Proof. The flow is defined for all time because \(M\) is compact. Suppose that there were points \(q_j\) and times \(t_j>0\) with \[t_j\longrightarrow0,\qquad \Psi_{t_j}(q_j)=q_j,\qquad X(q_j)\ne0.\] After passing to a subsequence, assume that \(q_j\to q\). Choose coordinates near \(q\), and a Euclidean ball about its coordinate image whose closure lies inside the coordinate domain. By continuity of the flow, all sufficiently late trajectory segments \(\{\Psi_s(q_j):0\le s\le t_j\}\) lie in this ball. The coordinate vector field \(V\) is Lipschitz there with a constant \(L\ge0\), since its derivative is bounded on the closed ball.

Write \(u_j(s)\) for the coordinate trajectory and set \[C_j=\max_{0\le s\le t_j}|\dot u_j(s)|>0.\] For any \(s,r\in[0,t_j]\), convexity of the ball and the Lipschitz estimate give \[\begin{split} |\dot u_j(s)-\dot u_j(r)| &=|V(u_j(s))-V(u_j(r))|\\ &\le L|u_j(s)-u_j(r)| \le LC_j|s-r| \le LC_jt_j. \end{split}\] The closed trajectory has zero mean coordinate velocity: \(\int_0^{t_j}\dot u_j(r)\,dr=0\). At a time \(s_j\) of maximum speed, it follows that \[C_j =\left| \frac1{t_j}\int_0^{t_j} \bigl(\dot u_j(s_j)-\dot u_j(r)\bigr)\,dr \right| \le LC_jt_j.\] Dividing by \(C_j\) contradicts \(t_j\to0\). Thus all fixed points of a sufficiently short positive-time map are zeros of \(X\). Conversely, every zero of \(X\) is stationary and hence fixed by every \(\Psi_t\). ◻

Proof of Proposition 2. The first step is to extend \(f\) so that criticality along \(F\) is equivalent to ambient criticality at points of \(F\). This is the finite-group case of symmetric criticality [15], which we verify directly. Extend \(f\) smoothly to a function \(\widetilde f\) on \(M\): in submanifold charts, extend independently of the normal coordinates, and combine these local extensions using a partition of unity. Define \[K=\frac1m\sum_{j=0}^{m-1}\widetilde f\circ A^j.\] Then \(K\circ A=K\) and \(K|_F=f\).

Fix \(q\in F\), put \(D=dA_q\), and consider \[P_q=\frac1m\sum_{j=0}^{m-1}D^j.\] Since \(D^m=\mathrm{id}\), we have \(DP_q=P_q\). Moreover, \(P_qv=v\) whenever \(Dv=v\). Thus \(P_q\) is a projection onto \(\ker(D-\mathrm{id})=T_qF\). Differentiating the averaged extension gives, for every \(v\in T_qM\), \[dK_q(v)=d\widetilde f_q(P_qv)=df_q(P_qv).\] Consequently, \[ \mathop{\mathrm{crit}}(K)\cap F=\mathop{\mathrm{crit}}(f). \tag{4}\]

Because \(A\) is symplectic and \(K\circ A=K\), its differential carries the Hamiltonian vector field \(X_K\) to itself. Uniqueness of solutions therefore gives \(A\circ B_s=B_s\circ A\) for every \(s\). Apply Lemma 3 to \(X_K\), and choose \(0<\varepsilon<\delta/m\). We then have \[\phi_\varepsilon^m=B_{m\varepsilon}, \qquad \mathop{\mathrm{Fix}}(B_{m\varepsilon})=\mathop{\mathrm{crit}}(K).\] If \(\phi_\varepsilon(q)=q\), then \(q\) is fixed by \(B_{m\varepsilon}\), so \(dK_q=0\). It follows that \(B_\varepsilon(q)=q\), and hence \(A(q)=q\). Conversely, every point of \(\mathop{\mathrm{crit}}(K)\cap F\) is fixed by \(\phi_\varepsilon\). Equation 4 proves the asserted equality of fixed and critical sets.

