Toric nearly Kähler manifolds

Abstract

We show that 6-dimensional strict nearly Kähler manifolds admitting effective ${\mathbb {T}}^3$ actions by automorphisms are completely characterized in the neigborhood of each point by a function on ${\mathbb {R}}^3$ satisfying a certain Monge–Ampère-type equation.

1 Introduction

Nearly Kähler manifolds were originally introduced as the class ${\mathcal {W}}_1$ in the Gray–Hervella classification of almost Hermitian manifolds [7]. More precisely, an almost Hermitian manifold (M, g, J) is called nearly Kähle (NK in short) if $(\nabla _XJ)(X)=0$ for every vector field X on M, where $\nabla$ denotes the Levi-Civita covariant derivative of g. A NK manifold is called strict if $\nabla J\ne 0$.

In [12], it was shown that every NK manifold is locally a product of one of the following types of factors:

• Kähle manifolds;

• 3-symmetric spaces;

• twistor spaces of positive quaternion-Kähle manifolds;

• 6-dimensional strict NK manifolds.

It is thus crucial to understand the 6-dimensional case, to which we will restrict in the sequel. In dimension 6, strict NK are important for several further reasons: They admit real Killing spinors [5]; in particular, they are Einstein with positive scalar curvature, and they can be characterized in terms of exterior differential systems as manifolds with special generic 3-forms in the sense of Hitchin [8].

Until 2015, the only known examples of compact 6-dimensional strict NK manifolds were the 3-symmetric spaces $S^6=\mathrm {G}_2/\mathrm {SU}(3)$, $F(1,2)=\mathrm {SU}_3/S^1\times S^1$, $CP^3=\mathrm {Sp}_2/S^1\times \mathrm {Sp}_1$ and $S^3\times S^3=\mathrm {Sp}_1\times \mathrm {Sp}_1\times \mathrm {Sp}_1/\mathrm {Sp}_1$. Moreover, J.-B. Butruille has shown in [1] that these are the only homogeneous examples.

A breakthrough was achieved very recently by L. Foscolo and M. Haskins, who studied cohomogeneity one NK metrics and obtained the first examples of non-homogeneous NK structures on $S^6$ and $S^3\times S^3$, cf. [3, 4]. The corresponding metrics are shown to exist, but cannot be constructed explicitly. However, their isometry group is known and is equal to $\mathrm {SU}(2)\times \mathrm {SU}(2)$ in both cases.

It is easy to show that a torus acting by automorphisms of a NK structure $(M^6,g,J)$ has dimension at most 3 (Corollary 3.2), and if equality holds, then the corresponding commuting vector fields span a totally real distribution on a dense open set of M (cf. Lemma 3.4). In the present paper, we study 6-dimensional nearly Kähle manifolds whose automorphism group has maximal possible rank. We call them toric NK structures by analogy with the Kähle case.

Our main result is to give a local characterization of toric NK structures in terms of a single function of 3 real variables satisfying a certain Monge–Ampère-type equation. We conjecture that the only compact toric NK manifold is $S^3\times S^3$ with its 3-symmetric NK structure.

2 Structure equations

Let $M^6$ be an oriented manifold. An $\mathrm {SU}(3)$-structure on M is a triple $(g,J,\psi )$, where g is a Riemannian metric, J is a compatible almost complex structure (i.e., $\omega :=g(J\cdot ,\cdot )$ is a 2-form), and $\psi =\psi ^++i\psi ^-$ is a (3, 0) complex volume form satisfying

$$\begin{aligned} \psi \wedge {\bar{\psi }}=-\,8i\mathrm {vol}_g. \end{aligned}$$

(2.1)

Following Hitchin [8], it is possible to characterize $\mathrm {SU}(3)$-structures in terms of exterior forms only. If $\psi ^+$ is a 3-form on M, one can define $K\in \mathrm {End}(\mathrm {T}M)\otimes \Lambda ^6M$ by

Lemma 2.1

([8]) A non-degenerate 2-form $\omega$ on M, and a 3-form $\psi ^+\in \Lambda ^3M$ satisfying

• (i) $\omega \wedge \psi ^+=0$.

• (ii) $\mathrm {tr}K^2=-\frac{1}{6} (\omega ^3)^2\in (\Lambda ^6M)^{\otimes 2}$.

• (iii) $\omega (X,K(X))/\omega ^3>0$ for every $X\ne 0$.

define an $\mathrm {SU}(3)$-structure on M.

Proof

It is easy to check that

(2.2)

From (ii), we see that $J:=6K/\omega ^3$ is an almost complex structure on M. The tensor g defined by $g(\cdot ,\cdot ):=\omega (\cdot ,J\cdot )$ is symmetric by (i) and positive definite by (iii). Finally, it is straightforward to check that $\psi ^++i\psi ^-$ is a (3, 0) complex volume form satisfying (2.1), where $\psi ^{-}:=-\psi ^{+}(J\cdot , \cdot , \cdot )$. $\square$

Since $\mathrm {vol}_g=\frac{1}{6}\omega ^3$, (2.1) is equivalent to

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

(2.3)

Definition 2.1

A strict NK structure on $M^6$ is an $\mathrm {SU}(3)$-structure $(\psi ^{\pm }, \omega )$ satisfying

$$\begin{aligned} \mathrm {d}\omega =3 \psi ^{+} \end{aligned}$$

(2.4)

and

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

(2.5)

For an alternative definition and more details on NK manifolds, we refer to [6] or [10].

Let g denote the Riemannian metric induced by $(\psi ^{\pm }, \omega )$, with Levi-Civita covariant derivative $\nabla$, and let J denote the induced almost complex structure. From now on, we identify vectors and 1-forms, as well as skew-symmetric endomorphisms and 2-forms using g.

We then have the relations (cf. [10]):

$$\begin{aligned} JX\lrcorner \psi ^+= & {} (X\lrcorner \psi ^+)\circ J=- J\circ (X\lrcorner \psi ^+),\qquad \forall X\in \mathrm {T}\mathrm{M}, \end{aligned}$$

(2.6)

$$\begin{aligned} \nabla _XJ= & {} X\lrcorner \psi ^+,\qquad \forall X\in \mathrm {T}\mathrm{M}. \end{aligned}$$

(2.7)

3 Torus actions by automorphisms

Suppose that $(M^6,\psi ^{\pm }, \omega ,g,J)$ is a strict NK structure carrying a toric action by automorphisms. More precisely, we assume that there exists some positive integer $d\ge 1$ and k linearly independent Killing vector fields $\zeta _i,\ 1 \le i \le d$ such that $[\zeta _i,\zeta _j]=0$ for $1\le i,j\le d$, which are pseudo-holomorphic in the sense that $L_{\zeta _i}J=0$ for $1\le i\le d$. This last condition is equivalent with the requirement that

$$\begin{aligned} L_{\zeta _i} \psi ^{\pm }=0,\ L_{\zeta _i} \omega =0,\qquad 1 \le i \le d. \end{aligned}$$

(3.1)

Notice that if M is compact and not isometric with the standard sphere, (3.1) follows directly from the Killing condition (cf. [10], Proposition 3.1).

