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 (7) and (6).

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 (8) 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 (10). ◻

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 \[X(t)=\widetilde X(t) \begin{pmatrix}\tau\mathrm{Id}_k&0\\0&\mathrm{Id}_m\end{pmatrix}.\] 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 \[ \widetilde X(D)= \begin{pmatrix}-G&0\\N&M\end{pmatrix}, \qquad G>0,\quad M\text{ invertible}, \tag{15}\] for some matrix \(N\). In particular \(\widetilde X(D)\) is invertible. For \(\tau\ne0\) sufficiently small, \[A(t)=P(t) \begin{pmatrix}\tau^{-1}\mathrm{Id}_k&0\\0&\mathrm{Id}_m\end{pmatrix} \widetilde X(t)^{-1}.\] 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 \(D\), \[P_1(D)=\begin{pmatrix}\mathrm{Id}_k\\0\end{pmatrix}, \qquad U_1(D)=\begin{pmatrix}-G^{-1}&0\end{pmatrix}.\] Their product is \(-\mathop{\mathrm{diag}}(G^{-1},0)\). 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 (13) with \(B=G^{-1}>0\), including the absence of any off-diagonal residue.

For fixed \(z\in\mathbb R^m\), 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 \(\mathrm{Id}_d\) by (6); hence \(X'(0)=-\mathrm{Id}_d\) and \(P(0)=\mathrm{Id}_d\). Dividing all value columns by \(t\) and repeating the preceding calculation gives residue \(-\mathrm{Id}_d\). 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 \(-\mathrm{Id}_d\) there, proving (14). ◻

Proof. Lemma 7 and (12) 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 \[H'(t)=\tau^{-2}\mathrm{Id}_m+O(1),\qquad V_b'(t)=\tau^{-2}\mathrm{Id}_k+O(1),\qquad C'(t)=O(1).\] 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 \[ S(t)=\tau^{-2}\mathrm{Id}_k+O(1)\qquad(t\to0,T). \tag{19}\] At \(D\), both \(C'\) and \(H'\) are smooth and \(H'(D)\) is invertible. The middle pole therefore yields \[ S(t)=(t-D)^{-2}B+O(1)\qquad(t\to D). \tag{20}\] Taking determinant roots gives the positive limits \[\begin{array}{c|cc} &\text{left endpoint}&\text{right endpoint}\\ \hline h(t)\sin^2t\quad(0<t<T)&1&1\\ s_0(t)\sin^2(2t)/4\quad(0<t<D)&1&\det(B)^{1/k}\\ s_1(t)\sin^2(2t)/4\quad(0<t<D)&\det(B)^{1/k}&1. \end{array}\] 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 \[ \begin{pmatrix}\mathrm{Id}_k&-R\\0&\mathrm{Id}_m\end{pmatrix} (\Delta A) \begin{pmatrix}\mathrm{Id}_k&0\\-R^{\mathsf T}&\mathrm{Id}_m\end{pmatrix} =\begin{pmatrix}Q&0\\0&\Delta H\end{pmatrix}. \tag{28}\] 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 (8). The matrix \(Q\) has size \(k\times k\) and therefore has full nullity: it is the zero matrix. This proves (24), also in the reversed order. Near a diagonal point within the chart, Taylor expansion gives \[\Delta H=(y-x)H'(x)+O((y-x)^2),\quad \Delta C=(y-x)C'(x)+O((y-x)^2),\] and the corresponding expansion for \(\Delta V_b\). Invertibility of \(H'(x)\) then gives \[ Q(x,y)=(y-x)S(x)+O((y-x)^2). \tag{29}\] This establishes the zero extension needed at the endpoints of the integrations below.