For completeness, the composition is generated by a smooth Hamiltonian on the unit time interval. Choose a Hamiltonian path \(A_t\) from \(\mathrm{id}\) to \(A\), generated by \(H_t\), and a smooth nondecreasing function \(a:[0,1]\to[0,1]\) equal to \(0\) near \(0\) and to \(1\) near \(1\). The time-dependent Hamiltonian \[L_t(x)= \begin{cases} 2\varepsilon a'(2t)K(x),&0\le t\le\tfrac12,\\[2pt] 2a'(2t-1)H_{a(2t-1)}(x),&\tfrac12\le t\le1 \end{cases}\] is smooth, since both pieces vanish near the joining time. Its flow first follows \(B_{\varepsilon a(2t)}\), and then \(A_{a(2t-1)}\circ B_\varepsilon\), so its time-one map is \(\phi_\varepsilon\). ◻

Proof. In any standard affine chart of \(\mathbb C\mathrm P^4\), the equation defining \(M\) takes the form \(1+\sum_j w_j^2=0\). Its differential cannot vanish on its zero set: that would force all the \(w_j\) to be zero. Thus \(M\) is a smooth complex hypersurface, of complex dimension three. As a closed subset of projective space it is compact, and it has no boundary.

To see connectedness, write a nonzero isotropic vector as \(z=x+iy\) with \(x,y\in\mathbb R^5\). Its real and imaginary parts satisfy \[|x|=|y|>0,\qquad x\cdot y=0.\] After rescaling, they form an ordered orthonormal pair. Every such pair extends to a positively oriented orthonormal basis, so these pairs form one orbit of the connected group \(\mathrm{SO}(5)\). Their map \((x,y)\mapsto[x+iy]\) onto \(M\) proves that \(M\) is connected.

We construct the symplectic form with its normalization explicit. On the unit sphere \(S^9\subset\mathbb C^5\), put \[\alpha=\sum_{j=0}^4(x_j\,dy_j-y_j\,dx_j).\] The form \(d\alpha\) is invariant under scalar phase rotations. If \(v\in T_zS^9\), then \[d\alpha(iz,v) =-2\operatorname{Re}\sum_{j=0}^4\overline{z_j}v_j=0.\] Consequently it descends under the quotient map \(p:S^9\to\mathbb C\mathrm P^4\) to a two-form \(\sigma\), characterized by \(p^*\sigma=d\alpha\). Local sections show that \(\sigma\) is smooth and closed. A projective tangent vector has a horizontal lift \(v\in\mathbb C^5\) satisfying \(\sum_j\overline{z_j}v_j=0\); multiplication by \(i\) on these lifts induces the complex structure. Since \[d\alpha(v,iv)=2|v|^2,\] \(\sigma\) is positive on every complex tangent line. Its restriction \(\omega=\sigma|_M\) is therefore closed and nondegenerate.

We use a family of Hamiltonian rotations: equal speeds will give the involution, and unequal speeds will later give a function with four critical points on \(M\). For \(a,b,t\in\mathbb R\), let \(\Phi_t^{a,b}\) keep \(z_0\) fixed and act on \((z_1,z_2)^{\mathsf T}\) and \((z_3,z_4)^{\mathsf T}\) by \(R_{at}\) and \(R_{bt}\), respectively, where \[R_\theta= \begin{pmatrix} \cos\theta&-\sin\theta\\ \sin\theta&\cos\theta \end{pmatrix}.\] These real orthogonal, complex unitary transformations preserve the quadric and \(\alpha\). If \(Z_{a,b}\) is their generating vector field on \(S^9\), Cartan’s formula gives \[\iota_{Z_{a,b}}d\alpha=-d\bigl(\alpha(Z_{a,b})\bigr).\] The function \(\alpha(Z_{a,b})\) is phase invariant, because the rotations commute with scalar multiplication. It thus descends to a smooth function \(g_{a,b}\) on projective space. The displayed identity descends as well, so \(G_{a,b}=-g_{a,b}|_M\) generates the projected rotations on \(M\) with our convention \(\iota_{X_{G_{a,b}}}\omega=dG_{a,b}\). Explicitly, \[ G_{a,b}([z])= \frac{2a\,\operatorname{Im}(\overline{z_1}z_2) +2b\,\operatorname{Im}(\overline{z_3}z_4)} {\sum_{j=0}^4|z_j|^2}. \tag{5}\] In particular, \(A=\Phi_1^{\pi,\pi}\) is Hamiltonian and preserves \(\omega\).

