==== TABLE OF CONTENTS ====
The-metric-Blaschke-theorem//figures/attachment-square.tikz.tex
The-metric-Blaschke-theorem//figures/schur-paths.tikz.tex
The-metric-Blaschke-theorem//paper.tex
The-metric-Blaschke-theorem//sections/00-introduction.tex
The-metric-Blaschke-theorem//sections/01-geometric-inputs.tex
The-metric-Blaschke-theorem//sections/02-lagrangian.tex
The-metric-Blaschke-theorem//sections/03-weighted-comparison.tex
The-metric-Blaschke-theorem//sections/04-global-comparison.tex
The-metric-Blaschke-theorem//sections/05-incidence.tex
The-metric-Blaschke-theorem//sections/06-signature.tex
The-metric-Blaschke-theorem//sections/07-geodesic-space.tex
The-metric-Blaschke-theorem//sections/08-volume-algebra.tex
The-metric-Blaschke-theorem//sections/09-classification.tex
The-metric-Blaschke-theorem//references.bib
===========================
==== BEGIN The-metric-Blaschke-theorem//figures/attachment-square.tikz.tex ====
\begin{tikzpicture}[>=Stealth,font=\small]
\node (SU) at (-3,1.8) {$S(\rho^*E)$};
\node (SB) at (1,1.8) {$S(E)$};
\node (U) at (-3,-.2) {$U=SM$};
\node (B) at (1,-.2) {$B=U/S^1$};
\node (M) at (4,1.8) {$M$};
\draw[->] (SU)--node[above] {$\widehat\rho$} (SB);
\draw[->] (SB)--node[above] {$f$} (M);
\draw[->] (SU)--node[left,align=right] {$\widetilde p$\\$S^2\text{ fibers}$} (U);
\draw[->] (SB)--node[right,align=left] {$p$\\$S^2\text{ fibers}$} (B);
\draw[->] (U)--node[below] {$\rho\quad(S^1\text{ fibers})$} (B);
\node[fill=white] at (-1,.8) {pullback};
\node[align=center] at (-3,-1.15) {marked unit-speed\\oriented geodesic};
\node[align=center] at (1,-1.15) {oriented geodesic\\with no marked point};
\node[align=center] at (.2,-2.25)
{$f\circ\widehat\rho\simeq\pi\circ\widetilde p$
\quad (evaluation homotopy)};
\node[align=center] at (.2,-3.0)
{$x=f^*u,\quad z=\pi^*u,\quad
\widehat\rho^*x=\widetilde p^*z$};
\end{tikzpicture}
==== END The-metric-Blaschke-theorem//figures/attachment-square.tikz.tex ====
==== BEGIN The-metric-Blaschke-theorem//figures/schur-paths.tikz.tex ====
\begin{tikzpicture}[x=1.05cm,y=1.05cm,>=Stealth,font=\small]
\draw[->] (0,0)--(4.7,0) node[right] {$a$};
\draw[->] (0,0)--(0,4.6) node[above] {$b$};
\draw[gray,densely dashed] (0,0)--(4,4);
\draw[gray,dotted] (4,0)--(4,4)--(0,4);
\node[below left] at (0,0) {$0$};
\node[below] at (4,0) {$D$};
\node[left] at (0,4) {$D$};
\foreach \a/\b in {1/u,3/v} {
\draw (\a,-.06)--(\a,.06) node[below=3pt] {$\b$};
\draw (-.06,\a)--(.06,\a) node[left=3pt] {$\b$};
}
\draw[gray,dotted] (1,0)--(1,1);
\draw[gray,dotted] (0,3)--(1,3);
\draw[very thick,->] (1,1)--(1,3);
\draw[very thick,->] (1,3)--(3,3);
\fill (1,1) circle (1.7pt);
\fill (1,3) circle (1.7pt);
\fill (3,3) circle (1.7pt);
\node[below right] at (1,1) {$Q(u+iD,u+jD)=0$};
\node[anchor=east] (qtarget) at (-.3,3.55)
{$Q(u+iD,v+jD)$};
\draw[gray,thin] (qtarget.east)--(.9,3.08);
\node[right,align=left] at (3,3) {$Q(v+iD,v+jD)=0$};
\node[rotate=90,above] at (.89,2) {$\partial_yQ\ge S(y)$};
\node[above] at (2,3.02) {$-\partial_xQ\ge S(x)$};
\node[align=center] at (2,5.05)
{$x=a+iD,\quad y=b+jD,\quad i,j\in\{0,1\}$};
\node[align=center] at (2,-.9)
{$00$. For $x,y\in\mathcal I$,
\begin{equation}
f(x,y)=
\frac{|\det(A(y)-A(x))|}
{\sqrt{\det A'(x)\det A'(y)}}.
\label{eq:lag-graph-determinant}
\end{equation}
\end{proposition}
\begin{proof}
Changing either orthonormal frame multiplies the pairing matrix by an
orthogonal matrix. This leaves its absolute determinant unchanged.
Local smooth orthonormal frames exist, so $f$ is continuous even when
the pairing is singular. Skew-symmetry of $\omega$ replaces the matrix
by its negative transpose when the arguments are exchanged, proving
symmetry. Periodicity follows from
\eqref{eq:lag-periodicity} and \eqref{eq:lag-velocity}.
Evaluation of the pairing at $y$ gives
\[
\omega(Z_{x,a},Z_{y,b})
=\langle Z_{x,a}(y),Z'_{y,b}(y)\rangle.
\]
The vectors $Z'_{y,b}(y)$ form an orthonormal normal basis, and the
fields $Z_{x,a}$ have zero values and orthonormal derivatives at $x$.