We define the smooth functions $\mu _{ij}$ on M by setting $\mu _{ij}:=\omega (\zeta _i, \zeta _j)$.

Lemma 3.1

The following relations hold for every $i,j,k\in \{1,\ldots ,d\}$:

• (i) $\mathrm {d}\mu _{ij}=-3\zeta _i \lrcorner \zeta _j \lrcorner \psi ^{+}$.

• (ii) $\psi ^{+}(\zeta _i, \zeta _j, \zeta _k)=0$.

• (iii) $[\zeta _i, J\zeta _j]=0$.

• (iv) $[J\zeta _i, J\zeta _j]=4(J\zeta _j\lrcorner \zeta _i\lrcorner \psi ^+)^\sharp$.

Proof

(i) From (2.4) together with the Cartan formula, we get

$$\begin{aligned} 0=L_{\zeta _j} \omega =\zeta _j \lrcorner \mathrm {d}\omega +\mathrm {d}(\zeta _j \lrcorner \omega )= 3\zeta _j \lrcorner \psi ^{+}+\mathrm {d}(\zeta _j \lrcorner \omega ). \end{aligned}$$

Taking now the interior product with $\zeta _i$ yields

$$\begin{aligned} 0=3\zeta _i \lrcorner \zeta _j \lrcorner \psi ^{+}+\zeta _i \lrcorner \mathrm {d}(\zeta _j \lrcorner \omega ) \end{aligned}$$

and the claim follows by taking into account that

$$\begin{aligned} \zeta _i \lrcorner \mathrm {d}(\zeta _j \lrcorner \omega )=L_{\zeta _i}(\zeta _j \lrcorner \omega )-\mathrm {d}(\zeta _i \lrcorner \zeta _j \lrcorner \omega )= \mathrm {d}\mu _{ij}. \end{aligned}$$

(ii) Using (i), we can write

$$\begin{aligned} \psi ^{+}(\zeta _i, \zeta _j, \zeta _k)=-\frac{1}{3}\mathrm {d}\mu _{jk}(\zeta _i)= -\frac{1}{3}L_{\zeta _i}(\omega (\zeta _j,\zeta _k))=0. \end{aligned}$$

(iii) Follows directly from $L_{\zeta _i}J=0$ and the fact that the $\zeta _i$’s mutually commute.

(iv) On every almost Hermitian manifold, the Nijenhuis tensor

$$\begin{aligned} N(X,Y):=[X,Y]+J[X,JY]+J[JX,Y]-[JX,JY] \end{aligned}$$

can be expressed as

$$\begin{aligned} N(X,Y)=J(L_XJ)Y-(L_{JX}J)Y \end{aligned}$$

(3.2)

for all vector fields X, Y. On the other hand, (2.7) shows that on every NK manifold, the Nijenhuis tensor satisfies

$$\begin{aligned} N(X,Y)=J(\nabla _XJ)Y-J(\nabla _YJ)X-(\nabla _{JX}J)Y+(\nabla _{JY}J)X=-\,4Y\lrcorner JX\lrcorner \psi ^+. \end{aligned}$$

(3.3)

Applying (3.2) and (3.3) to $X=\zeta _i$, and using the fact that $L_{\zeta _i}J=0$ yields

$$\begin{aligned} (L_{J\zeta _i}J)=4J\zeta _i\lrcorner \psi ^+.\end{aligned}$$

(3.4)

This, together with (iii), finishes the proof. $\square$

Lemma 3.2

If $\xi$ is a Killing vector field, $J\xi$ cannot be Killing on any open set U.

Proof

From Corollary 3.3 and Lemma 3.4 in [10], we have

$$\begin{aligned} (\mathrm {d}J\xi )^{(2,0)}=\mathrm {d}J\xi =-\xi \lrcorner \mathrm {d}\omega =-3\xi \lrcorner \psi ^+\end{aligned}$$

and

$$\begin{aligned} (\mathrm {d}\xi )^{(2,0)}=-J\xi \lrcorner \psi ^+\end{aligned}$$

for every Killing vector field $\xi$. If $J\xi$ were Killing on some open set, the same relations applied to $J\xi$ would read

$$\begin{aligned} (\mathrm {d}\xi )^{(2,0)}=3J\xi \lrcorner \psi ^+\end{aligned}$$

and

$$\begin{aligned} (\mathrm {d}J\xi )^{(2,0)}=\xi \lrcorner \psi ^+, \end{aligned}$$

a contradiction. $\square$

Assume from now on that the dimension of the torus acting by automorphisms satisfies $d\ge 2$.

Lemma 3.3

For every $i\ne j$ in $\{1,\ldots ,d\}$, the vector fields $\{\zeta _i,\zeta _j,J\zeta _i,J\zeta _j\}$ are linearly independent on a dense open subset of M.

Proof

One can of course assume $i=1,j=2$. If the contrary holds, there exists some open set U on which $\zeta _1$ does not vanish and functions $a,b:U\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \zeta _2=a\zeta _1+bJ\zeta _1. \end{aligned}$$

(3.5)

We differentiate this relation on U with respect to the Levi-Civita covariant derivative $\nabla$ and obtain the following relation between endomorphisms of $\mathrm {T}M$:

$$\begin{aligned} \nabla \zeta _2= & {} \mathrm {d}a\otimes \zeta _1+a\nabla \zeta _1+\mathrm {d}b\otimes J\zeta _1+b\nabla J\zeta _1\\= & {} \mathrm {d}a\otimes \zeta _1+a\nabla \zeta _1+\mathrm {d}b\otimes J\zeta _1-b\zeta _1\lrcorner \psi ^++bJ\circ (\nabla \zeta _1). \end{aligned}$$

Taking the symmetric parts in this equation yields

$$\begin{aligned} 0=\mathrm {d}a\odot \zeta _1+\mathrm {d}b\odot J\zeta _1+b(J\circ (\nabla \zeta _1))^\mathrm{sym}. \end{aligned}$$

Since $\nabla \zeta _1$ is skew-symmetric, $(J\circ (\nabla \zeta _1))^\mathrm{sym}$ commutes with J, whence J commutes with $\mathrm {d}a\odot \zeta _1+\mathrm {d}b\odot J\zeta _1$. On the other hand, J commutes with $\mathrm {d}a\odot \zeta _1+J\mathrm {d}a\odot J\zeta _1$; thus, it commutes with $(\mathrm {d}b-J\mathrm {d}a)\odot J\zeta _1$. This implies $\mathrm {d}b=J\mathrm {d}a$. Differentiating this again with respect to $\nabla$ yields

$$\begin{aligned} \nabla \mathrm {d}b=\nabla (J\mathrm {d}a)=-\mathrm {d}a\lrcorner \psi ^++J\circ \nabla \mathrm {d}a. \end{aligned}$$

Taking the skew-symmetric part in this equality shows that

$$\begin{aligned} \mathrm {d}a\lrcorner \psi ^+=(J\circ \nabla \mathrm {d}a)^\mathrm{skew}. \end{aligned}$$

But the left-hand side anti-commutes with J, whereas the right- hand side commutes with J (since $\nabla \mathrm {d}a$ is symmetric). Thus $\mathrm {d}a=0$, so a and b are constants. From (3.5), we obtain that $J\zeta _1$ is a Killing vector field on U, which is impossible by Lemma 3.2. This contradiction concludes the proof. $\square$

Corollary 3.1

The vector fields $\{\zeta _1,\zeta _2,J\zeta _1,J\zeta _2,\zeta _1\lrcorner \zeta _2\lrcorner \psi ^+,J\zeta _1\lrcorner \zeta _2\lrcorner \psi ^+\}$ are linearly independent on a dense open subset of M.