A projective line fixed by \(A\) lies in one of the two eigenspaces of \(\operatorname{diag}(1,-1,-1,-1,-1)\). The positive eigenspace gives the single point \([1:0:0:0:0]\), which does not lie in \(M\). The negative eigenspace gives precisely \(F=M\cap\{z_0=0\}\). To identify \(F\), use the invertible complex-linear map \[(z_1,z_2,z_3,z_4)\longmapsto \begin{pmatrix} z_1+iz_2&z_3+iz_4\\ -z_3+iz_4&z_1-iz_2 \end{pmatrix}.\] Its determinant is \(z_1^2+z_2^2+z_3^2+z_4^2\). It therefore identifies \(F\) with the projectivized nonzero rank-one matrices. The map \[\mathbb C\mathrm P^1\times\mathbb C\mathrm P^1\longrightarrow F, \qquad ([u],[v])\longmapsto[uv^{\mathsf T}]\] is a diffeomorphism: a nonzero column and row recover the two projective factors, smoothly on the corresponding matrix charts. Finally, stereographic coordinates identify each \(\mathbb C\mathrm P^1\) with \(S^2\). Only this smooth identification is needed below.

For the tangent-space assertion, fix \(q\in F\) and choose an affine chart \(z_j=1\) with \(j>0\). In its normalized coordinates, \(A\) sends \(w_0\) to \(-w_0\) and leaves all other coordinates fixed. The equation of \(M\) is \[w_0^2+1+\sum_{\substack{k>0\\k\ne j}}w_k^2=0.\] At \(q\), where \(w_0=0\), its linearization places no constraint on the \(w_0\) direction. The remaining differential is nonzero, so its kernel in the other coordinates is exactly \(T_qF\). Thus \[T_qM=T_qF\oplus\mathbb C\frac{\partial}{\partial w_0},\] with the asserted \(+1\) and \(-1\) actions. This also verifies directly that \(F\) is a smooth complex surface. The positivity established above therefore makes \(\omega|_F\) symplectic. ◻

Proof. For a fixed \(x\ne p\), the \(y\)-dependence is a height function on the second sphere, with critical points \(y=\pm(x-p)/|x-p|\). At \(x=p\) that height function is constant, but criticality of the full function still requires its derivative in the \(x\) directions to vanish. We therefore separate this exceptional fiber from the two height branches.

The constrained critical-point equations are \[ e+y=\lambda x,\qquad x-p=\mu y, \qquad |x|=|y|=1, \tag{7}\] for real multipliers \(\lambda,\mu\). If \(x=p\), the first equation gives \(y=-e+\lambda p\). Its unit norm forces \(\lambda=0\), yielding the unique point \[q_0=(p,-e),\qquad f(q_0)=0.\]

Suppose now that \(x\ne p\) and set \(t=|x-p|>0\). The second equation in (7) gives \(y=s(x-p)/t\) for a sign \(s\in\{1,-1\}\). The first then becomes \[t e-s p=(\lambda t-s)x.\] Since the left side is nonzero, \(\lambda t-s\ne0\) and \(x\) lies in the plane spanned by \(e\) and \(p\). Write \(x=(u,0,w)\). The definition of \(t\) gives \(w=1-t^2/2\). Taking the first equation in (7) against the tangent vector \((w,0,-u)\) gives \[w+\frac{s u}{t}=0, \qquad\text{so}\qquad u=-s t w.\] Consequently the unit-norm condition reduces to \[ 0=(1-t^2/2)^2(1+t^2)-1 =\frac{t^4(t^2-3)}4. \tag{8}\] Because \(t>0\), this forces \(t=\sqrt3\), \(w=-1/2\), and \(u=s\sqrt3/2\). In particular, the other pole \(x=-p\), for which \(t=2\), is excluded. The two resulting candidates are \[ q_s= \left( \left(\frac{s\sqrt3}{2},0,-\frac12\right), \left(\frac12,0,-\frac{s\sqrt3}{2}\right) \right), \qquad s\in\{1,-1\}. \tag{9}\] Both satisfy (7) with \(\lambda=\mu=s\sqrt3\), and \[f(q_s)=s\,\frac{3\sqrt3}{2}.\] The cases \(x=p\) and \(x\ne p\) exhaust the domain, proving the count. ◻