This proves the claimed determinant interpretation. More explicitly,
an angular variation of a unit direction with derivative $e$ gives the
Jacobi field $J$ with $J(x)=0$ and $J'(x)=e$. Its value at $x+r$ is the
angular derivative of $\exp_{\gamma(x)}(r\,\cdot)$. Together with the
unit radial derivative and Gauss's lemma, this gives the polar density
$f(x,x+r)\,dr\,d\sigma$. There is no additional power of $r$ in this
polar formula; the Cartesian normal-coordinate density would be
$f(x,x+r)/r^d$.
In the stated coordinates,
\[
\omega((q,p),(\widetilde q,\widetilde p))
=q^{\mathsf T}\widetilde p-p^{\mathsf T}\widetilde q,
\qquad L_0=\{q=0\}.
\]
Equation~\eqref{eq:lag-intersections} makes $L_t$ transverse to $L_0$
for $t\in\mathcal I$, proving the graph representation. Isotropy is
equivalent to $A=A^{\mathsf T}$. For the graph frame
$E_tq=(q,A(t)q)$,
\[
G_t(E_tq,E_tr)=q^{\mathsf T}A'(t)r,
\qquad
\omega(E_xq,E_yr)=q^{\mathsf T}(A(y)-A(x))r.
\]
Positivity of $G_t$ gives $A'(t)>0$. The frames
$E_tA'(t)^{-1/2}$ are orthonormal for $G_t$, so taking determinants in
the last identity proves \eqref{eq:lag-graph-determinant}.
\end{proof}
The coordinate basis is now fixed. Split both $q$ and $p$ into their
first $k$ and last $m$ coordinates and write
\begin{equation}
A(t)=
\begin{pmatrix}
V_b(t)&C(t)\\ C(t)^{\mathsf T}&H(t)
\end{pmatrix},
\qquad
S(t)=V_b'(t)-C'(t)H'(t)^{-1}C'(t)^{\mathsf T}.
\label{eq:lag-blocks}
\end{equation}
The subscript in $V_b$ distinguishes this matrix from the volume of the
manifold. Here $H$ is an $m\times m$ matrix and $S$ is a $k\times k$
matrix. Positive definiteness of $A'$ and block elimination give
\begin{equation}
H'(t)>0,\qquad S(t)>0,\qquad
\det A'(t)=\det H'(t)\det S(t)
\quad(t\in\mathcal I).
\label{eq:lag-velocity-factorization}
\end{equation}
For comparison, on the standard model the adapted chart has
\[
A_{\mathrm{mod}}(t)
=\diag\bigl(-2\cot(2t)\Id_k,-\cot(t)\Id_m\bigr).
\]
Thus $H'_{\mathrm{mod}}(t)=\csc^2(t)\Id_m$ and
$S_{\mathrm{mod}}(t)=4\csc^2(2t)\Id_k$.
These formulas motivate the scalar reference profiles below.
\subsection{The chart poles and determinant-root profiles}
The curve $L_t$ remains smooth at $0,D,T$; it is the projection from
$L_t$ to the fixed $q$-coordinate space that ceases to be invertible.
Its kernel has dimension $d$ at $0,T$ and dimension $k$ at $D$.
The following expansion shows how this loss of transversality enters
$A$: at $D$ the pole lies entirely in the first $k$ coordinates,
while the complementary block $H$ remains regular. Smooth remainders
are needed because the comparison uses derivatives of the graph matrices.
\begin{lemma}[Pole expansions]
\label{lag:poles}
There is a positive definite $k\times k$ matrix $B$ such that the graph
matrix has the expansions
\begin{align}
A(t)&=-\frac1{t-D}
\begin{pmatrix}B&0\\0&0\end{pmatrix}+R_D(t),
&&\text{near }D,\label{eq:lag-middle-pole}\\
A(t)&=-\frac1{t-t_0}\Id_d+R_{t_0}(t),
&&\text{near }t_0=0,T,\label{eq:lag-end-poles}
\end{align}
where each remainder extends smoothly across its indicated point.
In particular, $C$ and $H$ extend smoothly across $D$, and
$H'(D)>0$.
\end{lemma}
\begin{proof}
Put $\tau=t-D$. Choose a local smooth frame of $L_t$, written as the
columns of
\[
F(t)=\begin{pmatrix}X(t)\\P(t)\end{pmatrix},
\]
whose first $k$ columns at $D$ are the vertical vectors $(0,e_a)$ for
$1\le a\le k$. These are exactly the chosen coordinate basis of
$L_0\cap L_D$. Isotropy with these vectors forces every vector in the
image of $X(D)$ to have its first $k$ coordinates zero. The kernel of
the value projection on $L_D$ is $L_0\cap L_D$, so that image has
dimension $m$ and is the entire last coordinate space. Choose the
remaining frame columns so that their value projections give a basis
of this image. In blocks, their last $m$ value coordinates form an
invertible matrix $M$.
The first $k$ columns of $X(D)$ vanish. Dividing these columns by
$\tau$ gives a smooth matrix $\widetilde X(t)$ satisfying
\[
X(t)=\widetilde X(t)
\begin{pmatrix}\tau\Id_k&0\\0&\Id_m\end{pmatrix}.
\]
Let $G$ be the matrix of $G_D$ on the first $k$ frame vectors. For
$1\le a,b\le k$, the definition of the velocity form in the fixed
coordinates gives
\[
G_{ab}=\omega((0,e_a),\dot F_b(D))
=-e_a^{\mathsf T}X_b'(D).
\]
Consequently
\begin{equation}
\widetilde X(D)=
\begin{pmatrix}-G&0\\N&M\end{pmatrix},
\qquad G>0,\quad M\text{ invertible},
\label{eq:lag-divided-frame}
\end{equation}
for some matrix $N$. In particular $\widetilde X(D)$ is invertible.