Proof

This follows from Lemma 3.3 using the fact that the vectors $\zeta _1\lrcorner \zeta _2\lrcorner \psi ^+$ and $J\zeta _1\lrcorner \zeta _2\lrcorner \psi ^+$ are orthogonal to $\zeta _1,\zeta _2,J\zeta _1$ and $J\zeta _2$, and they both are non-vanishing at each point where $\{\zeta _1,\zeta _2,J\zeta _1,J\zeta _2\}$ are linearly independent. $\square$

From now on, we assume that $d\ge 3$.

Lemma 3.4

For every mutually distinct $1\le i,j,k\le d$, the 6 vector fields $\zeta _i,$$\zeta _j,$$\zeta _k$, $J\zeta _i,$$J\zeta _j,$$J\zeta _k$ are linearly independent on a dense open subset $M_0$ of M.

Proof

We may assume that $i=1$, $j=2$ and $k=3$. Like before, if the statement does not hold, there exists some open set U on which $\zeta _1$ does not vanish and functions $a_1,b_1,a_2,b_2:U\rightarrow {\mathbb {R}}$ such that

$$\begin{aligned} \zeta _3=a_1\zeta _1+b_1J\zeta _1+a_2\zeta _2+b_2J\zeta _2. \end{aligned}$$

(3.6)

By Lemma 3.3, one may assume that $\{\zeta _1,\zeta _2,J\zeta _1,J\zeta _2\}$ are linearly independent on U. Taking the Lie derivative with respect to $J\zeta _1$ in (3.6) and using Lemma 3.1 (iii) and (iv) yields

$$\begin{aligned} 0=J\zeta _1(a_1)\zeta _1+J\zeta _1(b_1)J\zeta _1+J\zeta _1(a_2)\zeta _2+ J\zeta _1(b_2)J\zeta _2+4b_2J\zeta _2\lrcorner \zeta _2\lrcorner \psi ^+. \end{aligned}$$

From Corollary 3.1, we get $b_2=0$. Similarly, taking the Lie derivative with respect to $J\zeta _2$ in (3.6), we get $b_1=0$. Therefore, (3.6) becomes

$$\begin{aligned} \zeta _3=a_1\zeta _1+a_2\zeta _2.\end{aligned}$$

(3.7)

Differentiating this equation with respect to $\nabla$ and taking the symmetric part yields

$$\begin{aligned} 0=\mathrm {d}a_1\odot \zeta _1+\mathrm {d}a_2\odot \zeta _2. \end{aligned}$$

Since $\zeta _1$ and $\zeta _2$ are linearly independent on U, this implies $\mathrm {d}a_1=c\zeta _2$ and $\mathrm {d}a_2=-c\zeta _1$ for some function $c:U\rightarrow {\mathbb {R}}$. On the other hand, taking the Lie derivative with respect to $\zeta _2$ in (3.7) yields $0=\zeta _2(a_1)\zeta _1+\zeta _2(a_2)\zeta _2$; thus, $\zeta _2(a_1)=0$, so finally $c|\zeta _2|^2=g(\mathrm {d}a_1,\zeta _2)=\zeta _2(a_1)=0$, whence $c=0$. This shows that $a_1$ and $a_2$ are constant, contradicting the hypothesis that $\zeta _1,\zeta _2$ and $\zeta _3$ are linearly independent Killing vector fields. This proves the lemma. $\square$

Corollary 3.2

The rank d of the automorphism group of M is at most 3.

Proof

Assume for a contradiction that $d\ge 4$, there exist 4 linearly independent mutually commuting Killing vector fields $\zeta _1,\ldots ,\zeta _4$ on M preserving the almost complex structure J. From Lemma 3.4, there exist functions $a_i$ and $b_i$ ($i=1,2,3$) on $M_0$ such that

$$\begin{aligned} \zeta _4=\sum _{j=1}^3a_j\zeta _j+b_jJ\zeta _j. \end{aligned}$$

(3.8)

From Lemma 3.1 (ii), we get $\psi ^{+}(\zeta _1, \zeta _2, \zeta _3)=\psi ^{+}(\zeta _1, \zeta _2, \zeta _4)=0$. Using (3.8) together with the fact that $\psi ^+(X,JX,\cdot )=0$ for every X, we get $b_3\psi ^{+}(\zeta _1, \zeta _2, J\zeta _3)=0$.

Assume that $b_3$ is not identically zero on M. Then $\psi ^{+}(\zeta _1, \zeta _2, J\zeta _3)=0$ on some non-empty open set U. On the other hand, the 1-form $\psi ^{+}(\zeta _1, \zeta _2,\cdot )$ vanishes when applied to $\zeta _1$, $J\zeta _1$, $\zeta _2$, $J\zeta _2$ and $\zeta _3$; so, by Lemma 3.4, $\psi ^{+}(\zeta _1, \zeta _2,\cdot )$ vanishes on the non-empty open set $U\cap M_0$. This contradicts Corollary 3.1. Consequently $b_3\equiv 0$, and similarly $b_2= b_1 \equiv 0$. We thus get

$$\begin{aligned} \zeta _4=\sum _{j=1}^3a_j\zeta _j. \end{aligned}$$

(3.9)

Taking the Lie derivative in (3.9) with respect to $\zeta _i$ and $J\zeta _i$ for $i=1,2,3$ and using Lemma 3.1 (iii) we obtain $\zeta _i(a_j)=J\zeta _i(a_j)=0$ for every $i,j\in \{1,2,3\}$, so $a_j$ are constant on $M_0$, thus showing that $\zeta _4$ is a linear combination of $\zeta _1,\zeta _2,\zeta _3$, a contradiction. $\square$

4 Toric NK structures

In view of Corollary 3.2 we can now introduce the following:

Definition 4.1

A 6-dimensional strict NK manifold is called toric if its automorphism group has rank 3, or equivalently, if it carries 3 linearly independent mutually commuting pseudo-holomorphic Killing vector fields $\zeta _1,\zeta _2,\zeta _3$.

