆) 1 (Re iu ,a )2 = − h Λ,∆ pΛi ̺(∆) = ̺(∆)p, p Re iaΛ,iuΛ,∆ Re iuΛ,∆,aΛ aΛ Re iaΛ, wΛ,1 = h − h i i h i ̺(∆) Re a ,u Re iu ,a Re ia ,a = h Λ Λ,∆i−p h Λ,∆ Λi h Λ Λi ̺(∆) =0 . p It follows that [vol (Λ)h (u ) Re ia , Re iu , ] (w , w ) 1 ∆ Λ,∆ h Λ −i ∧ h Λ,∆ −i Λ,1 Λ,2 = vol (Λ)h (u )[ Re iu , Re ia , ] (w , w ) 1 ∆ Λ,∆ − h Λ,∆ −i ∧ h Λ −i Λ,1 Λ,2 vol (Λ)h (u ) =2̺(∆) 1 ∆ Λ,∆ ( 1) 2 − vol (Λ)h (u ) =2̺(∆) 1 ∆ Λ,∆ [ Re w , Re w , ] (w , w ) 2 − h Λ,1 −i ∧ h Λ,2 −i Λ,1 Λ,2 vol 1(Λ)h∆(uΛ,∆) =2̺(∆) υE′ (wΛ,1, wΛ,2) , 2 ∆ which implies (77). Observe that changing the orientation of ∆, and hence of Λ, aﬀects by a minus sign the vectors a , w and the forms υ ′ ,υ ′ , however it changes neither the vectors u , w nor the Λ Λ,1 EΛ E∆ Λ,∆ Λ,2 preceding computations. By summing the relations (77) as Λ runs in (∆, 1), the equality (6) yields the desired relation (78). The lemma is thus completely proved. B Remark 9.2. Remark that the basis u ,a , w , w ,b ,...,b and a , u , w , w ,b ,...,b Λ,∆ Λ − Λ,1 Λ,2 5 2n Λ − Λ,∆ − Λ,1 Λ,2 5 2n n ⊥ ⊥ ⊥ are both positively oriented, so C gives EΛ the orientation induced by aΛ and E∆ gets from EΛ the orientation induced by u . This means that u orients E⊥ in the right way. − Λ,∆ − Λ,∆ ∆ Theorem 9.2. Let Γ (Cn), then ∈P c ∧2 (dd hΓ) =2 ̺(∆) vol 2(∆)λ∆ . (79) ∆∈BX(Γ,2) c ∧2 In particular Supp (dd hΓ) = Σ2.

Proof. By Theorem 9.1 we already know that

c dd hΓ = vol 1(Λ)λΛ , (80) Λ∈BX(Γ,1) so, for any (n 2,n 2)-test form ς, − − (ddch )∧2, ς = ddc(h ddch ), ς hh Γ ii hh Γ Γ ii = h ddch , ddcς hh Γ Γ ii = ddch ,h ddcς hh Γ Γ ii = vol (Λ) λ ,h ddcς 1 hh Λ Γ ii Λ∈BX(Γ,1) c = vol (Λ) υ ′ h dd ς 1 EΛ Γ KΛ ∧ Λ∈BX(Γ,1) Z c = lim vol 1(Λ) υE′ hΓdd ς ε→0 Λ KΛ∩Supp ς\(Σ2)ε ∧ Λ∈BX(Γ,1) Z c = lim vol 1(Λ) υE′ hΓdd ς , ε→0 Λ SΛ,ε ∧ Λ∈BX(Γ,1) Z

53 where S = K (K Supp ς (Σ ) ) , i.e. the intersection of K with the (ε/2)-neighbourhood Λ,ε Λ ∩ Λ ∩ \ 2 ε ε/2 Λ of K Supp ς (Σ ) . Unlike K Supp ς (Σ ) , the set S has a smooth relative boundary. Λ ∩ \ 2 ε Λ ∩ \ 2 ε Λ,ε ′ ⊥ c c The 1-form υE′ is a closed on E E , moreover the (2n 2)-forms dhΓ d ς and d hΓ dς Λ Λ ⊂ Λ − − ∧ ∧ have the same (n 1,n 1)-parts on E⊥C so that on K one has − − Λ Λ c c c d(vE′ hΓd ς) = dυE′ hΓd ς υE′ d(hΓd ς) Λ ∧ Λ ∧ − Λ ∧ c c =0 υE′ dhΓ d ς υE′ hΓdd ς − Λ ∧ ∧ − Λ ∧ c c = υE′ d hΓ dς υE′ hΓdd ς . Λ ∧ ∧ − Λ ∧ c Since on KΛ the 1-form d hΓ has constant coeﬃcients, it follows that

c c c υE′ hΓdd ς = d(υE′ d hΓ ς) d(υE′ hΓd ς) , Λ ∧ Λ ∧ ∧ − Λ ∧ whence

c ∧2 c (dd hΓ) , ς = lim vol 1(Λ) υE′ d hΓ ς (81) ε→0 Λ hh ii relbd SΛ,ε ∧ ∧ Λ∈BX(Γ,1) Z c lim vol 1(Λ) υE′ hΓd ς . (82) ε→0 Λ − relbd SΛ,ε ∧ Λ∈BX(Γ,1) Z

For any ﬁxed side Λ, if Supp ς relbd KΛ = ∅, one has KΛ Supp ς (Σ2)ε ( SΛ,ε ( KΛ and both K Supp ς (Σ ) and S approach∩ the6 set K Supp ς∩, as ε \0+. The relative boundary of Λ ∩ \ 1 ε Λ,ε Λ ∩ → SΛ,ε can be decomposed into three parts: the pieces belonging to ∂ Supp ς (on which ς is zero), the spherical pieces over the singular points of ∂[KΛ Supp ς (Σ2)ε], (which shrink to the corresponding singular points of ∂[K Supp ς (Σ ) ] as ε 0∩) and the\ ﬂat pieces parallel and near to the (2n 2)- Λ ∩ \ 2 ε → − dimensional parts of relbd KΛ, if any. The latter are the only parts of relbd SΛ,ε which really count. The (2n 2)-dimensional parts of relbd KΛ are nothing but the cones which are dual to the 2-faces ∆ of Γ admitting− Λ as a side. ε Let us denote XΛ(∆) the subset of relbd SΛ,ε parallel and near to K∆ and observe, by the way, that Xε approaches K Supp ς as ε 0. As a part of the manifold relbd S , the set Xε has the Λ(∆) ∆ ∩ → Λ,ε Λ(∆) orientation of relbd SΛ,ε, namely the orientation induced by the outer unit normal vector to SΛ,ε. On Xε the outer unit normal vector to S equals the vector pointing outside K Supp ς (Σ ) i.e. Λ(∆) Λ,ε Λ ∩ \ 2 ε uΛ,∆, which, by Remark 9.2, is coherent with the choice of the basis made in the proof of Lemma 9.4. −The situation is depicted in Figure 11 for a real 3-dimensional cube.3

3 3 3 Of course the picture refers to the intersection of KΛ ∩ Supp ς \ (Σ2)ε with the 3-dimensional subspace EΓ = R ⊂ C . ς K K In fact Supp can be 6-dimensional while Λ1 and Λ2 are 4-dimensional, so there are three more dimensions that cannot be represented in the picture.

54 u∆,Γ u∆,Γ

uo,Λ1

uΛ2,∆ uΛ1,∆ ε/2 u q,Λ2 ε

R3 K ∩ ∆ R3 K R3 K ∩ Λ2 ∩ Λ1

Xε Xε Λ2(∆) Λ1(∆) K S K S Λ2 ∩ \ (Σ2)ε Λ1 ∩ \ (Σ2)ε

ε ε/2 o Λ2 ∆ Λ1 q

Figure 11: S = Supp ς is the transparent cylinder with brown contour, Γ is a cube with a vertex in the origin, ∆ is the red face, Λ1 is the orange edge of ∆ with a vertex in the origin o, whereas Λ2 is the green 3 edge parallel to Λ1. The space C is spanned by the orthonormal basis (u∆,Γ, uΛ1,∆, uo,Λ1 ) or ⊥ (u∆,Γ, uΛ2,∆, uq,Λ2 ) and three more unit vectors w1,w2,w3 spanning EΓ thus getting in both cases 3 positively oriented basis of C . This choice implies that E∆ is oriented by (uΛ1,∆, uo,Λ1 ) or

(uΛ2,∆, uq,Λ1 ). The set KΛ1 ∩ Supp ς \ (Σ2)ε is the blue slice (on the right) of Supp ς along

KΛ1 \ (Σ2)ε while SΛ1,ε is the set bounded by the smooth path all around KΛ1 ∩ Supp ς \ (Σ2)ε. ε XΛ1(∆) is the red segment on the boundary of SΛ1,ε. Along the black part of the boundary of SΛ1,ε the form ς is zero because this part is included in ∂ Supp ς. On the blue circular (in fact

cylindrical!) pieces of ∂SΛ1,ε, over the singular points of KΛ1 ∩ Supp ς \ (Σ2)ε, the integral vanishes in the limit. The same considerations are valid for Λ2.