For $\tau\ne0$ sufficiently small,
\[
A(t)=P(t)
\begin{pmatrix}\tau^{-1}\Id_k&0\\0&\Id_m\end{pmatrix}
\widetilde X(t)^{-1}.
\]
To identify its residue, let $P_1(t)$ denote the first $k$ columns of
$P(t)$ and let $U_1(t)$ denote the first $k$ rows of
$\widetilde X(t)^{-1}$. Then the only possibly singular summand is
$\tau^{-1}P_1(t)U_1(t)$. At $D$,
\[
P_1(D)=\begin{pmatrix}\Id_k\\0\end{pmatrix},
\qquad U_1(D)=\begin{pmatrix}-G^{-1}&0\end{pmatrix}.
\]
Their product is $-\diag(G^{-1},0)$. The difference
$P_1(t)U_1(t)-P_1(D)U_1(D)$ divided by $\tau$ is smooth, as is the
summand from the other frame columns. This proves
\eqref{eq:lag-middle-pole} with $B=G^{-1}>0$, including the absence of
any off-diagonal residue.
For fixed $z\in\R^m$, the graph vectors
\[
E_z(t)=\left(\begin{pmatrix}0\\z\end{pmatrix},
\begin{pmatrix}C(t)z\\H(t)z\end{pmatrix}\right)
\]
extend smoothly through $D$. Their limits lie in $L_D$ by continuity
of $L_t$, and are nonzero if $z\ne0$. The velocity identity in the
fixed symplectic coordinates, now also at $D$, gives
\[
z^{\mathsf T}H'(D)z
=\omega(E_z(D),\dot E_z(D))
=G_D(E_z(D),E_z(D))>0.
\]
This proves strict positivity of the extended derivative.
At $t_0=0$, take a frame whose value at $0$ is the whole vertical
coordinate basis. All $X$-columns then vanish at $0$. The velocity
metric on this basis is $\Id_d$ by \eqref{eq:lag-velocity}; hence
$X'(0)=-\Id_d$ and $P(0)=\Id_d$. Dividing all value columns by $t$
and repeating the preceding calculation gives residue $-\Id_d$.
At $t_0=T$, periodicity identifies both $L_T$ and its velocity metric
with their values at zero in these same fixed coordinates. The same
argument gives residue $-\Id_d$ there, proving
\eqref{eq:lag-end-poles}.
\end{proof}
\begin{proposition}[Scalar profiles]
\label{lag:profiles}
Define
\begin{equation}
\begin{aligned}
h(t)&=\det(H'(t))^{1/m} &&(00$ throughout
$(0,T)$,
\begin{equation}
\Delta H=H(y)-H(x)=\int_x^y H'(t)\,dt
\label{eq:lag-H-difference}
\end{equation}
is positive definite for $y>x$ and negative definite for $yv$,
and both positive-semidefinite lower bounds still apply.}
\label{fig:schur-endpoint-paths}
\end{figure}
Both moving endpoints stay in one open half-period, so neither path
passes through a chart pole. In particular, when $i=1,j=0$ the actual
endpoint order is $u+D>v$. The first path starts at $(u+D,u)$ and the
second ends at $(v+D,v)$; the same zero values and the same derivative
inequalities apply. Thus neither matrix inequality reverses in this
case. Finally, the integrals on their right sides are positive
definite, so $Q(u+iD,v+jD)>0$.
\end{proof}
\subsection{Scalar lower bounds for the two factors}
For distinct $x,y\in\mathcal I$, define
\begin{equation}
f_H(x,y)=
\frac{|\det(H(y)-H(x))|}
{\sqrt{\det H'(x)\det H'(y)}},\qquad
f_K(x,y)=
\frac{|\det Q(x,y)|}
{\sqrt{\det S(x)\det S(y)}}.
\label{eq:lag-two-factors}
\end{equation}
These auxiliary factors are symmetric under exchanging the actual
endpoints. Combining \eqref{eq:lag-graph-determinant},
\eqref{eq:lag-velocity-factorization}, and
\eqref{eq:lag-Schur-factorization} gives
\begin{equation}
f(x,y)=f_H(x,y)f_K(x,y).
\label{eq:lag-factorization}
\end{equation}
The factors are positive away from the cut pairs $|y-x|=D$; at those
pairs $f_K=0$.
\begin{proposition}[Scalar block bounds]
\label{lag:block-bounds}
With the profiles of Proposition~\ref{lag:profiles}, every pair of
distinct chart endpoints satisfies
\begin{equation}
f_H(x,y)^{1/m}\ge
\frac{\left|\int_x^y h(t)\,dt\right|}{\sqrt{h(x)h(y)}}.
\label{eq:lag-H-scalar-bound}
\end{equation}
For $00\\\det P=1}}\frac1r\tr(PM)
\label{eq:lag-determinant-variational}
\end{equation}
follows by applying the arithmetic--geometric mean inequality to the
eigenvalues of $M^{1/2}PM^{1/2}$. Equality is attained at
$P=\delta_r(M)M^{-1}$. The formula proves monotonicity under the
positive-semidefinite order. It also proves, for any continuous
positive definite matrix function on a compact interval,
\begin{equation}
\delta_r\left(\int_a^b M(t)\,dt\right)
\ge\int_a^b\delta_r(M(t))\,dt.
\label{eq:lag-Minkowski}
\end{equation}
Indeed, for every admissible $P$ the integral of
$\tr(PM(t))/r$ is at least the right side, and one then takes the
infimum over $P$. This is the integral form of Minkowski's determinant
inequality; it retains the degree-one homogeneity of the determinant
root.