Proof. Suppose that \(h:M\to\mathbb R\) has at most \(r\) critical points. List its distinct critical values as \(c_1<\cdots<c_k\), where \(1\le k\le r\); the absolute minimum and maximum are among them. For a fixed Riemannian metric let \(g_t\) be the negative-gradient flow of \(h\), which is complete by compactness, and put \(P_b=\{h\le b\}\). We first cover \(M\) by \(k\) open sets, each diffeomorphic to a finite disjoint union of coordinate balls. Every closed positive-degree form will be exact on each of these sets; the cover will therefore force every product of \(k\) positive-degree cohomology classes to vanish.

Around the critical points at \(c_i\), choose an open set \(U_i\) that is a union of disjoint coordinate balls, and a smaller neighborhood \(V_i\) with \(\overline V_i\subset U_i\). A bound on the speed of the flow and the positive distance between \(M\setminus U_i\) and \(\overline V_i\) give \(T_i>0\) such that trajectories starting outside \(U_i\) stay outside \(V_i\) for \(0\le t\le T_i\).

Choose a closed level band \(\{|h-c_i|\le\delta_i\}\) narrow enough that it contains no critical points outside \(V_i\). On the part outside \(V_i\), compactness gives a positive lower bound \(b_i\) for \(|\nabla h|^2\); if that set is empty, any positive \(b_i\) will do. Choose \(0<\epsilon_i<\delta_i\) so that the intervals \([c_i-\epsilon_i,c_i+\epsilon_i]\) are separated and \(2\epsilon_i<T_i b_i\). Along the negative-gradient flow, \[\frac{d}{dt}h(g_t(q))=-|\nabla h(g_t(q))|^2.\] A point of \(P_{c_i+\epsilon_i}\setminus U_i\) either already lies below \(c_i-\epsilon_i\) or decreases at rate at least \(b_i\) until it reaches that level. It follows that \[ g_{T_i}(P_{c_i+\epsilon_i}\setminus U_i) \subset P_{c_i-\epsilon_i}. \tag{10}\] For \(i>1\), a further fixed flow time \(S_i>0\) carries \(P_{c_i-\epsilon_i}\) into \(P_{c_{i-1}+\epsilon_{i-1}}\): the compact band between these levels contains no critical points and has a positive gradient-norm lower bound.

Since \(c_1\) is the absolute minimum, \(P_{c_1-\epsilon_1}\) is empty, and (10) shows that \(P_{c_1+\epsilon_1}\subset U_1\). Inductively, cover \(P_{c_i+\epsilon_i}\) by \(U_i\) and the preimages under \(g_{T_i+S_i}\) of the \(i-1\) sets covering the preceding upper sublevel. Equation (10) proves that these sets cover. Each \(g_t\) is a diffeomorphism of \(M\), so every member remains diffeomorphic to a finite disjoint union of coordinate balls. The last upper sublevel is all of \(M\), since \(c_k\) is the absolute maximum.

Write the resulting cover as \(O_1,\ldots,O_k\). Choose closed forms \(\alpha_j\) representing \(a_j\) for \(1\le j\le k\). Since these forms have positive degree, the Poincaré lemma on each ball gives \(\alpha_j=d\lambda_j\) on \(O_j\). Choose a smooth partition of unity \(\psi_1,\ldots,\psi_k\) subordinate to the cover. Let \(\chi:[0,1]\to[0,1]\) be smooth, zero on \([0,1/(4k)]\) and one on \([1/(2k),1]\), and put \(\rho_j=\chi(\psi_j)\). At every point some \(\psi_j\ge1/k\), so some \(\rho_j\) is identically one nearby. Each \(\rho_j\lambda_j\) extends smoothly by zero outside \(O_j\).

