Global properties of toric nearly Kähler manifolds

Abstract

We study toric nearly Kähler manifolds, extending the work of Moroianu and Nagy. We give a description of the global geometry using multi-moment maps. We then investigate polynomial and radial solutions to the toric nearly Kähler equation.

1 Introduction

A nearly Kähler manifold is an almost Hermitian manifold (M, g, J) such that $\nabla J$ is skew symmetric: $(\nabla _X J)X =0$ for every vector field X on M. Each of these can be decomposed as a Riemannian product of nearly Kähler manifolds which are either Kähler, 6-dimensional, homogeneous, or twistor spaces over quaternionic Kähler manifolds of positive scalar curvature [14]. We will focus on the case of 6-dimensional nearly Kähler manifolds that are strict in the sense that they are not Kähler. These are characterized by being the links of metric cones with holonomy $G_2$, which makes them Einstein with positive scalar curvature [2].

A main challenge is to construct complete examples of 6-dimensional strictly nearly Kähler manifolds (which will be referred to simply as nearly Kähler manifolds in the rest of the paper). There are homogeneous nearly Kähler structures on $S^6$, $S^3\times S^3$, ${\mathbb {C}}P^3$, and the flag manifold $\mathrm {SU}(3)/T^2$. In [3], these are shown to be the only homogeneous examples. In [6], cohomogeneity one examples are constructed on $S^6$ and $S^3\times S^3$. No other complete examples are known. The cohomogeneity two case has been studied in [9], which shows that the infinitesimal symmetry group must be ${\mathfrak {u}}(2)$.

We will skip to cohomogeneity three in exchange for having an abelian symmetry group by studying nearly Kähler manifolds which are toric in the sense that the automorphism group contains a 3-torus. The homogeneous nearly Kähler structure on $S^3\times S^3$ is the only known example. The general case has been studied in [12], where the local theory is shown to be equivalent to a Monge–Ampère type equation which we will refer to as the toric nearly Kähler equation. The current paper represents the author’s efforts to build on this work. One of the main ingredients is the idea of a multi-moment map, which was introduced in [11] as a multi-symplectic generalization of the moment map associated to a symplectic form. A toric nearly Kähler manifold (M, g, J, T) is equipped with two multi-moment maps,

$$\begin{aligned} \mu :M\rightarrow {\varLambda }^2{\mathfrak {t}} \cong {\mathbb {R}}^3, \qquad \varepsilon :M\rightarrow {\varLambda }^3{\mathfrak {t}}\cong {\mathbb {R}}, \end{aligned}$$

which are defined in Sect. 2, where ${\mathfrak {t}}$ is the Lie algebra of the torus T acting on M. A foundational result in (symplectic) toric geometry [1, 7] states that the moment map of a compact toric manifold has connected fibres, and its image is a convex polytope. The first condition implies that the moment map induces a homeomorphism from the space of orbits onto its image. We prove a nearly Kähler analogue of this:

Theorem 1

Let M be a complete toric nearly Kähler manifold with the action of a torus $T^3$ with Lie algebra ${\mathfrak {t}}$. Then the multi-moment maps induce a homeomorphism

$$\begin{aligned} ({\bar{\mu }},{\bar{\varepsilon }}):M/T\cong S^3. \end{aligned}$$

The T action is free away from a finite number of orbits in the equator $\varepsilon ^{-1}(0)$. Moreover the two orbits in $\mu ^{-1}(0)$ are Lagrangian.

This theorem generalizes previous work by the author in [5], which describes multi-moment maps of the homogeneous nearly Kähler structure on $S^3\times S^3$. In that case, $\mu \left( \varepsilon ^{-1}(0)\right)$ is Cayley’s nodal cubic surface, whose 4 nodal singular points correspond to the singular T orbits. By studying the topological consequences of this theorem, we prove the following:

Corollary 1

Any complete toric nearly Kähler manifold has at least 4 torus orbits where the action is not free.

As a consequence of this, radial solutions to the toric nearly Kähler equation cannot give complete metrics. By studying the corresponding ODE, we see that the singularity that forms must occur at the Lagrangian orbit.

We also study the case when a hypothetical solution to the toric nearly Kähler equation is polynomial in the natural multi-moment map coordinates. The homogeneous nearly Kähler structure on $S^3\times S^3$ corresponds to a cubic solution $\varphi _0$ shown in Equation (4). Using an old theorem of Hesse [8], we prove:

Theorem 2

Every polynomial solution of the toric nearly Kähler equation with degree at most 5 is equivalent to the cubic solution $\varphi _0$ up to coordinate transformation.

The toric nearly Kähler equation restricted to the space of polynomials is overdetermined for polynomials of degree greater than three, so it is unlikely that there will be other polynomial solutions. However, to show this explicitly is computationally difficult even in the quintic case.

2 Local theory

In this section we review the local theory of toric nearly Kähler manifolds from [12], although we will use a coordinate invariant treatment in order to make clear the invariance properties of the expressions.

First we introduce $\mathrm {SU}(3)$ structures, which are a convenient framework for studying nearly Kähler manifolds:

Definition 1

An $\mathrm {SU}(3)$ structure $(\omega ,\psi = \psi ^+ + i \psi ^-)$ on a 6-manifold M is a pair of forms $\omega \in {\varOmega }^2(M)$ and $\psi \in {\varOmega }^3(M,{\mathbb {C}})$ satisfying

$$\begin{aligned} 2\omega ^3 = 3\psi ^+\wedge \psi ^-, \qquad \omega \wedge \psi = 0. \end{aligned}$$

We will refer to these equations as the SU(3) structure equations.

Theorem 3

([4]) A nearly Kähler structure is equivalent to an $\mathrm {SU}(3)$ structure $(\omega ,\psi ^+)$ satisfying

$$\begin{aligned} {{\,\mathrm{d}\,}}\omega = 3\psi ^+, \qquad {{\,\mathrm{d}\,}}\psi ^- = -2\omega \wedge \omega . \end{aligned}$$

We will refer to these equations as the nearly Kähler structure equations.

Definition 2

A toric nearly Kähler manifold $(M,\omega ,\psi ,T)$ is a 6-manifold M equipped with a nearly Kähler structure $\mathrm {SU}(3)$ structure $(\omega , \psi = \psi ^++i\psi ^-)$ which is invariant under the effective action of a 3-torus T. We will denote by (g, J) the associated Hermitian structure.

Let $K:{\mathfrak {t}}\rightarrow {\varGamma }(TM)$ be the linear map which sends elements of the Lie algebra ${\mathfrak {t}}$ of T to the vector field generating the corresponding action. Let $\mathfrak t_M$ denote the image of K. The exact forms ${{\,\mathrm{d}\,}}\omega$ and ${{\,\mathrm{d}\,}}\psi ^-$ are T-invariant, so by [11] admit natural multi-moment maps

$$\begin{aligned} \mu :=\omega \circ ({\varLambda }^2 K):M\rightarrow {\varLambda }^2{\mathfrak {t}}^*, \qquad \varepsilon :=\psi ^-\circ ({\varLambda }^3 K):M\rightarrow {\varLambda }^3{\mathfrak {t}}^*. \end{aligned}$$

Let $\mathring{M}:=M\backslash \varepsilon ^{-1}(0).$ By the definition of $\varepsilon$, K is injective on $\mathring{M}$. Thus we can define $\theta \in {\varOmega }^1(\mathring{M},{\mathfrak {t}})$ such that $\theta \circ K = \mathrm {Id}$ and $\theta |_{{\mathfrak {t}}_{\mathring{M}}^\perp }=0.$ Define

$$\begin{aligned} \gamma :=J\theta = -\theta \circ J\in {\varOmega }^1(\mathring{M},{\mathfrak {t}}). \end{aligned}$$

Since $\psi$ is a (3, 0)-form, $\mathring{M}$ is also described as the set of points where the image ${\mathfrak {t}}_M$ of K intersects transversally with $J{\mathfrak {t}}_M$. This allows us to write $T\mathring{M} = {\mathfrak {t}}_M\oplus J{\mathfrak {t}}_M$ with frame $(\theta ,\gamma )\in {\varOmega }^1(\mathring{M},{\mathfrak {t}}\oplus {\mathfrak {t}})$.

