The metric Blaschke theorem

Proof. Compactness and positive dimension give 0<D<\infty, and every pointwise injectivity radius is at least D. For 0<t<D, the exponential map at a point p is injective and nonsingular on a tangent ball containing the sphere of radius t. A minimizing geodesic to \exp_p(tv) exists by completeness and has length at most t. Its initial vector lies in that ball and maps to the same endpoint as tv, so injectivity identifies the two vectors. This proves both minimality and uniqueness. Taking t\uparrow D and using continuity of distance proves minimality at D. If a segment of length greater than D minimized, the distance between its endpoints would exceed the diameter.□

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 (3.3) and (3.2).

Evaluation of the pairing at y gives

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,

Equation (3.4) 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),

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 (3.6).□

Proof. Put \tau=t-D. Choose a local smooth frame of L_t, written as the columns of

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

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

Consequently

for some matrix N. In particular \widetilde X(D) is invertible. For \tau\ne0 sufficiently small,

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,

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 (3.9) with B=G^{-1}>0, including the absence of any off-diagonal residue.

For fixed z\in\R^m, the graph vectors

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

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 (3.2); 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 (3.10).□

Proof. Lemma 3.2 and (3.8) show that H' is positive throughout (0,T), including D. Near t_0=0,T, put \tau=t-t_0. Differentiating the smooth-remainder expansion gives

Here and below O(1) denotes a matrix bounded in operator norm as the indicated time is approached. Since H'(t)^{-1}=O(\tau^2), the Schur complement satisfies

At D, both C' and H' are smooth and H'(D) is invertible. The middle pole therefore yields

Taking determinant roots gives the positive limits

Each ratio thus extends continuously and positively to its compact closed interval. Its minimum and maximum supply the stated bounds.□

Proof. With R=(\Delta C)(\Delta H)^{-1}, block congruence gives

This proves the determinant identity and identifies the nullity of \Delta A with that of Q. Exchanging the endpoints changes the sign of each difference, and hence changes Q to -Q.

For a matched pair (u,u+D), the kernel of \Delta A parametrizes L_u\cap L_{u+D} by the common q coordinate. It has dimension k by (3.4). The matrix Q has size k\times k and therefore has full nullity: it is the zero matrix. This proves (3.20), also in the reversed order. Near a diagonal point within the chart, Taylor expansion gives

and the corresponding expansion for \Delta V_b. Invertibility of H'(x) then gives

This establishes the zero extension needed at the endpoints of the integrations below.

Direct differentiation of (3.18), using \partial_y(\Delta H)^{-1}=-(\Delta H)^{-1}H'(y)(\Delta H)^{-1}, gives

The identical calculation with -\partial_x gives the same expression with A'(x). For either endpoint a=x or a=y, completing the square gives the exact identity

The second summand is positive semidefinite. This proves (3.21). The calculation used invertibility of \Delta H, but did not require its positive definiteness.

To obtain (3.22), fix the first endpoint u+iD and move the second from u+jD to v+jD. The starting value of Q is zero by the diagonal extension if i=j, and by (3.20) otherwise. Integrating \partial_yQ\ge S(y) on this interval gives the result. For (3.23), fix the second endpoint v+jD and move the first from u+iD to v+iD. The terminal value is zero, so integration of -\partial_xQ\ge S(x) gives the second result. When i=j, one can first integrate on a shortened interval and pass to the diagonal endpoint using (3.25).

Figure 1 displays these integrations with the half-period indices held fixed.

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.□

Proof. We first record the determinant inequality used in both blocks. For a positive definite r\times r matrix M, write \delta_r(M)=\det(M)^{1/r}. The elementary variational formula

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,

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 (3.32) to H' on the interval between x and y. Equation (3.17) gives

For y<x this uses the positive matrix -\Delta H. Division by

proves (3.29).

Now put Q=Q(u+iD,v+jD)>0. Each of the two matrix bounds in Proposition 3.4, followed by determinant monotonicity and (3.32), gives

Thus \delta_k(Q) dominates their maximum and therefore their geometric mean. The denominator in the kth root of f_K is exactly

which proves (3.30) with the displayed powers.

Finally, every endpoint in \mathcal I has a unique expression u+iD with 0<u<D and i\in\{0,1\}. For a pair whose folded parameters differ, order those parameters as u<v, exchanging the actual endpoints if necessary. The preceding bounds apply for every choice of the two half-period indices. Equal folded parameters mean either a diagonal pair or a cut pair. The remaining excluded pairs have an endpoint at D. These are finitely many lines, and symmetry of both factors justifies the endpoint exchange everywhere else.□

Proof. Direct integration gives

The measure \nu_{xy} is a probability measure. Both its average of g and \log\int e^g\,d\nu_{xy}=\log(\int p/\int p_0) have absolute value at most \|g\|_\infty. Since W is essentially bounded, the asserted finiteness follows already from these bounds. Jensen’s inequality makes the integrand on the second line of (4.1) nonnegative. It remains to establish the equality of the integrated linear terms.

Set K(z)=W(z)/\sin(\lambda z) and I_W=\int_0^b W(z)\,dz, and define

Here the two-variable kernel q(x,y)=K(y-x)\sin(\lambda x)\sin(\lambda y) is essentially bounded. Fubini gives

Thus F is absolutely continuous on [0,b], F(0)=0, and, for almost every t,

For the second equality substitute b-z in the subtracted integral: K(b-z)=K(z) and \sin(\lambda(t-b+z))=-\sin(\lambda(t+z)). For the last equality, symmetry gives \int_0^b K(z)\cos(\lambda z)\,dz=0, while \int_0^b K(z)\sin(\lambda z)\,dz=I_W. Integration therefore yields

This argument has not differentiated the measurable weight.