Apply \eqref{eq:lag-Minkowski} to $H'$ on the interval between $x$ and
$y$. Equation~\eqref{eq:lag-H-difference} gives
\[
|\det(\Delta H)|^{1/m}
\ge\left|\int_x^y\det(H'(t))^{1/m}\,dt\right|.
\]
For $y0$. Each of the two matrix bounds in
Proposition~\ref{lag:schur}, followed by determinant monotonicity and
\eqref{eq:lag-Minkowski}, gives
\[
\delta_k(Q)\ge\int_u^v s_i(t)\,dt,
\qquad
\delta_k(Q)\ge\int_u^v s_j(t)\,dt.
\]
Thus $\delta_k(Q)$ dominates their maximum and therefore their
geometric mean. The denominator in the $k$th root of $f_K$ is exactly
\[
\bigl(\det S(u+iD)\det S(v+jD)\bigr)^{1/(2k)}
=\sqrt{s_i(u)s_j(v)},
\]
which proves \eqref{eq:lag-K-scalar-bound} with the displayed powers.
Finally, every endpoint in $\mathcal I$ has a unique expression
$u+iD$ with $00$, $\lambda=\pi/b$, and
\[
p_0(t)=\frac{\lambda^2}{\sin^2(\lambda t)},\qquad 0y$; the bound in
Proposition~\ref{lag:block-bounds} includes this order.
All three separations have the same absolute fast model factor
$B_0(z)$.
The four weights $w_{ij}(z)=w(|z+(j-i)D|)$ form the matrix
\begin{equation}
(w_{ij}(z))_{i,j=0}^1
=B_0(z)^k
\begin{pmatrix}
\sin^m z&\cos^m z\\
\cos^m z&\sin^m z
\end{pmatrix}.
\label{weighted:weight-matrix}
\end{equation}
Every row and every column sums to
\[
W_D(z)=(\sin^m z+\cos^m z)B_0(z)^k.
\]
This is symmetric under $z\mapsto D-z$, and
\[
\frac{W_D(z)}{\sin(2z)}
=2^{-k}(\sin^m z+\cos^m z)\sin^{k-1}(2z)
\]
is bounded for every $k\geq1$. When $k=1$, its limit at either endpoint
is $1/2$, so this case also satisfies Lemma~\ref{weighted:scalar}.
Multiply \eqref{weighted:folded-bound} by $w_{ij}(z)$ and sum over
$i,j$. The row sums control the terms indexed by $i$, and the column
sums control those indexed by $j$, giving exactly
\begin{align}
\sum_{i,j=0}^1 w_{ij}(z)\frac{L_K(u+iD,v+jD)}k
&\geq W_D(z)\sum_{i=0}^1
\left[A_i(u,v)-\frac{g_i(u)+g_i(v)}2\right].
\label{weighted:folded-sum}
\end{align}
Lemma~\ref{weighted:scalar}, applied on $(0,D)$ separately to $s_0$
and $s_1$ with weight $W_D$, shows that the right side has nonnegative
integral over $\Delta_D$.
For completeness, the four images of $\Delta_D$ used above consist
precisely of pairs whose first endpoint has smaller residue modulo $D$
than the second. Their transposes contain precisely the opposite
residue order. These eight disjoint regions cover $\mathcal R\setminus E$;
each coordinate change has Jacobian one. Since $w_{xy}L_K(x,y)$ is
symmetric, it follows that
\begin{align}
&\int_{\mathcal R}w_{xy}L_K(x,y)\,dx\,dy\notag\\
&\qquad=2k\int_{\Delta_D}\sum_{i,j=0}^1
w_{ij}(v-u)\frac{L_K(u+iD,v+jD)}k\,du\,dv
\ \geq\ 0.
\label{weighted:K-log}
\end{align}
Here the factor $2$ accounts for endpoint exchange and the factor $k$
restores the logarithm of $f_K$ from that of its $k$th root.
\paragraph{Comparison and equality.}
Adding \eqref{weighted:H-log} and \eqref{weighted:K-log} gives
\[
\int_{\mathcal R}w_{xy}\log(f/w_{xy})\,dx\,dy\geq0.
\]
For $r>0$, the strictly convex function
\[
\Phi(r)=r-1-\log r
\]
is nonnegative and vanishes exactly at $r=1$. On
$\mathcal R\setminus E$ put $r=f/w_{xy}$. All terms in the identity
\begin{equation}
\int_{\mathcal R}(f-w_{xy})\,dx\,dy
=\int_{\mathcal R}w_{xy}\log r\,dx\,dy
+\int_{\mathcal R}w_{xy}\Phi(r)\,dx\,dy
\label{weighted:strict-gap}
\end{equation}
are finite by the preceding integrability argument. Both terms on
the right are nonnegative, proving \eqref{weighted:comparison}.
If equality holds, the last integral is zero. Since $w_{xy}>0$ off
the diagonal and cut lines, $\Phi(r)=0$ almost everywhere, hence
$f=w_{xy}$ almost everywhere in $\mathcal R$.
Continuity of both functions on $[0,T]^2$ extends this identity to
every pair, including the exceptional lines and boundary. Conversely,
that pointwise identity gives equality in \eqref{weighted:comparison}.
\end{proof}
==== END The-metric-Blaschke-theorem//sections/03-weighted-comparison.tex ====
==== BEGIN The-metric-Blaschke-theorem//sections/04-global-comparison.tex ====
\section{Global comparison and the equality case}
\label{rig:section}
We retain all intermediate model types and the normalization of
Section~\ref{geom:section}. The preceding geodesic comparison
now gives the global volume bound. Its nonnegative defect will
also retain enough information to identify the equality metric.
