Lemma 5.3.48907196C3513

Proof. We first show that, for a given tuple $A_3,\ldots ,A_ n$, the second assertion follows from the first. Let $H$ be positive definite. Then $q(H,H)>0$ by lemma 5.3.3(3), so the space of Hermitian matrices is the $q$-orthogonal direct sum of $\mathbb RH$ and $H^\perp =\{ M:q(M,H)=0\} $. The restriction of $q$ to $H^\perp $ is nondegenerate, since a vector of $H^\perp $ orthogonal to $H^\perp $ would be orthogonal to the whole space. By Sylvester's law of inertia it has signature $(0,n^2-1)$; that is, it is negative definite.

We prove the signature assertion by induction on $n$. For $n=2$ there are no fixed matrices, and expanding $\det (t_1M+t_2N)$ gives $q(M,N)=(\operatorname {tr}M\operatorname {tr}N-\operatorname {tr}(MN))/2$. The decomposition of a matrix into a scalar matrix and a trace-free matrix is $q$-orthogonal, $q(I,I)=1$, and $q(M,M)=-\operatorname {tr}(M^2)/2<0$ for a nonzero trace-free $M$, since $\operatorname {tr}(M^2)$ is the sum of the squares of the real eigenvalues of $M$. Hence the signature is $(1,3)$.

Let $n\geq 3$, and assume both assertions in dimension $n-1$. We first show that $q$ is nondegenerate for every tuple of positive definite matrices. Replacing every matrix $X$ by $G^*XG$ multiplies $D$ by $|\det G|^2$, by the definition of $D$ through the determinant, and preserves positive definiteness; since $X\mapsto G^*XG$ is a linear automorphism of the Hermitian matrices, we may take $G=A_ n^{-1/2}$ and assume that $A_ n=I$. For a Hermitian matrix $X$, let $X^{(i)}$ be the $(n-1)\times (n-1)$ matrix obtained by deleting the $i$th row and column, and let $P_ i$ be the diagonal matrix with entry $1$ in position $(i,i)$ and zeros elsewhere. Expanding along the $i$th row, $\det (tP_ i+Y)=\det Y+t\det Y^{(i)}$; comparing coefficients of $t_1\cdots t_ n$ in $\det (t_1P_ i+\sum _{j\geq 2}t_ jB_ j)$ gives, for Hermitian matrices $B_2,\ldots ,B_ n$,

where the factor $1/n=(n-1)!/n!$ comes from the normalizations of $D$ and of the mixed discriminant $D_{n-1}$ of size $n-1$. Put

the matrices $A_ j^{(i)}$ and $I^{(i)}$ are positive definite, as principal submatrices of positive definite matrices. Using the last formula, and writing $I=\sum _ iP_ i$ in the last slot of $q(M,M)$, we obtain

Now let $M$ lie in the kernel of $q$. Then $q(M,P_ i)=0$, so $M^{(i)}$ lies in the hyperplane $\{ q_ i(\, \cdot \, ,I^{(i)})=0\} $, on which $q_ i$ is negative definite by the induction hypothesis. Since also $q(M,M)=0$, every term of the sum vanishes, and $M^{(i)}=0$ for every $i$. As $n\geq 3$, every entry of $M$ lies in some $M^{(i)}$, so $M=0$.

Finally, the set of tuples $A=(A_3,\ldots ,A_ n)$ of positive definite matrices is convex, hence connected, and we write $q_ A$ for the corresponding form. By Sylvester's law of inertia, the number $k(A)$ of positive squares of $q_ A$ is the largest dimension of a subspace on which $q_ A$ is positive definite. We show that $k$ is locally constant. Fix a tuple $A^0$. Since $q_{A^0}$ is nondegenerate, the space of Hermitian matrices is a direct sum $E_+\oplus E_-$, where $q_{A^0}$ is positive definite on $E_+$, negative definite on $E_-$, and $\dim E_+=k(A^0)$. The function $(A,M)\mapsto q_ A(M,M)$ is continuous, and it is positive on $\{ A^0\} \times S_+$ and negative on $\{ A^0\} \times S_-$, where $S_\pm $ is the unit sphere of $E_\pm $ for some norm. These spheres are compact, so for all tuples $A$ near $A^0$ the form $q_ A$ is still positive definite on $E_+$ and negative definite on $E_-$. Then $k(A)\geq \dim E_+$, and $k(A)\leq \dim E_+$ because a subspace of larger dimension meets $E_-$ in a nonzero vector. Hence $k(A)=k(A^0)$ near $A^0$, and $k$ is constant on the connected set of tuples.

It therefore suffices to compute the signature at $A_3=\cdots =A_ n=I$. With $s=t_3+\cdots +t_ n$, the coefficient of $s^{n-2}$ in $\det (t_1M+t_2N+sI)$ is the sum of the principal $2\times 2$ minors of $t_1M+t_2N$, that is, $((\operatorname {tr}X)^2-\operatorname {tr}(X^2))/2$ for $X=t_1M+t_2N$, in which $t_1t_2$ has the coefficient $\operatorname {tr}M\operatorname {tr}N-\operatorname {tr}(MN)$; moreover, the coefficient of $t_3\cdots t_ n$ in $s^{n-2}$ is $(n-2)!$. Hence

and, as in the case $n=2$, the decomposition into scalar and trace-free matrices shows that the signature is $(1,n^2-1)$. $\square$