The closed forms \[\alpha'_j=\alpha_j-d(\rho_j\lambda_j)\] represent the same classes as \(\alpha_j\). Near every point, at least one of these forms vanishes, and hence \(\alpha'_1\wedge\cdots\wedge\alpha'_k=0\). Thus \(a_1\smile\cdots\smile a_k=0\), contradicting the nonvanishing of \(a_1\smile\cdots\smile a_r\) because \(k\le r\). ◻

Proof. The class \([\omega]^n\) is nonzero, since \(\int_M\omega^n>0\) in the symplectic orientation. Apply Lemma 7 with each \(a_j=[\omega]\). ◻

Proof. Let \(h=c_1(\mathcal O_{Q^3}(1))\in H^2(Q^3;\mathbb Z)\) be the hyperplane class. Since \(Q^3\) has degree two in \(\mathbb C\mathrm P^4\), \[h^3=2[\mathrm{pt}],\] where \([\mathrm{pt}]\) evaluates to one on the fundamental class. This product remains nonzero over \(\mathbb Q\), giving \(\operatorname{cuplength}(Q^3;\mathbb Q)\ge4\).

For the matching upper bound, take the Hamiltonian \(G_{1,2}\) of (5). Its critical points are precisely the zeros of its Hamiltonian vector field. That field is induced on \(Q^3\) by the complex-linear generator \[T=\operatorname{diag}(0,J,2J), \qquad J=\begin{pmatrix}0&-1\\1&0\end{pmatrix}.\] The induced projective vector field vanishes at \([z]\) exactly when \(Tz\in\mathbb Cz\). The eigenvalues \(0,\pm i,\pm2i\) of \(T\) are distinct, so its projective zeros are its five eigenlines. The zero eigenline \([1:0:0:0:0]\) misses \(Q^3\), whereas the other four lines \[[0:1:\pm i:0:0],\qquad [0:0:0:1:\pm i]\] all satisfy the quadric equation. Thus \(G_{1,2}\) has exactly four critical points on \(Q^3\). Together with (11), this gives \[4\le\operatorname{cuplength}(Q^3;\mathbb Q) \le\mathop{\mathrm{Crit}}(Q^3)\le4.\] ◻

Proof of Theorem 1. Take the symplectic quadric \((M,\omega)\) and Hamiltonian involution \(A\) from Proposition 4. Its fixed submanifold \(F\) is diffeomorphic to \(S^2\times S^2\). Transport the function of Proposition 5 to \(F\); it has exactly three critical points.

Proposition 2 provides a smooth \(A\)-invariant extension \(K\colon M\to\mathbb R\) with \(K|_F=f\). Let \(B_s\) denote its Hamiltonian flow. For all sufficiently small \(\varepsilon>0\), the map \[\phi=A\circ B_\varepsilon\] is a Hamiltonian diffeomorphism and satisfies \(\mathop{\mathrm{Fix}}(\phi)=\mathop{\mathrm{crit}}(f)\), with the critical points viewed as points of \(F\). The exact comparison in Corollary 9 now gives \[\#\mathop{\mathrm{Fix}}(\phi)=3<4=\mathop{\mathrm{Crit}}(M) =\operatorname{cuplength}(M;\mathbb Q).\] It remains to verify the asserted degeneracy. Write \(\omega_F=\omega|_F\), which is symplectic, and let \(X_f\) be the Hamiltonian vector field of \(f\) on \((F,\omega_F)\). Invariance of \(K\) makes \(X_K\) tangent to \(F\), and restricting its defining equation to \(TF\) gives \(X_K|_F=X_f\). Indeed, at \(q\in F\) invariance gives \(dA_qX_K(q)=X_K(q)\), and the fixed tangent space is \(T_qF\). Thus \(B_s\) preserves \(F\), and its restriction is the Hamiltonian flow of \(f\) on \(F\). At the critical point \(q_0\) of Remark 6, differentiating \(\iota_{X_f}\omega_F=df\) gives \[\omega_F\bigl(DX_f(q_0)v,w\bigr) =\operatorname{Hess}_{q_0}f(v,w).\] Thus any nonzero vector \(v\) in the Hessian kernel satisfies \(DX_f(q_0)v=0\). Since \(A|_F=\mathrm{id}\) and \(q_0\) is an equilibrium, \[d\phi_{q_0}v =\exp\bigl(\varepsilon DX_f(q_0)\bigr)v =v.\] The fixed point \(q_0\) therefore has eigenvalue \(1\) in its linearization. This argument takes place entirely in the invariant subspace \(T_{q_0}F\); no assumption on the normal Hessian of the extension \(K\) is needed. ◻