Assume from now on that $(M^6,g,J,\zeta _1,\zeta _2,\zeta _3)$ is a toric NK manifold and consider on the dense open subset $M_0$ given by Lemma 3.4 the basis $\{ \theta ^1, \theta ^2, \theta ^3, \gamma ^1,\gamma ^2,\gamma ^3 \}$ of $\Lambda ^1M_0$ dual to $\{\zeta _1,\zeta _2,\zeta _3, J\zeta _1, J\zeta _2, J\zeta _3 \}$, together with the function

$$\begin{aligned} \varepsilon :=\psi ^{-}(\zeta _1, \zeta _2, \zeta _3). \end{aligned}$$

(4.1)

For further use, let us also introduce the symmetric $3\times 3$ matrix

$$\begin{aligned} C:=(C_{ij})=(g(\zeta _i,\zeta _j)). \end{aligned}$$

(4.2)

As a direct consequence of Lemma 3.4, we have that $\zeta +J\zeta =\mathrm {T}M_0$, where $\zeta$ is the 3-dimensional distribution spanned by $\zeta _k, 1 \le k \le 3$. This enables us to express $\psi ^+$, and $\psi ^-$ in terms of the basis $\{ \theta ^i, \gamma ^j \}$ and of the function $\varepsilon$, simply by checking that the two terms are equal when applied to elements of the basis $\{\zeta _i, J\zeta _j\}$ of $\mathrm {T}M_0$:

$$\begin{aligned} \begin{aligned} \psi ^{+}&=\varepsilon \Big (\gamma ^{123}- \theta ^{12} \wedge \gamma ^3-\theta ^{31} \wedge \gamma ^2-\theta ^{23} \wedge \gamma ^1 \Big ),\\ \psi ^{-}&= \varepsilon \Big (\theta ^{123}-\gamma ^{12} \wedge \theta ^3-\gamma ^{31} \wedge \theta ^2-\gamma ^{23} \wedge \theta ^1\Big ), \end{aligned} \end{aligned}$$

(4.3)

where here and henceforth the notation $\gamma ^{123}$ stands for $\gamma ^1\wedge \gamma ^2\wedge \gamma ^3$, etc. Recalling the definition of $\mu _{ij}:=\omega (\zeta _i, \zeta _j)$, the fundamental 2-form $\omega :=g(J\cdot ,\cdot )$ can be expressed by the formula:

$$\begin{aligned} \omega =\sum \limits _{1 \le i < j \le 3 }\mu _{ij}(\theta ^{ij}+\gamma ^{ij})+\sum \limits _{i=1}^3 \theta ^i \wedge c^i \end{aligned}$$

(4.4)

where the 1-forms $c^i$ in $\Lambda ^1(J\zeta ^*)$ are given by $c^i=\sum \limits _{j=1}^{3} C_{ij}\gamma ^j$. A short computation yields

$$\begin{aligned} \omega ^3=-\,6 \theta ^{123} \wedge c^{123}+6 \theta ^{123}\wedge c\wedge \eta , \end{aligned}$$

(4.5)

where $\eta$ in $\Lambda ^2(J\zeta ^*)$ is given by

$$\begin{aligned} \eta :=\sum \limits _{1 \le i<j\le 3} \mu _{ij} \gamma ^{ij} \end{aligned}$$

and c in $\Lambda ^1(J\zeta ^*)$ is given by

$$\begin{aligned} c:=\mu _{23}c^1+\mu _{31}c^2+\mu _{12}c^3. \end{aligned}$$

Therefore, from the compatibility relations (2.3), it follows that

$$\begin{aligned} c^{123}={\varepsilon ^2}\gamma ^{123}+c\wedge \eta , \end{aligned}$$

(4.6)

which is equivalent to

$$\begin{aligned} \det C=\varepsilon ^2+^t\!V CV, \end{aligned}$$

(4.7)

where we denote by

$$\begin{aligned} V:=\left( \begin{array}{c} \mu _{23}\\ \mu _{31}\\ \mu _{12} \end{array} \right) . \end{aligned}$$

(4.8)

Lemma 4.1

The following relations hold:

• (i) $\mathrm {d}\mu _{12}=-3\varepsilon \gamma ^3,\ \mathrm {d}\mu _{31}=-3\varepsilon \gamma ^2,\ \mathrm {d}\mu _{23}=-3\varepsilon \gamma ^1$;

• (ii) $\mathrm {d}\varepsilon =4c.$

Proof

(i) Using (2.4), (4.3) and the Cartan formula, we can write

$$\begin{aligned} \mathrm {d}\mu _{12}=\mathrm {d}(\zeta _2\lrcorner \zeta _1\lrcorner \omega )=\zeta _2\lrcorner \zeta _1\lrcorner \mathrm {d}\omega =3\zeta _2\lrcorner \zeta _1\lrcorner \psi ^+=-3\varepsilon \gamma ^3. \end{aligned}$$

The other formulas are similar.

(ii) Using (2.5), (4.4) and the Cartan formula again, we get

$$\begin{aligned} \mathrm {d}\varepsilon= & {} \mathrm {d}(\zeta _3\lrcorner \zeta _2\lrcorner \zeta _1\lrcorner \psi ^-)=-\zeta _3\lrcorner \zeta _2\lrcorner \zeta _1\lrcorner \mathrm {d}\psi ^-\\= & {} 2\zeta _3\lrcorner \zeta _2\lrcorner \zeta _1\lrcorner \omega ^2=4(\mu _{23}c^1+\mu _{31}c^2+\mu _{12}c^3). \end{aligned}$$

$\square$

We will now show that Eq. (2.5) is equivalent to some exterior system involving the 1-forms $\theta ^i$.

Lemma 4.2

Equation (2.5) holds if and only if the forms $\theta _i, 1 \le i \le 3$ satisfy the differential system:

$$\begin{aligned} \begin{aligned} \frac{1}{4}\varepsilon \mathrm {d}\theta ^1&= c^2 \wedge c^3-\mu _{23} \eta \\ \frac{1}{4}\varepsilon \mathrm {d}\theta ^2&= c^3 \wedge c^1-\mu _{31} \eta \\ \frac{1}{4}\varepsilon \mathrm {d}\theta ^3&= c^1 \wedge c^2-\mu _{12} \eta \end{aligned} \end{aligned}$$

(4.9)

Proof

Assume that (2.5) holds. By (4.3)

$$\begin{aligned} \zeta _2 \lrcorner \zeta _1 \lrcorner \psi ^{-}=\varepsilon \theta ^3. \end{aligned}$$

(4.10)

Since $\zeta _k, 1 \le k \le 3$ are commuting Killing vector fields preserving the whole $\mathrm {SU}(3)$-structure, (4.4) yields

$$\begin{aligned} \mathrm {d}(\zeta _2 \lrcorner \zeta _1 \lrcorner \psi ^{-})=\zeta _2 \lrcorner \zeta _1 \lrcorner \mathrm {d}\psi ^{-}=-2 \zeta _2 \lrcorner \zeta _1 \lrcorner (\omega \wedge \omega )=-\,4\theta ^3\wedge c-4\mu _{12}\eta +4c^1\wedge c^2. \end{aligned}$$

hence by (4.10) and Lemma 4.1 (ii), we get

$$\begin{aligned} \frac{1}{4}\varepsilon \mathrm {d}\theta ^3=\frac{1}{4}\mathrm {d}(\varepsilon \theta ^3)-\frac{1}{4}\mathrm {d}\varepsilon \wedge \theta ^3 =-\theta ^3\wedge c-\mu _{12}\eta +c^1\wedge c^2-c\wedge \theta ^3=c^1\wedge c^2-\mu _{12}\eta . \end{aligned}$$

The proof of the two other relations is similar.