Direct differentiation of (22), using \(\partial_y(\Delta H)^{-1}=-(\Delta H)^{-1}H'(y)(\Delta H)^{-1}\), gives \[\begin{align*} \partial_yQ &=V_b'(y)-C'(y)R^{\mathsf T}-R C'(y)^{\mathsf T} +R H'(y)R^{\mathsf T}\\ &=\begin{pmatrix}\mathrm{Id}_k&-R\end{pmatrix}A'(y) \begin{pmatrix}\mathrm{Id}_k\\-R^{\mathsf T}\end{pmatrix}. \end{align*}\] 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 \[\begin{align*} &\begin{pmatrix}\mathrm{Id}_k&-R\end{pmatrix}A'(a) \begin{pmatrix}\mathrm{Id}_k\\-R^{\mathsf T}\end{pmatrix} \\ &\quad=S(a)+ \bigl(R-C'(a)H'(a)^{-1}\bigr)H'(a) \bigl(R-C'(a)H'(a)^{-1}\bigr)^{\mathsf T}. \tag{30}\end{align*}\] The second summand is positive semidefinite. This proves (25). The calculation used invertibility of \(\Delta H\), but did not require its positive definiteness.

To obtain (26), 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 (24) otherwise. Integrating \(\partial_yQ\ge S(y)\) on this interval gives the result. For (27), 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 (29).

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 \[ \delta_r(M)= \inf_{\substack{P>0\\\det P=1}}\frac1r\mathop{\mathrm{tr}}(PM) \tag{35}\] 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, \[ \delta_r\left(\int_a^b M(t)\,dt\right) \ge\int_a^b\delta_r(M(t))\,dt. \tag{36}\] Indeed, for every admissible \(P\) the integral of \(\mathop{\mathrm{tr}}(PM(t))/r\) 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 (36) to \(H'\) on the interval between \(x\) and \(y\). Equation (21) gives \[|\det(\Delta H)|^{1/m} \ge\left|\int_x^y\det(H'(t))^{1/m}\,dt\right|.\] For \(y<x\) this uses the positive matrix \(-\Delta H\). Division by \[\bigl(\det H'(x)\det H'(y)\bigr)^{1/(2m)} =\sqrt{h(x)h(y)}\] proves (33).

Now put \(Q=Q(u+iD,v+jD)>0\). Each of the two matrix bounds in Proposition 9, followed by determinant monotonicity and (36), 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 (34) 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. The full cut map \[c:SM\longrightarrow M\times M,\qquad c(\xi)=(\pi(\xi),\exp_{\pi(\xi)}(D\xi))\] has constant rank \(4r+(4r-4)=8r-4\). Its fibers are the connected great three-spheres from Proposition 5. 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 \[ N_pC_{q_1}\cap N_pC_{q_2}=\{0\}: \tag{51}\] 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 (Hatcher 2002, Theorem 4D.1), using \(1,u_2,\ldots,u_2^{r-1}\) on the fibers, and disjointness from the diagonal give the integral presentation \[ H^*(F;\mathbb Z)=\mathbb Z[u_1,u_2]/(u_1^{r+1},\Delta(u_1,u_2)), \qquad \Delta(a,b)=\sum_{j=0}^r a^jb^{r-j}. \tag{52}\] 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 \(H^4(F;\mathbb Z)\), we have \(p_1(\mathcal V)=t(u_1+u_2)\) for an integer \(t\). Also \(H^2(F;\mathbb Z/2)=0\), 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 (2010, secs. 2.1–2.2). For an oriented spin four-plane bundle, its two complex half-spin bundles have second Chern classes \[\frac{-p_1(\mathcal V)+2e(\mathcal V)}4,\qquad \frac{-p_1(\mathcal V)-2e(\mathcal V)}4.\] 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 (51) have direct sum inside the pullback of \(TM\) from \(p\). Choose a complementary bundle \(W\) of rank \(4(r-2)\). Set \[D_s(v,x)=1+2s(v+x)+(v-x)^2, \qquad p_0(v)=\frac{(1+2sv+v^2)^{r+1}}{1+4sv}.\] The rank-four identity \(p_2=e^2\) (Milnor and Stasheff 1974, Corollary 15.8) and Proposition 14 give \[ p(W)=\frac{p(TM)(v)}{D_s(v,x_1)D_s(v,x_2)}. \tag{56}\] Its polynomial degree is at most \(t_0=2r-4\).

