Lemma 14.13.5.DBC2334A4FC0 Diagonalize $g$ at a point and order its eigenvalues as $\lambda _1\geq \cdots \geq \lambda _ n$. Put
\[ \mu _ i=\frac{\sum _{j\ne i}\lambda _ j}{n-1}>0, \quad t_ i=\mu _ i^{-1},\quad f_ i=\frac1{n-1}\sum _{j\ne i}t_ j, \quad \mathcal F=\sum _ i f_ i=\sum _ i t_ i. \]
Then $\mu _1\leq \cdots \leq \mu _ n$, $f_1\leq \cdots \leq f_ n$, and
051408386360
14.13.6
\begin{gather} \sum _ i f_ i\lambda _ i=n,\qquad \frac{\mathcal F}{n(n-1)}\leq f_ k\leq \frac{\mathcal F}{n-1} \quad (k>1),\qquad t_ i\leq (n-1)f_1\quad (i>1), \\ f_ n\geq \mathcal F/n. \notag \end{gather}
Fix $\tau ,D>0$ and assume $Q_0\geq \tau \gamma $ and $\det P_\gamma (g)\leq D$. For $\lambda _1$ larger than a constant depending only on $n,\tau ,D$,
2B456F0E5D2B
14.13.7
\begin{equation} \sum _ i f_ i(\chi _{i\bar i}-\lambda _ i) \geq \frac\tau 2\mathcal F. \end{equation}
There is also a uniform positive lower bound for $\mathcal F$ depending only on $n,D$.
Proof.
The ordering follows from the formula for $\mu _ i$, and then from that for $f_ i$. The identity $\sum _ if_ i\lambda _ i=n$, which reflects the relation $\log \det P_\gamma (sg)=\log \det P_\gamma (g)+n\log s$, can be checked directly:
\[ \sum _ i f_ i\lambda _ i =\sum _ j\frac{t_ j}{n-1}\sum _{i\ne j}\lambda _ i =\sum _ jt_ j\mu _ j=n. \]
For $k>1$, the sum defining $f_ k$ contains the largest weight $t_1\geq \mathcal F/n$. The upper bound for $f_ k$ and the estimate for $t_ i$ are immediate, and the largest weight $f_ n$ is at least the average $\mathcal F/n$.
The map $P_\gamma $ is self-adjoint for the matrix trace pairing. Consequently
\[ \sum _ i f_ i\chi _{i\bar i} =\sum _ i t_ i(Q_0)_{i\bar i}\geq \tau \mathcal F. \]
Moreover $\sum _ i\lambda _ i=\sum _ i\mu _ i>\lambda _1>0$, where the strict inequality follows from $\mu _1>0$. Hence $\mu _ n>\lambda _1/n$. Since the other $n-1$ numbers $\mu _ i$ are at least $\mu _1$,
\[ \mu _1^{n-1}\mu _ n\leq \prod _ i\mu _ i\leq D, \qquad \mathcal F\geq \mu _1^{-1} \geq \left(\frac{\lambda _1}{nD}\right)^{1/(n-1)}. \]
For $\lambda _1$ sufficiently large this is at least $2n/\tau $, so $n\leq \tau \mathcal F/2$. Subtracting $\sum _ i f_ i\lambda _ i=n$ therefore proves (14.13.7). Finally, the arithmetic–geometric mean inequality applied to the $t_ i$ gives $\mathcal F\geq nD^{-1/n}$.
$\square$
Pome
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