Since $\theta +i\gamma$ gives a framing of ${\varLambda }^{(1,0)}\mathring{M}$, we can write $\psi = i\varepsilon {\varLambda }^3 (\theta +i\gamma )$, so that

$$\begin{aligned} \psi ^+ = \varepsilon ({\varLambda }^3\gamma -\gamma \wedge {\varLambda }^2 \theta ), \qquad \psi ^- = \varepsilon ({\varLambda }^3\theta -\theta \wedge {\varLambda }^2\gamma ). \end{aligned}$$

Similarly, the rest of the structures can be given in terms of the multi-moment maps $(\mu ,\varepsilon )$, the frame $(\theta ,\gamma )$, and a matrix

$$\begin{aligned} C=g\circ (\odot ^2 K):M\rightarrow \mathrm {Sym}^2{\mathfrak {t}}^*. \end{aligned}$$

For example,

$$\begin{aligned} \omega = \mu ({\varLambda }^2\theta +{\varLambda }^2\gamma ) + \theta \wedge \mathbf {c}, \end{aligned}$$

where $\mathbf {c} = \gamma \lrcorner C \in {\varOmega }^1(\mathring{M},{\mathfrak {t}}^*).$

Lemma 1

([12]) Using this framework, the $\mathrm {SU}(3)$ structure equations are equivalent to

$$\begin{aligned} \det \, C = \varepsilon ^2 + C(V,V), \end{aligned}$$

(1)

while the nearly Kähler structure equations are equivalent to

$$\begin{aligned} {{\,\mathrm{d}\,}}(\varepsilon \mathbf {c}) = 0, \qquad \tfrac{1}{4}\varepsilon {{\,\mathrm{d}\,}}\theta = \mathbf {c}\wedge \mathbf {c} - \mu \mu \cdot (\gamma \wedge \gamma ), \end{aligned}$$

where $V\in {\varGamma }(\mathring{M},{\varLambda }^3{\mathfrak {t}}^*\otimes {\mathfrak {t}})$ is the element corresponding to $\mu$ via the natural isomorphism $\natural :{\varLambda }^3{\mathfrak {t}}^*\otimes {\mathfrak {t}}\cong {\varLambda }^2{\mathfrak {t}}^*$.

Here, by $\det \, C\in {\varGamma }\left( \mathring{M},({\varLambda }^3{\mathfrak {t}}^*)^{\otimes 2}\right)$, we mean the square of the volume form on ${\mathfrak {t}}$ induced by C. This agrees with the usual determinant in coordinates.

Since the functions $\varepsilon$, V, and C are T-invariant, they descend to $\mathring{M}/ T$, which can be locally identified with ${\varLambda }^2{\mathfrak {t}}^*$ via $\mu$. Since we can think of $\mu$ as giving coordinates on ${\varLambda }^2{\mathfrak {t}}^*$, we can think of $\varepsilon$, V, and C as functions locally given in these coordinates on some $U\subset {\varLambda }^2{\mathfrak {t}}^*$.

These coordinates allow explicit computations of several expressions in terms of a potential function:

Theorem 4

([12]) There exists a function $\varphi :U\rightarrow ({\varLambda }^3{\mathfrak {t}}^*)^2$ whose Hessian in $\mu$ coordinates is C. We also have

$$\begin{aligned} \varepsilon ^2 = \tfrac{8}{3}(1-\partial _r)\varphi , \qquad C(V,V) = (\partial _r^2-\partial _r)\varphi , \end{aligned}$$

where $\partial _r$ is the Euler vector field for ${\varLambda }^2{\mathfrak {t}}^*$ (so that in coordinates $\partial _r = \mu ^i\partial _{\mu _i}$).

Combining this with Equation (1) gives the Monge–Ampère type equation

$$\begin{aligned} \det \circ {{\,\mathrm{Hess}\,}}\varphi = \left( \tfrac{8}{3}-\tfrac{11}{3}\partial _r+\partial _r^2\right) \varphi , \qquad \qquad \qquad (\star ) \end{aligned}$$

which we will refer to as the toric nearly Kähler equation or just $(\star )$.

Note that with respect to the frame $(\theta ,\gamma )$, g is represented by the matrix

$$\begin{aligned} D:=\begin{pmatrix} {{\,\mathrm{Hess}\,}}\varphi &{} -\mu \\ \mu &{} {{\,\mathrm{Hess}\,}}\varphi \end{pmatrix}\in {\varGamma }\left( U,\mathrm {Sym}^2({\mathfrak {t}}\oplus {\mathfrak {t}})^*\right) , \end{aligned}$$

where $\mu \in {\varGamma }(U,{\varLambda }^2{\mathfrak {t}}^*)$ is the inclusion (identity) map.

The above theorem has a partial converse:

Theorem 5

([12]) Every solution of the toric nearly Kähler equation on some open set U of ${\varLambda }^2{\mathfrak {t}}^*$ defines in a canonical way a nearly Kähler structure with 3 linearly independent commuting infinitesimal automorphisms on $U_0\times T^3$, where

$$\begin{aligned} U_0 = \left\{ x\in U: (1-\partial _r)\varphi > 0 \text { and } D \text { is positive definite}\right\} . \end{aligned}$$

Note that if $\varphi$ is given by a toric nearly Kähler structure, then $(1-\partial _r)\varphi$ is proportional to the $\varepsilon ^2$, and D is the expression of g in the frame $(\theta ,\gamma )$. Now consider the following set with an a priori weaker constraint than $U_0$:

$$\begin{aligned} {\hat{U}}_0 := \left\{ x\in U: (1-\partial _r)\varphi > 0 \text { and } {{\,\mathrm{Hess}\,}}\varphi \text { is positive definite}\right\} . \end{aligned}$$

However, this constraint is not weaker:

Lemma 2

$U_0 = {\hat{U}}_0.$

Proof

Since D being positive definite implies that ${{\,\mathrm{Hess}\,}}\varphi$ is positive definite, we find that $U_0\subseteq {\hat{U}}_0$. It remains to show that D has no null vectors in ${\hat{U}}_0$, which implies the reverse inclusion.

Let $C={{\,\mathrm{Hess}\,}}\varphi$ and $\varepsilon ^2 = \tfrac{8}{3}(1-\partial _r)\varphi$. Defining $j = C^{-1}\mu$, we find that any null vector for D is of the form $(v,w)\in {\mathfrak {t}}\oplus {\mathfrak {t}}$ at some point $p\in {\hat{U}}_0$ with

$$\begin{aligned} jw = v, \qquad jv = -w. \end{aligned}$$

Thus v and w are eigenvectors of $j^2$ at p with eigenvalue $-1$. Thus it suffices to show that $j^2\in {\varGamma }\left( U_0,{{\,\mathrm{End}\,}}({\mathfrak {t}})\right)$ never attains an eigenvalue $-1$.

