Let $\Gamma\subset\mathbb C^n$ be a convex polytope and let $h_\Gamma(z)=\max_{v\in\Gamma}{\rm Re}\langle z,v\rangle$ be its support function with respect to the standard scalar product on $\mathbb C^n$ obtained as the real part of the standard hermitian one.

If ${\rm d}^c=i(\bar\partial-\partial)$, $B_m$ is the unit $m$-dimensional ball and $\varkappa_m$ its $m$-dimensional Lebesgue measure, Boris Yakovlevich Kazarnovskii, in the paper entitled "On the zeros of exponential sums" (Soviet Math. Dokl. Vol. 23, (1981), no. 2, 347-351), claims that the wedge product of positive currents ${\rm dd}^c h_\Gamma \wedge ({\rm dd}^c h_{B_{2n}})^{\wedge(n-1)}$ yields a positive measure $\mu_1$ such that $\mu_1(B_{2n})$ equals, up to a dimensional coefficient, the sum, as $\Delta$ runs in the set ${\mathcal B}(\Gamma,1)$ of sides of $\Gamma$, of $\varkappa_{2n-1}^{-1}vol_1(\Delta)vol_{2n-1}(K_\Delta\cap B_{2n})$, where $vol_1(\Delta)$ is the length of $\Delta$, $K_\Delta$ its dual cone and $vol_{2n-1}(K_\Delta\cap B_{2n})$ the $(2n-1)$-Lebesgue measure of $K_\Delta\cap B_{2n}$.

The paper provides a more general statement about the measure $\mu_k=({\rm dd}^c h_\Gamma)^{\wedge k} \wedge ({\rm dd}^c h_{B_{2n}})^{\wedge(n-k)}$ and, although I understand the reason why $\mu_k$ is a well defined positive measure, I am confused about the way to prove both statements.

I can prove that the value $\langle\!\langle {\rm dd}^c h_\Gamma,\varsigma\rangle\!\rangle$ of the current ${\rm dd}^c h_\Gamma$ on any $(n-1,n-1)$-test form is given by the following expression:

\begin{equation} \langle\!\langle {\rm dd}^c h_\Gamma,\varsigma\rangle\!\rangle=\sum_{\Delta\in{\mathcal B}(\Gamma,1)} vol_1(\Delta)\int_{K_\Delta} \iota^*_\Delta(\upsilon_{E_\Delta^\prime}\wedge \varsigma) \end{equation}

where $E_\Delta$ is the line issuing from the origin and parallel to $\Delta$, $E_\Delta^\prime=i E_\Delta$, $\upsilon_{E_\Delta^\prime}$ is the volume form on $E_\Delta^\prime$ and $\iota_\Delta:K_\Delta\to\mathbb C^n$ is the inclusion mapping.

I think that the preceding representation of ${\rm dd}^c h_\Gamma$ can be of some help in proving Kazarnovskii's claims, however my computation does not seem to give the expected answer.

Could anybody help me to understand how to prove Kazarnovskii's claim? Thank you in advance.

Edit: The general equality claimed by Kazarnovskii can be proved by a differential geometric argument (cf. https://arxiv.org/pdf/1910.03099.pdf Theorem 8.1), however I am still searching for an alternative proof using the theory of currents. Here is an attempt on which I am stuck. Thanks a lot to everybody will find time to look at my computation.