As a last remark, observe that the form υ ′ may vanish on K , indeed this happens precisely when EΛ ∆ E′ is orthogonal to K E⊥ and since the ortho-complement of E⊥ is E , it follows that the pullback Λ ∆ ⊂ ∆ ∆ ∆ of υ ′ to K is zero if and only if E′ E = 0 , which means that ia = u . We deduce that EΛ ∆ Λ ∆ Λ Λ,∆ such a pullback is non-zero if and only if∩ ∆ is6 equidimensional.{ } Whence ±

c ∧2 c (dd h ) , ς = lim vol (Λ) υ ′ d h ς Γ 1 EΛ Γ hh ii ε→0 Xε ∧ ∧ Λ∈B(Γ,1) ∆∈B(Γ,2) Z Λ(∆) X ΛX≺∆

c lim vol (Λ) υ ′ h d ς 1 EΛ Γ − ε→0 Xε ∧ Λ∈B(Γ,1) ∆∈B(Γ,2) Z Λ(∆) X ΛX≺∆

c = vol (Λ) υ ′ d h ς 1 EΛ Γ K∆∩Supp ς ∧ ∧ ∆∈BXed(Γ,2) Λ∈BX(∆,1) Z c vol (Λ) υ ′ h d ς . 1 EΛ Γ − K∆∩Supp ς ∧ ∆∈BXed(Γ,2) Λ∈BX(∆,1) Z

55 Now, for every ∆ (Γ, 2), by Lemma 9.3 and Lemma 9.4 ∈Bed

c vol (Λ) υ ′ d h ς 1 EΛ Γ K∆∩Supp ς ∧ ∧ Λ∈BX(∆,1) Z

c = vol (Λ)υ ′ d h ς 1 EΛ Γ K∆∩Supp ς  ∧  ∧ Z Λ∈BX(∆,1)   =2̺(∆) vol 2(∆) υE′ ς ∆ ∧ ZK∆∩Supp ς =2̺(∆) vol (∆) λ , ς , 2 hh ∆ ii whereas by Lemma 9.2

c vol (Λ) υ ′ h d ς 1 EΛ Γ K∆∩Supp ς ∧ Λ∈BX(∆,1) Z

c = vol (Λ)υ ′ h d ς 1 EΛ Γ K∆∩Supp ς   ∧ Z Λ∈BX(∆,1) =0 .  

It follows that

c ∧2 (dd hΓ) =2 ̺(∆) vol 2(∆)λ∆ =2 ̺(∆) vol 2(∆)λ∆ (83) ∆∈BXed(Γ,2) ∆∈BX(Γ,2) c ∧2 and, consequently, Supp (dd hΓ) = Σ2, which completes the proof.

Corollary 9.1. Let A , A (Cn) be such that A A is convex. Then 1 2 ∈ K 1 ∪ 2 c ∧2 c ∧2 c ∧2 c ∧2 (dd hA1 ) + (dd hA2 ) = (dd hA1∪A2 ) + (dd hA1∩A2 ) . (84)

In particular ddch ddch = ddch ddch . (85) A1 ∧ A2 A1∪A2 ∧ A1∩A2 Proof. Since the space of 4-currents is Hausdorﬀ, by virtue of Theorem 2.2, it is enough to show that, for every hyperplane H Cn and any polytope Γ, ⊂ c ∧2 c ∧2 c ∧2 c ∧2 (dd hΓ∩H ) + (dd hΓ) = (dd hΓ∩H+ ) + (dd hΓ∩H− ) . (86)

Remark that all the 2-dimensional faces of Γ H are also 2-faces of Γ H+ and Γ H−. For each such ∩ ∩ ∩ + common 2-face ∆ one has a disjoint union K∆,Γ∩H+ K∆,Γ∩H− = K∆,Γ∩H . Moreover, Γ H and Γ H− may respectively have 2-faces ∆ and ∆ such∪ that ∆ ∆ is a 2-face of Γ not included∩ in ∩ 1 2 1 ∪ 2 + H. In this case, ̺(∆1)= ̺(∆2)= ̺(∆1 ∆2), vol 2(∆1) + vol 2(∆2) = vol 2(∆1 ∆2) and K∆1,Γ∩H = − ∪ + − ∪ K∆2,Γ∩H = K∆1∪∆2,Γ. Any other 2-face of Γ H and Γ H which does not intersect H is also a 2-face of Γ. The preceding analysis and Theorem∩ 9.2 imply (∩86). Moreover, by (66)

c c ∧2 c c ∧2 (dd hA1 + dd hA2 ) = (dd hA1∪A2 + dd hA1∩A2 ) , i.e.

(ddch )∧2 + 2 ddch ddch + (ddch )∧2 A1 A1 ∧ A2 A2 = (ddch )∧2 + 2 ddch ddch + (ddch )∧2 , A1∪A2 A1∪A2 ∧ A1∩A2 A1∩A2 which, thanks to (84), yields (85).

9.3 The case k > 2.

c ∧k For k> 2, the valuation property of the current (dd hA) can be proved by adapting the argument used in the case k = 2, however the following corollary shows that the general case is indeed a consequence of the case k =2.

56 Corollary 9.2. Let A , A (Cn) be such that A A is convex. Then 1 2 ∈ K 1 ∪ 2 c ∧k c ∧k c ∧k c ∧k (dd hA1 ) + (dd hA2 ) = (dd hA1∪A2 ) + (dd hA1∩A2 ) . (87)

c Proof. We prove the statement by induction on k. For the sake of notation let us set a = dd hA1 , c c c b = dd hA2 , c = dd hA1∪A2 and d = dd hA1∩A2 . By(66) and Corollary 9.1, we know that a + b = c + d, a∧2 + b∧2 = c∧2 + d∧2 and a b = c d (whence (a b)∧ℓ = (c d)∧ℓ, for every ℓ N). By induction, suppose that, for any 1 s <∧ k, one∧ has a∧s + b∧s =∧c∧s + d∧s.∧ It then follows, for∈1 ℓ k 1, that ≤ ≤ ≤ − a∧(k−ℓ) b∧ℓ + a∧ℓ b∧(k−ℓ) ∧ ∧ (a b)∧ℓ a∧(k−2ℓ) + b∧(k−2ℓ) if ℓ < k ℓ = ∧ ∧ − (a b)∧(k−ℓ) a∧(2ℓ−k) + b∧(2ℓ−k) if ℓ > k ℓ ( ∧ ∧ − (c d)∧ℓ c∧(k−2ℓ) + d∧(k−2ℓ) if ℓ < k ℓ = ∧ ∧ − (c d)∧(k−ℓ) c∧(2ℓ−k) + d∧(2ℓ−k) if ℓ > k ℓ ( ∧ ∧ − = c∧(k−ℓ) d∧ℓ + c∧ℓ d∧(k−ℓ) . ∧ ∧ The preceding computation show that

k−1 k a∧(k−ℓ) b∧ℓ ℓ ∧ Xℓ=1 (k/2)−1 k ∧k/2 k ∧(k−ℓ) ∧ℓ ∧ℓ ∧(k−ℓ) (a b) + a b + a b , if k 2 0  k/2 ∧ ℓ ∧ ∧ ≡ ℓ=1 = ⌊k/2⌋ X  k  a∧(k−ℓ) b∧ℓ + a∧ℓ b∧(k−ℓ) , if k 1 ℓ ∧ ∧ ≡2  ℓ=1  X  (k/2)−1  k ∧k/2 k ∧(k−ℓ) ∧ℓ ∧ℓ ∧(k−ℓ) (c d) + c d + c d , if k 2 0  k/2 ∧ ℓ ∧ ∧ ≡ ℓ=1 = ⌊k/2⌋ X  k  c∧(k−ℓ) d∧ℓ + c∧ℓ d∧(k−ℓ) , if k 1 ℓ ∧ ∧ ≡2  Xℓ=1 k−1  k =  c∧(k−ℓ) d∧ℓ , ℓ ∧ Xℓ=1 then, by (a + b)∧k = (c + d)∧k, it follows that a∧k + b∧k = c∧k + d∧k.

n Corollary 9.3. Let m 1,...,n and, if m

gm(A)= Qn(A[m], Am+1,...,An) , is a continuous, translation invariant and unitarily invariant valuation. In particular Pn is a continuous, translation invariant and unitarily invariant valuation.

c ∧m Proof. By Corollary 9.2, the mapping A (dd hA) is a valuation, then so is the mapping A (ddch )∧m ddch ... ddch . It7→ follows that g is a valuation too. Continuity as well7→ A ∧ Am+1 ∧ ∧ An m as translation and unitary invariance follow from the analogous properties of Qn. In particular, by choosing m = n one gets the statement for Pn.

n Lemma 9.5. Let Γ (C ), ∆ (Γ, k) and Λ ed(∆, k 1). Then ∆ is equidimensional if and only if E′ E = 0∈. P ∈ B ∈ B − Λ ∩ ∆ { } ′ ⊥ Proof. If EΛ E∆ = 0 , the outer normal vector uΛ,∆, spanning E∆ EΛ , does not belong to ′ ∩ { }′ ∩ EΛ. Observe that dimR E∆ = k 1 since otherwise linC E∆ would have odd real dimension. Since ′ 6 −′ ⊥ ′ ′ ⊥ ′ EΛ linC EΛ linC E∆ = E∆ E∆ = EΛ (E∆ EΛ ) E∆, it follows that EΛ (E∆ EΛ ) E∆, ⊂ ⊆′ ⊕ ′ ⊕ ∩ ′ ⊕ ′ ⊆ ⊥ ∩ ′ ⊕ whence dimR E∆ = k 2 or dimR E∆ = k. If dimR E∆ = k 2, then EΛ = (E∆ EΛ ) E∆, but this − ′ − ∩ ⊕ ′ means that uΛ,∆ belongs to EΛ, which is contrary to our assumption. It follows that dimR E∆ = k, i.e. ∆ is equidimensional. On the other hand, suppose E′ E = 0 . As E′ E⊥ and since u spans Λ ∩ ∆ 6 { } Λ ⊂ Λ Λ,∆

57 E⊥ E , it follows that u E′ E . Let v ,...,v an orthonormal basis of E , by Lemma 4.1 Λ ∩ ∆ Λ,∆ ∈ Λ ∩ ∆ 1 k−1 Λ the vectors t ,...,t deﬁned by t = iv k−1 Re iv , v v , for 1 ℓ k 1, provide a basis of E′ . 1 k−1 ℓ ℓ − s=1 h ℓ si s ≤ ≤ − Λ Since u E′ , we have u = k−1 c t , for some non trivial choice of coeﬃcients c ,...,c R. Λ,∆ Λ Λ,∆ ℓ=1 ℓ ℓP 1 k−1 This representation∈ yields the equality ∈ P k−1 k−1 k−1 u + c Re iv , v v = c iv , Λ,∆ ℓ h ℓ si s ℓ ℓ s=1 Xℓ=1 X Xℓ=1 where the vector on the left hand side belongs to E∆ and that on the right hand side belongs to iEΛ 0 . If a denotes such a vector, then ia E iE E iE , whence ̺(∆) = 0. \{ } ∈ Λ ∩ ∆ ⊂ ∆ ∩ ∆ Lemma 9.6. Let ∆ (Cn) be an equidimensional oriented k-polytope. Then ∈P

∗ ι ̺(Λ) vol k−1(Λ)υE′ =0 , (88) ∆  Λ  Λ∈BX(∆,k−1)   where ι : K Cn is the inclusion mapping and every Λ (∆, k 1) is oriented by the unit outer ∆ ∆ → ∈B − normal vector uΛ,∆.