Conversely, we notice that (2.5) holds if and only if

$$\begin{aligned} {\left\{ \begin{array}{ll} \zeta _i \lrcorner \zeta _j \lrcorner \mathrm {d}\psi ^{-}=-2\zeta _i \lrcorner \zeta _j \lrcorner \omega ^2,\qquad \forall \ 1\le i,j\le 3,\\ J\zeta _1\lrcorner J\zeta _2\lrcorner J\zeta _3\lrcorner \mathrm {d}\psi ^{-}=-2J\zeta _1\lrcorner J\zeta _2\lrcorner J\zeta _3\lrcorner \omega ^2. \end{array}\right. } \end{aligned}$$

The first relation was just shown to be equivalent to (4.9). It remains to check, by a straightforward calculation, that the second relation is automatically fulfilled. $\square$

We finally interpret Eq. (2.4) in terms of the frame $\{c^i\}$.

Lemma 4.3

Equation (2.4) holds if and only if (4.6) holds and the forms $\varepsilon c^k$ are closed for $1 \le k \le 3$.

Proof

Taking the interior product with $\zeta _1$ in (2.4) and using (4.3), (4.4) and Lemma 4.1 (i) yields

$$\begin{aligned} 3\varepsilon (-\theta ^2\wedge \gamma ^3+\theta ^3\wedge \gamma ^2)= & {} 3\zeta _1\lrcorner \psi ^+=\zeta _1\lrcorner \mathrm {d}\omega =-\mathrm {d}(\zeta _1\lrcorner \omega )=-\mathrm {d}(\mu _{12}\theta ^2-\mu _{31}\theta ^3+c^1)\\= & {} 3\varepsilon \gamma ^3\wedge \theta ^2 -\mu _{12}\mathrm {d}\theta ^2 -3\varepsilon \gamma ^2\wedge \theta ^3 +\mu _{31}\mathrm {d}\theta ^3-\mathrm {d}c^1, \end{aligned}$$

whence

$$\begin{aligned} \mathrm {d}c^1=\mu _{31}\mathrm {d}\theta ^3 -\mu _{12}\mathrm {d}\theta ^2. \end{aligned}$$

From Lemma 4.2 and 4.1 (ii), we thus obtain

$$\begin{aligned} \mathrm {d}(\varepsilon c^1)= & {} 4c\wedge c^1+4 \Big [\mu _{31}(c^1 \wedge c^2-\mu _{12}\eta )- \mu _{12}(c^3 \wedge c^1-\mu _{31}\eta )\Big ]\\= & {} 4(\mu _{23}c^1+\mu _{31}c^2+\mu _{12}c^3)\wedge c^1+4(\mu _{31}c^1 \wedge c^2- \mu _{12}c^3 \wedge c^1)=0. \end{aligned}$$

Conversely, we notice that (2.4) holds if and only if

$$\begin{aligned} {\left\{ \begin{array}{ll} \zeta _i \lrcorner \mathrm {d}\omega =3\zeta _i \lrcorner \psi ^+,\qquad \forall \ 1\le i\le 3,\\ J\zeta _1\lrcorner J\zeta _2\lrcorner J\zeta _3\lrcorner \mathrm {d}\omega =3J\zeta _1\lrcorner J\zeta _2\lrcorner J\zeta _3\lrcorner \psi ^+. \end{array}\right. } \end{aligned}$$

We have just shown that the first equation is equivalent to $\varepsilon c^k$ being closed. The component of $\mathrm {d}\omega =3\psi ^{+}$ on $\Lambda ^3 J\zeta$ is given by

$$\begin{aligned} \mathrm {d}\eta +\sum \limits _{k=1}^3 \mathrm {d}\theta ^k \wedge c^k=3\varepsilon \gamma ^{123}, \end{aligned}$$

so using (4.9), the second equation is equivalent to (4.7). $\square$

Let us now consider the 3-dimensional quotient $U:=M_0\slash \zeta$ of the open set $M_0$ by the action of the 3-dimensional torus generated by the Killing vector fields $\zeta _i$. Clearly, the natural projection $\pi :M\rightarrow U$ is a submersion. We shall now interpret the geometry of the situation down on U. Since $\zeta _i(\mu _{jk})=0$, there exist functions $y_i$ on U such that $\pi ^*y_1=\mu _{23},\pi ^*y_2=\mu _{31},\pi ^*y_3=\mu _{12}$. Moreover, since $\varepsilon$ does not vanish on $M_0$, Lemma 4.1 (i) shows that $\{y_i\}$ define a global coordinate system on U. From now on, we will identify the projectable functions or exterior forms on M with their projection on U. Since everything is local, we may suppose that U is contractible.

Remark 4.1

By Lemma 3.1 (i), it follows that the map $\mu :M\rightarrow \Lambda ^2{\mathbb {R}}^3\cong {{\mathfrak {s}}}{{\mathfrak {o}}}(3)$ defined by

$$\begin{aligned} \mu := \begin{pmatrix} 0 &{} \mu _{12} &{} \mu _{13} \\ \mu _{21} &{} 0 &{} \mu _{23} \\ \mu _{31} &{} \mu _{32} &{} 0 \end{pmatrix}=\pi ^*\begin{pmatrix} 0 &{} y_3 &{} -y_2 \\ -y_3 &{} 0 &{} y_1 \\ y_2 &{} -y_1 &{} 0 \end{pmatrix} \end{aligned}$$

is the multi-moment map of the strong geometry $(M,\psi ^+)$ defined by Madsen and Swann in [9] and studied further by Dixon [2] in the particular case where $M=S^3\times S^3$. Similarly, the function $\varepsilon$ can be seen as the multi-moment map associated with the closed 4-form $\mathrm {d}\psi ^-$. These maps will play an important role in Sects. 5 and 6 below.

Proposition 4.1

There exists a function $\varphi$ on U (defined up to an affine function) such that $\mathrm{Hess}(\varphi )=C$ in the coordinates $\{y_i\}$.