For the sake of simplicity, let us set $n=2$ and in $\mathbb C^2$ consider the coordinates $z_1=x_1+iy_1$ e $z_2=x_2+iy_2$. If the convex polytope $\Gamma\subset\mathbb C^2$ is not reduced to a single point, its support function $h_\Gamma(z)=\max_{u\in\Gamma}{\mathop{\rm Re\,}\nolimits}\langle z,u\rangle$ is continuous, convex and 1-homogeneous, however it is not differentiable. The support function of $B_4$ is just the euclidean norm and the wedge product of $\mathrm {dd}^{\mathrm c} h_\Gamma$ and $\mathrm {dd}^{\mathrm c} h_{B_4}$ is defined as \begin{equation} \langle\!\langle \mathrm {dd}^{\mathrm c} h_\Gamma\wedge\mathrm {dd}^{\mathrm c} h_{B_4},\varsigma \rangle\!\rangle = \langle\!\langle \mathrm {dd}^{\mathrm c}(h_{B_4}\mathrm {dd}^{\mathrm c} h_\Gamma),\varsigma \rangle\!\rangle = \langle\!\langle \mathrm {dd}^{\mathrm c} h_\Gamma,h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma \rangle\!\rangle.\tag{1} \end{equation} for any test function $\varsigma$ on $\mathbb C^2$. Since $h_\Gamma$ is convex, $\mathrm {dd}^{\mathrm c} h_\Gamma$ is a current of measure type, so its value against the continuous form with compact support $h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma$ is well defined. Such a form is not smooth because $h_{B_4}$ is singular at the origin. The current $\mu_1=\mathrm {dd}^{\mathrm c} h_\Gamma\wedge\mathrm {dd}^{\mathrm c} h_{B_4}$, is just a positive measure and in order to compute $\mu_1(B_4)$ there are at least two different ways. The first one would involve a convolution of $h_\Gamma$ and $h_{B_4}$ with some regularizing kernel $\varrho_\epsilon$ thus getting $$ \mu_1(B_4)=\lim_{\varepsilon\to0}\int_{B_4}\mathrm {dd}^{\mathrm c}(h_\Gamma*\varrho_\varepsilon)\wedge \mathrm {dd}^{\mathrm c}(h_{B_4}*\varrho_\varepsilon). $$ This first method seems quite cumbersome to me, but I am possibly wrong. A second possibility makes use of a regularization of the characteristic function of $B_4$ together with the representation \begin{equation}\label{c} \mathrm {dd}^{\mathrm c} h_\Gamma = \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,\lambda_\Delta\tag{2} \end{equation} (cf. [1] Theorem 9.1), where ${\mathcal B}(\Gamma,1)$ is the set of the edges of $\Gamma$ and, for every such edge $\Delta$, ${\mathop{\rm vol\,}\nolimits}_1\Delta$ is the length of $\Delta$, whereas $\lambda_\Delta$ is the positive current acting against a generic $(1,1)$-test form $\varphi$ as $$ \langle\!\langle \lambda_\Delta,\varphi \rangle\!\rangle = \int_{K_\Delta} \iota_\Delta^*(\upsilon_{E'_\Delta}\wedge\varphi). $$ Here $K_\Delta$ is the dual cone to the face $\Delta$ (i.e. the relatively open convex polyhedral cone spanned by the rays issuing from the origin and orthogonal to the affine hulls of the cells of $\Gamma$ admitting $\Delta$ as an edge), $\iota_\Delta:K_\Delta\to \mathbb C^2$ is the inclusion and $\upsilon_{E'_\Delta}$ is the volume form on the line $E'_\Delta=iE_\Delta$, where $E_\Delta$ denotes the line parallel to $\Delta$ and passing through the origin.