Proof. Since ∆ is equidimensional, any of its facets is such. As ∆ is oriented each facet gets the orientation corresponding to the outer unit normal vector. For every Λ (∆, k 1), we have E⊥ = ∈ B − ∆ u⊥ E⊥ and by (35), on E⊥, ̺(Λ)υ ′ = ( 1)(k−1)(k−2)/2υ . Since ι∗ υ = ι∗ ( d Re iu , ), Λ,∆ Λ Λ EΛ iEΛ ∆ iEΛ ∆ Λ,∆ by linearity,∩ the statement is a consequence of− (6). Observe that changing the orientation∗ ofh ∆ changes−i ∗ ∗ the orientation of each of its facets, so that ι∆υiEΛ = ι∆ ( d Re iuΛ,∆, ), however this change of sign does not aﬀect the equality (88). − ∗ h −i

Lemma 9.7. Let Γ (Cn) and ∆ (Γ, k). If ∆ is oriented, then ∈P ∈Bed

∗ c ι ̺(Λ) vol k−1(Λ)υE′ d hΓ ∆  Λ ∧  Λ∈BX(∆,k−1)   ∗ = ι ̺(Λ) vol k−1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, , ∆  Λ ∧ h −i Λ∈BX(∆,k−1)   where ι : K Cn is the inclusion and every Λ (∆, k 1) is oriented by the unit outer normal ∆ ∆ → ∈ B − vector uΛ,∆.

Proof. Let Λ (∆, k 1) be fixed. Recall that E⊥ affR Λ= p and E⊥ affR ∆= p . Then, ∈B − Λ ∩ { Λ} ∆ ∩ { ∆} by virtue of (8), pΛ = p∆ + h∆(uΛ,∆)uΛ,∆ and

c υE′ d hΓ = υE′ Re ipΛ, Λ ∧ Λ ∧ h −i = υE′ Re ip∆, + h∆(uΛ,∆)υE′ Re iuΛ,∆, , Λ ∧ h −i Λ ∧ h −i By Lemma 9.6,

∗ ι ̺(Λ) vol k−1(Λ)υE′ Re ip∆, ∆  Λ ∧ h −i Λ∈BX(∆,k−1)   ∗ ∗ = ι ( Re ip∆, ) ι ̺(Λ) vol k−1(Λ)υE′ =0 ∆ − h −i ∧ ∆  Λ  Λ∈BX(∆,k−1)   whence

∗ ∗ c ι ̺(Λ) vol k−1(Λ)υE′ ι d hΓ ∆  Λ ∧ Λ  Λ∈BX(∆,k−1)   ∗ = ι ̺(Λ) vol k−1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, ∆  Λ ∧ h −i Λ∈BX(∆,k−1)   which proves the lemma.

58 Lemma 9.8. Let Γ (Cn) and ∆ (Γ, k). If ∆ is oriented and Λ (∆, k 1) is oriented by ∈ P ∈ Bed ∈ B − the unit outer normal vector uΛ,∆, then ∗ ι ̺(Λ) vol k−1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, ∆ Λ ∧ h −i = ̺(∆) vol (Λ)h (u )υ ′ . k−1 ∆ Λ,∆ E∆ In particular

∗ ι ̺(Λ) vol k−1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, ∆  Λ ∧ h −i Λ∈BX(∆,k−1) = k̺(∆) vol (∆)υ ′ .  k E∆

′ ⊥ ⊥ ∗ ∗ Proof. By (35), on E E E we have ι ̺(Λ)υE′ = ι (υiE ), then ∆ ⊂ ∆ ⊂ Λ ∆ Λ ∆ Λ ∗ ι ̺(Λ)υE′ Re iuΛ,∆, ∆ Λ ∧ h −i (k−1)(k−2)/2 ∗ = ( 1) ι (υiEΛ Re iuΛ,∆, ) − ∆ ∧ h −i = ( 1)(k−1)(k−2)/2( 1)(k−1)ι∗ (Re iu , υ ) − − ∆ h Λ,∆ −i ∧ iEΛ = ( 1)k(k−1)/2ι∗ (v ) − ∆ iE∆ = ̺(∆)v ′ . E∆ As ∆ is oriented, each of its facets gets the orientation induced by the outer unit normal vector and, by (7), it follows that

∗ ι ̺(Λ) vol k−1(Λ)h∆(uΛ,∆)υE′ Re iuΛ,∆, ∆  Λ ∧ h −i Λ∈BX(∆,k−1)   vol k−1(Λ)h∆(uΛ,∆) = k̺(∆) vE′  k  ∆ Λ∈BX(∆,k−1) = k̺(∆) vol (∆)v ′ .  k E∆ The proof is thus complete. Theorem 9.3. Let Γ (Cn) and 1 < k n, then ∈P ≤ c ∧k (dd hΓ) = k! ̺(∆) vol k(∆)λ∆ . (89) ∆∈BX(Γ,k) c ∧k In particular Supp (dd hΓ) = Σk.

Proof. We prove the theorem by induction on k. The case k =1 and k =2 have been already proved by Theorem 9.1 and Theorem 9.2, respectively. By induction, suppose that (ddch )∧(k−1) = (k 1)! ̺(Λ) vol (Λ)λ . Γ − k−1 Λ Λ∈BX(Γ,k−1) Then, for any (n k,n k)-test form ς, − − (ddch )∧k, ς hh Γ ii = ddc h (ddch )∧(k−1) , ς hh Γ Γ ii c ∧(k−1) c = hΓ(dd hΓ) , dd ς hh ii = (k 1)! ̺(Λ) vol (Λ) λ ,h ddcς − k−1 hh Λ Γ ii Λ∈BedX(Γ,k−1) c = (k 1)! ̺(Λ) vol (Λ) υ ′ h dd ς k−1 EΛ Γ − KΛ ∧ Λ∈BedX(Γ,k−1) Z c = (k 1)! lim ̺(Λ) vol k−1(Λ) υE′ hΓdd ς ε→0 Λ − KΛ∩Supp ς\(Σk)ε ∧ ∆∈BedX(Γ,k−1) Z c = (k 1)! lim ̺(Λ) vol k−1(Λ) υE′ hΓdd ς , ε→0 Λ − SΛ,ε ∧ ∆∈BedX(Γ,k−1) Z

59 where S = (K (K Supp ς (Σ ) ) , i.e. the (ε/2)-neighbourhood of K Supp ς (Σ ) . Λ,ε Λ ∩ Λ ∩ \ k ε ε/2 Λ ∩ \ k ε Since υ ′ is a closed (k 1)-form on E′ E⊥ and as the (2n 2k + 2)-forms dh dcς and dch dς EΛ Λ Λ Γ Γ have the same (n k +1−,n k + 1)-parts,⊂ it follows that on− the manifold S − ∧ ∧ − − Λ,ε c c k−1 c d(vE′ hΓd ς) = dυE′ hΓd ς + ( 1) υE′ d(hΓd ς) Λ ∧ Λ ∧ − Λ ∧ k−1 c k−1 c =0+( 1) υE′ dhΓ d ς + ( 1) υE′ hΓdd ς − Λ ∧ ∧ − Λ ∧ k c k−1 c = ( 1) υE′ d hΓ dς + ( 1) υE′ hΓdd ς . − Λ ∧ ∧ − Λ ∧ Observe that, on K , the 1-form dch has constant coeﬃcients, so the k-form υ ′ dch is closed on Λ Γ EΛ Γ k c c ∧ KΛ and ( 1) υE′ d hΓ dς = d(υE′ d hΓ ς). It follows that − Λ ∧ ∧ Λ ∧ ∧ k c c c ( 1) υE′ hΓdd ς = d(υE′ d hΓ ς) d(υE′ hΓd ς) , − Λ ∧ Λ ∧ ∧ − Λ ∧ whence