Using (4.2) and then (4.3), the integrated interval averages are

The exchange is absolutely integrable: replacing g by |g| in the left side bounds it by \|g\|_\infty\int_{\Delta_b}W(y-x)\,dx\,dy. In particular, no integral of p_0 over the entire interval is taken. The endpoint terms have exactly the same value, since

The final equality again uses W(b-z)=W(z). Subtracting these equal finite quantities proves the identity and hence the lemma.□

Proof. All constants in this proof may depend on the chosen geodesic. Write w_{xy}=w(|y-x|). We first work off the null set

There f,w_{xy}, and the two determinant factors f_H,f_K of Proposition 3.5 are positive, and f=f_Hf_K. Define

Set their values on E equal to zero for integration. Both are symmetric in the actual endpoint slots x,y.

Finiteness of the logarithmic integrals. The profile bounds in Proposition 3.3 give constants 0<\alpha\leq\beta<\infty such that

Set \eta=\alpha/\beta\leq1. For x<y, the first bound in Proposition 3.5 and the chord identity (4.2) imply

Symmetry gives this lower bound for either endpoint order. Similarly, put B_0(z)=\sin(2z)/2 for 0<z<D. For 0<u<v<D and i,j\in\{0,1\}, the second block bound gives

In these coordinates B_0(v-u)=|\sin(2|y-x|)/2|. Every pair outside E has this form, possibly after exchanging endpoints. Consequently

With a^+=\max(a,0), the elementary estimate

and continuity of f on the compact square prove integrability of w_{xy}L^+. The individual negative parts are bounded by (4.5); also

Thus w_{xy}L_H, w_{xy}L_K, and w_{xy}L are all absolutely integrable. The following additions of logarithmic integrals therefore involve only finite quantities.

The H factor on the full square. Apply Lemma 4.1 with b=T, \lambda=1, p=h, and W=w. Its assumptions hold by Proposition 3.3 and

which is bounded because m\geq1. Writing g_H=\log(h/\csc^2t), the first block bound and (4.2) give, for 0<x<y<T outside E,

After multiplication by w(y-x), Lemma 4.1 makes the integral of the right side nonnegative. Multiplication by m and endpoint symmetry then yield

The K factor and the four folded weights. Put p_0(t)=4\csc^2(2t) on (0,D) and define

For (u,v)\in\Delta_D, z=v-u, x=u+iD, and y=v+jD, the second block bound gives the precise logarithmic inequality

Indeed, (4.2) with \lambda=2 converts the reference interval integral and its endpoint factors into B_0(z). The actual separation is z if i=j, D+z if (i,j)=(0,1), and D-z if (i,j)=(1,0). In the last case x>y; the bound in Proposition 3.5 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

Every row and every column sums to

This is symmetric under z\mapsto D-z, and

is bounded for every k\geq1. When k=1, its limit at either endpoint is 1/2, so this case also satisfies Lemma 4.1.

Multiply (4.7) 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

Lemma 4.1, 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

Here the factor 2 accounts for endpoint exchange and the factor k restores the logarithm of f_K from that of its kth root.

Comparison and equality. Adding (4.6) and (4.10) gives

For r>0, the strictly convex function

is nonnegative and vanishes exactly at r=1. On \mathcal R\setminus E put r=f/w_{xy}. All terms in the identity

are finite by the preceding integrability argument. Both terms on the right are nonnegative, proving (4.4). 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 (4.4).□

Proof. The open injectivity ball maps diffeomorphically onto the complement of the cut locus. The latter is a smooth submanifold of dimension d-k<n, hence has zero Riemannian volume. Polar coordinates therefore prove the first formula. Periodicity of the Jacobi data and reversal of the geodesic give

The sign in the reversed initial derivative disappears under the absolute determinant. The substitution t\mapsto T-t in the second half-period, followed by the area-preserving antipodal map on S_pM, proves the second formula.□

Proof of Theorem 1.2. The circle, sphere-type, and real-projective-type cases are already covered in Section 2. In an intermediate case use the normalization (2.1). Theorem 4.2 says that

Time translation along a geodesic gives

The possible choice of endpoint for the residue modulo T has no effect on any integral. Fubini’s theorem, translation on the time circle, flow invariance of \mu, and Lemma 5.1 yield

All integrands are continuous and bounded on the compact domains in question. Since w(T-t)=w(t), its periodic extension gives likewise

Integrating (5.1) and dividing by 2TV>0 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 4.2 now gives

In Cartesian normal coordinates at any point, the volume density is J(\xi,t)/t^d. By (5.2) it depends only on t. At the origin it extends smoothly with value one, as also follows directly from the normal-coordinate volume form. Thus M is harmonic. It is compact and simply connected by Proposition 2.2. The compact simply connected case of the Lichnerowicz conjecture, proved by Szabó [25], identifies its metric with a compact rank-one symmetric metric up to scale. Its type identifies the model, and the fixed diameter identifies the scale. For the radial-density definition and the concluding metric classification, see also the author’s original preprint [24, pp. 2–4 and Lemmas 4.9–4.10].

Conversely, a model is isometric to itself and has volume V_0. Finally both sides of the volume comparison scale by the same factor under a constant metric rescaling. Undoing (2.1) proves the assertion at every diameter.□

Proof of Corollary 1.3. The upper bound and Theorem 1.2 give equal volumes, so its equality case applies.□

Proof. The full cut map

has constant rank 4r+(4r-4)=8r-4. Its fibers are the connected great three-spheres from Proposition 2.2. Here constant rank and compactness give an embedded image, as follows. Constant-rank charts give local image germs through any image point. Charts overlapping along its fiber have the same germ, so connectedness of that fiber makes all these germs agree. Cover the compact fiber by finitely many such charts. Properness excludes additional inverse images outside their union over a sufficiently small neighborhood of the image point. Thus the whole image in that neighborhood is the one smooth submanifold germ. This proves the assertion. The first projection is a submersion because its composition with c is \pi; symmetry gives the same conclusion for the second. Proper submersions are smooth fiber bundles.

