Three fixed points on the symplectic quadric threefold

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

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

For any s,r\in[0,t_j], convexity of the ball and the Lipschitz estimate give

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

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 2.1. 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

Then K\circ A=K and K|_F=f.

Fix q\in F, put D=dA_q, and consider

Since D^m=\id, we have DP_q=P_q. Moreover, P_qv=v whenever Dv=v. Thus P_q is a projection onto \ker(D-\id)=T_qF. Differentiating the averaged extension gives, for every v\in T_qM,

Consequently,

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 2.2 to X_K, and choose 0<\varepsilon<\delta/m. We then have

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 \CritSet(K)\cap F is fixed by \phi_\varepsilon. Equation 2.1 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 \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

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 \CP^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\R^5. Its real and imaginary parts satisfy

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\C^5, put

The form d\alpha is invariant under scalar phase rotations. If v\in T_zS^9, then

Consequently it descends under the quotient map p:S^9\to\CP^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\C^5 satisfying \sum_j\overline{z_j}v_j=0; multiplication by i on these lifts induces the complex structure. Since

\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\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

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

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,

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

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

is a diffeomorphism: a nonzero column and row recover the two projective factors, smoothly on the corresponding matrix charts. Finally, stereographic coordinates identify each \CP^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

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

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

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

Suppose now that x\ne p and set t=|x-p|>0. The second equation in (4.2) gives y=s(x-p)/t for a sign s\in\{1,-1\}. The first then becomes

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 (4.2) against the tangent vector (w,0,-u) gives

Consequently the unit-norm condition reduces to

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

Both satisfy (4.2) with \lambda=\mu=s\sqrt3, and

The cases x=p and x\ne p exhaust the domain, proving the count.□

Proof. Suppose that h:M\to\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,

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

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 (5.1) 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 (5.1) 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

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 5.1 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 \CP^4,

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 (3.1). Its critical points are precisely the zeros of its Hamiltonian vector field. That field is induced on Q^3 by the complex-linear generator

The induced projective vector field vanishes at [z] exactly when Tz\in\C z. 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

all satisfy the quadric equation. Thus G_{1,2} has exactly four critical points on Q^3. Together with (5.2), this gives

□

Proof of Theorem 1.1. Take the symplectic quadric (M,\omega) and Hamiltonian involution A from Proposition 3.1. Its fixed submanifold F is diffeomorphic to S^2\times S^2. Transport the function of Proposition 4.1 to F; it has exactly three critical points.

Proposition 2.1 provides a smooth A-invariant extension K\colon M\to\R with K|_F=f. Let B_s denote its Hamiltonian flow. For all sufficiently small \varepsilon>0, the map

is a Hamiltonian diffeomorphism and satisfies \Fix(\phi)=\CritSet(f), with the critical points viewed as points of F. The exact comparison in Corollary 5.3 now gives

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 4.2, differentiating \iota_{X_f}\omega_F=df gives

Thus any nonzero vector v in the Hessian kernel satisfies DX_f(q_0)v=0. Since A|_F=\id and q_0 is an equilibrium,

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.□