(ddch )∧k, ς hh Γ ii k c = (k 1)! lim ( 1) ̺(Λ) vol k−1(Λ) υE′ d hΓ ς (90) ε→0 Λ − − relbd SΛ,ε ∧ ∧ Λ∈BedX(Γ,k−1) Z k c (k 1)! lim ( 1) ̺(Λ) vol k−1(Λ) υE′ hΓd ς . (91) ε→0 Λ − − − relbd SΛ,ε ∧ Λ∈BedX(Γ,k−1) Z Let us first consider the term (91). For any fixed equidimensional (k 1)-face Λ such that Supp ς − ∩ relbd KΛ = ∅, the only parts of relbd SΛ,ε which really count are the flat parts parallel and near to 6 c the (2n k)-dimensional pieces of relbd KΛ on which hΓd ς and υE′ are linearly independent. Such − Λ (2n k)-dimensional pieces of relbd KΛ are nothing but the cones which are dual to the k-faces ∆ of − ε Γ admitting Λ as a facet. Let us denote XΛ(∆) the subset of relbd SΛ,ε parallel and near to K∆ and ε observe, by the way, that X approaches K∆ Supp ς as ε 0. The form υE′ may vanish on K∆, Λ(∆) ∩ → Λ indeed this happens precisely when E′ admits a non-zero vector that is orthogonal to K E⊥. Since Λ ∆ ⊂ ∆ the ortho-complement of E⊥ is E , it follows that the pullback of υ ′ to K is zero if and only if ∆ ∆ EΛ ∆ E′ E = 0 , which, by Lemma 9.5, means that ∆ is not equidimensional. Λ ∩ ∆ 6 { } It follows that (91) becomes

k−1 c ( 1) (k 1)! lim ̺(Λ) vol (Λ) υ ′ h d ς . (92) k−1 EΛ Γ − − ε→0 Xε ∧ Λ∈Bed(Γ,k−1) ∆∈Bed(Γ,k) Z Λ(∆) X ΛX≺∆

Let us now ﬁx ∆ ed(Γ, k) and give it an orientation. This provides each of its facets Λ (∆, k 1) the orientation induced∈B by the outer unit normal vector u E⊥ E . In particular, if v ∈B,...,v − Λ,∆ ∈ Λ ∩ ∆ Λ,1 Λ,k−1 spans EΛ and u , v ,...,v , ( 1)(k−1)k/2w ,...,w Λ,∆ Λ,1 Λ,k−1 − ∆,1 ∆,k is the positive basis of linC E∆ deﬁned in Remark 4.2, up to completing it to a positive basis

u , v ,...,v , ( 1)(k−1)k/2w ,...,w ,b ,...,b Λ,∆ Λ,1 Λ,k−1 − ∆,1 ∆,k 2k+1 2n of Cn, it is clear that the basis

v ,...,v , ( 1)k−1u , ( 1)(k−1)k/2w ,...,w ,,b ,...,b Λ,1 Λ,k−1 − Λ,∆ − ∆,1 ∆,k 2k+1 2n is positive too and the subspace E⊥ is oriented by the vector ( 1)k−1u . On the other hand Xε Λ − Λ,∆ Λ(∆) has the orientation coming from relbd SΛ,ε, this orientation corresponds to the unit normal vector to Xε pointing outside K Supp ς (Σ ) , i.e. u , so that the orientation of Xε is ( 1)k times Λ(∆) Λ ∩ \ k ε − Λ,∆ Λ(∆) − that of K∆ as a subset of KΛ. Thus the expression (92) becomes

c ( 1)(k 1)! ̺(Λ) vol (Λ) υ ′ h d ς k−1 EΛ Γ − − K∆∩Supp ς ∧ ∆∈BXed(Γ,k) Λ∈BedX(∆,k−1) Z

c = ( 1)(k 1)! ̺(Λ) vol (Λ)υ ′ h d ς . k−1 EΛ Γ − − K∆∩Supp ς   ∧ ∆∈BXed(Γ,k) Z Λ∈BedX(∆,k−1)  

60 By Lemma 9.6, the latter term is zero. We now turn our attention to the term (90). Again, for any fixed equidimensional (k 1)-face Λ, ∗ c − if Supp ς relbd KΛ = ∅, the only parts of relbd SΛ,ε on which ι (υE′ d hΓ ς) gives a non-zero ∩ 6 Λ Λ ∧ ∧ contribution are the flat pieces on which ι∗ ς = 0 and ι∗ (υ ′ dch ) is a non-zero k-form. The first Λ Λ EΛ Γ condition produces the same conclusions as in6 the case of the∧term (91), i.e. the integral in (90) has ε to be performed on the subsets of the form XΛ(∆). The second condition requires that ∆ has to be an ∗ c ′ equidimensional k-face and ι (υE′ d hΓ) a volume form on E . So Λ Λ ∧ ∆

k c (k 1)! lim ( 1) ̺(Λ) vol (Λ) υ ′ d h ς k−1 EΛ Γ − ε→0 − Xε ∧ ∧ Λ∈Bed(Γ,k−1) ∆∈Bed(Γ,k) Z Λ(∆) X ΛX≺∆

c = (k 1)! ̺(Λ) vol (Λ) υ ′ d h ς k−1 EΛ Γ − K∆∩Supp ς ∧ ∧ ∆∈BXed(Γ,k) Λ∈BedX(∆,k−1) Z

c = (k 1)! ̺(Λ) vol (Λ)υ ′ d h ς . k−1 EΛ Γ − K∆∩Supp ς  ∧  ∧ ∆∈BXed(Γ,k) Z Λ∈BX(Λ,k−1)   ∗ c By Lemma 9.8, ι ̺(Λ) vol k−1(Λ)υE′ d hΓ = k̺(∆) vol k(∆)vE′ , hence ∆ Λ∈B(∆,k−1) Λ ∧ ∆

cP ∧k (dd hΓ) = k! ̺(∆) vol k(∆)λ∆ = k! ̺(∆) vol k(∆)λ∆ . ∆∈BXed(Γ,k) ∆∈BX(Γ,k) c ∧k Of course, Supp (dd hΓ) = Σk and the proof is thus complete.

Corollary 9.4. Let 1 k n, Γ ,..., Γ, k (Cn) and Γ= k Γ . Then ≤ ≤ 1 ∈P ℓ=1 ℓ ddch ... ddch = k! ̺(∆)VP(∆ ,..., ∆ )λ , (93) Γ1 ∧ ∧ Γk k 1 k ∆ ∆∈BX(Γ,k) k where, for every ∆ (Γ, k) and any 1 ℓ k, ∆ℓ 4 Γℓ is such that ∆ = ℓ=1 ∆ℓ. In particular c ∈ Bc ≤ ≤ Supp dd hΓ ... dd hΓ = Σk,Γ. 1 ∧ ∧ k P k Proof. Let t1,...,tk > 0. By applying Theorem 9.3 to the polytope Γ= ℓ=1 tℓΓℓ one gets

∧k P k k e c e tℓdd hΓℓ = k! ̺(∆) vol k tℓ∆ℓ λ∆ , ! e e ! Xℓ=1 ∆∈BX(Γ,k) Xℓ=1 e k where ∆ = ℓ=1 tℓ∆ℓ is the representation of ∆ as a sum of faces ∆ℓ 4 Γℓ, for 1 ℓ k. If k k ≤ ′≤ ′ Γ= Γ and ∆= ∆ , with the same face summands as ∆, then E e = E whence E e = E , ℓ=1 ℓ ℓ=1 ℓ ∆ ∆ ∆ ∆ e P e ̺(∆) = ̺(∆) and υE′ = υE′ . By comparing the coeﬃcients of the monomial t1 ...tk on both sides one P ∆e P ∆ obtains e k e c 2 k! dd hΓℓ = (k!) ̺(∆)Vk (∆1,... ∆k) λ∆ , ℓ^=1 ∆∈BX(Γ,k) whence ddch ... ddch = k! ̺(∆)V (∆ ,... ∆ ) λ . Γ1 ∧ ∧ Γk k 1 k ∆ ∆∈BX(Γ,k) The statement about the support of ddch ... ddch is a consequence of Theorem 9.3. Γ1 ∧ ∧ Γk Remark 9.3. For k = n, Corollary 9.4 yields a proof of Corollary 8.1 not depending on Theorem 8.1. Indeed, in the notation of Corollary 9.4, for every ∆ (Γ,n) one has E′ = E⊥, so if ς is a ∈ Bed ∆ ∆ m

61 regularization of χB2n 1 Qn(Γ1,..., Γn) = lim ̺(∆)Vn (∆1,... ∆n) λ∆, ςm m→∞ κn hh ii ∆∈BXed(Γ,n) 1 = lim ̺(∆)Vn (∆1,... ∆n) ςmυE⊥ m→∞ ∆ κn K∆ ∆∈BXed(Γ,n) Z 1 = ̺(∆)Vn (∆1,... ∆n) υ ⊥ E∆ κn K∆∩B2n ∆∈BXed(Γ,n) Z 1 = ̺(∆)Vn (∆1,... ∆n) vol n(K∆ B2n) κn ∩ ∆∈BXed(Γ,n) = ̺(∆)Vn (∆1,... ∆n) ψΓ(∆) ∆∈BXed(Γ,n) ̺ = Vn (Γ1,..., Γn) .

By continuity, the same conclusion holds true for arbitrary convex bodies.

c ∧k c ∧ℓ 9.4 The current (dd hΓ) ∧ (dd hB2n )

c ∧k c ∧ℓ In the present section we will show that the current (dd hΓ) (dd hB2n ) is an absolutely continuous ∧ c ∧k measure with respect to the (measure) coeﬃcients of the current (dd hΓ) . In order to do so, we c ∧k will show that the representation of the current (dd hΓ) given in the preceding section yields a corresponding representation of (ddch )∧k (ddch )∧ℓ. Γ ∧ B2n n ℓ If Γ (C ) is ﬁxed, for every 1 k n, 1 ℓ (n k) and any ∆ ed(Γ, k), let µ∆ be the (k +∈ℓ, P k + ℓ)-current of measure type≤ on≤ Cn acting≤ ≤ on any− continuous and∈ compactly B supported (n k ℓ,n k ℓ)-form ς as − − − −

ℓ ∗ ′ c ∧ℓ µ , ς = ι υE (dd hB n ) ς , (94) hh ∆ ii ∆ ∆ ∧ 2 ∧ ZK∆ where ι : K Cn is the inclusion, υ ′ = ι∗ (Re w , ... Re w , ) and w ,...,w is an ∆ ∆ E∆ ∆ 1 k 1 k → ′ h −i ∧ ∧ h −iℓ orthonormal basis of E∆. Let us point out a few remarks about the current µ∆.