For (a,b)\in \mathcal V_{(p,q)}\subset T_pM\oplus T_qM, orthogonality to (T_pC_q,0) gives a\in N_pC_q. If a=0, surjectivity of the second projection on T_{(p,q)}F forces b=0. The projection to N_pC_q is therefore injective and, both spaces having dimension four, an isomorphism. First variation for the constant distance D shows that the initial directions from p to q are normal to C_q. Their great sphere has the full dimension of S(N_pC_q), proving the last assertion. In particular, if q_1,q_2\in C_p are distinct, then

a common nonzero direction would reach both points at time D.

The cut loci and M are simply connected, so the fibration makes F simply connected. Its normal bundle is orientable. Leray–Hirsch [9, Theorem 4D.1], using 1,u_2,\ldots,u_2^{r-1} on the fibers, and disjointness from the diagonal give the integral presentation

To justify completeness, the displayed relation is the restriction of the diagonal class in M\times M. It is monic of degree r in u_2, so the proposed quotient has exactly the Leray–Hirsch basis. The dual of F in M\times M restricts on either factor slice to \pm u. Its restriction to the disjoint diagonal is zero. Its two factor coefficients are consequently opposite, and after an orientation choice its restriction to F is u_1-u_2. The self-intersection formula gives the asserted Euler class.

The interchange of factors preserves the unoriented normal bundle. Since u_1,u_2 are an integral basis of H^4(F;\Z), we have p_1(\mathcal V)=t(u_1+u_2) for an integer t. Also H^2(F;\Z/2)=0, so \mathcal V is spin. We use the classical identification \operatorname{Spin}(4)=\operatorname{SU}(2)\times\operatorname{SU}(2) and its two complex two-dimensional half-spin representations; see Moore [17, Sections 2.1–2.2]. For an oriented spin four-plane bundle, its two complex half-spin bundles have second Chern classes

This follows directly from the formal half-spin weights \pm(t_1+t_2)/2 and \pm(t_1-t_2)/2, where p_1=t_1^2+t_2^2 and e=t_1t_2. Integrality in the basis u_1,u_2 implies t\equiv2\pmod4. Writing t=2s proves the proposition.□

Proof. Over Z, the two actual normal four-planes in (6.2) have direct sum inside the pullback of TM from p. Choose a complementary bundle W of rank 4(r-2). Set

The rank-four identity p_2=e^2 [16, Corollary 15.8] and Proposition 6.1 give

Its polynomial degree is at most t_0=2r-4.

We will prove that the formal characteristic class

has degree at most t_0 and a degree-t_0 component that is nonzero modulo v. The freeness in (6.5) will then show that any first discrepancy between p(TM)(v) and p_0(v) violates the rank bound on W. For the degree bound we use only the algebra (6.4), without presuming the answer for TM. Successive monic polynomial divisions factor

in an extension of A obtained by adjoining roots of monic polynomials. Give each root degree one; the monic relations are homogeneous for this grading. Each adjoining algebra is free with a basis containing 1, so this extension is injective. The positive-degree elementary symmetric functions of all r+1 roots vanish. Consequently, the product of D_s(v,z) over all these roots is (1+2sv+v^2)^{r+1}. The factor at z=v is 1+4sv. Dividing these constant-one series therefore gives

There are r-2 factors, each of degree at most two, as required.

We also need the top component (\mathcal W_0)_{t_0} to be nonzero modulo v. For r=2 it is 1. For r\ge3, let

The recursion a_0=1, a_1=-2s and a_j=-2sa_{j-1}-a_{j-2} gives

Indeed the left side is unchanged on replacing j by j-1, and its initial value is one. Modulo v the algebra can also be presented as

In degree 2r-4 the surviving monomials after the first two truncations are

The remaining ideal gives exactly their sum as a relation: multiplying its generator by any monomial of degree r-3 gives this same sum. Their coefficients in \mathcal W_0 are, in order, a_{r-3}a_{r-1},a_{r-2}^2,a_{r-3}a_{r-1}. Equation (6.8) makes the middle coefficient differ from either outside coefficient by one. Thus the component modulo v is the nonzero class of x_1^{r-2}x_2^{r-2}, one of the monomials in (6.5).

Finally suppose that p(TM)(v)/p_0(v) first differs from 1 in degree i, where 1\le i\le r. Write its first coefficient as b\ne0. In degree t_0+i, modulo v^{i+1}, (6.7) is b v^i(\mathcal W_0)_{t_0}. This is nonzero by (6.5) and the preceding calculation: in a free module over \mathbb Q[v]/(v^{r+1}), a coefficient nonzero modulo v remains nonzero after multiplication by v^i, modulo v^{i+1}, for i\le r. It contradicts the rank bound on W. Therefore there is no such i, which proves the theorem.□

Proof. Keep the choices s\ge1 and \int_Mu^r=1. Set

Hirzebruch’s signature theorem [16, Signature Theorem 19.4], applied to Theorem 6.2 gives

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

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 (7.1) gives

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

Then 0<\tau\le3-\sqrt8<0.172 and \sqrt2<b<1.435. Substitute u=-t^2 in (7.1). Since f(-a^2t^2)=at/\tan(at) and a_+a_-=1, taking the coefficient by a residue and then putting z=a_+t gives

The change of variables contributes the factor 1/(ca_+)=\tau/b; this fixes the normalization of the residue.