We will prove that the formal characteristic class \[\mathcal W_0=\frac{p_0(v)}{D_s(v,x_1)D_s(v,x_2)}\in A\] has degree at most \(t_0\) and a degree-\(t_0\) component that is nonzero modulo \(v\). The freeness in (54) 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 (53), without presuming the answer for \(TM\). Successive monic polynomial divisions factor \[T^{r+1}=(T-v)(T-x_1)(T-x_2) \prod_{j=4}^{r+1}(T-z_j)\] 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 \[\mathcal W_0=\prod_{j=4}^{r+1}D_s(v,z_j).\] 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 \[(1+2sx+x^2)^{-1}=\sum_{j\ge0}a_jx^j.\] The recursion \(a_0=1\), \(a_1=-2s\) and \(a_j=-2sa_{j-1}-a_{j-2}\) gives \[ a_j^2-a_{j-1}a_{j+1}=1\qquad(j\ge1). \tag{57}\] 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 \[\mathbb Q[x_1,x_2]/ \left(x_1^r,x_2^r,\sum_{j=0}^{r-1}x_1^jx_2^{r-1-j}\right).\] In degree \(2r-4\) the surviving monomials after the first two truncations are \[x_1^{r-3}x_2^{r-1},\qquad x_1^{r-2}x_2^{r-2},\qquad x_1^{r-1}x_2^{r-3}.\] 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 (57) 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 (54).

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 \(v^{i+1}\), (56) is \(b v^i(\mathcal W_0)_{t_0}\). This is nonzero by (54) 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. 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 \(\mu(U)=|S^d|\mathop{\mathrm{Vol}}(M)\). Since \(d+1=4r\), this proves (65).

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 (65) gives \[N_0=\frac{(4r-1)!}{r(2r+1)((2r-1)!)^2} =\frac{(4r)!}{(2r)!(2r+1)!},\] which is (66). ◻

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 \[e(TM)=(r+1)u^r.\] The integral Gysin sequence of the oriented sphere bundle \(\pi\) has the form \[H^{q-n}(M)\xrightarrow{\ e(TM)\smile\ }H^q(M) \xrightarrow{\ \pi^*\ }H^q(U) \xrightarrow{\ \pi_!\ }H^{q-n+1}(M) \xrightarrow{\ e(TM)\smile\ }H^{q+1}(M).\] It gives the complete additive calculation \[ H^q(U;\mathbb Z)= \begin{cases} \mathbb Z\{z^a\},&q=4a,\quad 0\le a<r,\\ \mathbb Z/(r+1)\{z^r\},&q=n,\\ \mathbb Z\{y_a\},&q=n-1+4a,\quad 1\le a\le r,\\ 0,&\text{otherwise}. \end{cases} \tag{71}\] 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 (71).

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 \[ 0\longrightarrow H^{2j-2}(B) \xrightarrow{\ e\smile\ }H^{2j}(B) \xrightarrow{\ \rho^*\ }H^{2j}(U) \longrightarrow0. \tag{72}\] For \(2j=2\) this says \(H^2(B)=\mathbb Ze\). 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 (72) 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 (67).

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 \(H_i(B;\mathbb Z)\) as torsion in \(H^{i+1}(B;\mathbb Z)\), 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 \(H^n(U)=\mathbb Z/(r+1)\), and is precisely (69). The Poincaré pairing \[H^{n-2}(B;\mathbb Z)\times H^n(B;\mathbb Z)\longrightarrow\mathbb Z\] is unimodular, since these two groups are free. Composing one factor with the injection of index \(r+1\) in (69) proves the determinant assertion (70). Finally \(H^{n+1}(U)=H^{n+2}(U)=0\) by (71). 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 \[\sum_{j=0}^{2r-1}\bigl(\lfloor j/2\rfloor+1\bigr) =2\sum_{a=0}^{r-1}(a+1)=r(r+1).\] Poincaré duality supplies another \(r(r+1)\) dimensions above degree \(n-1\), and the middle odd group is zero. This proves (68). ◻