ℓ The current µ∆ is a kind of restriction to the manifold K∆ of the current induced by the form • ′ c ∧ℓ 0 υE (dd hB n ) . Of course, µ = λ∆. ∆ ∧ 2 ∆ ℓ The definition of µ∆ makes sense if and only if ∆ is equidimensional. Indeed, the degree of the • ∗ ′ c ∧ℓ form ι (υE (dd hB n ) ς) equals the dimension of K∆ if and only if ∆ is equidimensional. ∆ ∆ ∧ 2 ∧ ′ ⊥C n ⊥ ′ ⊥C For any equidimensional face ∆ of Γ, one has E∆ E∆ E∆ = C and K∆ E∆ = E∆ E∆ , • ′ ⊕ ⊥C⊕ ⊂ ⊕ so the orientation of K∆ comes from that of E∆ E∆ . Since the mutually orthogonal C-linear ′ ⊥C ⊕ ⊥ subspaces E∆ E∆ and E∆ are positively oriented, it follows that the orientation of E∆ merely ⊕ ′ depends on the orientation of E∆. As shown in Remark 4.2, choosing an orientation on E∆ implies a corresponding choice of orientation on E′ yielding the positive orientation on E E′ . The ∆ ∆ ⊕ ∆ volume form υ ′ involved in the definition of the current µℓ conforms to the chosen orientation E∆ ∆ ′ on E∆. This way the integral in (94) is not affected by the choice of such a basis and the current ℓ µ∆ is thus well defined. c ∧ℓ −ℓ The coefficients of the form (dd hB2n ) ς are O( z ), as z 0. Since ℓ n k dimR K∆ • 1=2n k 1, the integral involved in (∧94) is convergent.k k → ≤ − ≤ − − − c c The only terms of (dd h )∧ℓ ς giving a non zero contribution to ι∗ (υ ′ (dd h )∧ℓ ς) are B2n ∆ E∆ B2n • ∧ ⊥C ∧ ∧ those involving the differentials of the coordinate functions of E∆ , nevertheless the coefficients of such terms do depend on the full array of coordinates of K E⊥. ∆ ⊂ ∆ The support of the current µℓ is the closure of K , but it concentrated on K . • ∆ ∆ ∆ Corollary 9.5. Let Γ (Cn) and 1 k,ℓ n such that k + ℓ n. Then ∈P ≤ ≤ ≤ λ (ddch )∧ℓ = µℓ . (95) ∆ ∧ B2n ∆

62 In particular (ddch )∧k (ddch )∧ℓ = k! ̺(∆) vol (∆)µℓ . (96) Γ ∧ B2n k ∆ ∆∈BXed(Γ,k)

c ∧ℓ Proof. Both the currents λ∆ and (dd hB2n ) are positive, so their wedge product is legitimate. Since n hB2n is not diﬀerentiable on the whole C , let us consider, for ε> 0, the regularisation hB2n ϕε. For any (n k ℓ,n k ℓ)-test form ς, ∗ − − − − c ∧ℓ c ∧ℓ λ∆ (dd hB ) , ς = lim λ∆ (dd (hB ϕε)) , ς hh ∧ 2n ii ε→0hh ∧ 2n ∗ ii ∗ c ∧ℓ = lim ι v ′ (dd (h ϕ )) ς (97) ∆ E∆ B2n ε ε→0 K ∧ ∗ ∧ Z ∆ ∗ ′ c ∧ℓ = ι vE (dd hB n ) ς (98) ∆ ∆ ∧ 2 ∧ ZK∆ = µℓ , ς , hh ∆ ii where the equality (98) is due to the dominated convergence in (97). The equality (96) is a consequence of (95) and Theorem 9.3.

Remark 9.4. A second longer proof of (96) can be carried on by induction on ℓ by using the results of the preceding section. Moreover (96) with ℓ = n k can be used to obtain an alternative proof of Theorem 8.1. −

10 Examples

n Example 10.1. The n-pseudovolume of the unit full-dimensional ball B2n of C . As ρ (z)= 1+ z 2 = 1+ n z z¯ , we have ρ (z) =2 z =2 for every z ∂B , so B2n − k k − ℓ=1 ℓ ℓ k∇ B2n k k k ∈ 2n P n n ∧(n−1) in 1 P (B )= z d¯z z¯ dz dz d¯z n 2n n!κ 2 ℓ ℓ ℓ ℓ ℓ ℓ n ∂B2n − ! ∧ ∧ ! Z Xℓ=1 Xℓ=1 n ∧n in = dz d¯z n!κ ℓ ℓ n B2n ∧ ! Z Xℓ=1 2n = υ κ 2n n ZB2n 2nκ = 2n κn 2nΓ (1/2)nΓ (1 + (n/2)) = . n! Table 1 collects the values of P (B ) for 1 n 10. n 2n ≤ ≤ Table 1: Pseudovolume of full-dimensional unit balls of Cn.

n 1234 5 6 7 8 9 10 2 3 3 4 4 5 5 2 4π π 8π π 16π π 32π Pn(B n) π 2π π 2 3 2 15 6 105 24 945

The computation through the Levi form is easier. Indeed K 1 on ∂B , so ∂B2n ≡ 2n n−1 n 2 (n 1)! 2 κ2n Pn(B2n)= − 2nκ2n = . n!κn κn Finally the computation of P (B ) as a convex body is possible too. In fact, h (z)= z so n 2n B2n k k ∂2 z (ddch )∧n =4nn! det k k υ , B2n ∂z ∂z¯ 2n ℓ k

63 and Lemma 11.1 yields ∂2 z det k k =2−(n+1) z −n . ∂z ∂z¯ k k ℓ k By using (100), it follows that 1 1 (ddch )∧n = 2n−1 z −n υ n!κ B2n κ k k 2n n ZB2n n ZB2n n−1 1 2π 2n−2 π 2 n−1 2n−ℓ−1 = ρ dρ dϑ2n−1 (sin ϑℓ) dϑℓ κn 0 0 0 Z Z ℓY=1 Z n 2n−2 π/2 2 π 2n−ℓ−1 = 2 (sin ϑℓ) dϑℓ nκn 0 ℓY=1 Z 2n−2 2nπ Γ (2n ℓ)/2 = Γ (1/2) − nκn Γ (2n ℓ + 1)/2 ℓ=1 Y − 2nπ πn−1 = nκ (n 1)! n − 2nκ = 2n , κn just as expected. Since 2/κ1 = 1, it follows that the surface area κ2 of the unit disk in C equals its n 1-dimesional pseudovolume. However 2 /κn > 1 for n> 1, so in higher dimension the 2n-dimensional volume κ2n of B2n is strictly smaller that its n-dimensional pseudovolume. Moreover Pn(B2n) increases with n for n 6, then it decreases to 0 as n + . ≤ → ∞ Example 10.2. The n-pseudovolume of odd-dimensional unit balls B Cn. 2n−1 ⊂ n If E C is a real hyperplane, the intersection E B2n is a (2n 1)-unit ball B2n−1 in E. Up to a ⊂ n ∩ n − unitary transformation of C , we may suppose that E = z C Re z1 =0 . In this case { ∈ | } 2 n z1 z¯1 2 h − (z)= − + z . B2n 1 v 2i | ℓ| u ℓ=2 u X t If n =1, B = [ i,i] C then, by (57), P (B )=2. For n> 1, Lemma 11.2 shows that 1 − ⊂ 1 1 2 n 2 ∂ hB2n−1 ℓ=2 zℓ det = | n|+2 , ∂z ∂z¯ (2h − ) ℓ k PB2n 1 which, by virtue of (99), in polar coordinates becomes

2 2 2 2 ρ (sin ϑ1) (sin ϑ2) (sin ϑ2) n+2 n+2 n+2 = n+2 n n . 2 ρ (sin ϑ1) 2 ρ (sin ϑ1)

It follows that Pn(B2n−1) equals

2 2n−2 1 n (sin ϑ2) 2n−1 2n−ℓ−1 4 n! n+2 n n ρ (sin ϑℓ) dρ dϑ n!κn B2n 2 ρ (sin ϑ1) ∧ Z ℓY=1 i.e. n−2 2n−2 2 n−1 n−2 2n−1 2n−ℓ−1 ρ (sin ϑ1) (sin ϑ2) (sin ϑℓ) dρ dϑ , κn B2n ∧ Z ℓY=3 where, for the sake of notation, dϑ = dϑ ... dϑ . By(100) 1 ∧ ∧ 2n−1 π Γ (1/2)Γ ((n 1)/2) (sin ϑ )n−2dϑ = − , 1 1 Γ (n/2) Z0 π Γ (1/2)(n 1)! (sin ϑ )2n−1dϑ = − , 2 2 Γ (n + (1/2)) Z0 2n−2 π 2n−4 2n−ℓ−1 Γ (1/2) sin ϑℓdϑℓ = , 0 (n 2)! ℓY=3 Z −

64 so that 2n−22πΓ (1/2)2n−2Γ ((n 1)/2)(n 1)! P (B )= − − n 2n−1 κ nΓ (n/2)Γ (n + (1/2))(n 2)! n − 2n−2Γ (1/2)nΓ ((n 1)/2)(n 1) = − − . Γ (n + (1/2))

The table 2 collects the values of Pn(B2n−1) and Pn(B2n), for 1 n 10. An easy proof by induction ∗ ≤ ≤ shows that Pn(B2n−1) < Pn(B2n) for every n N . Moreover, Pn(B2n−1) increases with n for n 6, then it decreases to 0, as n + . ∈ ≤ → ∞

Table 2: n-pseudovolume of B2n−1 and B2n.

n 1234 5 6 7 8 9 10 4π 32π 32π2 1024π2 256π3 16384π3 2048π4 1048576π4 16384π5 Pn(B n− ) 2 2 1 3 15 35 945 693 45045 19305 11486475 692835 2 3 3 4 4 5 5 2 4π π 8π π 16π π 32π Pn(B n) π 2π π 2 3 2 15 6 105 24 945

Example 10.3. The n-pseudovolume of the standard 2n-cube of Cn. n Consider the standard 2n-cube I2n C . I2n is the cartesian product of n copies of the convex subset conv ( 1+ i, 1+ i, 1 i, 1 i ) ⊂ C. As follows from Example 2.2, the n-th component of the face { − − − − } ⊂ vector f(I2n) is given by 2n f (I )=2n . n 2n n 2n 2n In R there are n n-dimensional coordinate subspaces and the n-faces of I2n are organized into 2n groups, each group consisting of faces parallel to a same real n-dimensional coordinate subspace of R2n Cn. As shown in Example 4.3, of such real subspaces only 2n are equidimensional (and in ≃ n fact real similar), so that the cardinal number of ed(I2n,n) is 4 . For each ∆ ed(I2n,n), one has n −n B n ∈ B ̺(∆) = 1, vol n(∆) = 2 and ψΓ(∆) = 2 , whence Pn(I2n)=4 . Notice that Pn(I2n) = vol 2n(I2n), quite an uncommon circumstance!