Contour representation. Take the rectangle with vertical sides \Re z=\pm X, X=3\pi/4, and horizontal sides \Im z=\pm4. Apart from zero, the only poles of the integrand in this rectangle are at z=\pm\rho. Zeros of \tan z occur only at multiples of \pi, zeros of \tan(\tau z) only at multiples of \pi/\tau, and the next poles of \tan(bz) are outside because 3\pi/(2b)>X. 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 (7.2) as

Symmetry of the four sides and the triangle inequality bound |\mathcal E|/L by the sum of

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

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

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

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, (7.6) now implies

Here the integral equals

On a horizontal side the identity for the modulus of tangent gives

Since \tau/\tanh(4\tau) is increasing,

Substitution in (7.5), with X/\pi=3/4, therefore gives

The numerical trigonometric inequalities used here can be checked without numerical approximation: use

for \tan\rho and the smaller real arguments. For the hyperbolic inequalities use

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 (7.3) and r+1\ge5 imply

Combining these estimates yields

contradicting either possible cohomological signature. Thus s=1. Restricting (6.1) to a fixed-p fiber gives e(\mathcal V)=-u_2 and p_1(\mathcal V)=2u_2 there. Proposition 6.1 identifies this restriction with the actual tautological four-plane bundle, which proves the last assertion as well.□

Proof. Integration over the circle fibers gives

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

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 (8.2).

The model’s polar density on (0,D) is \sin^{4r-1}t\cos^3t. With y=\sin^2t,

Using |S^{4r-1}|=2\pi^{2r}/(2r-1)! and T=\pi in (8.2) gives

which is (8.3).□

Proof. The sphere fiber of \pi has dimension n-1\ge7. Its homotopy sequence and simple connectivity of M give \pi_1(U)=0. The homotopy sequence of the circle bundle then gives \pi_1(B)=0.

The Euler number of TM is \chi(M)=r+1. The chosen orientation therefore gives

The integral Gysin sequence of the oriented sphere bundle \pi has the form

It gives the complete additive calculation

Here the upper generators may be chosen with \pi_!y_a=u^a and y_a=z^{a-1}y_1; the latter assertion follows from the projection formula and the fact that \pi_! is an isomorphism in these upper degrees. At degree n-1, the Gysin map from H^0(M) to H^n(M) is multiplication by the nonzero integer r+1, so H^{n-1}(U)=0. This also explains the sole torsion group in (8.8).

Apply next the integral Gysin sequence of \rho. The vanishing of odd groups of U below degree n inductively gives H^{2j-1}(B)=0 for 2j-1<n. For positive even 2j<n the remaining terms give

For 2j=2 this says H^2(B)=\Z e. For 2j=4, the last group is freely generated by z, so choose an integral lift v. If j is odd, the last group in (8.9) vanishes and multiplication by e gives the next basis. If j=2a is even, the last group is generated by z^a, and the class v^a lifts that generator. Induction proves exactly the integral basis (8.4).

In all degrees at most n-2, the homology groups of B are free and have the same ranks as its cohomology groups. Indeed the universal coefficient theorem detects torsion in H_i(B;\Z) as torsion in H^{i+1}(B;\Z), and the latter groups are free through degree n-1. Poincaré duality on the oriented (2n-2)-manifold B now supplies all its cohomology in degrees at least n. It is free in even degrees and zero in odd degrees; together with the low-degree calculation this proves the asserted global statement.

The circle Gysin sequence at degree n uses H^{n-1}(U)=H^{n-1}(B)=0 and H^n(U)=\Z/(r+1), and is precisely (8.6). The Poincaré pairing

is unimodular, since these two groups are free. Composing one factor with the injection of index r+1 in (8.6) proves the determinant assertion (8.7). Finally H^{n+1}(U)=H^{n+2}(U)=0 by (8.8). The next portion of the circle Gysin sequence therefore makes multiplication by e from degree n to degree n+2 an isomorphism.

In every even degree q\ge n, the group H^q(U;\mathbb Q) vanishes. The rational circle Gysin sequence therefore makes e\smile:H^{q-2}(B;\mathbb Q)\to H^q(B;\mathbb Q) surjective. The low-degree bases now show that the full rational ring is generated by e,v. Below degree n, the basis ranks sum to

Poincaré duality supplies another r(r+1) dimensions above degree n-1, and the middle odd group is zero. This proves (8.5).□

Proof. We give the index, attachment, and integral-coefficient details.

Critical manifolds and indices. The positive critical levels are j^2T/2, j\ge1; the critical manifold at each is a copy of U, parametrized by w\mapsto(t\mapsto\gamma_w(jt)). These are all the nonconstant critical loops because every geodesic has prime length T.

For the jth iterate, reparametrize variations by arclength on [0,jT]. The kernel of the periodic Hessian consists of all periodic Jacobi fields. All normal Jacobi fields are T-periodic by the differential of \phi_T=\Id, so their dimension is 2d. A tangential Jacobi field is an affine scalar multiple of the velocity; the periodic boundary conditions force its scalar to be constant. The kernel therefore has dimension 2d+1=\dim U. Variations of the initial unit tangent vector, including the time-shift direction, realize precisely these fields. Thus the Hessian kernel is the tangent space to the critical manifold, as required by the Morse–Bott condition.