By virtue of a result of Kazarnovskii's (stated in [2] and proved in [1] Theorem 8.1) one has \begin{equation}\label{k} \mu_1(B_4)= 2 \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) {\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4)\,.\tag{3} \end{equation} The preceding formula reveals an interesting fact: $\mu_1(B_4)$ is invariant under orthogonal transformations of $\Gamma$ as a subset of $\mathbb R^4$. If $m\in\mathbb N^*$, let \begin{equation*} \varsigma_m(z)= \begin{cases} 1 & \textrm{if}\quad \Vert z\Vert \leq 1\,,\\ \exp\left(1+\dfrac{1}{m^2(1-\Vert z\Vert)^2-1}\right) &\textrm{if}\quad1< \Vert z\Vert <1+\dfrac{1}{m}\,,\\ 0 &\textrm{if}\quad \Vert z\Vert \geq 1+\dfrac{1}{m}\,. \end{cases} \end{equation*} The function $\varsigma_m$ is radial, smooth, its support is the full-dimensional ball of radius $1+(1/m)$ about the origin, it takes on the value 1 on $B_4$, $0\leqslant\varsigma_m\leqslant 1$ and it converges (point-wise) to the characteristic function of $B_4$. So \begin{equation} \mu_1(B_4) = \lim_{m\to\infty} \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m)\label{s}\tag{4} \end{equation} and in view of (4), it would be enough to prove that, for any $\Delta\in{\mathcal B}(\Gamma,1)$, \begin{equation} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c} \varsigma_m) = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\label{u}\tag{5} } \end{equation} Let $\Delta$ be fixed; up to an orthogonal trasformation, we may suppose that $E_\Delta=\{z\in\mathbb C^2\mid {\mathop{\rm Im\,}\nolimits} z_1=z_2=0\}$. In such a case, $K_\Delta\subset E_\Delta^\perp=\{z\in\mathbb C^2\mid x_1=0\}$, $\iota_\Delta^*(\upsilon_{iE_\Delta})={\mathrm d}y_1$ and \begin{align} &\iota_\Delta^*(\upsilon_{iE_\Delta}\wedge h_{B_4}\mathrm {dd}^{\mathrm c}\varsigma_m)\tag{6} \\ &= {\mathrm d}y_1\wedge \iota_\Delta^*\left(h_{B_4}2i\dfrac{\partial^2\varsigma_m}{\partial z_2\partial\bar z_2}{\mathrm d}z_2\wedge{\mathrm d}\bar z_2\right)\tag{7} \\ &= {\mathrm d}y_1\wedge\iota_\Delta^*\left(h_{B_4}2i\left[\dfrac{1}{2}\left(\dfrac{\partial}{\partial x_2}-i\dfrac{\partial}{\partial y_2}\right)\dfrac{1}{2}\left(\dfrac{\partial\varsigma_m}{\partial x_2}+i\dfrac{\partial\varsigma_m}{\partial y_2}\right)\right](-2i){\mathrm d}x_2\wedge{\mathrm d}y_2\right)\tag{8} \\ &= \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right]\tag{9} \end{align} so that (5) becomes \begin{equation}\label{w} { \lim_{m\to\infty} \int_{K_\Delta} \iota_\Delta^*\left[h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\right] = 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4).\tag{10} } \end{equation} In order to prove this equality let us pass to spherical coordinates in $E_\Delta^\perp$. Set $y_1=\rho\cos\vartheta_1$, $x_2=\rho\sin\vartheta_1\cos\vartheta_2$, $y_2=\rho\sin\vartheta_1\sin\vartheta_2$, with $\rho\geqslant0$, $\vartheta_1\in[0,\pi)$ and $\vartheta_2\in[0,2\pi)$. In these coordinates, on $K_\Delta$ one has $h_{B_4}=\rho$, \begin{equation} \dfrac{\partial \varsigma_m}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{\partial\rho}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_1} \dfrac{\partial\vartheta_1}{\partial x_2} + \dfrac{\partial\varsigma_m}{\partial \vartheta_2} \dfrac{\partial\vartheta_2}{\partial x_2} = \dfrac{\partial\varsigma_m}{\partial \rho} \dfrac{x_2}{\rho} = \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\cos\vartheta_2\tag{11} \end{equation} and similarly \begin{align} \dfrac{\partial \varsigma_m}{\partial y_2} &= \dfrac{\partial\varsigma_m}{\partial \rho} \sin\vartheta_1\sin\vartheta_2\tag{12} \end{align} whence \begin{equation} \dfrac{\partial^2 \varsigma_m}{\partial x_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\cos^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\cos^2\vartheta_2}{\rho} \right)\tag{13} \end{equation} and \begin{align} \dfrac{\partial^2 \varsigma_m}{\partial y_2^2} = \dfrac{\partial^2 \varsigma_m}{\partial \rho^2} \sin^2\vartheta_1\sin^2\vartheta_2 + \dfrac{\partial \varsigma_m}{\partial \rho} \left( \dfrac{1-\sin^2\vartheta_1\sin^2\vartheta_2}{\rho} \right)\tag{14} \end{align} so that \begin{align} & \lim_{m\to\infty} \int_{K_\Delta} h_{B_4}\left(\dfrac{\partial^2\varsigma_m}{\partial x_2^2}+\dfrac{\partial^2\varsigma_m}{\partial y_2^2}\right) {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge {\mathrm d}y_2\tag{15} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho \left[ \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^2\vartheta_1 + \dfrac{\partial \varsigma_m}{\partial \rho}\left(\dfrac{2-\sin^2\vartheta_1}{\rho}\right) \right] \rho^2\sin\vartheta_1{\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{16} \\ &= \lim_{m\to\infty} \int_{K_\Delta} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{1}\tag{17} \\ &+ \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{2}\tag{18} \\ &- \lim_{m\to\infty} \int_{K_\Delta} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\sin^3\vartheta_1 {\mathrm d}\rho\wedge{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\label{3}.\tag{19} \end{align} For computing the latter integrals, let us observe that the derivatives of $\varsigma_m$ vanish at every $\rho\in[0,1]\cup[1+(1/m),+\infty)$ and that $$ 0\leqslant\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\leqslant\left(1+\dfrac{1}{m}\right)\cdot1\cdot\left(1+\dfrac{1}{m}-1\right)=\dfrac{1}{m}+\dfrac{1}{m^2} $$ so \begin{equation}\label{z} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho\,\varsigma_m\,{\mathrm d}\rho=0\,.\tag{20} \end{equation} Now by Fubini' theorem, (17) equals \begin{equation} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{21} \end{equation} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} \rho^3 \dfrac{\partial^2 \varsigma_m}{\partial \rho^2}{\mathrm d}\rho &= \lim_{m\to\infty} \left[\rho^3\dfrac{\partial\varsigma_m}{\partial\rho}\right]_1^{1+(1/m)}-3\int_1^{1+(1/m)}\rho^2\dfrac{\partial\varsigma_m}{\partial\rho}\,{\mathrm d}\rho\tag{22} \\ &= \lim_{m\to\infty} -3\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}+6\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{23} \\ &= 3\,,\tag{24} \end{align} next (18) equals \begin{align} \lim_{m\to\infty} \int_{K_\Delta} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{25} \end{align} with \begin{align} \lim_{m\to\infty} \int_1^{1+(1/m)} 2\rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho &= \lim_{m\to\infty} 2\left[\rho^2\varsigma_m\right]_1^{1+(1/m)}-2\int_1^{1+(1/m)}\rho\,\varsigma_m\,{\mathrm d}\rho\tag{26} \\ &= -2\tag{27} \end{align} and finally (19) equals \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin^3\vartheta_1 {\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{28} \end{align} with \begin{align} \lim_{m\to\infty} -\int_1^{1+(1/m)} \rho^2 \dfrac{\partial \varsigma_m}{\partial \rho}\,{\mathrm d}\rho =1.\tag{29} \end{align} It follows that the left hand side of (10) is equal to \begin{align} \int_{K_\Delta\cap\partial B_4} (4\sin^3\vartheta_1-2\sin\vartheta_1){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{30} \end{align} whereas the right hand one equals \begin{align} 2{\mathop{\rm vol\,}\nolimits}_3(K_\Delta\cap B_4) &= 2\int_{K_\Delta\cap B_4} {\mathrm d}y_1\wedge{\mathrm d}x_2\wedge{\mathrm d}y_2\tag{31} \\ &= 2\int_0^1 \rho^2{\mathrm d}\rho \cdot \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\tag{32} \\ &= \dfrac{2}{3} \int_{K_\Delta\cap\partial B_4} \sin\vartheta_1{\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2\,,\tag{33} \end{align} hence (10) is equivalent to \begin{align} { \int_{K_\Delta\cap\partial B_4} \left(4\sin^3\vartheta_1-\dfrac{8}{3}\sin\vartheta_1\right){\mathrm d}\vartheta_1\wedge{\mathrm d}\vartheta_2 =0\label{x}\tag{34} } \end{align} where $K_\Delta\cap \partial B_4$ is a subset of the unit 2-dimensional sphere about the origin. In fact $\partial B_4$ is the unit 3-dimensional sphere about the origin, $K_\Delta\cap\partial B_4\subseteq E_\Delta^\perp\cap \partial B_4$ and $\dim E_\Delta^\perp\cap \partial B_4=2$, since $E_\Delta^\perp$ is a hyperplane transverse to $\partial B_4$.