Choosing a basis for ${\mathfrak {t}}$ so that $C_p$ is diagonal at any chosen $p\in {\hat{U}}_0$ allows one to verify that $C^{-1}\mu C^{-1} = \left( \frac{CV}{\det \, C}\right) ^\natural$, where we abuse notation by using $\natural$ to also denote the isomorphism ${\varLambda }^3{\mathfrak {t}}\otimes {\mathfrak {t}}^*\cong {\varLambda }^2{\mathfrak {t}}$. Then

$$\begin{aligned} j^2 = C^{-1}\mu C^{-1}\mu = \left( \frac{CV}{\det \, C}\right) ^\natural V^\natural = \frac{-C(V,V){{\,\mathrm{Id}\,}}+ (CV)\otimes V}{\det \, C}, \end{aligned}$$

where throughout this computation we’ve been using juxtaposition to denote ‘matrix multiplication’, or contraction of a single ${\mathfrak {t}}, {\mathfrak {t}}^*$ index pair. Since $V\in \mu ^{-1}(0)\subseteq j^{-1}(0)$, we find that $j^2$ has eigenvalues 0 with multiplicity 1 and $-\frac{C(V,V)}{\det \, C}$ with multiplicity 2. By the toric nearly Kähler equation, $-1$ is an eigenvalue only when $\varepsilon =0$, which is impossible on ${\hat{U}}_0$ by definition. $\square$

This lemma can be used to interpret what goes wrong when trying to find a completion of a local toric nearly Kähler manifold. If some connected $\mathring{M}$ is maximal in the sense that it is not properly contained in a toric nearly Kähler manifold where $\varepsilon$ doesn’t vanish, what is happening at the boundary? Using $\mu$, we can interpret this boundary as a set of points in ${\varLambda }^2{\mathfrak {t}}^*$. By (1), C is going to remain positive definite as long as $\varepsilon$ doesn’t vanish. Thus the previous lemma shows that if $\varepsilon$ does not limit to 0 at the boundary point, then the local solution $\varphi$ to the toric nearly Kähler equation cannot be extended to the boundary point. In Sect. 6, we show that local radial solutions can be extended to have the radius defined between 0 and some finite $r_0$. The differential equation is singular at 0, while $\varepsilon$ vanishes when the radius is $r_0$.

3 Relation to toric $G_2$ manifolds

For a strict nearly Kähler manifold $(M,\omega ,\psi ^++i\psi ^-)$ with metric g, consider the Riemannian cone $\left( N=M\times (0,\infty ),g_N = r^2 g + {{\,\mathrm{d}\,}}r^2\right)$, where $r\in (0,\infty )$ is the radial coordinate. It is well known that N admits a parallel $G_2$ structure given by

$$\begin{aligned} \phi :={{\,\mathrm{d}\,}}\left( \frac{r^3\omega }{3}\right) , \quad *\phi :=-{{\,\mathrm{d}\,}}\left( \frac{r^4\psi ^-}{4}\right) . \end{aligned}$$

If M is toric, then the torus action lifts to a multi-Hamiltonian action on N with respect to the forms $\phi$ and $*\phi$. This makes N a toric $G_2$ manifold as studied in [10]. The corresponding multi-moment maps for $\phi$ and $*\phi$, respectively, are

$$\begin{aligned} \nu _N:=\tfrac{1}{3}r^3\mu :N\rightarrow {\varLambda }^2{\mathfrak {t}}^*,\quad \varepsilon _N:=-\tfrac{1}{4}r^4\varepsilon :N\rightarrow {\varLambda }^3{\mathfrak {t}}^*. \end{aligned}$$

(2)

From [10], $\nu _N\oplus \varepsilon _N$ maps the set of singular orbits S of N to a graph in ${\varLambda }^2{\mathfrak {t}}^*\oplus {\varLambda }^3{\mathfrak {t}}^*\cong {\mathbb {R}}^3\oplus {\mathbb {R}}$. Moreover, $\varepsilon _N$ is constant on each connected component of S. In the case when $N=M\times (0,\infty )$ is the cone over a toric nearly Kähler manifold, then the radial symmetries of (2) imply that $\varepsilon _N$ vanishes on the graph, and moreover each edge of the graph is a radial ray shining out from the origin in ${\varLambda }^2{\mathfrak {t}}^*$. Since points on the edge of the graph correspond to torus orbits where a single circle collapses, we immediately find

Corollary 2

On a toric nearly Kähler manifold, the torus action is free away from a finite set of orbits where a single circle collapses and $\varepsilon$ vanishes.

4 Global properties

Let $(M,\omega ,\psi , T)$ be a connected complete toric nearly Kähler 6-manifold. In this section we will prove the properties of the multi-moment maps claimed in Theorem 1. Recall that we define $\mathring{M} = M\backslash \varepsilon ^{-1}(0)$.

Lemma 3

$\mu |_{\mathring{M}}$ is a submersion.

Proof

Lemma 4.1(i) in [12] gives $d\mu |_{\mathring{M}} = -4\varepsilon \cdot \gamma$. The result follows since $\gamma$ has full rank and $\varepsilon$ does not vanish on $\mathring{M}$. $\square$

Now we can show that $\mu (\mathring{M})$ is star-shaped around 0:

Lemma 4

Every $p\in \mathring{M}$ is contained in some path $\ell$ on $\mathring{M}$ such that $\mu |_\ell :\ell \rightarrow {\varLambda }^2{\mathfrak {t}}^*$ is an injective map whose image is a line segment between 0 and $\mu (p)$.

Proof

If $\mu (p)=0$, there is nothing to show. Otherwise, there exists some maximal connected subset L of the line segment $\overline{0\mu (p)}$ which lifts to a path $\ell$ on $\mathring{M}$ through p. By Lemma 3, L is non-empty and open. If L is closed, then $L =\overline{0\mu (p)}$ as required. Otherwise, since M is complete, $\ell$ has a limiting point $p'\in M\backslash \mathring{M}$. Hence L is a radial line segment travelling outward from $\mu (p')$, a point at which $\varepsilon$ vanishes by the definitions of $p'$ and $\mathring{M}$. However, using Theorem 4, we compute the radial directional derivative

$$\begin{aligned} \partial _r(\varepsilon ^2)=\tfrac{8}{3}\partial _r(1-\partial _r)\varphi =-\tfrac{8}{3} C(V,V)\le 0. \end{aligned}$$

This implies that $\varepsilon ^2$ is non-positive along L, contradicting $\varepsilon \ne 0$ at $p\in \mathring{M}$. $\square$

Since $\mu$ and $\varepsilon$ are T-invariant, they induce maps ${\bar{\mu }}:M/ T\rightarrow {\varLambda }^2{\mathfrak {t}}^*$ and ${\bar{\varepsilon }}:M/ T\rightarrow {\varLambda }^3{\mathfrak {t}}^*$, which are called orbital multi-moment maps.

Lemma 5

For any connected component $M_0$ of $\mathring{M}$, ${\bar{\mu }}|_{M_0/ T}$ is injective.

Proof

For any $p\in M_0$, let $\ell$ be the path between p and some $p'\in \mu ^{-1}(0)$ guaranteed by the previous lemma. The map

$$\begin{aligned} F:M_0/ T\rightarrow {\bar{\mu }}^{-1}(0): T p\mapsto T p' \end{aligned}$$

is clearly well defined and continuous. Since $M_0/ T$ is connected and ${\bar{\mu }}^{-1}(0)$ is discrete, the image of F is a single orbit which we will denote by $o_0\in M_0/ T$.

Now let $o_1,o_2\in M_0/ T$ with ${\bar{\mu }}(o_1)={\bar{\mu }}(o_2)$. For each $i\in \{1,2\}$, the previous lemma can be used to construct a path $\ell _i$ between $o_0$ and $o_i$ in $M_0/ T$ which is a lift of the line segment L between 0 and ${\bar{\mu }}(o_1)={\bar{\mu }}(o_2)$. Since $\mu |_{\mathring{M}}$ is a submersion, ${\bar{\mu }}|_{\mathring{M}/ T}$ is a local homeomorphism. In particular, $T_{o_0}\ell _1 = {\bar{\mu }}^{-1}(T_0 L )= T_{o_0}\ell _2$, implying that $\ell _1=\ell _2$. Thus $o_1=o_2$ as required. $\square$

By this lemma, ${\bar{\mu }}$ gives global coordinates on $M_0/ T$. This allows us to view the multi-moment image $(\mu ,\varepsilon )(M_0)$ as the graph of ${\bar{\varepsilon }}|_{M_0/ T}$. Since $\varepsilon$ vanishes on $\partial M_0$, we find that M/T is recovered by gluing together two of these graphs. We will first need a lemma:

Lemma 6

Along the set $\{\varepsilon = 0\}$, $d\varepsilon$ only vanishes on a finite set of orbits.