Example 10.4. The 2-pseudovolume of the standard unit 4-crosspolytope of C4. Consider the standard unit 2n-crosspolytope Θ Cn. As shows Example 2.3, the n-th component of 2n ⊂ the face vector f(Θ2n) is given by 2n f (Θ )=2n+1 . n 2n n +1 In the case n =2, the 4-crosspolytope Θ4 has only equidimensional 2-faces, 16 of which are real similar whereas the remaining ones have a coeﬃcient of area distortion equal to 2/3. The outer angle of any 2-face equals 1/6 and the area of any 2-face is √3/2, so that P2(Θ4) = (20/9)√3. Observe by the way that the 3-dimensional crosspolytope, (i.e. the octahedron) Θ C Re C has 8 real similar facets, so 3 ⊂ × that P2(Θ3)=2√3.

Example 10.5. The 2-mixed pseudovolume of B3 and B4.

The computation of the derivatives of hB4 and hB3 carried on in the proofs of Lemma 11.1 and Lemma 11.2 show that ddch ddch equals B4 ∧ B3 1 − ( z 2 +2 z 2)(2h2 z 2)+ z 2(2 z 2 + z 2 (Re z )2) 4h3 h3 1 2 B3 2 2 1 2 1 B4 B3 | | | | − | | | | | | | | − times dz d¯z dz d¯z which in polar coordinates becomes 1 ∧ 1 ∧ 2 ∧ 2 2ρ sin ϑ sin ϑ + ρ sin ϑ (sin ϑ )3 + ρ(sin ϑ )3(sin ϑ )3 ρ(sin ϑ )3(sin ϑ )5 1 2 1 2 1 2 − 1 2 times dρ dϑ dϑ dϑ . By using (100), it follows easily that Q (B ,B ) = 248/45. ∧ 1 ∧ 2 ∧ 3 2 4 3

65 Example 10.6. If n> 1, the Kazarnovskiˇı n-dimensional pseudovolume is not orthogonally invariant. n n n 2n Let In R C be the standard unit n-cube. We know that Pn(In) = vol n(In)=2 . If F : R 2n ⊂ ⊂ → R is the orthogonal mapping considered in Remark 4.5, it is easy to realize that the image of F (In) n C via F is a n-cube in C such that E = 0 , so that Pn(F (In)) = 0. In particular, the present F (In) 6 { } example implies that the 1-form α∂A cannot be orthogonally invariant.

Example 10.7. The n-dimensional Kazarnovskiˇımixed psudovolume is not rational on lattice polytopes. Given n polytopes Γ ,..., Γ Cn whose vertices have coordinates in the ring of Gauss’ integers, then 1 n ⊂ n!Qn(Γ1,..., Γn) is, generally, not an integer number and (2n)!Qn(Γ1,..., Γn) need not be either. If n = 1, consider the polytope Γ C with vertices 1, 1, i, i; then 1!P1(Γ) = 2√2 / Z and 2!P1(Γ) = 4√2 / Z. The example can be generalized⊂ to Cn. − − ∈ ∈

11 Useful constructions and computations

We collect here some constructions and computations which have been used in the preceding sections.

11.1 Polar coordinates

Consider the following coordinate transformation in Rm

ℓ−1 ξ = ρ cos ϑ , ξ = ρ(cos ϑ ) sin ϑ , if 2 ℓ m 1 1 1 ℓ ℓ j ≤ ≤ − j=1 Y and m−1 ξm = ρ sin ϑj , j=1 Y where ρ (0, + ), ϑm−1 (0, 2π) and ϑℓ (0, π), for every 1 ℓ m 2. The corresponding jacobian is given by∈ ∞ ∈ ∈ ≤ ≤ − m−2 m−1 m−ℓ−1 ρ (sin ϑℓ) . ℓY=1 Observe that, for every 1 k m, ≤ ≤ 2 m k−1 ξ2 = ρ sin ϑ . (99) ℓ  j  j=1 Xℓ=k Y   Recall also that, for every n N, the n-th Wallis’ integral can be computed via Euler’s gamma function as follows ∈ π Γ (1/2)Γ ((n + 1)/2) (sin ϑ)ndϑ = . (100) Γ ((n + 2)/2) Z0 11.2 Hessians

Lemma 11.1. For every n N∗ and every z Cn 0 , det HessC h =2−(n+1) z −n. ∈ ∈ \{ } B2n k k Proof. An easy computation shows that, for every 1 ℓ, k n, ≤ ≤ ∂2h 2 z 2δ z z¯ B2n = k k ℓ,k − k ℓ , ∂z ∂z¯ 4 z 3 ℓ k k k so that we are led to show that

2n 3n n−1 2n 2 z det HessC h =2 z . (101) k k B2n k k 2 2 Consider the matrix M = (2 z δℓ,k zkz¯ℓ)ℓ,k and observe that it admits the eigenvalue λ1 = 2 z . Indeed, the matrix M λ I k =k ( z −z¯ ) has rank 1 because, for every 1 k n, the k-th columnk k − 1 n − k ℓ ℓ,k ≤ ≤

66 t equals zk (¯z1,..., z¯n). It follows that the dimension of the corresponding eigenspace is n 1. A direct − 2 − computation shows that the other eighenvalue of M is λ2 = z . Indeed, let aℓ denote the ℓ-th row of M λ I and suppose z = 0, then a = a ( z /z k).k Since the columns a , with ℓ = k, are − 2 n k 6 k ℓ6=k ℓ − ℓ k ℓ 6 linearly independent, it follows that M λ2In has rank n 1. − P − Lemma 11.2. For every n N 0, 1 and every z z Cn Re z =0 ∈ \{ } ∈{ ∈ | 6 } n 2 zℓ ℓ=2 C | | det Hess hB2n−1 = n+2 . (2XhB2n−1 )

Proof. We already know that

2 n z1 z¯1 2 h − (z)= − + z B2n 1 v 2i | ℓ| u ℓ=2 u X t and an easy computation shows that, for every 2 ℓ, k n, ≤ ≤ 2 2 ∂ hB − 2hB − δℓ,k zkz¯ℓ 2n 1 = 2n 1 − , ∂z ∂z¯ 4h3 ℓ k B2n−1

2 2 ∂ hB − (¯z z )¯z ∂ hB − (z z¯ )z 2n 1 = 1 − 1 ℓ , 2n 1 = 1 − 1 k , ∂z ∂z¯ 8h3 ∂z ∂z¯ 8h3 ℓ 1 B2n−1 1 k B2n−1 n 2 z 2 ∂ hB − ℓ 2n 1 = ℓ=2 | | , ∂z ∂z¯ 4h3 1 1 X B2n−1 so it’s enough to show that

n n 3n n−2 2 n−1 2 4 h det HessC hB − =2 (h ) zℓ . B2n−1 2n 1 B2n−1 ℓ=2 | | 3 2 X C − Let Mn =4hB2n−1 Hess hB2n 1 . Observe that λ1 = 2hB2n−1 is an eigenvalue of Mn, in fact, for every 2 ℓ n, the ℓ-th row of M λ I is equal to ≤ ≤ n − 1 n z¯ z z¯ 1 − 1 ,z ,...,z , − ℓ 2 2 n i.e. a multiple of v = ((¯z1 z1)/2,z2,...,zn). Since none of the last n 1 rows is identically zero on B 0 and as the ﬁrst− row of M λ I is not a multiple of v, it follows− that M λ I has rank 2n−1 \{ } n − 1 n n − 1 n 2, i.e. Mn admits a diagonal form with λ1 occurring on the last n 2 diagonal elements. The remaining eigenvalues are −

2 z1 z¯1 2 z1 z¯1 λ2 = h hB − − and λ3 = h + hB − − . B2n−1 − 2n 1 2i B2n−1 2n 1 2i

Indeed, for j =2, 3, the ﬁrst row of the matrix Mn λj In is a linear combination of the remaining ones, (which are all non zero). For j =2, 3 and 2 ℓ −n, the coeﬃcient of the the ℓ-th row in such a linear combination is ≤ ≤

−1 −1 z1 z¯1 z1 z¯1 iz h − + − or iz h − − , ℓ B2n 1 2i − ℓ B2n 1 − 2i respectively. It follows that

2 2 n−2 2 2 z1 z¯1 det Mn = (2h ) h h − B2n−1 B2n−1 B2n−1 − 2i " # n n−2 2 n−1 2 =2 (h ) zℓ B2n−1 ℓ=2 | | X as required.