For $\xi\in U$, write $\gamma_\xi$ for the geodesic with initial
velocity $\xi$ and $f_\xi$ for the determinant of
Proposition~\ref{lag:determinant}. Put
\[
J(\xi,t)=f_\xi(0,t).
\]
This is the absolute determinant of the normal Jacobi tensor
with initial value zero and derivative $\Id$. It is continuous
in $(\xi,t)$, positive for $00$
proves $V\ge V_0$.
If $V=V_0$, the nonnegative function $\delta$ has integral zero.
It is continuous, because the underlying Jacobi tensors depend
smoothly on the initial velocity and taking the absolute
determinant preserves continuity. Full support of $\mu$ forces
$\delta(\xi)=0$ for every $\xi$. The equality assertion in
Theorem~\ref{weighted:geodesic} now gives
\begin{equation}\label{rig:radial-density}
J(\xi,t)=w(t),\qquad \xi\in U,\quad 00,\qquad c=2\sqrt{s},
\quad f(y)=\frac{\sqrt y}{\tanh\sqrt y}.
\]
Hirzebruch's signature theorem
\citep[Signature Theorem~19.4]{MilnorStasheff1974}, applied to
Theorem~\ref{inc:pontryagin} gives
\begin{equation}\label{sig:formula}
\sigma_r=[u^r]
\frac{(f(a_+^2u)f(a_-^2u))^{r+1}}{f(c^2u)}.
\end{equation}
Here $f$ is its power series at zero; the two factors reflect
$1+2su+u^2=(1+a_+^2u)(1+a_-^2u)$.
The integral cohomology ring of $M$ gives
\[
\sigma_r=\begin{cases}1,&r\text{ even},\\0,&r\text{ odd}.
\end{cases}
\]
Indeed for even $r$ the middle pairing has the positive basis
$u^{r/2}$, and for odd $r$ its middle cohomology is zero.
Expanding $f(y)=1+y/3-y^2/45+2y^3/945+\cdots$ in
\eqref{sig:formula} gives
\[
\sigma_2-1=\frac{8(s^2-1)}{15},\qquad
\sigma_3=-\frac{32s(s^2-1)}{105}.
\]
These prove the assertion for $r=2,3$.
We give uniform estimates for all $r\ge4$. Since $s$ is odd,
the contrary case is $s\ge3$. We express the signature as a residue
and show that the two nearest nonzero poles give a main term that
dominates the remaining contour integral, forcing
$(-1)^r\sigma_r>1$. Put
\[
\tau=\frac{a_-}{a_+}=s-\sqrt{s^2-1},\qquad
b=\frac c{a_+}=\sqrt{2(1+\tau^2)},\qquad
Q(z)=\tan z\tan(\tau z),\qquad \rho=\frac\pi{2b}.
\]
Then $0<\tau\le3-\sqrt8<0.172$ and
$\sqrt2X$. The poles of $\tan z$ at $\pm\pi/2$
give zeros of the reciprocal denominator, not poles.
Both residues at $\pm\rho$ equal
$-b^{-1}Q(\rho)^{-r-1}$, with $Q(\rho)>0$.
The residue theorem therefore expresses \eqref{sig:residue} as
\begin{equation}\label{sig:main-term}
(-1)^r\sigma_r=L+\mathcal E,\qquad
L=\frac{2\tau}{b^2}Q(\rho)^{-r-1}.
\end{equation}
Symmetry of the four sides and the triangle inequality bound
$|\mathcal E|/L$ by the sum of
\begin{align}
E_v&=\frac b\pi\int_0^4
|\tan(b(X+iy))|
\left(\frac{Q(\rho)}{|Q(X+iy)|}\right)^{r+1}\,dy,
\label{sig:vertical}\\
E_h&=\frac b\pi\int_0^X
|\tan(b(x+4i))|
\left(\frac{Q(\rho)}{|Q(x+4i)|}\right)^{r+1}\,dx.
\label{sig:horizontal}
\end{align}
\paragraph{Uniform bounds.}
The same contour works for every $r\ge4$ once both boundary ratios
are bounded by constants below one: their fifth powers then control
every exponent $r+1\ge5$. We retain the $\tau$ dependence in the
numerator and denominator so these bounds remain uniform as
$\tau\downarrow0$. All finite decimals below
denote the exact indicated rational numbers. Using
$3.14<\pi<3.142$ gives
\[
\rho<1.111,\qquad \tan\rho<2.022,\qquad
\tau\rho<0.192,\qquad Q(\rho)<2.30\tau.
\]
For the last inequality, the increase of $\tan t/t$ and the
Taylor bounds on $[0,0.192]$ give $\tan t/t<1.013$,
and $2.022\cdot1.111\cdot1.013<2.30$.
The same monotonicity gives
\begin{equation}\label{sig:ratio}
\frac{Q(\rho)}{\tan(\tau X)}
\le\frac{\rho\tan\rho}{X}<0.955.
\end{equation}
On the vertical sides, $|\tan(X+iy)|=1$.
Also $\cos(2bX)\ge0$, since
$2bX=3b\pi/2\in(2\pi,5\pi/2)$, and hence
$|\tan(b(X+iy))|\le1$.
The elementary complex tangent identity gives
\begin{align*}
\frac{|\tan(\tau(X+iy))|^2}{\tan^2(\tau X)}
&=\frac{1+\sinh^2(\tau y)/\sin^2(\tau X)}
{1+\sinh^2(\tau y)/\cos^2(\tau X)}\\
&\ge 1+0.08y^2\qquad(0\le y\le4).
\end{align*}
For this last bound, the numerator is at least $1+y^2/X^2$,
whereas $\cos(\tau X)>0.917$ and
$\sinh(\tau y)\le1.082\tau y$ bound the denominator by
$1+Ay^2$ with $A=(1.082\cdot0.172/0.917)^2$.