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

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: \[p^*E\cong\mathbb R\oplus T_{\mathrm{vert}}S(E).\] 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 \(H^3(B;\mathbb Z)=0\), 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 \[H^*(S(E);\mathbb Q) =p^*H^*(B;\mathbb Q) \oplus\xi\,p^*H^{*-2}(B;\mathbb Q).\] Since \(p_!x=e\), there is a unique rational class \(Y\) with \(x=p^*Y+p^*e\,\xi\). The integral class \[2x-p^*e\,e(T_{\mathrm{vert}}S(E))\] has zero sphere integral. Integral Gysin exactness expresses it as \(p^*Y_2\) for \(Y_2\in H^4(B;\mathbb Z)\). 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 19. The class \(\rho^*e\) vanishes, as it does for the Euler class pulled to its own circle bundle. Equation (73) now gives \[\widetilde p^*(\rho^*Y)=\widehat\rho^*x =\widetilde p^*z.\] 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 18. Since \(\rho^*v=z\) is a primitive generator of \(H^4(U;\mathbb Z)\), pulling back gives \(a=2\). This proves (80).

Similarly, integrality of \(p_1(E)\) and (75) give \[p_1(E)=A e^2+4v,\qquad A\in\mathbb Z.\] 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 \[L=\frac{A-2\beta b}{4}X^2+\beta XY.\] Thus \(4\alpha=A-2\beta b\) is integral, proving (81).

The first equality in (82) 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 \(H^{n-2}(B;\mathbb Z)\) from (67) is \[b_i=eX^{r-1-i}v^i,\qquad 0\le i<r.\] The matrix of (70) in this basis has entries \[\int_Be b_i b_j =\int_B eX^{2r-1-i-j}v^{i+j}.\] 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 18 proves the first assertion in (83). 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 18 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 18. Therefore \(Q=0\).

Suppose now that \(\alpha=0\). We use the nondegenerate moment pairing in (85) to obtain a contradiction. Write \[q(y)=Q(1,y),\qquad a(y)=R(1,y),\qquad A_0=\mathbb C[y]/(q).\] 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 \(F:A_0\to\mathbb C\). 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 (85). The relation \(R=0\) similarly gives \[ F(a(y)y^j)=0\qquad(0\le j\le r-2). \tag{88}\]

Let \(\varepsilon:A_0\to\mathbb C\) 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 (88) 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 \[c_\eta=\frac{\eta-1}{\eta+1},\qquad y_0=\frac{\beta}{c_\eta^2},\qquad Z_0=c_\eta y_0.\] 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 \(\mathbb C[X,Y]\) 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 (89). On setting \(X=0\), the quotient becomes \(\mathbb C[Y]/(Y^r)\). Thus the degree-\(r\) cokernel of multiplication by \(X\) vanishes. The two dimensions read from (89) are equal, which proves (90).

For the assertion about the whole cohomology ring, Lemma 18 gives generation of \(H^*(B;\mathbb Q)\) by \(e,v\) and total dimension \(2r(r+1)\). Since \(Y=v+(b/2)e^2\) by (80), the same ring is generated by \(e,Y\).

The polynomial algebra \(\mathbb C[e,Y]\) is free on \(1,e\) over \(\mathbb C[X,Y]\) under \(X=e^2\). Quotienting by the two displayed relations therefore gives \(\mathcal A\oplus e\mathcal A\), of dimension \(2r(r+1)\) by (89). Its surjection to \(H^*(B;\mathbb C)\) 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 \[R=0,\qquad ZQ=0.\] It follows at once that multiplying either \(R\) or \(Q\) by \(4Z^2\) gives zero, so (92) annihilates the top-degree multiples of both relations.

We next prove nondegeneracy, uniformly in the complex parameter \(\beta\). The coefficient functional on \[D_0=\mathbb C[a_1,a_2]/(a_1^{r+1},a_2^{r+1})\] extracting \(a_1^ra_2^r\) has a perfect product pairing, since every monomial pairs with its complementary monomial. The quadratic relation in (91) 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 \[\mathbb C[X,Y]\oplus Z\mathbb C[X,Y].\] The ideal generated by \(R,ZQ\) has even part \((R,LQ)\) and odd part \(Z(R,Q)\). Consequently \[ \mathcal D^-\simeq Z\mathcal A \quad\hbox{as a graded module, with a shift by one}. \tag{95}\] This follows by taking the two summands of the ideal; no cancellation of \(Z\) in the quotient is used.