67 11.3 Linear groups

For every n N∗, let M(n, C), (resp. M(2n, R)), denote the set of n-by-n matrices (resp. 2n-by-2n matrices) with∈ entries in C (resp. R). As usual GL(n, C)= M M(n, C) det M =0 , { ∈ | 6 } GL(2n, R)= M M(2n, R) det M =0 , { ∈ | 6 } U(n)= M M(n, C) M tM = I , { ∈ | n} O(2n)= M M(2n, R) M tM = I , { ∈ | n} SO(2n)= M O(2n) det M =1 . { ∈ | } Lemma 11.3. The mapping ψ : M(1, C)= C M(2, R) deﬁned, for m C, by 1 → ∈ Re m Im m ψ (m)= , 1 Im m −Re m has the following properties for every m,m′ C: ∈ ′ ′ 1. ψ1(mm )= ψ1(m)ψ1(m ), 2. det ψ (m)= m 2, 1 | | t 3. ψ1(m)= ψ1(m).

In particular, ψ1(U(1)) = SO(2, R).

Proof. The proof is straightforward. For the last statement, observe that, if m U(1), then ∈ t ψ1(m) ψ1(m)= ψ1(m)ψ1(m)= ψ1(mm)= ψ1(1) = I2 so that ψ1(m) SO(2, R), i.e. ψ1(U(1)) SO(2, R). Conversely, an element from SO(2, R) can always be written as ∈ ⊆ cos θ sin θ , sin θ −cos θ for some θ R, and such a matrix is nothing but ψ (eiθ). ∈ 1 Lemma 11.4. If n N 0, 1 , the mapping ψn : M(n, C) M(2n, R) deﬁned, for M = (mℓj ) M(n, C), by ∈ \{ } → ∈

ψ1(m11) ... ψ1(m1n) . . . ψn(M)=  . .. .  , ψ (m ) ... ψ (m )  1 n1 1 nn    has the following properties for every M,M ′ M(n, C): ∈ ′ ′ 1. ψn(MM )= ψn(M)ψn(M ), 2. det ψ (M)= det M 2, n | | t t 3. ψn(M)= ψn( M).

In particular ψ (U(n)) SO(2n, R), but the inclusion is strict. n ⊂ ′ ′ Proof. The equality ψn(MM )= ψn(M)ψn(M ) is straightforward. If Re M = (Re mℓ,j) and Im M = (Im mℓ,j ), then M = Re M + i Im M and

det ψn(M) Re m Im m ... Re m Im m 11 − 11 1n − 1n ......  . . . . .  − n(n 1) Re mn1 Im mn1 ... Re mnn Im mnn = ( 1) 2 det  − −  −  Im m11 Re m11 ... Im m1n Re m1n     . . . . .   ......     Im m Re m ... Im m Re m   n1 n1 nn nn  Re M Im M  = det , Im M −Re M

68 whence Re M Im M Re M + i Im M Im M + i Re M det = det Im M −Re M Im M − Re M Re M + i Im M 0 = det Im M Re M i Im M − so that M 0 det = (det M)(det M) Im M M = (det M)(det M) = det M 2 . | | Also t t ψ1(m11) ... ψ1(mn1) t . . . ψn(M)=  . .. .  tψ (m ) ... tψ (m )  1 1n 1 nn    ψ1(m11) ... ψ1(mn1) . . . =  . .. .  ψ (m ) ... ψ (m )  1 1n 1 nn   t  = ψn( M) .

t t t so that M U(n) implies I2n = ψn(In)= ψn(M M)= ψn(M)ψn( M)= ψn(M) ψn(M), i.e. ψn(M) ∈ 2 ∈ O(2n, R). Since M is unitary, det ψn(M) = det M = 1, then ψn(M) SO(2n, R). If A M(2, R) denotes the matrix | | ∈ ∈ 0 1 A = , 1 0 when n = 2k, the matrix with 2k diagonal blocks equal to A provides an element of SO(4k, R) \ ψ2k(U(2k)). If n =2k +1, the matrix with 2k diagonal blocks equal to A and a block equal to I2, yields an element of SO(4k +2, R) ψ (U(2k + 1)). \ 2k+1 ∗ 2n Remark that, for every n N , ψn(iIn) is the matrix of the operator of R deﬁning the usual complex structure. ∈

12 Open questions

In this section we present some open questions about Qn as well as some partial answers. The open questions are essentially three.

1. Non vanishing (or non degeneracy) condition for Qn: prove or disprove Corollary 5.2 in the case of arbitrary convex bodies;

2. Find monotonicity conditions for Qn;

3. Prove or disprove Alexandroﬀ-Fenchel inequality for Qn.

We discuss the questions of the preceding list following the same enumeration.

12.1 Non vanishing condition

In the polytopal case a non vanishing condition has been already stated in [Ka1] and proved in [Ka10]; our Corollary 5.2 gives an alternative proof. This non vanishing condition in the polytopal case is necessary and suﬃcient. However, in the general case, it is just suﬃcient and it is not known if it is also necessary.

Lemma 12.1. Let A ,...,A (Cn) and A = A + ... + A their Minkowski sum. If dimC A

69 (m) Proof. For 1 ℓ n, let (Γℓ )m be a sequence of polytopes with vertices belonging to ∂Aℓ that ≤ ≤ (m) (m) converges to Aℓ for the Hausdorﬀ metric. As Γℓ Aℓ, it follows that, for every m N, dimC(Γ1 + (m) ⊆ m m ∈ ... +Γn ) dimC A 0 for every m N, but n 1 ∈ (m) (m) + lim Qn(Γ ,..., Γ )= Qn(A1,...,An)=0 . m→∞ 1 n

12.2 Monotonicity

The question of monotonicity of Qn (with respect to inclusion in each argument) is also very interesting and it seems related to non degeneracy and to the valuation property. Let us start with a general monotonicy statement.

n Lemma 12.2. Let A1,...,Am,K2,...,Kn (C ) and A = A1 + ... + Am. For every 1 ℓ m and 1 k n, ∈ K ≤ ≤ ≤ ≤ m Q (A [k],K ,...,K ) Q (A [k],K ,...,K ) (102) n ℓ k+1 n ≤ n j k+1 n j=1 X Q (A[k],K ,...,K ) . (103) ≤ n k+1 n In particular, if 0 A A then A A and ∈ ℓ ∩ ℓ ⊆ Q (A [k],K ,...,K ) Q (A[k],K ,...,K ) . n ℓ k+1 n ≤ n k+1 n Proof. The last statement is a monotonicity result and is a straightforward consequence of the (102), so it is enough to prove the latter inequalities. By direct computation: 1 Q (A [k],K ,...,K )= (ddch )∧k ddch ... ddch n ℓ k+1 n n!κ Aℓ ∧ Kk+1 ∧ ∧ Kn n ZBn 1 m (ddch )∧k ddch ... ddch ≤ n!κ Aj ∧ Kk+1 ∧ ∧ Kn n Bn j=1 Z X m = Qn(Aj [k],Kk+1,...,Kn) j=1 X and m Qn(Aj [k],Kk+1,...,Kn) j=1 X 1 m = (ddch )∧k ddch ... ddch n!κ Aj ∧ Kk+1 ∧ ∧ Kn n Bn j=1 Z X m 1 k! c ∧kℓ c c (dd hAℓ ) dd hKk+1 ... dd hKn ≤ n!κn Bn k1! km! ∧ ∧ ∧ Z 0≤k1,...,km≤k ··· ℓ=1 k1+...X+km=k ^ 1 = (ddch + ... + ddch )∧k ddch ... ddch n!κ A1 Am ∧ Kk+1 ∧ ∧ Kn n ZBn 1 = (ddch )∧k ddch ... ddch n!κ A ∧ Kk+1 ∧ ∧ Kn n ZBn = Qn(A[k],Kk+1,...,Kn) . The lemma is thus proved.

Observe that, by symmetry, Lemma 12.2 is still valid in any of the arguments of Qn. The following Example 12.1 shows that without further assumptions, Lemma 12.2 seems the only general monotonicity result for convex bodies in Cn.

70 Example 12.1. The Kazarnovskiˇı n-dimensional pseudovolume is, generally, not monotonically in- cresing. Consider the the square

∆ = conv 2+2i, 2+2i, 2 2i, 2 2i C = C 0 C2 { − − − − }⊂ ×{ }⊂ and let, for every λ> 0, Γλ = conv ∆; (0, 2λ) C2. The polytope Γλ is a 3-dimensional pyramid on the base ∆ with apex in the point (0{, 2λ). Let}⊂ also

Kλ = conv (i, 0), ( i, 0), (i, λ), ( i, λ) C2 , { − − }⊂ then Kλ Γλ, for every λ> 0. Figure 12 depicts both Kλ and Γλ when λ =1/4. ⊂

Re z2

Im z1

Re z1

λ λ Figure 12: The rectangle K and the pyramid Γ for λ = 1/4.

λ λ λ It is very easy to show that P2(K )=2λ. In order to compute P2(Γ ) notice that Γ has ﬁve 2-dimensional faces, namely

∆ = conv (2+2i, 0), ( 2+2i, 0), (0, 2λ) , 1 { − } ∆ = conv (2 2i, 0), ( 2 2i, 0), (0, 2λ) , 2 { − − − } ∆ = conv (2+2i, 0), (2 2i, 0), (0, 2λ) , 3 { − } ∆ = conv ( 2+2i, 0), ( 2 2i, 0), (0, 2λ) . 4 { − − − } λ and ∆5 = ∆. The face ∆5 does not contribute to P2(Γ ) because ̺(∆5)=0, whereas

λ2 1 ̺(∆ )= , vol (∆ )=4 1+ λ2 , ψ(∆ )= , ℓ 1+ λ2 2 ℓ ℓ 2 p for every ℓ 1, 2, 3, 4 , so that ∈{ } 2 λ 8λ P2(Γ )= . √1+ λ2 λ λ This computation shows that P2(Γ ) < P2(K ) as soon as 0 <λ< 1/√15. The preceding example is a variation of one provided by B. Ya. Kazarnovskiˇı(in a private communication) and can be generalized to higher dimensions. Observe that dimR Kλ < dimR Γλ, i.e. Kλ is dimensionally smaller than Γλ, however it can have a bigger pseudovolume than Γλ.