The inequality follows from
$X^{-2}\ge A+0.08+16(0.08)A$.
Since $r+1\ge5$, \eqref{sig:ratio} now implies
\begin{align*}
E_v
&<\frac{1.435}{3.14}(0.955)^5
\int_0^4(1+0.08y^2)^{-5/2}\,dy\\
&<0.783.
\end{align*}
Here the integral equals
\[
\frac{4(1+2\cdot0.08\cdot16/3)}{(1+0.08\cdot16)^{3/2}}
<2.154.
\]
On a horizontal side the identity for the modulus of tangent
gives
\[
|Q(x+4i)|\ge\tanh4\,\tanh(4\tau),\qquad
|\tan(b(x+4i))|\le\coth(4b)<1.001.
\]
Since $\tau/\tanh(4\tau)$ is increasing,
\[
\frac{Q(\rho)}{\tanh4\,\tanh(4\tau)}
<\frac{2.30\cdot0.172}{0.999\cdot0.595}.
\]
Substitution in \eqref{sig:horizontal}, with $X/\pi=3/4$,
therefore gives
\[
E_h<\frac{3\cdot1.435}{4}\,1.001
\left(\frac{2.30\cdot0.172}{0.999\cdot0.595}\right)^5
<0.142.
\]
The numerical trigonometric inequalities used here can be checked
without numerical approximation: use
\[
\sin t\le t-t^3/6+t^5/120,\qquad
\cos t\ge1-t^2/2+t^4/24-t^6/720
\quad(0\le t\le1.111)
\]
for $\tan\rho$ and the smaller real arguments. For the
hyperbolic inequalities use
\[
\frac{\sinh t}{t}\le1+\frac{t^2}{6}
+\frac{t^4/120}{1-t^2/42}\quad(0\le t\le0.688).
\]
The first seven terms of the exponential series at $1.376$
give $\tanh(0.688)>0.595$, while $e^8>2000$ gives
$\tanh4>0.999$. The same bound and $b>\sqrt2$ imply
$\coth(4b)<1.001$.
Finally $Q(\rho)<2.30\tau<1$, so \eqref{sig:main-term}
and $r+1\ge5$ imply
\[
L>\frac{\tau}{1+\tau^2}(2.30\tau)^{-5}
\ge\frac{1}{(1+0.172^2)\,2.30^5\,0.172^4}>17.
\]
Combining these estimates yields
\[
(-1)^r\sigma_r>(1-0.783-0.142)17=1.275>1,
\]
contradicting either possible cohomological signature. Thus $s=1$.
Restricting \eqref{inc:normal-classes} to a fixed-$p$ fiber
gives $e(\mathcal V)=-u_2$ and $p_1(\mathcal V)=2u_2$ there.
Proposition~\ref{inc:normal} identifies this restriction with
the actual tautological four-plane bundle, which proves the
last assertion as well.
\end{proof}
==== END The-metric-Blaschke-theorem//sections/06-signature.tex ====
==== BEGIN The-metric-Blaschke-theorem//sections/07-geodesic-space.tex ====
\section{The geodesic space and its integral classes}
\label{geo:section}
Throughout this section, $M$ has quaternionic model type, so that
\[
n=4r,\qquad r\ge2,\qquad d=4r-1,\qquad k=3,
\qquad H^*(M;\Z)=\Z[u]/(u^{r+1}),\quad |u|=4.
\]
We retain the normalization $D=\pi/2$ and $T=2D=\pi$.
Choose the sign of $u$ as permitted by Theorem~\ref{sig:coefficient},
so that the actual four-plane bundle $\mathcal V$ on cut incidence
$F$ satisfies
\begin{equation}\label{geo:incidence-input}
p_1(\mathcal V)=2(u_1+u_2).
\end{equation}
Orient $M$ by $\int_Mu^r=1$.
Only the characteristic-class conclusion
\eqref{geo:incidence-input} and the bundle identification in
Proposition~\ref{inc:normal} are used from the incidence argument.
Let $U=SM$, let $\pi:U\to M$ be the unit tangent projection, and
write $z=\pi^*u$. The prime period is $T$, so the action
\[
(\R/\Z)\times U\longrightarrow U,
\qquad (a,w)\longmapsto\phi_{Ta}(w)
\]
is free. Its quotient and quotient map will be denoted by
\[
B=U/S^1,\qquad \rho:U\longrightarrow B.
\]
Thus $B$ is the closed smooth manifold of oriented, unmarked
geodesics, of dimension $2n-2$.
\subsection{The contact normalization and the two Gysin sequences}
The contact calculation is the volume method for periodic geodesic
flows associated with Weinstein \citep{Weinstein1974}; we include
it to fix the period, Euler sign and angular-area factors exactly.
The ensuing integral Gysin sequences, universal coefficient theorem
and Poincar\'e duality are used in their standard forms
\citep{Hatcher2002,MilnorStasheff1974}.
Yang's complex-projective volume calculation also uses the Gysin
sequences of the unit-tangent sphere bundle and the geodesic-circle
bundle \citep[pp.~381--382]{Yang1991}. Here the degree-four
generator leads to a different middle-degree lattice.
Let $\lambda$ be the unit tangent contact form,
\[
\lambda_w(\zeta)=\langle w,d\pi_w\zeta\rangle.
\]
Its Reeb flow is the unit-speed geodesic flow. Consequently
$\theta=\lambda/T$ is a connection for the above $\R/\Z$ action:
it is invariant and integrates to one on each orbit. Its curvature
is the pullback of a symplectic form $\omega_B$ on $B$.