Proof

Let $p\in \varepsilon ^{-1}(0)$, so that $\psi ^-|_{{\varLambda }^3{\mathfrak {t}}_M}$ vanishes at p. Note that by T invariance of the closed form $d\mu =({\varLambda }^2 K)\lrcorner (3\psi ^+)$, we have $\psi ^+|_{{\varLambda }^3{\mathfrak {t}}_M} =0$. Thus $\psi |_{{\varLambda }^3{\mathfrak {t}}_M}$ vanishes at p. Since $\psi$ is a complex volume form, this implies that ${\varLambda }^3{\mathfrak {t}}_M|_p \subseteq T^{1,1}_p M\wedge T_p M$. Thus for any $0\ne {\varXi }\in {\varLambda }^3{\mathfrak {t}}$, we can write ${\varXi }_M|_p = P\wedge X\in T^{1,1}_p M\wedge T_p M$. We compute

$$\begin{aligned} d\varepsilon ({\varXi })_p ={\varXi }_M|_p\lrcorner d\varphi ^-_p =(P\wedge X)\lrcorner (-2\omega \wedge \omega ) = -2 \omega (P) (X\lrcorner \omega ). \end{aligned}$$

By the non-degeneracy of $\omega$, this only vanishes when ${\varXi }_M$ does, which are the points where the torus action is not free. By Corollary 2, this vanishing is at a finite set of orbits. $\square$

Theorem 6

$({\bar{\mu }},{\bar{\varepsilon }}):M/ T\rightarrow {\varLambda }^2{\mathfrak {t}}^*\oplus {\varLambda }^3{\mathfrak {t}}^*$ is injective with image a 3-sphere. Moreover, the component of $({\varLambda }^2{\mathfrak {t}}^*\oplus {\varLambda }^3{\mathfrak {t}}^*)\backslash (\mu ,\varepsilon )(M)$ containing 0 is star-shaped about 0.

Proof

Let $M_+$ be a connected component of $\mathring{M}$. By Lemma 4, $U:=\mu (M_+)$ is star shaped around the origin. In particular, $\partial M_+$ has one connected component. Since $\varepsilon \ne 0$ on $\mathring{M}$, $\varepsilon$ has a sign on $M_+$. Since $\varepsilon$ vanishes on the 2-dimensional $\partial M_+\subseteq \partial \mathring{M}$, the previous lemma implies that $\varepsilon$ changes sign when crossing this boundary. Thus there is some other connected component $M_-$ of $\mathring{M}$ with the opposite sign of $\varepsilon$ and $\partial M_+\subseteq \partial M_-$. But $\mu (M_-)$ must also be star shaped around 0 with boundary $\partial U$, so $\mu (M_-)=U=\mu (M_+)$. Thus $\partial M_-=\partial M_+$, so that we have $M={\overline{M}}_+\cup {{\overline{M}}}_-$. In particular, ${\bar{\mu }}:M\rightarrow U$ is a double cover ramified over $\partial U$, with the sign of ${\bar{\varepsilon }}$ distinguishing the points in each ${\bar{\mu }}$ fibre. Thus $({\bar{\mu }},{\bar{\varepsilon }})$ is injective. Since U is diffeomorphic to a 3-ball, the image of $({\bar{\mu }},{\bar{\varepsilon }})$ is a 3-sphere.

By relabelling if necessary, we can assume that $\pm \varepsilon$ is positive on $M_\pm$. The component of $({\varLambda }^2{\mathfrak {t}}^*\oplus {\varLambda }^3{\mathfrak {t}}^*)\backslash (\mu ,\varepsilon )(M)$ containing 0 can be written as $D_+\cup D_-$, where

$$\begin{aligned} D_\pm =\left\{ (x,y)\in {\varLambda }^2{\mathfrak {t}}^*\oplus {\varLambda }^3{\mathfrak {t}}^*:x\in {\bar{\mu }}\left( \overline{M_\pm }\right) ,\pm y\in \left[ 0,\pm \varepsilon (y)\right] \right\} . \end{aligned}$$

Now each $D_\pm$ is star-shaped around 0, since ${\bar{\mu }}(\overline{M_\pm })$ is and $\pm \varepsilon$ is decreasing in radial directions. Thus $D_+\cup D_-$ is star-shaped around 0 as required. $\square$

We can now wrap up the proof of the main theorem:

Proof of Theorem 1

Corollary 2 combined with Theorem 6 gives most of the claim. T orbits in $\mu ^{-1}(0)$ must be Lagrangian by definition, and there are two of them, since ${\bar{\mu }}$ is a double cover ramified at $\varepsilon ^{-1}(0)$. $\square$

5 Some topology

We apply the results from the previous section to prove Corollary 1. The obstruction we use to prove this comes from Myers’ theorem [13], which asserts that if a complete Riemannian manifold has Ricci curvature positive and bounded away from zero, then the diameter must be bounded. Since the same must be true for the universal cover, the fundamental group must be finite. In particular, the first Betti number must vanish.

Proposition 1

Let $(M,\omega ,\psi , T)$ be a connected complete toric nearly Kähler 6-manifold. Then the action of T is not free.

Proof

Assume that the action of T is free, so that M is a $T^3$ bundle over $S^3$. It follows that we have the Wang long exact sequence [16]

$$\begin{aligned} H_\bullet ( T^3)\rightarrow H_\bullet (M)\rightarrow H_{\bullet -3}( T^3)\xrightarrow {[-1]}, \end{aligned}$$

which shows that $H_1(M)\cong H_1( T^3)\cong {\mathbb {Z}}^3$ has positive rank. This contradicts Myers’ theorem. $\square$

Consider the set $S\in M/ T$ of T orbits where the action is not free. By Corollary 2, each orbit $s\in S$ has a one-dimensional isotropy group whose Lie algebra is given by a line $\iota (s)\in {\mathbb {P}}{\mathfrak {t}}$. We will need the following lemma:

Lemma 7

The map $\iota :S\rightarrow {\mathbb {P}}{\mathfrak {t}}$ is at most two to one.

Proof

Since C is the metric on the torus orbits, S is the vanishing locus of $\det \, C$. Thus (1) implies that the non-negative functions $\varepsilon ^2$ and C(V, V) both vanish on S. Thus $S\subset \partial \mathring{M}/ T$, and for any $s\in S$, $\iota (s) = (\mathbb PV)(s)$, where $\mathbb PV\in {\varGamma }(M/ T,{\mathbb {P}}{\mathfrak {t}})$ is the projectivication of V. Note that the identification ${\varLambda }^2{\mathfrak {t}}^*\cong {\varLambda }^3{\mathfrak {t}}^*\otimes {\mathfrak {t}}$ induces an identification ${\mathbb {P}}{\varLambda }^2{\mathfrak {t}}^*\cong {\mathbb {P}}{\mathfrak {t}}$ under which ${\mathbb {P}}\mu$ is identified with $\mathbb PV$. But by Theorem 6, ${\bar{\mu }}$ is injective on $\partial \mathring{M}/ T\supset S$. Thus it suffices to show that ${\bar{\mu }}(S)\subset \mu (\partial \mathring{M})$ intersects the line ${\mathbb {P}}\mu (s)$ at most two times. But $\mu (M)$ is star-shaped around 0, so $\mu (\partial \mathring{M})$ intersects the line ${\mathbb {P}}\mu (s)$ at two antipodal points. $\square$

Consider the case where S has k orbits. We will consider decompositions $M = A\cup B$, where A and B are both unions of T-orbits and $A\cap B\cap S=\emptyset$. To see that such a decomposition exists, note that if H is a hyperplane in ${\varLambda }^2{\mathfrak {t}}^*\oplus {\varLambda }^3{\mathfrak {t}}^*$ disjoint from $(\mu ,\varepsilon )(S)$, then there exists a neighbourhood U of H also disjoint from $(\mu ,\varepsilon )(S)$. Now it is clear that we can find A and B as claimed with $(\mu ,\varepsilon )(A\cap B) = U$. Moreover, we see that no two orbits in $S\cap A$ (respectively $S\cap B$) correspond to the same element of ${\mathfrak {t}}$, since by the previous lemma, they would correspond to antipodal points in $\mu \left( \varepsilon ^{-1}(0)\right)$, which are avoided by this construction.