Proof

From Lemma 4.3, there exist functions $f_i$ on U such that $\mathrm {d}f_i=\varepsilon c^i$ for $1\le i\le 3$. Notice that by Lemma 4.1 (i), this is equivalent to

$$\begin{aligned} \frac{\partial f_i}{\partial y_j}=-3C_{ij}. \end{aligned}$$

(4.11)

From Lemma 4.1 (i), we get

$$\begin{aligned} \mathrm {d}\left( \sum _{i=1}^3f_i\mathrm {d}y_i\right) = \sum _{i=1}^3\mathrm {d}f_i\wedge \mathrm {d}y_i=-3 \sum _{i=1}^3\varepsilon c^i\wedge \varepsilon \gamma ^i= \sum _{i,j=1}^3\varepsilon ^2C_{ij}\gamma ^j\wedge \gamma ^i=0, \end{aligned}$$

so there exists some function $\varphi$ such that

$$\begin{aligned} \mathrm {d}\varphi =-\frac{1}{3}\sum _{i=1}^3f_i\mathrm {d}y_i. \end{aligned}$$

This means that $\frac{\partial \varphi }{\partial y_i}=-\frac{1}{3}f_i$, which together with (4.11) finishes the proof. $\square$

Let us introduce the operator $\partial _r$ of radial differentiation, acting on functions on U by

$$\begin{aligned} \partial _rf:=\sum _{i=1}^3 y_i\frac{\partial f}{\partial y_i}. \end{aligned}$$

Proposition 4.2

The function $\varphi$ can be chosen in such a way that

$$\begin{aligned} \varepsilon ^2=\frac{8}{3}(\varphi -\partial _r\varphi ) . \end{aligned}$$

(4.12)

Proof

It is clearly enough to show that the exterior derivatives of the two terms coincide. Since

$$\begin{aligned} \frac{\partial (\partial _r\varphi )}{\partial y_j}=\sum _{i=1}^3\frac{\partial ^2\varphi }{\partial y_i\partial y_j}y_i+\frac{\partial \varphi }{\partial y_j}, \end{aligned}$$

Lemma 4.1 yields

$$\begin{aligned} -\frac{8}{3}\mathrm {d}(\partial _r\varphi -\varphi )=-\frac{8}{3}\sum _{i,j=1}^3 C_{ij}y_i\mathrm {d}y_j=8\sum _{i,j=1}^3 C_{ij}y_i\varepsilon \gamma ^j=8\varepsilon c=\mathrm {d}(\varepsilon ^2). \end{aligned}$$

$\square$

Summing up, we get the following result:

Corollary 4.1

The function $\varphi$ given in the previous proposition satisfies the equation

$$\begin{aligned} \det (\mathrm{Hess}(\varphi ))=\frac{8}{3}\varphi -\frac{11}{3}\partial _r\varphi +\partial _r^2\varphi . \end{aligned}$$

(4.13)

Proof

We have

$$\begin{aligned} \partial _r^2\varphi =\partial _r\bigg (\sum _{i=1}^3 y_i\frac{\partial \varphi }{\partial y_i}\bigg )=\sum _{i=1}^3 y_i\frac{\partial \varphi }{\partial y_i}+\sum _{i,j=1}^3 y_iy_j\frac{\partial ^2\varphi }{\partial y_i\partial y_j} =\partial _r\varphi +^t\!VCV,\end{aligned}$$

(4.14)

so (4.13) is a consequence of (4.7) and (4.12). $\square$

5 The inverse construction

In this section, we will show that conversely, every solution $\varphi$ of Eq. (4.13) on some open set $U\subset {\mathbb {R}}^3$ defines a NK structure with 3 linearly independent commuting Killing vector fields on $U_0\times {\mathbb {T}}^3$, where $U_0$ is some open subset of U. More precisely, let $y_1,y_2,y_3$ be the standard coordinates on U and let $\mu$ be the $3\times 3$ skew-symmetric matrix

$$\begin{aligned} \mu := \begin{pmatrix} 0 &{} y_3 &{} -y_2 \\ -y_3 &{} 0 &{} y_1 \\ y_2 &{} -y_1 &{} 0 \end{pmatrix}. \end{aligned}$$

(5.1)

Define the $6\times 6$ symmetric matrix

$$\begin{aligned} D:=\begin{pmatrix}\mathrm{Hess}(\varphi )&{}-\mu \\ \mu &{}\mathrm{Hess}(\varphi )\end{pmatrix}. \end{aligned}$$

Let $U_0$ denote the open set

$$\begin{aligned} U_0:= \Big \{x\in U\ |\ \varphi (x)-\partial _r\varphi (x)>0\ \hbox {and}\ D\ \hbox {is positive definite}\Big \}. \end{aligned}$$

(5.2)

The next result is straightforward:

Lemma 5.1

The matrix D is positive definite if and only if

• (i) $C=\mathrm{Hess}(\varphi )$ is positive definite and

• (ii) $\langle \mu a,b\rangle ^2 <\langle Ca,a \rangle \langle Cb,b \rangle$ for all $(a,b)\in ({\mathbb {R}}^3\times {\mathbb {R}}^3)\setminus {(0,0)}$.

On $U_0$ we define a positive function $\varepsilon$ by (4.12), 1-forms $\gamma ^i$ by $\mathrm {d}y_i=-3\varepsilon \gamma ^i$ and a 2-form $\eta :=y_1\gamma ^2\wedge \gamma ^3+y_2\gamma ^3\wedge \gamma ^1+y_3\gamma ^1\wedge \gamma ^2$. We denote as before by C the Hessian of $\varphi$ and define $c^i:=\sum _{j=1}^3 C_{ij}\gamma ^j$.

Lemma 5.2

The following hold:

• (i) The 1-forms $\varepsilon c^i$ are exact.

• (ii) The 2-forms $\tau _1:=(c^2 \wedge c^3-y_1 \eta )/\varepsilon$, $\tau _2:=(c^3 \wedge c^1-y_2 \eta )/\varepsilon$ and $\tau _3:=(c^1 \wedge c^2-y_3 \eta )/\varepsilon$ are closed.

Proof

(i) We have:

$$\begin{aligned} \mathrm {d}\bigg (-\frac{1}{3}\frac{\partial \varphi }{\partial y_i}\bigg )=-\frac{1}{3}\sum _{j=1}^3\frac{\partial ^2 \varphi }{\partial y_i\partial y_j}\mathrm {d}y_j =-\frac{1}{3}\sum _{j=1}^3C_{ij}\mathrm {d}y_j=\varepsilon c^i. \end{aligned}$$

(ii) We first compute using (i):

$$\begin{aligned} \mathrm {d}(\varepsilon ^3\tau _1)= & {} \mathrm {d}(\varepsilon ^2(c^2 \wedge c^3-y_1 \eta ))= -\mathrm {d}(y_1\varepsilon ^2\eta )\\= & {} -\mathrm {d}(y_1^2\varepsilon ^2\gamma ^{23}+y_1y_2\varepsilon ^2\gamma ^{31}+y_1y_3\varepsilon ^2\gamma ^{12})=12y_1\varepsilon ^3\gamma ^{123}. \end{aligned}$$

On the other hand,

$$\begin{aligned} \mathrm {d}(\varepsilon ^3)\wedge \tau _1= & {} 3\varepsilon ^2 \mathrm {d}\varepsilon \wedge \tau _1=12\varepsilon \left( \sum _{j=1}^3y_jc^j\right) \wedge (c^2 \wedge c^3-y_1 \eta )\\= & {} 12\varepsilon y_1\left( \det C-\sum _{i,j=1}^3C_{ij}y_iy_j\right) \gamma ^{123}=12y_1\varepsilon ^3\gamma ^{123}, \end{aligned}$$

the last equality (which is the converse to (4.7)) following from (4.12), (4.13) and (4.14). These two relations show that $\tau _1$ is closed. The proof that $\mathrm {d}\tau _2=\mathrm {d}\tau _3=0$ is similar. $\square$

By replacing $U_0$ with a smaller open subset if necessary, one can find 1-forms $\sigma _i$ such that $\mathrm {d}\sigma _i=4\tau _i$. Consider now the 6-dimensional manifold $M:=U_0\times {\mathbb {T}}^3$ with coordinates $y_1,y_2,y_3$ and $x_1,x_2,x_3$ (locally defined). The 1-forms $\theta ^i:=\mathrm {d}x_i+\sigma _i$ satisfy the differential system (4.9). We define $\psi ^\pm$ and $\omega$ by (4.3) and (4.4) and we claim that they determine a strict NK structure on M whose automorphism group contains a 3-torus.