We fix the Euler-class sign convention by
\[
\rho^*\omega_B=d\theta,\qquad e=[\omega_B]\in H^2(B;\Z),
\]
and give $B$ its symplectic orientation.
\begin{proposition}[Contact volume]\label{geo:contact}
With this orientation and Euler class,
\begin{equation}\label{geo:contact-formula}
N:=\int_B e^{4r-1}
=\frac{(4r-1)!\,|S^{4r-1}|\,\Vol(M)}{T^{4r}}>0.
\end{equation}
For the standard quaternionic model at the same diameter, this
number is
\begin{equation}\label{geo:model-number}
N_0=\frac{1}{2r+1}\binom{4r}{2r}.
\end{equation}
\end{proposition}
\begin{proof}
Integration over the circle fibers gives
\[
\int_Be^d
=\int_U\theta\wedge(d\theta)^d
=T^{-(d+1)}\int_U\lambda\wedge(d\lambda)^d,
\]
where $U$ has the orientation induced by the positive circle
direction and the chosen orientation of $B$.
To compute the last form as a density, use the connection splitting
of $T_wU$ into horizontal and vertical directions. Choose an
orthonormal frame $(w,e_1,\ldots,e_d)$ at $\pi(w)$.
The contact form reads off the horizontal $w$ coordinate, and
$d\lambda$ pairs each horizontal $e_i$ coordinate with its vertical
$e_i$ coordinate, up to one common sign convention.
In the $d$-fold exterior power each pairing occurs once in each of
$d!$ orders. Hence
\[
\big|\lambda\wedge(d\lambda)^d\big|=d!\,d\mu,
\]
where $d\mu$ is base Riemannian volume times unit tangent sphere
area. The chosen orientation makes the contact integral positive,
and $\mu(U)=|S^d|\Vol(M)$. Since $d+1=4r$, this proves
\eqref{geo:contact-formula}.
The model's polar density on $(0,D)$ is
$\sin^{4r-1}t\cos^3t$. With $y=\sin^2t$,
\[
\int_0^{\pi/2}\sin^{4r-1}t\cos^3t\,dt
=\frac12\int_0^1 y^{2r-1}(1-y)\,dy
=\frac{1}{4r(2r+1)}.
\]
Using $|S^{4r-1}|=2\pi^{2r}/(2r-1)!$ and $T=\pi$ in
\eqref{geo:contact-formula} gives
\[
N_0=\frac{(4r-1)!}{r(2r+1)((2r-1)!)^2}
=\frac{(4r)!}{(2r)!(2r+1)!},
\]
which is \eqref{geo:model-number}.
\end{proof}
Thus the desired upper volume bound is $N\le N_0$. The integral
cohomology calculation below computes the determinant of a middle
pairing whose entries are top-degree products, including this
characteristic number.
\begin{lemma}[Integral Gysin information]\label{geo:gysin}
The manifolds $U$ and $B$ are simply connected. There is an
integral class $v\in H^4(B;\Z)$ with $\rho^*v=z$.
For $2j0.
\label{alg:middle-input}
\end{equation}
Our goal is
$N\le\binom{4r}{2r}/(2r+1)$, the model value computed in
Proposition~\ref{geo:contact}. The attachment gives polynomial
relations; the middle determinant fixes the scale of the top
functional; an explicit coefficient sum then gives the bound.
We do not prescribe the cohomology ring of the geodesic quotient
in advance.
\subsection{The power relations and the nonzero parameter}
Introduce a formal variable $Z$ with $Z^2=L$ and define homogeneous
polynomials in $X,Y$ by
\begin{align}
Q(X,Y)&=\frac{(Y+Z)^{r+1}-(Y-Z)^{r+1}}{2Z}
=\sum_{j\ge0}\binom{r+1}{2j+1}Y^{r-2j}L^j,
\label{alg:Q}\\
R(X,Y)&=\frac{(Y+Z)^{r+1}+(Y-Z)^{r+1}}2
=\sum_{j\ge0}\binom{r+1}{2j}Y^{r+1-2j}L^j.
\label{alg:R}
\end{align}
Only the terms with valid binomial indices are included. The quotient
in~\eqref{alg:Q} is a polynomial identity; it involves no division by a
cohomology class. The degrees of $Q,R$ are $r,r+1$, respectively.
\begin{lemma}
\label{alg:relations}
The classes $Q(X,Y)$ and $R(X,Y)$ vanish in
$H^*(B;\mathbb Q)$. Moreover $\alpha\ne0$, and consequently
$|4\alpha|\ge1$.
\end{lemma}
\begin{proof}
On the sphere bundle $p:S(E)\to B$, the attachment class is
$x=Y+e\xi$ and satisfies $x^{r+1}=0$. The fiber antipode sends $\xi$
to $-\xi$, so $(Y-e\xi)^{r+1}=0$ as well. Taking the sum gives
$p^*R=0$. The rational Leray--Hirsch decomposition with basis $1,\xi$
makes $p^*$ injective, hence $R=0$. Taking the difference and integrating
over the sphere fiber, using $p_!\xi=1$, gives $eQ=0$. The class $Q$ has
cohomological degree $4r=n$, and multiplication by $e$ is injective from
$H^n(B;\mathbb Q)$ to $H^{n+2}(B;\mathbb Q)$ by
Lemma~\ref{geo:gysin}. Therefore $Q=0$.
Suppose now that $\alpha=0$. We use the nondegenerate moment pairing
in~\eqref{alg:middle-input} to obtain a contradiction. Write
\[
q(y)=Q(1,y),\qquad a(y)=R(1,y),\qquad
A_0=\C[y]/(q).
\]
The polynomial $q$ has degree $r$ and leading coefficient $r+1$.