Now we have $A\cong D_i$ and $B\cong D_j$ where $i+j = k$ and $D_i$ is a $T^3$ fibration over the three-ball $D^3$ with i orbits where circles collapse. Moreover, these circles are different, in the sense that the collapsing directions correspond to different vectors in ${\mathfrak {t}}$. Using $\simeq$ to denote homotopy equivalence, we will compute these for the first few $D_i$:

Lemma 8

$D_0\simeq T^3$, $D_1\simeq T^2$, $D_2\simeq S^3\times S^1$.

Proof

Since every bundle over $D^3$ is trivial, $D_0 \cong T^3\times D^3\simeq T^3$.

$D_1$ is a neighbourhood of the collapsed orbit. Thus $D_1\cong T^2\times {\mathbb {R}}^4$ by identifying ${\mathbb {R}}^4$ with $D_1/ T$ and $T^2$ the quotient of T with the circle that collapses.

$D_2/ T$ is a neighbourhood of a curve C connecting the two collapsing orbits. Thus $D_2$ retracts to some ${{\tilde{D}}}_2$ such that ${{\tilde{D}}}_2/ T\cong C$. Since the circles that collapse are different, we can write ${{\tilde{D}}}_2\cong (S^1\times S^1\times [0,1])/\sim ,$ where $\sim$ collapses the first circle at 0 and the second circle at 1. Now

$$\begin{aligned} {{\tilde{D}}}_2\rightarrow {\mathbb {C}}^2\times S^1:[\theta _1,\theta _2,\theta _3,x]\mapsto \left( \sin \left( \tfrac{\pi }{2}x \right) e^{i\theta _1}, \cos \left( \tfrac{\pi }{2} x\right) e^{i\theta _2},\theta _3\right) \end{aligned}$$

identifies ${{\tilde{D}}}_2\cong S^3\times S^1$. $\square$

Before we proceed to applying Mayer–Vietoris to the decomposition $M=A\cup B$, we still need to understand the equatorial region $E:=A\cap B$.

Lemma 9

$h_1(E)=h_4(E)\in [2,3]$.

Proof

E must be a $T^3$ bundle over $U\simeq S^2$. Thus E retracts to a $T^3$ bundle ${{\bar{E}}}$ over $S^2$. Since ${{\bar{E}}}$ is compact, we have the duality $h_1({{\bar{E}}}) =h_4({{\bar{E}}})$. Part of the Wang sequence is

$$\begin{aligned} \rightarrow H_{2-2}( T^3)\rightarrow H_1( T^3)\rightarrow H_1({{\bar{E}}})\rightarrow 0. \end{aligned}$$

Thus $h_1(E) = h_1({{\bar{E}}})\in h_1( T^3) - \left[ 0,h_0( T^3)\right] = 3 - [0,1] = [2,3]$. $\square$

We can now work with the Mayer–Vietoris sequence with respect to the decomposition $M = A\cup B = D_i\cup D_j$. Since $A\cap B = E$, this sequence is

$$\begin{aligned} H_\bullet (E)\rightarrow H_\bullet (D_i)\oplus H_\bullet (D_j)\rightarrow H_\bullet (M)\xrightarrow {[-1]}. \end{aligned}$$

We are now ready to prove the main result of this section:

Proof of Corollary 1

The Mayer–Vietoris sequence at $\bullet = 1$ gives

$$\begin{aligned} h_1(D_i)+h_1(D_j)\le h_1(E)+h_1(M)\le 3, \end{aligned}$$

where the second inequality uses the previous lemma and Myers’ theorem. But by Lemma 8, $h_1(D_i)=3-i$ or $i\in \{0,1,2\}$. In particular, for $k=i+j<3$, we have $h_1(D_i)+h_1(D_j) = 6-(i+j) > 3$, contradicting our upper bound.

For $k=3$, choose $A\cong D_1$ and $B\cong D_2$. Since $h_1(M)=0$, the Mayer–Vietoris sequence at $\bullet = 1$ gives $h_1(E)\ge h_1(D_1)+h_1(D_2) = 3$. Combining this with the previous lemma gives $h_1(E)=h_4(E) = 3$. Since $h_5(M)=h_1(M)=0$, the Mayer–Vietoris sequence at $\bullet = 4$ gives the contradiction

$$\begin{aligned} 3 = h_4(E) \le h_4(D_1)+h_4(D_2) = 1. \end{aligned}$$

$\square$

6 Radial solutions

In this section, we study solutions of the form $\varphi (\mu ) = x(t)$, where $t = \tfrac{1}{2}\Vert \mu \Vert ^2$ is a radial coodinate, and $\Vert \cdot \Vert$ is the Euclidean metric on ${\varLambda }^2{\mathfrak {t}}^*$. These were studied in [12], where they show that the nearly toric equation simplifies to the ODE

$$\begin{aligned} 0 = {\mathcal {D}}(x):=3(x'^2-2t)(x'+2tx'')-8(x-2tx') \end{aligned}$$

subject to the constraint

$$\begin{aligned} x>2tx'>2t\sqrt{2t}, \end{aligned}$$

(3)

where the derivatives are taken with respect to t. The main result is that such a radial solution cannot be complete:

Theorem 7

If $(M,\omega ,\psi , T)$ is a connected complete toric nearly Kähler 6-manifold corresponding to a solution $\varphi$ to the toric nearly Kähler equation, then $\varphi$ is not radially symmetric.

Proof

Assume that $\varphi$ is radially symmetric. Combining this symmetry with Theorem 6, $\mu (M)$ must be a closed 3-disc ${\varDelta }$ centred at the origin. Now consider the set of points S in M where the torus action is not free. By Corollary 2, $\mu (S)$ is a discrete set of points in $\partial {\varDelta }$. But by radial symmetry, $\mu (S)$ must be either empty or all of $\partial {\varDelta }$. But $\mu (S)$ is a discrete set, so it can’t be $\partial {\varDelta }$. Thus $\mu (S)$, and hence S is empty. This contradicts Proposition 1. $\square$

We now investigate what goes wrong with the ODE to prevent completeness. Local existence of solutions to ODE’s will give a local solution x(t) to ${\mathcal {D}}(x)=0$ near any prescribed initial 1-jet $\left( t_0,x(t_0),x'(t_0)\right)$ satisfying the constraints (3). Let $(t_-,t_+)$ be the maximal open interval on which the solution can be extended while satisfying the constraints.

$t_\pm$ must be either a point where x(t) blows up or a boundary point of the constraints. By Lemma 2, $\varepsilon ^2 = \tfrac{8}{3}(x-2tx')>0$ implies the other constraint $x'>\sqrt{2t}$. Thus the boundary condition is simply $\varepsilon ^2=0$.

Lemma 10

x(t) does not blow up at $t_+ < \infty$.

Proof

First note that $\varepsilon ^2>0$ implies that $(\log x)'<\frac{1}{2t}$. Integrating this implies that $x(t) < x_0\sqrt{\tfrac{t}{t_0}}$. Since x(t) is also positive, it cannot blow up in finite time.

On the other hand, integrating $x'>\sqrt{2t}$ gives $x-x_0 > \frac{\sqrt{2t}^3-\sqrt{2t_0}^3}{3}$. This lower bound for x grows faster as t increases than the upper bound for x in the previous paragraph. Thus $t_+$ is finite. $\square$

We compute

$$\begin{aligned} \varepsilon ^2 = \tfrac{8}{3} (x-2t x'), \qquad 2tx'' = \frac{\varepsilon ^2}{x'^2-2t}-x', \qquad (\varepsilon ^2)' = -\frac{8}{3}\frac{\varepsilon ^2}{x'^2-2t}. \end{aligned}$$

Note that by Lemma 2, the constraints can be rewritten as $0<\varepsilon ^2\propto x-2tx'$.

Thus the constraints imply that $x'^2-2t>0$, so the ODE is regular when the constraints hold and $t>0$.

Lemma 11

$t_-=0$.