The periodic index equals the fixed-endpoint index. To see this without an index correction formula, evaluation at the marked initial point maps the periodic kernel onto T_{\pi(w)}M: arbitrary normal initial values are possible, and the constant tangent field supplies the remaining direction. Choose a linear right inverse to this evaluation. Subtracting the corresponding kernel field from a periodic test field makes both endpoint values zero. The index form is unchanged, including its mixed terms, because a kernel field pairs to zero with every periodic test field by integration by parts. This subtraction is injective on every negative definite subspace. Conversely, fixed-endpoint test fields form a subspace of the periodic domain. The two indices are consequently equal.

Interior conjugate instants on [0,jT] occur at the j odd multiples of D, with multiplicity three, and the j-1 positive multiples of T, with multiplicity d. The fixed-endpoint Morse index theorem [15, Theorem 15.1, p. 83] gives

In particular \lambda_1=3, while every later index is at least d+6=n+5\ge13. The index count is unchanged on completing the piecewise smooth zero-endpoint fields in H^1: the index form is continuous, and any finite negative Gram matrix remains negative after sufficiently close smooth approximation.

Negative bundles and attachment. Let \nu\to U be the negative spectral bundle of the periodic L^2 Jacobi operator

This self-adjoint operator has domain the periodic H^2 fields, and its quadratic form is the index form on periodic H^1 fields. Its negative eigenvalues form an isolated finite cluster. The constant index and nullity, together with compactness of U, make the corresponding spectral projections locally smooth in the initial data. Elliptic regularity makes their eigenfields smooth, so \nu is a genuine rank-three bundle. It represents the negative bundle for the Hilbert-space Morse–Bott attachment as well. Indeed, projection from the negative subspace of the bounded H^1-Riesz Hessian to \nu_w is injective: a vector in its kernel would have nonnegative index form in the L^2 spectral decomposition. Equal negative ranks make this a bundle isomorphism. Each fixed parameter rotation intertwines these operators and projections with those at the rotated geodesic. The resulting action on \nu is continuous and covers the free smooth circle action on U. Hence \nu descends to a real bundle E\to B, with \nu\cong\rho^*E. Since B is simply connected, E is orientable.

We use the Morse–Bott disk-attachment theorem for the energy on the H^1 loop space of a closed Riemannian manifold. The energy satisfies the Palais–Smale condition. Its Hessian is nondegenerate normal to each compact critical manifold just described, and has finite negative index. The bounded H^1 Hessian is Fredholm; its identified kernel therefore gives an invertible restriction normal to the critical manifold, as in the formulation cited below. Thus sublevels bracketing one critical value differ, up to homotopy, by the disk and sphere bundles of its negative bundle. This construction also respects parameter rotation. Indeed each fixed rotation is a smooth isometry for the natural loop-space metric, preserves energy, commutes with the gradient deformation, and preserves the Hessian splitting. The local normal form and the contraction of its positive directions may be constructed from these invariant metrics and splittings. Joint differentiability of the circle action on the whole H^1 manifold is not needed. For this equivariant negative-bundle formulation see Ottosen and Bökstedt [18, Section 7, pp. 2190–2192], following the Hilbert-space attachment of Klingenberg [12, Theorem 2.4.10]. The associated Borel vector bundle and Thom quotient are described in Ottosen and Bökstedt [18, Lemma 5.1, pp. 2185–2186].

Here is the attaching map and its evaluation property explicitly. Start a sufficiently small negative disk at every first-level geodesic by pointwise exponential of the negative fields. Its boundary lies strictly below the first positive critical value, uniformly over U. That lower sublevel deforms equivariantly to the constants. This follows from the noncritical energy deformation between the boundary level and a small positive level, followed by contraction of small loops to constants. For the latter, embed M in Euclidean space, average a sufficiently short loop, and project the straight contraction to a tubular neighborhood of M. Small energy gives uniformly small length, so this is defined and continuous, preserves the constants, and commutes with rotation. The projection has bounded derivative, so choosing the initial small level sufficiently small keeps this contraction below the first critical value.

Insert the boundary deformation in an outer collar of the small negative disk bundle. This produces an equivariant map of pairs

whose zero section is the first critical manifold and which represents its Morse–Bott attachment. On the boundary it takes values in the constants. Because the action on the constants is trivial, this boundary map descends to f:S(E)\to M. Evaluation at time zero composed with \mathcal F_1 is homotopic, by radial contraction of the disk fibers, to \pi composed with their bundle projection. Restrict this homotopy to S(\nu)\cong S(\rho^*E). Its boundary value is f\circ\widehat\rho, which proves (8.10).

Choose regular sublevels \Lambda_- and \Lambda_+ immediately below and above the first positive critical value. The preceding attachment gives, over \Z or any field \mathbb F,

These are Thom isomorphisms; in the second one the free first critical action identifies the Borel negative bundle with E up to homotopy. One can take finite-dimensional free approximations to ES^1 throughout, apply the invariant attachment, and then pass to the limit. For every later critical level, the Borel negative bundle still exists over the Borel critical manifold even though the action has finite stabilizers. Its relative cohomology vanishes below its rank \lambda_j, with any necessary orientation local system. The same statement holds without the Borel construction. Equation (8.13) therefore shows that subsequent attachments do not change cohomology in degrees at most five. The exhaustion by sublevels causes no extra term in these degrees: the restriction systems in the relevant degrees are eventually constant, so their inverse-limit obstruction vanishes.

We have constructed the attachment and isolated the degrees in which no later attachment can alter it. What remains is to fix its coefficient integrally and identify its actual negative bundle.

The coefficient is a unit over every prime field. Since H^2(B;\Z)=\Z e, write p_!x=c e with c\in\Z. Under (8.15), the connecting homomorphism

sends the class u from M to p_!x, up to the Thom orientation sign. Indeed the connecting map first restricts to the attaching sphere, where that class is x, and then uses the boundary map for (D(E),S(E)); the inverse Thom isomorphism identifies this last map with integration over the oriented sphere fiber.