By the arbitrary choice of $\Gamma$, equality (34) should always hold true, i.e. for every subset of the 2-sphere of the form $K_\Delta\cap \partial B_4$. The simplest case is when $\Gamma$ is a segment, so $\Gamma=\Delta$, $K_\Delta=E_\Delta^\perp$ and $K_\Delta\cap\partial B_4$ is the whole unit 2-sphere about the origin. The second simplest case is when $\Gamma$ is a polygon and, for each of its edges $\Delta$, $K_\Delta$ is a half-space and $K_\Delta\cap\partial B_4$ is a half of the unit 2-sphere about the origin. If $\Gamma$ is a polyhedron, for any of its edges $\Delta$, $K_\Delta\cap\partial B_4$ is a spherical polygon with two edges only, smaller than a hemisphere. If $\Gamma$ is a polychoron, for any of its edges $\Delta$, $K_\Delta$ is a 3-dimensional convex polyhedral cone with apex in the origin not including lines through the origin and $K_\Delta\cap\partial B_4$ is a spherical polygon.

However (34) does not seem to be correct in general. Indeed by Stokes' theorem (34) becomes \begin{align}\label{y} { \int_{\partial(K_\Delta\cap\partial B_4)} -\dfrac{4}{3} \sin^2\vartheta_1\cos\vartheta_1{\mathrm d}\vartheta_2 =0\,.\tag{35} } \end{align} and the latter equality does not hold true for the spherical triangle with vertices of spherical coordinates $A_0=(1,0,0)$, $A_1=(1,\pi/4,0)$ e $A_2=(1,\pi/4,\pi/2)$. In fact, the integral along $A_0A_1$ and $A_2A_0$ vanish, but that along $A_1A_2$ is strictly negative as $\vartheta_2$ increases from $0$ to $\pi/2$ and the function $(-4/3) \sin^2\vartheta_1\cos\vartheta_1$ is negative in the intevall $[\pi/4,\pi/2]$.


[1] J. Silipo Kazarnovskii pseudovolume, a clever use of support function https://arxiv.org/abs/1910.03099

[2] B. J. Kazarnovskii On the zeros of exponential sums Dokl. Akad. Nauk SSSR (Russian) 257 (4), 804–808 (1981).


1 Answer 1


I finally found a purely current theoretical proof of the equality \begin{equation*} \mu_1(B_{2n}) = 2^{n-1}(n-1)! \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_{1}(\Delta) {\mathop{\rm vol\,}\nolimits}_{2n-1}({K_\Delta\cap B_{2n-1}}). \end{equation*} Indeed, the following equality (cf. [1], Theorem 8.1) $$ \mathrm{dd}^{\mathrm c} h_\Gamma = \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,[K_\Delta]\wedge \upsilon_{E_\Delta^\prime}\,, $$ where $[K_\Delta]$ denotes the integration current on $K_\Delta$ and $\upsilon_{E_\Delta^\prime}$ is a conveniently chosen volume form on $E_\Delta^\prime=iE_\Delta$, implies that the trace measure $\nu_1$ of $\mathrm{dd}^{\mathrm c} h_\Gamma$ equals \begin{align*} &\dfrac{1}{4^{n-1}(n-1)!}\, \mathrm{dd}^{\mathrm c} h_\Gamma \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-1)} \\ &= \dfrac{1}{4^{n-1}(n-1)!} \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,[K_\Delta]\wedge \upsilon_{E_\Delta^\prime}\wedge\left(2i\sum_{\ell=1}^n \mathrm{d} z_\ell\wedge \mathrm{d}\bar z_\ell\right)^{\wedge(n-1)} \\ &= \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,[K_\Delta]\wedge \upsilon_{E_\Delta^\prime}\wedge \upsilon_{E_\Delta^{\perp_\mathbb C}} \\ &= \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \,[K_\Delta]\wedge \upsilon_{E_\Delta^\perp} \,, \end{align*} where $E_\Delta^{\perp_\mathbb C}$ and $E_\Delta^\perp$ are the orthocomplements of $E_\Delta$ with respect to the standard hermitian and scalar products $\langle\,,\rangle$ and $ \mathrm{Re}\langle\,,\rangle$ on $\mathbb C^n$, whereas $\upsilon_{E_\Delta^{\perp_\mathbb C}}$ and $\upsilon_{E_\Delta^\perp}=\upsilon_{E_\Delta^\prime}\wedge \upsilon_{E_\Delta^{\perp_\mathbb C}}$ are conveniently chosen volume forms on $E_\Delta^{\perp_\mathbb C}$ and $E_\Delta^\perp$, respectively. It follows that $$ \nu_1(B_{2n}) = \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) \int_{K_\Delta\cap B_{2n}} \upsilon_{E_\Delta^\perp} = \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_1(\Delta) {\mathop{\rm vol\,}\nolimits}_{2n-1}(K_\Delta\cap B_{2n})\,. $$ However, a regularization of $h_\Gamma$ and of the characteristic function of $B_{2n}$ together with Stokes' theorem imply that \begin{align*} \int_{B_{2n}} \mathrm{dd}^{\mathrm c} h_\Gamma \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-1)} &= \int_{\partial B_{2n}} \mathrm{dd}^{\mathrm c} h_\Gamma \wedge 2h_{B_{2n}} (\mathrm{d}^{\mathrm c} h_{B_{2n}})^{\wedge (n-2)} \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-2)} \\ &= 2 \int_{\partial B_{2n}} \mathrm{dd}^{\mathrm c} h_\Gamma \wedge \mathrm{d}^{\mathrm c} h_{B_{2n}} \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-2)} \\ &= 2 \int_{B_{2n}} \mathrm{dd}^{\mathrm c} h_\Gamma \wedge \mathrm{dd}^{\mathrm c} h_{B_{2n}} \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-2)}. \end{align*} An iteration of the same argument $(n-2)$ more times yields $$ \int_{B_{2n}} \mathrm{dd}^{\mathrm c} h_\Gamma \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-1)} = 2^{n-1} \int_{B_{2n}} \mathrm{dd}^{\mathrm c} h_\Gamma \wedge (\mathrm{dd}^{\mathrm c} h_{B_{2n}})^{\wedge (n-1)}\,, $$ so \begin{align*} \mu_1(B_{2n}) &= 2^{n-1}(n-1)! \nu_1(B_{2n}) \\ &= 2^{n-1}(n-1)! \sum_{\Delta\in{\mathcal B}(\Gamma,1)} {\mathop{\rm vol\,}\nolimits}_{1}(\Delta) {\mathop{\rm vol\,}\nolimits}_{2n-1}({K_\Delta\cap B_{2n-1}}) \end{align*} as claimed.