Proof

Since $t_-$ is the boundary point of a maximal domain of an ODE subject to the constraint $\varepsilon ^2>0$, at $t_-$ either the ODE is singular and the solution x(t) becomes unbounded or $\varepsilon ^2$ vanishes. Since the ODE is singular at $t=0$, we must have $t_-\ge 0$. Since x is positive and increasing, it must be bounded in $(t_-,t_+)$. Since $\varepsilon ^2$ is decreasing, it cannot vanish at $t_-$. Thus $t_-$ must a singular point of the ODE, in particular the only one: 0. $\square$

By Theorem 7, there must be some singularity for x(t) in $[0,t_+]$, and by the previous two lemmas it must be at $t=0$.

Note that the estimate in Lemma 10 doesn’t essentially require radial symmetry: it only uses $\varepsilon ^2\propto \varphi -\partial _r\varphi >0$. In particular, continuing the discussion following Lemma 2, $\varphi$ should not become unbounded as one tries to extend solutions in radial directions away from the origin.

7 Polynomial solutions

In this section we will try to understand polynomial solutions to the toric nearly Kähler equation. As described in [12], the toric nearly Kähler structure on $S^3\times S^3$ corresponds to the solution of the toric nearly Kähler equation

$$\begin{aligned} \varphi _0 := 3 + \sum _j \mu _j^2 + \frac{1}{\sqrt{3}}\prod _j\mu _j, \end{aligned}$$

(4)

where $\left\{ \mu _j\right\} _{j=1}^3$ are coordinates on ${\varLambda }^2{\mathfrak {t}}^*$ induced by the multi-moment map $\mu$. We will prove Theorem 2 by treating each degree of polynomial separately. First we will introduce some notation. If E is an equation or expression, and m is a monomial in ${\mathbb {R}}\left[ \mu _1,\mu _2,\mu _3\right] ,$ then [m]E and (m)E will refer, respectively, to the coefficient of m in E, and the part of E which is a multiple of m. We will use $\nabla$ to denote the gradient in $\left\{ \mu _j\right\} _{j=1}^3$ coordinates, and abuse notation by not distinguishing it from its transpose, or a restricted gradient to an context-appropriate subset of the coordinates. Similarly, $\nabla ^2$ will denote the Hessian, where the set of coordinates may depend on context.

Proposition 2

Every cubic solution to the toric nearly Kähler equation is equivalent to $\varphi _0$ up to linear changes in coordinates.

Proof

Let $\varphi$ be some cubic solution of the toric nearly Kähler equation $(\star )$. Write $\varphi = \sum _{j=0}^3\varphi ^k$, where each $\varphi ^k$ is a degree k homogeneous polynomial in $\left\{ \mu _j\right\} _{j=1}^3$. As noted in [12], $\varphi ^1$ can be chosen to be 0. The degree 3 term of $(\star )$ gives

$$\begin{aligned} \left| \nabla ^2\varphi ^3\right| = \left( \tfrac{8}{3}-\tfrac{11}{3}\partial _r+\partial _r^2\right) \varphi ^3 = \left( \tfrac{8}{3}-{11}+9\right) \varphi ^3 =\tfrac{2}{3}\varphi ^3. \end{aligned}$$

Since $\left| \nabla ^2\varphi ^3\right| \propto \varphi ^3$, the plane algebraic curve $V\left( \varphi ^3\right) \subset {\mathbb {C}}P^2$ is a union of lines [15], so that $\varphi ^3$ is a product of linear factors. We can choose coordinates along these lines so that $\varphi ^3=\lambda \prod _{j=1}^3\mu _j$ for some $\lambda \in {\mathbb {R}}$. We can compute $\left| \nabla ^2\varphi ^3\right| =2\lambda ^2\varphi ^3,$ so we must have $\lambda ^2 = \tfrac{1}{3}$. Again, by changing coordinates, we may assume that $\lambda = \tfrac{1}{\sqrt{3}}$.

Writing $\varphi ^2 = \varphi ^2_d + \varphi ^2_o$, where $\nabla ^2\varphi ^2_d$ is a diagonal matrix, while $\nabla ^2\varphi ^2_o$ vanishes on the diagonal, one can compute that the degree 2 term of $\left| \nabla ^2\varphi \right|$ is $\frac{2}{3}\left( \varphi ^2_o-\varphi ^2_d\right)$. Thus the degree 2 term of $(\star )$ is

$$\begin{aligned} -\frac{2}{3}\varphi ^2 = \frac{2}{3}\left( \varphi ^2_o-\varphi _d\right) , \end{aligned}$$

so that $\varphi _o=0$. Thus $\nabla ^2\varphi ^2$ is a diagonal matrix. We still have the freedom to scale the coordinates so that $\nabla ^2\varphi ^2=2{{\,\mathrm{Id}\,}}$, or equivalently $\varphi ^2=\sum _{j=1}^3\mu _j^2.$ The degree 0 term of $(\star )$ now gives $\tfrac{8}{3}\varphi ^0=\left| \nabla ^2\varphi ^2\right| =8$, so that $\varphi ^0=3$. Thus $\varphi =\varphi _0$. $\square$

For higher degree polynomial solutions, the toric nearly equation becomes overdetermined, since $\left| \nabla ^2\varphi \right|$ is formally a polynomial of degree $3\left( \deg (\varphi )-2\right)$. This suggests that the cubic solution might be the only polynomial solution. By diagonalizing $\nabla ^2\varphi ^2$, we can always choose a basis so that $\varphi ^2=\sum _{j=1}^3\mu _j^2$. As discussed in the previous proof, we will also get $\varphi ^0=3$ and $\varphi ^1=0$. The degree 1 term of $(\star )$ tells us that $\varphi ^3$ is harmonic.

Our main tool will be the following theorem, which we will refer to as Hesse’s theorem since Hesse originally claimed the result:

Theorem 8

([8]) If f is a homogeneous polynomial in 4 or less variables over an algebraically closed field, then $\det \circ {{\,\mathrm{Hess}\,}}(f)=0$ if and only if f is independent of one of the variables after a suitable homogeneous coordinate change.

Remark 1

More geometrically, this theorem shows that $\det \circ {{\,\mathrm{Hess}\,}}(f)$ vanishes whenever the directional derivative of f vanishes along some vector X. The coordinate change corresponds to any linear map which sends X to a generator of a coordinate axis. In particular, we can choose such a linear map to be orthogonal, meaning that the quadratic function $\sum _{j=1}^3\mu _j^2$ is preserved. For the rest of the paper, whenever we use Hesse’s theorem, we will use such an orthogonal coordinate change.

Proposition 3

There are no quartic solutions to the toric nearly Kähler equation.

Proof

Let $\varphi =\sum _{j=0}^4\varphi ^j$ be a solution to $(\star )$, where each $\varphi ^j$ is a homogeneous polynomial of degree j. Assume that $\varphi ^4\ne 0$, so that $\varphi$ is quartic. We can choose coordinates so that the quadratic part of $\varphi$ is $3+x^2+y^2+z^2$, where (x, y, z) is a relabelling of the coordinates $\left( \mu _j\right) _{j=1}^3$.

The degree 6 term of $(\star )$ is $\left| \nabla ^2\varphi _4\right| =0$. By Hesse’s theorem, this is equivalent to $\varphi _4$ being a function of two variables. Thus, we can choose coordinates so that $\varphi ^4 = \varphi ^4(x,y)$.

Lemma 12

After some rotation of the $y-z$ coordinates, $\varphi ^3_{zz}=0$.

Proof

Assume that the claim is false, so that $\varphi ^3_{zz}\ne 0$. The degree 5 term of $(\star )$ gives $\left| \nabla ^2_{xy}\varphi ^4\right| \varphi ^3_{zz}=0$, so that $\left| \nabla ^2_{xy}\varphi ^4\right| =0$. Using Hesse’s theorem, we can choose coordinates so that $\varphi ^4$ depends only on x. Now the degree 4 term of $(\star )$ gives

$$\begin{aligned} 4\varphi ^4=\varphi ^4_{xx}\left| \nabla ^2_{yz}\varphi ^3\right| . \end{aligned}$$

Write $\varphi ^3 = x^3\,B^0+ x^2\,B^1+x\,B^2+B^3$, where each $B^j$ is a degree j homogeneous polynomial in y and z. The $x^2$ term of the above equation gives $0=\left| \nabla ^2 B^3\right|$. Using a similar argument as in Remark 1, we can use Hesse’s theorem to rotate the $y-z$ coordinates (preserving $y^2+z^2$ while fixing x) so that $B^3\propto y^3$. Note that our assumption still gives $\varphi ^3_{zz}\ne 0$, hence $B^2_{zz}\ne 0$. The $x^3$ term of the above equation then gives $B^3_{yy}B^2_{zz}=0$, implying $B^3=0$. The $x^4$ term of the above equation gives $\left| \nabla ^2B^2\right| = \tfrac{1}{3}$.