Under suitable assumptions, special monotonicy results can be obtained by using the valuation property. The following lemma is an example of this kind.

Lemma 12.3. Let A (Cn) such that dimR A =2n 1 and let H Cn a real aﬃne hyperplane such C ∈ K − ⊂+ − that EA EH = EA. Then Pn(A H)=0 and Pn(A) = Pn(A H )+ Pn(A H ). In particular P (A) ∩max P (A H+), P (A ∩H−) . ∩ ∩ n ≥ { n ∩ n ∩ }

71 Proof. The hyperplane H cannot be included in aﬀR A, since otherwise (for dimensional reasons) C it should coincide with aﬀR A, then EA = EH and EH = EA = EA, which is impossible because C dimR EH =2n 1 whereas dimR EA =2n 2. The convex body A H has zero pseudovolume. In fact, any sequence of− polytopes with vertices on− the relative boundary∩ of A H and converging to A H has the property that each polytope of the sequence has a complex dimension∩ that is smaller than∩ n. By Corollary 5.1, Corollary 8.1 and Remark 8.2, it follows that Pn(A H)=0. Thanks to Corollary 9.3, P (A)= P (A H+)+ P (A H−), then P (A) max P (A H∩+), P (A H−) . n n ∩ n ∩ n ≥ { n ∩ n ∩ } + − Remark 12.1. By Lemma 12.3, one gets Pn(A) > 0 as soon as Pn(A H ) > 0 or Pn(A H ) > 0. Unless A is a polytope, the latter inequalities are hardly checked, so Lemma∩ 12.3 does not∩ imply any non vanishing criterion for Pn(A).

12.3 Alexandroﬀ-Fenchel inequality

An Alexandroﬀ-Fenchel inequality for Kazarnovskˇıi mixed pseudovolume would read as follows:

Q (A , A , A ,...,A )2 Q (A [2], A ,...,A )Q (A [2], A ,...,A ) (104) n 1 2 3 n ≥ n 1 3 n n 2 3 n for every A , A , A ,...,A (Cn). 1 2 3 n ∈ K Although I currently have no proof of (104), I do believe it should hold true. Indeed two possible proofs seem natural:

one inspired by Alexandroﬀ’s second proof of (16), [A2]; • the other inspired by Ewald’s proof of (16), [Ewa2]. •

Both approaches require approximation of A1, A2, A3,...,An, by smooth strictly convex bodies in the first case and by strictly combinatorially isomorphic polytopes in the second one. The first way could benefit from the the theory of mixed discriminants of hermitian matrices and should involve a variation of Hilbert’s parametrix. A first problem to face in this first approach deals with (50). Indeed, by writing (14) as

Vn(A1,...,An)= Dn(hA1 In, HessR hA2 ,..., HessR hAn )υ∂Bn , (105) Z∂Bn a comparison between (105) and (50) immediately shows that, unlike hA1 In, the matrix M in (50) is not a multiple of In, which complicates the adaption of Hilbert’s method to the case of mixed pseudovolume. The second idea would involve Alexandroﬀ-Fenchel inequality for mixed volume and could eventually hold true even for the more general case of mixed ϕ-volumes in the polytopal case. If ϕ is continuous, the usual continuity argument would possibly yield the result for arbitrary convex bodies. In particular such a result would imply the Alexandroﬀ-Fenchel inequality for mixed pseudovolume. The feasibility of both attempts need further analysis and the question is deferred to a forthcoming paper.

72 List of notation

Notation Meaning Page

χA Characteristic function of the set A, χ(x)=1 if x A and χ(x)=0 if x / A 5 ( , ) Standard scalar product in Rn ∈ ∈ 5 affR A Affine subspace spanned by the subset A of Rn or Cn over R 5 EA Linear subspace parallel to affR A 5 relint A Relative interior of A 5 relbd A Relative boundary of A 5 n n k(R ) Grassman manifold of k-dimensional linear subspaces of R 5 G(Rn) Topological sum of (Rn) 5 G Gk E⊥ Orthocomplement of the linear subspace E Rn with respect to the standard 5 scalar product ( , ) ⊂ (Rn) Convex subsets of Rn 5,24 K(Rn) Non empty compact subsets of Rn 5 C (Rn) Convex polytopes of Rn 5,24 P dA Real dimension of EA 5 [v, w] Segment with v and w as endpoints 5 hA Support function of the convex subset A 5 n Bn Full-dimensional unit ball in R about the origin 5 HA(v) Supporting hyperplane of the convex subset A in the direction v 5 4 , , < , Order relations for faces 5 (Γ)≺ ≻ Boundary complex of the polytope Γ 6 B(Γ, k) Set of k-dimensional faces of the polytope Γ 6 fB(Γ) Face vector of the polytope Γ 6 u∆k−1,∆k Outer unit normal vector to the face ∆k−1 7 K∆,Γ Dual cone to the face ∆ of the polytope Γ 8 Σk,Γ k-star of the polytope Γ 9 ψΓ(∆) Outer angle of the face ∆ 4 Γ with respect to Γ 9 ℓ vol ℓ ℓ-dimensional Lebesgue measure in R 9 n κℓ ℓ-dimensional Lebesgue measure of the ℓ-dimensional closed unit ball Bℓ in R 9 Σ(Γ) Normal fan of the polytope Γ 10 (A)ε ε-neighborhood of A 10 ∞(Rn) Neighborhoods of polytopes in Rn 10,24 P ∆n Standard n-simplex 10 In Standard n-cube 10 Θn Standard n-crosspolytope 11 Subd hA Subdifferential of hA 11 k(Rn) k-regular subsets of Rn 11 S∞(Rn) Smooth subsets of Rn 11 S ρA Defining function of A 11 k(Rn) k-regular convex subsets of Rn 11 K∞(Rn) Smooth convex subsets of Rn 24 K n n 1(R ) Strictly convex subsets of R 11 K∞ n n 1 (R ) Smooth strictly convex subsets of R 11,24 K n ℓ n ℓ(R ) Convex bodies with (R 0 ) support function 11,24 Kk n C \{ }ℓ n ℓ (R ) k-regular convex bodies with (R 0 ) support function 11,24 hK Haussdorf metric C \{ } 12 RεA ε-regularization of A 12 ϕε Cut-off function 12 n υn Standard volume n-form on R 12 A[k] Sequence of k subsets equal to A 13 vk k-th intrinsic volume 13 Vn n-dimensional (Minkowski) mixed volume 13

υ∂Bn Standard volume form on ∂Bn 14 HessR h Real Hessian matrix of h 14

73 Notation Meaning Page

Dn−1 Mixed discriminant 14 vϕ k k-th intrisic ϕ-volume 15 ϕ Vk k-dimensional mixed ϕ-volume 15 ̺(E1, E2) Coefficient of volume distortion under the projection of E1 on E2 19 dimR E Dimension of the R-linear subspace E Cn 19 ⊆ dimC E Complex dimension of the C-linear subspace linC E Cn 19 EC Maximal complex subspace of the R-linear subspace⊆E Cn 19 E⊥ Orthocomplement of the linear subspace E Cn with respect⊆ to the 19 standard scalar product Re , ⊆ E⊥C Orthocomplement of the linearh i subspace E Cn with respect to the 19 standard hermitian product , ⊆ h i E′ E⊥ linC E 19 (Cn, R) Set∩ of R-linear subspaces of Cn 19 ̺G(E) Coefficient of volume distortion under the projection of E on iE′ 20 4 Equivalence modulo 4 20 ≡ n υE Volume form on the R-linear subspace E C 27 (Cn) Convex subsets of Cn ⊂ 24 K(Cn) Convex polytopes of Cn 24 P(Cn) Smooth compact subsets of Cn 24 S ′ ̺(A) Coefficient of volume distortion under the projection of EA on iEA 24 ed(Γ, k) Set of k-dimensional equidimensional faces of the polytope Γ 24 B n υ2n Standard volume 2n-form on C 26 dc i(∂¯ ∂) 26 ddc The− Bott-Chern operator 2i∂∂¯ 26 Hodge -operator 27 υ∗ Volume∗ form on the oriented real linear subspace E Cn 27 E ⊂ ν∂A Gauss map of ∂A 28 υ∂A Volume form on ∂A 28 α∂A Differential 1-form defined on ∂A 29 Pn n-dimensional Kazarnovskiˇıpseudovolume 29 II∂A Second quadratic form of ∂A 31 IIC II ∂A R-bilinear symmetric form associated to ∂A 32 ∂A Levi form of ∂A 33 KL Product of the eigenvalues of 32 ∂A L∂A Sn Symmetric group over the set 1,...,n 37 σ { } mix (M1,...,Mn) mixed version of matrices 37 mixσ(M1,...,Mn) mixed version of matrices 37 HessC h Complex hessian matrix of h 38 Dn Mixed discriminant 37 M [j,k] sub-matrix of M 38 T Current 43 ς Test form 43 T, ς Pairing between the current T and the testform ς 43 hh ii λ∆ (k, k)-current associated to the equidimensional k-face ∆ of a poly- 45 tope Γ ι Inclusion mapping K Cn 45 ∆ ∆ →

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