By (90), multiplication by \(X\) carries \(\mathcal D^-_r\) isomorphically onto \(\mathcal D^-_{r+1}\). In the basis \[(a_1-a_2)X^{r-1-i}Y^i,\qquad 0\le i<r,\] 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 \[A_1=Y+X/2,\qquad A_2=Y-X/2.\] For \(p+q=2r-1\), the coefficient of \(X\) in the quadratic remainder of \(A_1^pA_2^q\) is the polynomial \[\frac{a_1^pa_2^q-a_1^qa_2^p}{2(a_1-a_2)}.\] The numerator is divisible by \(a_1-a_2\). Multiplying by \((a_1-a_2)^2\) and extracting \(a_1^ra_2^r\) gives \[ \Phi_{1/4,0}(A_1^pA_2^q)= \begin{cases} -1,&(p,q)=(r,r-1),\\ 1,&(p,q)=(r-1,r),\\ 0,&\text{otherwise}. \end{cases} \tag{96}\] 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 (93) at \((\alpha,\beta)=(1/4,0)\).

At \(\beta=0\) for general \(\alpha\ne0\), odd-indexed \(J_j\) vanish. For \(j=2h\), quadratic reduction gives \[J_{2h}(\alpha,0) =(4\alpha)^{-(r-1-h)}J_{2h}(1/4,0).\] Every nonzero permutation term in the size-\(r\) determinant therefore has parameter exponent \[-\sum_{i=0}^{r-1}\left(r-1-\frac{i+\sigma(i)}2\right) =-\frac{r(r-1)}2.\] This proves (93) 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 (85) and (93) gives \[r+1=|K|^r(r+1)|4\alpha|^{-r(r-1)/2},\] which is (94). ◻

Proof. The quadratic relation in (91) is \[X^2=\mathfrak a X+\mathfrak b,\qquad \mathfrak a=-\frac{\beta}{2\alpha}(a_1+a_2),\quad \mathfrak b=\frac{(a_1-a_2)^2}{4\alpha}.\] 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 \[D_{2r-1}= \sum_{j=0}^{r-1}\binom{2r-2-j}{j}\mathfrak a^{2r-2-2j}\mathfrak b^j.\] Consequently, with ordinary polynomial coefficient extraction, \[ \Phi_{\alpha,\beta}(X^{2r-1}) =\sum_{j=0}^{r-1}\binom{2r-2-j}{j} \left(-\frac{\beta}{2\alpha}\right)^{2r-2-2j} (4\alpha)^{-j}c_j, \tag{98}\] where \[c_j=[a_1^ra_2^r] (a_1+a_2)^{2r-2-2j}(a_1-a_2)^{2j+2}.\] The exact sign of this central coefficient is \[ \operatorname{sign}(c_j)=(-1)^{j+1}. \tag{99}\] 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 21 and 23, 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 \[\begin{align*} |K|\left|\frac{\beta}{2\alpha}\right|^{2r-2-2j} |4\alpha|^{-j} &=2^{2r-2-2j}|4\alpha|^{j-3(r-1)/2}\\ &\le 2^{2r-2-2j}. \end{align*}\] 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 (99) shows that all the summands in (98) 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 \[\begin{align*} D_{2r-1} &=\frac{(\sqrt{a_1}+\sqrt{a_2})^{4r-2} -(\sqrt{a_1}-\sqrt{a_2})^{4r-2}} {4\sqrt{a_1a_2}}\\ &=\frac12\sum_{h=0}^{2r-2}\binom{4r-2}{2h+1} a_1^{2r-2-h}a_2^h. \end{align*}\] Thus multiplication by \((a_1-a_2)^2\) and central coefficient extraction give \[\left|\Phi_{-1/4,1}(X^{2r-1})\right| =\binom{4r-2}{2r-1}-\binom{4r-2}{2r-3} =\frac1{2r+1}\binom{4r}{2r}.\] Since \(N=\ell(X^{2r-1})>0\), the preceding termwise comparison proves (97). ◻