Now ${\varDelta }\varphi ^3=0$ implies that $B^1=0$. The degree 3 term of $\left| \nabla ^2\varphi \right|$ modulo $x^3$ is

$$\begin{aligned} \left| \nabla ^2(x\,B^2)\right| = -2x\,B^2\left| \nabla ^2 B^2\right| = -\tfrac{2}{3} x\,B^2, \end{aligned}$$

where the first equality can be verified directly since $B^2$ is a homogeneous quadratic polynomial. Thus the coefficient of x of the degree 3 term of $(\star )$ is $\tfrac{2}{3} B^2 = -\tfrac{2}{3}B^2$, so that $B^2=0$. This contradicts $\left| \nabla ^2B^2\right| = \tfrac{1}{3}$. $\square$

Since $\varphi ^3_{zz}=0$, we can write $\varphi ^3 = B^3(x,y) + z\,B^2(x,y).$ We have

$$\begin{aligned} \varphi = 3 + x^2+y^2+z^2 + B^3 + z\,B^2 + \varphi ^4. \end{aligned}$$

Since $\varphi ^3$ is harmonic, so are $B^2$ and $B^3$.

$$\begin{aligned} \nabla ^2\varphi = \begin{pmatrix} 2{{\,\mathrm{Id}\,}}+ \nabla ^2(B_3+\varphi ^4) &{} \nabla B^2 \\ \nabla B^2 &{} 2 \end{pmatrix}+z\begin{pmatrix} \nabla ^2 B^2 &{} 0 \\ 0 &{} 0 \end{pmatrix}. \end{aligned}$$

Now $\left[ z^2\right]$ $(\star )$ gives $2\left| \nabla ^2 B^2\right| =-\tfrac{2}{3}.$ Combining this with ${\varDelta } B^2=0$, we can rotate coordinates so that $B^2 = \tfrac{xy}{\sqrt{3}}$. Now (z) $(\star )$ gives

$$\begin{aligned} 0 = \tfrac{4}{\sqrt{3}}(B^3+\varphi ^4)_{xy}. \end{aligned}$$

Thus $0 = \varphi ^4_{xy} = B^3_{xy}$. Since $B^3$ is harmonic, it must vanish. Note that we now have $\varphi = \varphi _0 + \varphi ^4.$ Now it is easy to see that the degree 2 part of $(\star )$ gives ${\varDelta }\varphi ^4=0$, since it is the only term depending on $\varphi ^4$. Combining this with $\varphi ^4_{xy}=0$ shows that $\varphi ^4=0$, a contradiction. $\square$

Working quite a bit harder we can establish the quintic case:

Proposition 4

There are no quintic solutions to the toric nearly Kähler equation.

Proof

Let $\varphi =\sum _{j=0}^5\varphi ^j$ be a solution to $(\star )$ with $\varphi ^5\ne 0$. The degree 9 term of $(\star )$ gives $\left| \nabla ^2\varphi ^5\right| =0$. Using Hesse’s theorem, we can change variables so that $\varphi ^5_z=0$.

Lemma 13

$\varphi ^4_{zz}=0$

Proof

Assume $\varphi ^4_{zz}\ne 0$. The degree 8 term of $(\star )$ gives $0=\varphi ^4_{zz}\left| \nabla ^2_{xy}\varphi ^5\right| .$ Thus $\left| \nabla ^2_{xy}\varphi ^5\right| =0$. By Hesse’s theorem, we may assume that $\varphi ^5_y=0$. Write $\varphi ^4 = \sum _j x^j\,C^{4-j}$, where each $C^j$ is a degree j homogeneous polynomial in y and z. The degree 7 term of $(\star )$ gives

$$\begin{aligned} 0&= \left| \nabla ^2_{yz}\varphi ^4\right| = \left| \nabla ^2 C^4 + x\,\nabla ^2 C^3 + x^2 \,\nabla ^2 C^2\right| \\&= \left| \nabla ^2 C^4\right| + x\left\langle \nabla ^2 C^4,\nabla ^2 C^3\right\rangle \\&\quad + x^2\left( \left| \nabla ^2 C^3\right| +\left\langle \nabla ^2 C^4,\nabla ^2 C^2\right\rangle \right) \\&\quad + x^3 \left\langle \nabla ^2 C^3,\nabla ^2 C^2\right\rangle + x^4\left| \nabla ^2 C^2\right| , \end{aligned}$$

where $\langle N,M\rangle := 2{{\,\mathrm{adj}\,}}(N)\cdot M = [t]|N+tM|.$ If any $C^j$ vanishes, then we have

$$\begin{aligned} 0=\left| \nabla ^2 C^k\right| =\left| \nabla ^2 C^\ell \right| =\left\langle \nabla ^2 C^k,\nabla ^2 C^\ell \right\rangle , \end{aligned}$$

where $\{j,k,\ell \}=\{2,3,4\}$. This implies that $C^k$ and $C^\ell$ are both proportional to powers of the same linear term, which we may choose to be y by changing coordinates. This contradicts $\varphi ^4_{zz}\ne 0$. Thus no $C^j$ vanishes. Since $\left| \nabla ^2C^2\right| =0$, we may use Hesse’s theorem to choose coordinates so that $C^2\propto y^2$. Since $\left\langle \nabla ^2 C^3,\nabla ^2C^2\right\rangle =0$, we must have $C^3_{zz}=0$. Thus $C^4_{zz}=\varphi ^4_{zz}\ne 0$. Combining this with $\left| \nabla ^2C^4\right| =0$, we may choose coordinates so that $C^4\propto z^4$. Then $0=\left\langle \nabla ^2 C^4,\nabla ^2 C^3\right\rangle$ implies that $C^3_{yy}=0$. Thus $C^3= 0$, a contradiction $\square$

Lemma 14

$\varphi _{zzz}=0$.

Proof

Assume $\varphi _{zzz}\ne 0$. Write $\varphi = \alpha + z\,\beta + z^2\,\gamma + z^3\,\delta$, where $\alpha ,\beta ,\gamma ,$ and $\delta$ are polynomials in x and y of degrees 5, 3, 1,  and 0, respectively. Again, we will use exponents to denote the corresponding homogeneous parts. We are assuming $\delta \ne 0$. We have

$$\begin{aligned} \nabla ^2\varphi = \begin{pmatrix} \nabla ^2\alpha &{} \nabla \beta \\ \nabla \beta &{} 2\gamma \end{pmatrix}+z\begin{pmatrix} \nabla ^2\beta &{} 2\nabla \gamma \\ 2\nabla \gamma &{} 6\delta \end{pmatrix}. \end{aligned}$$

The degree 7 term of $(\star )$ with a factor of z gives $\left| \nabla ^2\alpha ^5\right| =0$. The degree 6 term with a factor $z^2$ gives $\left\langle \nabla ^2\alpha ^5,\nabla ^2\beta ^3\right\rangle =0$. The degree 5 term with a factor of $z^3$ gives $\left| \nabla ^2\beta ^3\right| =0$. Hesse’s theorem allows us to interpret this as $\alpha ^5$ and $\beta ^3$ each depending on only one variable, which must be the same due to the cross-term. Thus we can assume that $\alpha ^5$ and $\beta ^3$ are both functions of x.

The degree 6 term with a factor of z gives $0=\alpha ^5_{xx}\alpha ^4_{yy}\delta$, so that $\alpha ^4_{yy}=0$. Thus we have $\nabla ^2_{yz}\varphi ^4=0$. Note that we now have no distinguished direction in the $y-z$ plane.