Suppose that c vanished modulo a prime \ell, and put \mathbb F=\Z/\ell. Exactness would extend the pure class u\in H^4(M\times BS^1;\mathbb F) over (\Lambda_+)_{hS^1} and hence over (\Lambda M)_{hS^1}, by the higher-index gap. After forgetting equivariance, its restriction would be the evaluation class

To justify uniqueness here, (8.14) gives H^4(\Lambda_+,M;\mathbb F)=H^1(U;\mathbb F)=0. Higher attachments have no effect in this range, so restriction H^4(\Lambda M;\mathbb F)\to H^4(M;\mathbb F) is injective. The evaluation class is already one extension of u, and thus is the only one.

An ordinary class restricted from the Borel construction has zero action slant. For completeness, if Y is any circle space, fix e_0\in ES^1 and let i(y)=[e_0,y]\in Y_{hS^1}. For the action map A:S^1\times Y\to Y, the two maps i\circ A and i\circ\operatorname{pr}_Y are homotopic:

and the loop g\mapsto e_0g contracts in ES^1. Consequently the pullback of a Borel class under i\circ A has no component in H^1(S^1;\mathbb F)\otimes H^{*-1}(Y;\mathbb F). Equivalently its slant with [S^1] is zero. This is also the usual action description of the Borel d_2 differential.

The evaluation class a has a nonzero action slant over every prime field. The integral cohomology ring of M and the universal coefficient theorem give H_2(M;\Z)=H_3(M;\Z)=0. Simple connectivity and successive applications of the Hurewicz theorem [9, Theorem 4.32] make M three-connected; Hurewicz then identifies \pi_4(M) with H_4(M;\Z)=\Z. Choose a based map g:S^4\to M with \langle g^*u,[S^4]\rangle=1, and let \widehat g:S^3\to\Omega M\subset\Lambda M be its adjoint. Evaluation of the rotated based loops on S^1\times S^3 factors, up to the orientation convention, through the smash quotient

The quotient takes the cross-product fundamental class to the fundamental class of S^4, up to sign. Therefore

integrally, and remains nonzero after reduction modulo \ell. This contradicts the zero slant required of the hypothesized Borel extension. No prime divides c, so c=\pm1. Reverse the orientation of E if necessary to obtain (8.11).

The negative bundle is the cut-fiber tangent bundle. For a\in K_w, truncate J_a to the first half-period and extend it by zero on the second half:

The endpoint values agree, so this is an H^1 periodic test field. Integration by parts on [0,D], with the vanishing endpoint values and the Jacobi equation, gives zero index form on this entire three-dimensional space. Project these fields to the negative spectral space \nu_w. If a projected vector were zero, its zero index form and its membership in the nonnegative spectral subspace would force it into the Hessian kernel. It would then be a smooth periodic Jacobi field. But a nonzero J_a has nonzero derivative at D, since both zero value and zero derivative would make it identically zero. Its truncation has a derivative jump at D and cannot be a kernel field. The projection is thus injective, and equal ranks make it an isomorphism K_w\to\nu_w. The truncation depends continuously in the H^1 norm on w and a, and the spectral projections vary continuously as above. We obtain an isomorphism of real bundles K\cong\nu\cong\rho^*E. The truncations are used in the H^1 form domain, not in the H^2 operator domain; their derivative jump is the reason that a nonzero truncation cannot lie in the smooth kernel.

Finally define the full cut map

By the actual bundle identification in Proposition 6.1, its pullback of \mathcal V is the initial-direction four-plane. The marked unit direction w gives a trivial line in this plane; its orthogonal complement is exactly K_w, the tangent space of the great-three-sphere fiber. Hence

The two point maps from U to M in \mathfrak a are \pi and \pi\circ\phi_D. The geodesic flow provides a homotopy between them, so \mathfrak a^*u_1=\mathfrak a^*u_2=z. Taking first Pontryagin classes and using (8.1) proves (8.12).□

Proof. Give each unit sphere fiber the boundary orientation from E. The marked unit vector in p^*E spans a trivial line, with orthogonal complement the oriented vertical tangent two-plane:

For an oriented real two-plane bundle, the first Pontryagin class is the square of its Euler class. Thus \xi^2=p^*p_1(E)/4. On a sphere fiber the Euler class of its tangent bundle integrates to two, giving p_!\xi=1.

The Euler class of the oriented rank-three bundle E lies in H^3(B;\Z)=0, so it vanishes. The sphere-bundle Gysin sequence then gives, in every degree, a short exact sequence whose first map is p^* and whose last map is p_!. In particular p^* is injective. Over \mathbb Q, the classes 1,\xi restrict to a basis of each sphere’s cohomology, so Leray–Hirsch gives the module decomposition

Since p_!x=e, there is a unique rational class Y with x=p^*Y+p^*e\,\xi. The integral class

has zero sphere integral. Integral Gysin exactness expresses it as p^*Y_2 for Y_2\in H^4(B;\Z). Rational injectivity of p^* gives Y_2=2Y. This proves half-integrality without assuming that Y itself is integral.

Pull back to the square over U in Proposition 8.3. The class \rho^*e vanishes, as it does for the Euler class pulled to its own circle bundle. Equation (8.10) now gives

The pullback sphere bundle also has zero Euler class; its rational Gysin sequence makes \widetilde p^* injective. It follows that \rho^*Y=z. Write 2Y=a v+b e^2 in the integral basis supplied by Lemma 8.2. Since \rho^*v=z is a primitive generator of H^4(U;\Z), pulling back gives a=2. This proves (8.17).