The case of the measure $\mu_k=(\mathrm{dd}^{\mathrm c} h_\Gamma)^{\wedge k} \wedge (\mathrm{dd}^{\mathrm c} h_{B_{2n}})^{\wedge (n-k)} $ is handled in the same way but it is based on the equality $$ (\mathrm{dd}^{\mathrm c} h_\Gamma)^{\wedge k} = k! \sum_{\Delta\in{\mathcal B}(\Gamma,k)} \varrho(\Delta) {\mathop{\rm vol\,}\nolimits}_{k}(\Delta) [K_\Delta]\wedge\upsilon_{E_\Delta^\prime}\,, $$ where ${\mathcal B}(\Gamma,k)$ is the set of $k$-dimensional faces of $\Gamma$ and for each such face, $\varrho(\Delta)$ is a weight ascribed to $\Delta$ (which equals $1$ if $k=1$), ${\mathop{\rm vol\,}\nolimits}_{k}(\Delta)$ is the $k$-dimensional Lebesgue measure of $\Delta$, $K_\Delta$ is the dual cone to the face $\Delta$, $E_\Delta^\prime=E_\Delta^\perp\cap (E_\Delta+i E_\Delta)$ and $\upsilon_{E_\Delta^\prime}$ is a conveniently chosen volume form on $E_\Delta^\prime$ (cf. [1], Theorem 8.3). Now the trace measure $\nu_k$ of $(\mathrm{dd}^{\mathrm c} h_\Gamma)^{\wedge k}$ is $$ \dfrac{1}{4^{n-k}(n-k)!} (\mathrm{dd}^{\mathrm c} h_\Gamma)^{\wedge k} \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-k)} = k! \sum_{\Delta\in{\mathcal B}(\Gamma,k)} \varrho(\Delta) {\mathop{\rm vol\,}\nolimits}_{k}(\Delta) [K_\Delta]\wedge\upsilon_{E_\Delta^\perp}\,, $$ so that $$ \nu_k(B_{2n}) = k! \sum_{\Delta\in{\mathcal B}(\Gamma,k)} \varrho(\Delta) {\mathop{\rm vol\,}\nolimits}_{k}(\Delta) {\mathop{\rm vol\,}\nolimits}_{2n-k}(K_\Delta\cap B_{2n}). $$ Just like in the case $k=1$, one shows that $$ \int_{B_{2n}} (\mathrm{dd}^{\mathrm c} h_\Gamma)^{\wedge k} \wedge (\mathrm{dd}^{\mathrm c} h^2_{B_{2n}})^{\wedge (n-k)} = 2^{n-k} \int_{B_{2n}} (\mathrm{dd}^{\mathrm c} h_\Gamma)^{\wedge k} \wedge (\mathrm{dd}^{\mathrm c} h_{B_{2n}})^{\wedge (n-k)}\,, $$ whence \begin{align*} \mu_k(B_{2n}) &= 2^{n-k}(n-k)! \nu_k(B_{2n}) \\ &= 2^{n-k}k!(n-k)! \sum_{\Delta\in{\mathcal B}(\Gamma,k)} \varrho(\Delta) {\mathop{\rm vol\,}\nolimits}_{k}(\Delta) {\mathop{\rm vol\,}\nolimits}_{2n-k}(K_\Delta\cap B_{2n}). \end{align*} I still don't understand what was the problem with my previous attempt.


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