Let us first check that $(\psi ^\pm ,\omega )$ satisfy the conditions of Lemma 2.1. The relation (i) is straightforward, (ii) is equivalent to (4.7), and (iii) holds from the definition (5.2) of $U_0$.

In order to prove that $(\psi ^\pm ,\omega )$ defines a NK structure, we need to check (2.4) and (2.5). By Lemma 4.3, (2.4) is equivalent to $\varepsilon c^i$ being closed (Lemma 5.2 (i)) together with (4.7). Similarly, Lemma 4.2 shows that (2.5) is equivalent to the system (4.9) together with (4.7) again.

It remains to check that the automorphism group contains a 3-torus. This is actually clear: The action of ${\mathbb {T}}^3$ on $M=U_0 \times {\mathbb {T}}^3$ by multiplication on the first factor preserves the $\mathrm {SU}(3)$ structure. We have proved the following result:

Theorem 5.1

Every solution of the Monge–Ampère-type equation (4.13) on some open set U in ${\mathbb {R}}^3$ defines in a canonical way a NK structure with 3 linearly independent commuting infinitesimal automorphisms on $U_0\times {\mathbb {T}}^3$, where $U_0$ is defined by (5.2).

6 Examples

We will illustrate the above computations on a specific example of toric nearly Kähler manifold, namely the 3-symmetric space $S^3\times S^3$.

Let $K:=\mathrm {SU}_2$ with Lie algebra ${\mathfrak {k}}= {{\mathfrak {s}}}{{\mathfrak {u}}}_2$ and $G:=K\times K \times K$ with Lie algebra ${\mathfrak {g}}= {\mathfrak {k}}\oplus {\mathfrak {k}}\oplus {\mathfrak {k}}$. We consider the 6-dimensional manifold $M = G/K$, where K is diagonally embedded in G. The tangent space of M at $o=eK$ can be identified with

$$\begin{aligned} {\mathfrak {p}}= \{(X,Y,Z)\in {\mathfrak {k}}\oplus {\mathfrak {k}}\oplus {\mathfrak {k}}\,|\, X+Y+Z=0\} . \end{aligned}$$

Consider the invariant scalar product B on ${{\mathfrak {s}}}{{\mathfrak {u}}}_2$ such that the scalar product

$$\begin{aligned} \langle (X,Y,Z),(X,Y,Z)\rangle := B(X,X) + B(Y,Y) + B(Z,Z) \end{aligned}$$

defines the homogeneous nearly Kähler metric g of scalar curvature 30 on $M=S^3\times S^3$ (cf. [11], Lemma 5.4).

The G-automorphism $\sigma$ of order 3 defined by $\sigma (a_1,a_2,a_3) =(a_2,a_3,a_1)$ induces a canonical almost complex structure on the 3-symmetric space M by the relation

$$\begin{aligned} \sigma =\frac{-\mathrm {Id}+\sqrt{3} J}{2},\qquad \hbox {on}\ {\mathfrak {p}}, \end{aligned}$$

whence

$$\begin{aligned} J(X,Y,Z) = \tfrac{2}{\sqrt{3}} (Y,Z,X) + \tfrac{1}{\sqrt{3}}(X,Y,Z),\qquad \forall (X,Y,Z)\in {\mathfrak {p}}. \end{aligned}$$

(6.1)

Let $\xi$ be a unit vector in ${{\mathfrak {s}}}{{\mathfrak {u}}}_2$ with respect to B. The right-invariant vector fields on G generated by the elements

$$\begin{aligned} {\tilde{\zeta }}_1=(\xi ,0,0),\qquad {\tilde{\zeta }}_2=(0,\xi ,0),\qquad {\tilde{\zeta }}_3=(0,0,\xi ) \end{aligned}$$

of ${\mathfrak {g}}$, define three commuting Killing vector fields $\zeta _1$, $\zeta _2$, $\zeta _3$ on M.

Let us compute $g(\zeta _1,J\zeta _2)$ at some point $aK\in M$, where $a=(a_1,a_2,a_3)$ is some element of G. By the definition of J, we have

$$\begin{aligned} g(\zeta _1,J\zeta _2)_{aK}= & {} \left\langle \left( a ^{-1}{\tilde{\zeta }}_1 a\right) _{{\mathfrak {p}}},J\left( a ^{-1}{\tilde{\zeta }}_2 a\right) _{{\mathfrak {p}}}\right\rangle =\left\langle \left( a_1 ^{-1}\xi a_1,0,0\right) _{{\mathfrak {p}}},J\left( 0,a_2 ^{-1}\xi a_2,0\right) _{{\mathfrak {p}}}\right\rangle \\= & {} \frac{1}{9}\left\langle \left( 2a_1 ^{-1}\xi a_1,-a_1 ^{-1}\xi a_1,-a_1 ^{-1}\xi a_1\right) ,J\left( -a_2^{-1}\xi a_2,2a_2^{-1}\xi a_2,-a_2^{-1}\xi a_2\right) \right\rangle \\= & {} \frac{1}{9}\left\langle \left( 2a_1 ^{-1}\xi a_1,-a_1 ^{-1}\xi a_1,-a_1 ^{-1}\xi a_1\right) ,\sqrt{3} \left( a_2^{-1}\xi a_2,0,-a_2^{-1}\xi a_2\right) \right\rangle \\= & {} \frac{1}{\sqrt{3}}B\left( a_1 ^{-1}\xi a_1,a_2^{-1}\xi a_2\right) . \end{aligned}$$

We introduce the functions $y_1,y_2,y_3:G\rightarrow {\mathbb {R}}$ defined by

$$\begin{aligned} y_i(a_1,a_2,a_3)=\frac{1}{\sqrt{3}}B\left( a_j^{-1}\xi a_j,a_k^{-1}\xi a_k\right) ,\qquad \end{aligned}$$

for every permutation (i, j, k) of (1, 2, 3). The previous computation yields

$$\begin{aligned} g(\zeta _2,J\zeta _3)_{aK}=y_1(a),\qquad g(\zeta _3,J\zeta _1)_{aK}=y_2(a),\qquad g(\zeta _1,J\zeta _2)_{aK}=y_3(a), \qquad \forall a\in G. \end{aligned}$$

A similar computation yields

$$\begin{aligned} g(\zeta _i,\zeta _j)_{aK}=2\delta _{ij}+\frac{1}{\sqrt{3}}y_k(a) \end{aligned}$$

for every even permutation (i, j, k) of (1, 2, 3). In other words, the matrix C defined in (4.2) satisfies

$$\begin{aligned} C_{ij}=2\delta _{ij}+\frac{1}{\sqrt{3}}y_k, \end{aligned}$$

where by a slight abuse of notation we keep the same notations $y_i$ for the functions defined on M by the K-invariant functions $y_i$ on G.

