Lemma 5.3.33D25BAD0C21E

Proof. (1) The $l$th column of $\sum _ jt_ jA_ j$ is $\sum _ jt_ jA_ je_ l$. Expanding the determinant by multilinearity in its columns, we obtain

the sum running over all maps $\tau \colon \{ 1,\ldots ,n\} \to \{ 1,\ldots ,n\} $. The monomial $t_{\tau (1)}\cdots t_{\tau (n)}$ equals $t_1\cdots t_ n$ exactly when $\tau $ is a permutation, which gives the displayed formula. A permutation of $A_1,\ldots ,A_ n$ permutes the terms of this sum, so $D$ is symmetric. Each term is multilinear in $(A_1,\ldots ,A_ n)$, since each $A_ j$ occurs in exactly one column, so $D$ is multilinear. For $A_1=\cdots =A_ n=A$ all $n!$ terms equal $\det A$, so $D(A,\ldots ,A)=\det A$. Finally, $D$ is real: for real $t_ j$ the matrix $\sum _ jt_ jA_ j$ is Hermitian, so its determinant is real, being the product of its real eigenvalues; hence the polynomial $\det (\sum _ jt_ jA_ j)$ is real on $\mathbb R^ n$, and so are its coefficients, which are its partial derivatives at $t=0$ divided by factorials.

(2) In the expansion of $\omega _ A^ n$, the products $\bigwedge _ l(\mathrm dz_{j_ l}\wedge \mathrm d\bar z_{k_ l})$ vanish unless $(j_1,\ldots ,j_ n)$ and $(k_1,\ldots ,k_ n)$ are permutations of $(1,\ldots ,n)$. Since the $2$-forms $\mathrm dz_ j\wedge \mathrm d\bar z_ k$ commute, such a product equals $\bigwedge _ l(\mathrm dz_ l\wedge \mathrm d\bar z_{\pi (l)})$ for a permutation $\pi $, which arises from $n!$ ordered products, and $\bigwedge _ l(\mathrm dz_ l\wedge \mathrm d\bar z_{\pi (l)})=\operatorname {sign}(\pi )\bigwedge _ l(\mathrm dz_ l\wedge \mathrm d\bar z_ l)$, because the two products differ by a permutation of the factors $\mathrm d\bar z_ k$. Therefore

where the last equality is the case $A=I$ of the preceding one. Since $\mathrm i\mathrm dz_ l\wedge \mathrm d\bar z_ l=2\, \mathrm dx_ l\wedge \mathrm dy_ l$, we get $\omega _ I^ n=2^ nn!\, \mathrm dV_{2n}=2^ n\beta ^ n$ by (5.2.8). Now let $t_1,\ldots ,t_ n$ be real, and apply the first identity to the Hermitian matrix $\sum _ jt_ jA_ j$, for which $\omega _{\sum _ jt_ jA_ j}=\sum _ jt_ j\omega _{A_ j}$. Since $2$-forms commute, the coefficient of $t_1\cdots t_ n$ in $(\sum _ jt_ j\omega _{A_ j})^ n$ is $n!\, \omega _{A_1}\wedge \cdots \wedge \omega _{A_ n}$, while in $\det (\sum _ jt_ jA_ j)\, \omega _ I^ n$ it is $n!\, D(A_1,\ldots ,A_ n)\, \omega _ I^ n$. Two polynomials that agree on $\mathbb R^ n$ have the same coefficients, and we obtain the second identity.

(3) A real $(1,1)$-form $\omega _ A$ with $A$ positive semidefinite is strongly positive, by the discussion following proposition 5.3.2: a unitary diagonalization of $A$ writes it as a sum of generators with nonnegative coefficients. By proposition 5.3.2, the product $\omega _{A_1}\wedge \cdots \wedge \omega _{A_ n}$ is then strongly positive, hence weakly positive, that is, a nonnegative multiple of the volume form. By (2) and since $\omega _ I^ n$ is a positive multiple of the volume form, $D(A_1,\ldots ,A_ n)\geq 0$. If all $A_ j$ are positive definite, choose $\varepsilon >0$ smaller than the least eigenvalue of every $A_ j$, so that every $A_ j-\varepsilon I$ is positive semidefinite, and expand $D(A_1,\ldots ,A_ n)=D\bigl ((A_1-\varepsilon I)+\varepsilon I,\ldots ,(A_ n-\varepsilon I)+\varepsilon I\bigr )$ by multilinearity. Every term is the mixed discriminant of positive semidefinite matrices, hence nonnegative, and one of them is $D(\varepsilon I,\ldots ,\varepsilon I)=\varepsilon ^ n\det I=\varepsilon ^ n>0$. $\square$