Similarly, integrality of p_1(E) and (8.12) give

If the opposite sign of the incidence coefficient is retained, this reads p_1(E)=A e^2+4\beta v, with \beta=\pm1. In either convention substitution of v=Y-(b/2)e^2 gives

Thus 4\alpha=A-2\beta b is integral, proving (8.18).

The first equality in (8.19) is x^{r+1}=f^*(u^{r+1})=0. Fiber antipodal reflection covers the identity of B and reverses the orientation of the two-dimensional sphere fibers. Its pullback fixes base classes and sends the vertical Euler class, hence \xi, to its negative. Pulling back the first equality proves the second one.

It remains to identify the moment determinant. An integral basis of H^{n-2}(B;\Z) from (8.4) is

The matrix of (8.7) in this basis has entries

Replace v in these basis vectors by Y=v+(b/2)X. The change of basis over \mathbb Q is triangular with every diagonal entry equal to one. Hence its determinant is one and it does not change the determinant of the bilinear form. Lemma 8.2 proves the first assertion in (8.20). The second follows from eX^{2r-1}=e^{4r-1} and the definition of N. The final injectivity statements were proved above for p^* and in Lemma 8.2 for multiplication by e.□

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 8.2. Therefore Q=0.

Suppose now that \alpha=0. We use the nondegenerate moment pairing in (9.2) to obtain a contradiction. Write

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 (9.2). The relation R=0 similarly gives

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+j<r-1 and one when i+j=r-1, for 0\le i,j<r. Reversing the columns gives a triangular matrix with diagonal entries one. Hence there is a unique b\in A_0 such that F(f)=\varepsilon(bf) for all f. Nondegeneracy of the F pairing means that multiplication by b is invertible, so b is a unit. For the \varepsilon pairing the orthogonal complement of 1,y,\ldots,y^{r-2} consists exactly of the constants. To see this, write a remainder c_0+\cdots+c_{r-1}y^{r-1} and successively test against 1,y,\ldots,y^{r-2}; the tests force c_{r-1},c_{r-2},\ldots,c_1 to vanish in that order. Equation (9.5) therefore says that ba is constant. In particular any nonzero class a satisfying that equation must be a unit: its product with the unit b is a nonzero constant.

But the class of a=R(1,y) is nonzero and is not a unit. Both q and a vanish at y=0 when \alpha=0, so evaluation at zero precludes invertibility. To check that the class is nonzero, choose an (r+1)st root of unity \eta\ne1,-1, and put

These numbers are nonzero, Z_0^2=\beta y_0, and y_0+Z_0=\eta(y_0-Z_0) with y_0-Z_0\ne0. The formulas for Q,R give q(y_0)=0 but a(y_0)=(y_0-Z_0)^{r+1}\ne0. Thus a cannot be zero modulo q, a contradiction. We conclude that \alpha\ne0. Its stated integrality then gives |4\alpha|\ge1.□

Proof. At a common zero (X,Y) choose a complex square root Z of \alpha X^2+\beta XY. If Z\ne0, simultaneous vanishing of Q,R would force both (Y+Z)^{r+1} and (Y-Z)^{r+1} to vanish, which is impossible unless Y=Z=0. If Z=0, the formulas become Q=(r+1)Y^r and R=Y^{r+1}, so Y=0; then \alpha X^2=0 forces X=0 as well.

The two homogeneous polynomials are therefore relatively prime. Multiplication by Q in \C[X,Y] is injective, and multiplication by R is injective modulo Q: in the polynomial unique factorization domain, Q\mid Rf and relative primality imply Q\mid f. The two resulting multiplication exact sequences give (9.6). On setting X=0, the quotient becomes \C[Y]/(Y^r). Thus the degree-r cokernel of multiplication by X vanishes. The two dimensions read from (9.6) are equal, which proves (9.7).

For the assertion about the whole cohomology ring, Lemma 8.2 gives generation of H^*(B;\mathbb Q) by e,v and total dimension 2r(r+1). Since Y=v+(b/2)e^2 by (8.17), the same ring is generated by e,Y.

The polynomial algebra \C[e,Y] is free on 1,e over \C[X,Y] under X=e^2. Quotienting by the two displayed relations therefore gives \mathcal A\oplus e\mathcal A, of dimension 2r(r+1) by (9.6). Its surjection to H^*(B;\C) is an isomorphism by the equality of dimensions. The relations and maps are defined over \mathbb Q, giving the rational assertion as well.□

Proof. Put Y=(a_1+a_2)/2 and Z=(a_1-a_2)/2 inside \mathcal D. Its quadratic relation becomes Z^2=L and its other relations become

It follows at once that multiplying either R or Q by 4Z^2 gives zero, so (9.9) annihilates the top-degree multiples of both relations.

We next prove nondegeneracy, uniformly in the complex parameter \beta. The coefficient functional on

extracting a_1^ra_2^r has a perfect product pairing, since every monomial pairs with its complementary monomial. The quadratic relation in (9.8) is monic after division by 4\alpha, so \mathcal D=D_0\oplus XD_0. Its coefficient pairing is perfect as well. Indeed, if f+Xg annihilates every element of D_0, extraction shows that g=0; testing then against XD_0 shows that f=0. The grading makes this a perfect pairing between complementary degrees summing to 2r+1.

The involution exchanging a_1,a_2 preserves \varepsilon_{\mathcal D}. Its even and odd subspaces are therefore orthogonal, and the odd subspaces in complementary degrees pair perfectly. Their exact module description is particularly useful. Before imposing R,ZQ, the relation Z^2=L gives the direct sum

The ideal generated by R,ZQ has even part (R,LQ) and odd part Z(R,Q). Consequently

This follows by taking the two summands of the ideal; no cancellation of Z in the quotient is used.

