Proposition 12.6.36.02AE1B3B4027 On the unit-volume hypersurface $\mathscr S=\{ \theta \in \mathscr C:\mathcal V_\mu (\theta )=1\} $, the functions $U=\log b$ and $v$ are global coordinates, in which
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12.6.37
\begin{equation} \begin{aligned} g_{\mathscr S}& =m\, \mathrm{d}U^2+\rho (v)\, \mathrm{d}v^2,\\ \rho (v)& =\frac{Q''/Q-2(Q'/Q)^2}{n} =\frac2n\operatorname {Var}_{\nu _ v}(\ell _ v)>0. \end{aligned} \end{equation}
For any $v_*\in (-1/M,\infty )$, the coordinates
\[ Y=\sqrt m\, U,\qquad R(v)=\int _{v_*}^ v\sqrt{\rho (t)}\, \mathrm{d}t \]
define a global isometry onto $\mathbb R\times I$, where $I$ is an open interval with the Euclidean metric. The intrinsic distance is the Euclidean distance in $(Y,R)$.
Proof.
Consider an affine variation $\delta =(\delta b,\delta c,\delta e)$ of the coefficients, write $F=\delta L/L$, and note that $\sigma /(HL)=\mathrm{d}\nu _ v$ because $HL=Q(v)(1+va)$. Differentiating under the integral sign gives
\[ D^2\log H(\delta ,\delta ) =2\mathbb E_{\nu _ v} F^2-(\mathbb E_{\nu _ v} F)^2, \]
where $\mathbb E_{\nu _ v}$ denotes expectation with respect to $\nu _ v$. This is positive unless $\delta L=0$ everywhere, which forces $\delta c=\delta e=0$. Since $\log \lambda _\mu =(\log n+m\log b-\log H)/n$,
\[ g=-D^2\log \lambda _\mu =\frac mn\left(\frac{\mathrm{d}b}{b}\right)^2 +\frac1nD^2\log H \]
is positive definite. On the unit-volume hypersurface, the relation $H(c,e)=Q(v)/e=nb^ m$ gives
\[ e=\frac{Q(v)}{n\exp (mU)},\qquad c=ve. \]
To pull back the Hessian, write $t=\mathrm{d}e/e$ and $z=Q'/Q$. Along an affine path of coefficients one has $v''=-2tv'$, so differentiating $\log H=-\log e+\log Q(v)$ twice gives
\[ D^2\log H=t^2-2zt\, \mathrm{d}v +(Q''/Q-z^2)\, \mathrm{d}v^2. \]
On $\mathscr S$ we have $t=z\, \mathrm{d}v-m\, \mathrm{d}U$; substituting this and using $n=m+1$ gives (12.6.37). Since $Q'/Q=-\mathbb E_{\nu _ v}\ell _ v$ and $Q''/Q=2\mathbb E_{\nu _ v}\ell _ v^2$, the variance formula follows, and nonconstancy of $a$ makes this variance positive. In the coordinates $(Y,R)$ the metric is Euclidean on the convex set $\mathbb R\times I$, so straight segments prove the distance assertion.
$\square$
Pome
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