We write $\varphi = A + B + C + D$, where A, B, C,  and D have degree in [y, z], respectively 0, 1, 2,  and 3 and total degrees, respectively, 5, 4, 3 and 3. The degree 5 term of $(\star )$ with a factor of $x^3$ gives $0 = A^5_{xx}\left| \nabla ^2_{yz} D^3\right|$. Since $A^5=\varphi ^5\ne 0$, this shows that $\left| \nabla ^2_{yz}D^3\right| =0$. Using Hesse’s theorem, we can write $D^3$ as a function of y. This contradicts $\varphi _{zzz}\ne 0$. $\square$

We will continue to use the decomposition with Greek letters. The previous lemma shows that $\delta =0$.

Lemma 15

$\gamma = 1$.

Proof

Assume $\gamma \ne 1$, so that $\nabla \gamma \ne 0$. We can rotate the x-y coordinates to write $\gamma =1 + c x$ for some $0\ne c\in {\mathbb {R}}$. Thus $(z^3)$ $(\star )$ gives $0=-\beta _{yy}(2c)^2$, so that $\beta _{yy}=0$. We compute

$$\begin{aligned} \left( z^2\right) \left| \nabla ^2\varphi \right| =-4c^2\alpha _{yy}-2\gamma \left( \beta _{xy}\right) ^2+4c\beta _{xy}\beta _y =-4c^2\alpha _{yy}-2\left( \beta _{xy}\right) ^2+2c\beta _{xy}\left( 2-\partial _r\right) \beta _y. \end{aligned}$$

Thus $\left( z^2y\right)$ $(\star )$ gives $\alpha _{yyy}=0$. Now $\left[ z^2x^3\right]$ $(\star )$ gives $\alpha _{yy}^5=0$. We compute

$$\begin{aligned} (yz)\left| \nabla ^2\varphi \right| = 2\begin{vmatrix} 0&\alpha _{xyy}&\beta _{xy} \\ \beta _{xy}&\alpha _{yy}&\beta _y \\ 2c&\beta _y&2\gamma \end{vmatrix}. \end{aligned}$$

This is a polynomial in x, whose quartic term is proportional to $\left( \beta _{xxy}^3\right) ^3$. Thus $\left[ x^4yz\right]$ $(\star )$ gives $\beta ^3_y=0$. Now $\left[ x^2z^2\right]$ $(\star )$ gives $\alpha ^4_{yy}=0$. Now the degree 7 term of $(\star )$ gives $0=c\left( \alpha ^5_{xy}\right) ^2$, so that $\alpha ^5_y=0$.

Now $\left[ z^2\right]$ $(\star )$ gives $\tfrac{2}{3} = \beta ^2_{xy}+c^2$, so that $\left[ xz^2\right]$ $(\star )$ gives $0=\alpha ^3_{xyy}+c$. But $\varphi$ is harmonic, giving

$$\begin{aligned} 0 = {\varDelta }\varphi ^3 = {\varDelta }\alpha ^3 + 2c + z\,{\varDelta }\beta ^2 = \alpha ^3_{xx}+z\,\beta ^2_{xx}. \end{aligned}$$

Thus $\alpha ^3_{xx}=0=\beta ^2_{xx}$. Now $\left[ x^3\right]$ $(\star )$ gives $\alpha ^5=0$, contradicting $\varphi ^5\ne 0$. $\square$

Now that $\gamma =1$, $\left( z^2\right)$ $(\star )$ gives $2\left| \nabla ^2\beta \right| =-{2}/{3}$. Using Hesse’s theorem, we can change coordinates so that $\beta ^3= bx^3$, for some $b\in {\mathbb {R}}$. Combining these equations with ${\varDelta }\beta ^2=0$ (from $\nabla \varphi ^3=0$), we can choose coordinates so that $\beta ^2=\frac{xy}{\sqrt{3}}$, independent of whether or not b vanishes. We will show that b does indeed vanish. (z) $(\star )$ gives

$$\begin{aligned} \frac{2xyz}{3\sqrt{3}} + 4bzx^3 =&[z]\left| \nabla ^2\varphi \right| = z\left( 2\left\langle \nabla ^2\alpha ,\nabla ^2\beta \right\rangle +\begin{vmatrix} \nabla ^2\beta&\nabla \beta \\ \nabla \beta&0 \end{vmatrix}\right) \\ =&z\left( 12bx\,\alpha _{yy}-\frac{4}{\sqrt{3}}\alpha _{xy}-6bx\left( \beta _y\right) ^2+\frac{2}{\sqrt{3}}\beta _x\beta _y\right) \\ =&z\left( 12bx\,\alpha _{yy}-\frac{4}{\sqrt{3}}\alpha _{xy}-2bx^3+\frac{2}{3}x\left( 3bx^2+\frac{y}{\sqrt{3}}\right) \right) , \end{aligned}$$

so that

$$\begin{aligned} 0 = 12bx\,\alpha _{yy} - \frac{4}{\sqrt{3}}\alpha _{xy}-4bx^3. \end{aligned}$$

Taking the coefficients of $x^0$, $x^1$, and $x^2$, respectively, of this equation gives

$$\begin{aligned} \left[ x^1\right] \alpha _y=0, \qquad \left[ x^2\right] \alpha _y= \frac{{3}\sqrt{3}}{2}b\left[ x^0\right] \alpha _{yy}, \qquad \left[ x^3\right] \alpha _y=\sqrt{3} b\left[ x^1\right] \alpha _{yy}. \end{aligned}$$

In particular, $6\sqrt{3} b =\alpha ^3_{xxy} = - \alpha ^3_{yyy},$ where the second inequality comes from $\alpha ^3$ being harmonic. Now we compute

$$\begin{aligned} \left[ x^0z^0\right] \left| \nabla ^2\varphi \right| = \left[ x^0\right] \begin{vmatrix} \alpha _{xx}&0&\frac{y}{\sqrt{3}} \\ 0&\alpha _{yy}&0 \\ \frac{y}{\sqrt{3}}&0&2 \end{vmatrix} = \alpha _{yy}\left( 2\alpha _{xx}-\frac{y^2}{3}\right) . \end{aligned}$$

In particular, $\left[ y^2\right]$ $(\star )$ gives

$$\begin{aligned} 0 = \left[ y^2\right] \alpha _{yy}\alpha _{xx}=\alpha ^4_{yyyy}+\alpha _{yyy}\alpha _{xxy} +\alpha _{xxyy} = \alpha ^4_{yyyy} - 162b^2. \end{aligned}$$

Now $\left[ x^3yz\right]$ $(\star )$ gives

$$\begin{aligned} 0 = b\,\alpha _{xxyyy}\propto b^2\,\alpha ^4_{yyyy}\propto b^4, \end{aligned}$$

so that $b=0$. Now [z] $(\star )$ is simply $\alpha _{xy}=0$. The remainder of $(\star )$ is

$$\begin{aligned} \left( \tfrac{8}{3}-\tfrac{11}{3}\partial _r+\partial _r^2\right) \alpha = \begin{vmatrix} \alpha _{xx}&0&\frac{y}{\sqrt{3}} \\ 0&\alpha _{yy}&\frac{x}{\sqrt{3}} \\ \frac{y}{\sqrt{3}}&\frac{x}{\sqrt{3}}&2 \end{vmatrix} = 2\,\alpha _{xx}\alpha _{yy} + \frac{1}{3}\left( \partial _r-\partial _r^2\right) \alpha . \end{aligned}$$

Thus (xy) $(\star )$ gives $0={\alpha }_{xxx}{\alpha }_{yyy}$. Without loss of generality, by changing coordinates we have $\alpha _{yyy}=0$. Thus $(\star )$ becomes

$$\begin{aligned} 0 = \left( \tfrac{8}{3}-4\partial _r+\tfrac{4}{3}\partial _r^2\right) \alpha -4\,\alpha _{xx}. \end{aligned}$$

In particular, $0 = -\frac{32}{3}\alpha ^5$, contradicting $\varphi ^5=\alpha ^5\ne 0$. $\square$

Combining the previous three propositions gives Theorem 2.