By (9.7), multiplication by X carries \mathcal D^-_r isomorphically onto \mathcal D^-_{r+1}. In the basis

of \mathcal D^-_r, the bilinear pairing (f,g)\mapsto\varepsilon_{\mathcal D}(fXg) has matrix (J_{i+j}). It is nondegenerate because the odd complementary-degree pairing is perfect. This proves in particular that \Phi is nonzero.

For fixed nonzero \alpha, reduction of powers of X by its monic quadratic relation makes every J_j a polynomial in \beta. The determinant is thus a polynomial in \beta which, by the preceding argument, has no complex zero. It is a nonzero constant in \beta. It remains to evaluate it at \beta=0.

First set \alpha=1/4 as well, so that X^2=(a_1-a_2)^2. In \mathcal A use the linear coordinates

For p+q=2r-1, the coefficient of X in the quadratic remainder of A_1^pA_2^q is the polynomial

The numerator is divisible by a_1-a_2. Multiplying by (a_1-a_2)^2 and extracting a_1^ra_2^r gives

In the degree-(r-1) monomial basis A_1^{r-1-i}A_2^i, the matrix of the middle X=A_1-A_2 pairing has entries -2 for i+j=r-1, entries 1 for i+j=r-2 or r, and zero otherwise. Reversing columns gives the tridiagonal matrix with diagonal -2 and adjacent entries 1. Its size-r determinant is (-1)^r(r+1): expansion along the last row gives the recurrence d_r=-2d_{r-1}-d_{r-2} with d_0=1,d_1=-2. Column reversal has sign (-1)^{r(r-1)/2}. The linear change from X,Y to A_1,A_2 has determinant one, so its degree-(r-1) monomial change also has determinant one. We have proved (9.10) at (\alpha,\beta)=(1/4,0).

At \beta=0 for general \alpha\ne0, odd-indexed J_j vanish. For j=2h, quadratic reduction gives

Every nonzero permutation term in the size-r determinant therefore has parameter exponent

This proves (9.10) for \beta=0, and its independence of \beta proves the full identity.

Finally, \ell and \Phi are nonzero functionals on the same one-dimensional top degree of \mathcal A, so \ell=K\Phi. Taking absolute determinants in (9.2) and (9.10) gives

which is (9.11).□

Proof. The quadratic relation in (9.8) is

The coefficient of X in the remainder of X^q satisfies D_1=1, D_2=\mathfrak a, and D_q=\mathfrak a D_{q-1}+\mathfrak b D_{q-2}. Induction on this recurrence yields

Consequently, with ordinary polynomial coefficient extraction,

where

The exact sign of this central coefficient is

Indeed, substitute a_1=e^{it}, a_2=e^{-it} and average over a full period. The coefficient becomes (-1)^{j+1} times the strictly positive integral of a positive constant times \cos^{2r-2-2j}(t)\sin^{2j+2}(t).

By Lemmas 9.1 and 9.3, the parameter factor in the absolute value of each summand of K\Phi_{\alpha,\beta}(X^{2r-1}), apart from its binomial and central coefficients, is

Here |\beta|=1, |4\alpha|\ge1, and j-3(r-1)/2\le-(r-1)/2<0. The bound on the right is attained by the formal parameters \alpha=-1/4, \beta=1, for which the normalization factor |4\alpha|^{(r-1)/2} is one. At those parameters (9.16) shows that all the summands in (9.15) have the same, negative, sign. The sum of the termwise absolute bounds therefore equals the absolute value of that single formal evaluation.

We evaluate it explicitly. When \alpha=-1/4, \beta=1, the two formal roots of the quadratic in X are (\sqrt{a_1}+\sqrt{a_2})^2 and (\sqrt{a_1}-\sqrt{a_2})^2. Their divided difference gives the polynomial

Thus multiplication by (a_1-a_2)^2 and central coefficient extraction give

Since N=\ell(X^{2r-1})>0, the preceding termwise comparison proves (9.14).□

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 [9, Theorem 4.57]. A closed smooth 2r-manifold has the homotopy type of a finite 2r-dimensional CW complex, so cellular approximation [9, Theorem 4.8] 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 [9, Corollary 4.33] makes this map a homotopy equivalence. Yang’s volume theorem [28, Theorem, p. 380] therefore gives equality with the model at the same prime geodesic period, equivalently the same diameter. Corollary 1.3 applies.□

Proof. The projective-plane structure of such a Blaschke manifold is given by McKay [14, Lemma 5]. The classification of smooth topological projective planes [13, Corollary C] 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 [27, p. 282]. The diffeomorphism above supplies its homeomorphism hypothesis. Thus V=V_0, and this equality persists when the common rescaling is undone. Corollary 1.3 applies.□

Proof. For \HP^r with r\ge2, use Theorem 9.4 and Proposition 8.1. At the common period T=\pi, the contact formula has the same positive volume factor for M and its model, so

Here N_0 is the model number in (8.3), and the model volume computed there is

Theorem 9.4 gives N\le N_0, hence \Vol(M,g)\le\Vol(\HP^r). Corollary 1.3 proves the metric statement. The projective line is already the sphere case.□

Proof of Theorem 1.1. Lemma 2.1 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 2. The classical structure list leaves complex type, quaternionic type, and the Cayley plane. These are respectively Corollaries 10.1, 10.3, and 10.2. Each conclusion is a metric isometry with the model at equal diameter. Undoing the normalization therefore gives exactly the asserted classification up to scale.□