The relations $Q=0$ imply that the rule $y^j\mapsto I_j$, for
$0\le j\le2r-1$, agrees with reduction modulo $q$ followed by a
functional $F:A_0\to\C$. Indeed, every multiple of $q$ needed in
dividing a polynomial of this degree has multiplier of degree at most
$r-1$, and its homogenization is a top-degree multiple of $Q$.
Thus the matrix of $(f,g)\mapsto F(fg)$ in $1,y,\ldots,y^{r-1}$ is
precisely the nonsingular matrix in~\eqref{alg:middle-input}.
The relation $R=0$ similarly gives
\begin{equation}
F(a(y)y^j)=0\qquad(0\le j\le r-2).
\label{alg:orthogonal-R}
\end{equation}
Let $\varepsilon:A_0\to\C$ extract the coefficient of $y^{r-1}$
in the remainder of degree less than $r$. Its product pairing is
nondegenerate: the entry $\varepsilon(y^{i+j})$ is zero when
$i+j0$, the preceding termwise comparison proves
\eqref{alg:sharp-upper}.
\end{proof}
==== END The-metric-Blaschke-theorem//sections/08-volume-algebra.tex ====
==== BEGIN The-metric-Blaschke-theorem//sections/09-classification.tex ====
\section{Completion of the metric classification}
\label{completion:section}
Theorem~\ref{thm:comparison} reduces metric rigidity to a volume
upper bound at the same diameter. The intrinsic quaternionic bound
has now been proved. We first check the precise classical inputs
for the complex and plane cases, then assemble the model list.
For $\HP^2$, the classical plane argument is an alternative to the
intrinsic bound, which already includes every quaternionic rank $r\ge2$.
\begin{corollary}[Complex-projective type]\label{rig:complex}
A Blaschke manifold of complex-projective type is isometric
to its equally scaled standard complex projective model.
\end{corollary}
\begin{proof}
Write its real dimension as $2r$, with $r\ge2$; the projective
line is a sphere. Its integral cohomology ring is
$\Z[u]/(u^{r+1})$, where $u$ has degree two, and it is simply
connected. Represent $u$ by a map to $\CP^\infty$
\citep[Theorem~4.57]{Hatcher2002}. A closed
smooth $2r$-manifold has the homotopy type of a finite
$2r$-dimensional CW complex, so cellular approximation
\citep[Theorem~4.8]{Hatcher2002} places the map in the
$2r$-skeleton $\CP^r$. Pullback identifies the integral
cohomology generators and all their powers. The universal
coefficient theorem for these finite CW complexes then gives
an integral homology isomorphism. The homology Whitehead
theorem for simply connected CW complexes
\citep[Corollary~4.33]{Hatcher2002} makes this map a homotopy
equivalence. Yang's volume theorem
\citep[Theorem, p.~380]{Yang1991} therefore gives equality with the model at the
same prime geodesic period, equivalently the same diameter.
Corollary~\ref{thm:volume-reduction} applies.
\end{proof}
\begin{corollary}[Quaternionic and Cayley planes]\label{rig:planes}
A Blaschke manifold with model $\HP^2$ or $\OP^2$ is isometric
to that model at the same diameter.
\end{corollary}
\begin{proof}
The projective-plane structure of such a Blaschke manifold is
given by \citet[Lemma~5]{McKay2016}. The classification of smooth
topological projective planes
\citep[Corollary~C]{KramerStolz2007} makes $M$ diffeomorphic to
its standard model. Both metrics are Zoll: all their unit-speed
geodesics are simply closed with the same prime length $T=2D$.
After a common rescaling to $T=\pi$, the Riemannian weak Blaschke
volume theorem recorded by Wilking and attributed there to
Reznikov applies: a manifold with this common least period
that is homeomorphic to a compact rank-one symmetric space has
the same volume as its model \citep[p.~282]{Wilking2001}.
The diffeomorphism above supplies its homeomorphism hypothesis.
Thus $V=V_0$, and this equality persists when the common rescaling
is undone. Corollary~\ref{thm:volume-reduction} applies.
\end{proof}
\begin{corollary}[Quaternionic type]\label{rig:quaternionic}
A Blaschke manifold of quaternionic type is isometric to its
standard quaternionic model at the same diameter.
\end{corollary}
\begin{proof}
For $\HP^r$ with $r\ge2$, use Theorem~\ref{alg:upper} and
Proposition~\ref{geo:contact}. At the common period $T=\pi$, the
contact formula has the same positive volume factor for $M$ and its
model, so
\[
\frac{N}{N_0}=\frac{\Vol(M,g)}{\Vol(\HP^r)}.
\]
Here $N_0$ is the model number in \eqref{geo:model-number}, and the
model volume computed there is
\[
\Vol(\HP^r)=\frac{\pi^{2r}}{(2r+1)!}.
\]
Theorem~\ref{alg:upper} gives $N\le N_0$, hence
$\Vol(M,g)\le\Vol(\HP^r)$.
Corollary~\ref{thm:volume-reduction} proves the metric statement.
The projective line is already the sphere case.
\end{proof}
\begin{proof}[Proof of Theorem~\ref{thm:blaschke}]
Lemma~\ref{geom:cut-time} gives the common minimizing distance
from the stated global injectivity-radius equality. The
one-dimensional, sphere-type and real-projective-type cases
are covered by Section~\ref{geom:section}. The classical
structure list leaves complex type, quaternionic type, and the
Cayley plane. These are respectively
Corollaries~\ref{rig:complex}, \ref{rig:quaternionic},
and~\ref{rig:planes}. Each conclusion is a metric isometry
with the model at equal diameter. Undoing the normalization
therefore gives exactly the asserted classification up to scale.
\end{proof}
==== END The-metric-Blaschke-theorem//sections/09-classification.tex ====
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==== END The-metric-Blaschke-theorem//references.bib ====