The function $\varphi$ in the coordinates $y_i$ such that $\mathrm{Hess}(\varphi )=C$ is determined by

$$\begin{aligned} \varphi (y_1,y_2,y_3)=y_1^2+y_2^2+y_3^2+\frac{1}{\sqrt{3}}y_1y_2y_3+h, \end{aligned}$$

(6.2)

up to some affine function h in the coordinates $y_i$. On the other hand, since

$$\begin{aligned} \det (C)=-\frac{2}{3}\left( y_1^2+y_2^2+y_3^2\right) +\frac{2}{3\sqrt{3}}y_1y_2y_3+8, \end{aligned}$$

an easy computation shows that the function $\varphi$ given by (6.2) satisfies indeed the Monge–Ampère-type equation (4.13) for $h=3$. For the sake of completeness, we list the other functions involved in the previous section, in the particular case of the present situation:

$$\begin{aligned} \varepsilon ^2= -\frac{8}{3}\left( y_1^2+y_2^2+y_3^2\right) -\frac{16}{3\sqrt{3}}y_1y_2y_3+8, \\ \quad ^t\!VCV=2\left( y_1^2+y_2^2+y_3^2\right) +2\sqrt{3}y_1y_2y_3, \end{aligned}$$

where $\varepsilon$ is defined in (4.1) and V in (4.8).

6.1 Radial solutions

We search here particular solutions to Eq. (4.13), namely when $\varphi$ is a radial function on (some open subset of) ${\mathbb {R}}^3$ with coordinates $y_k, 1 \le k \le 3$. Let therefore $\varphi (y_1,y_2,y_3):=x(\frac{r^2}{2})$ where x is a function of one real variable and $r^2=y_1^2+y_2^2+y_3^2$. A direct computation yields

$$\begin{aligned} \begin{aligned} \mathrm{Hess}(\varphi )&=\,\begin{pmatrix} y_1^2 x^{\prime \prime }+x^{\prime }&{} y_1y_2 x^{\prime \prime } &{} y_1y_3 x^{\prime \prime }\\ y_1y_2 x^{\prime \prime } &{} y_2^2 x^{\prime \prime }+x^{\prime } &{} y_2y_3 x^{\prime \prime }\\ y_1y_3 x^{\prime \prime } &{} y_2y_3 x^{\prime \prime } &{} y_3^2 x^{\prime \prime }+x^{\prime } \end{pmatrix}\\&=\,x^{\prime }\mathrm {Id}+x^{\prime \prime }(\frac{r^2}{2}) V \cdot {}^{t}V \end{aligned} \end{aligned}$$

where $V:=\begin{pmatrix} y_1\\ y_2\\ y_3 \end{pmatrix}$. In particular,

$$\begin{aligned} \begin{aligned} \det \mathrm{Hess}(\varphi )&=\,(x^{\prime })^2 x^{\prime \prime }r^2+(x^{\prime })^3\\ \partial _r \varphi&=\,r^2 x^{\prime }, \ \partial _r^2 \varphi =r^4 x^{\prime \prime }+2r^2x^{\prime }, \end{aligned} \end{aligned}$$

whence after making the substitution $t:=\frac{r^2}{2}$ we get:

Proposition 6.1

Radial solutions to the Monge–Ampère-type equation (4.13) are given by solutions of the second-order ODE

$$\begin{aligned} x^{\prime \prime }=F(t,x,x^{\prime }) \end{aligned}$$

(6.3)

where $F(t,p,q):=\frac{8p-(10tq+3q^3)}{6(q^2t-2t^2)}$.

To decide which solutions to (6.3) yield genuine Riemannian metrics in dimension six, we observe that

Proposition 6.2

For a radial solution $\varphi =x(\frac{r^2}{2})$ to (4.13), the set $U_0$ defined in (5.2) is

$$\begin{aligned} U_0=\left\{ t>0\ |\ x(t)>2tx^{\prime }(t)> 2t\sqrt{2t} \right\} . \end{aligned}$$

Proof

Having $\varphi -\partial _r \varphi >0$ is equivalent with

$$\begin{aligned} 2tx^{\prime }(t)-x(t)<0. \end{aligned}$$

The matrix $\mathrm{Hess}(\varphi )$ has the eigenvalues $x^{\prime }(\frac{r^2}{2})$ with eigenspace $E:=\{a \in {\mathbb {R}}^3\ |\ \left\langle a,y\right\rangle =0\}$ and $x^{\prime }(\frac{r^2}{2})+r^2 x^{\prime \prime }(\frac{r^2}{2})$ with eigenvector y. Therefore, $\mathrm{Hess}(\varphi )>0$ if and only if

$$\begin{aligned} x^{\prime }(t)>0, \ x^{\prime }(t)+2tx^{\prime \prime }(t)>0. \end{aligned}$$

(6.4)

However, $x^{\prime }(t)+2tx^{\prime \prime }(t)=\frac{8(x-2tx^{\prime })}{3((x^{\prime })^2-2t)}$ from (6.3), thus showing that the system (6.4) is equivalent to $x^{\prime }(t) >\sqrt{2t}$. By Lemma 5.1, it remains to interpret the condition

$$\begin{aligned} \langle \mu a,b\rangle ^2 <\langle Ca,a \rangle \langle Cb,b \rangle \end{aligned}$$

(6.5)

for all $(a,b)\in ({\mathbb {R}}^3\times {\mathbb {R}}^3)\setminus {(0,0)}$.

We split $a=\lambda _1 y+v_1$, $b=\lambda _2 y+v_2$ with $v_1,v_2\in E$ and take into account that C preserves the orthogonal decomposition ${\mathbb {R}}^3={\mathbb {R}}y \oplus E$ and also that y belongs to $\ker \mu$. Then,

$$\begin{aligned} \langle Ca,a \rangle \langle Cb,b \rangle =\left( \lambda _1^2\langle Cy,y\rangle +\langle Cv_1, v_1 \rangle \right) \left( \lambda _2^2\langle Cy,y\rangle +\langle Cv_2, v_2 \rangle \right) \end{aligned}$$

and since $\mu$ is skew-symmetric,

$$\begin{aligned} \langle \mu a,b\rangle ^2=\langle \mu v_1, v_2\rangle ^2. \end{aligned}$$

Thus, (6.5) holds if and only if $\langle Cv_1, v_1 \rangle \langle Cv_2,v_2 \rangle > \langle \mu v_1,v_2\rangle ^2$ for all nonzero $v_1,v_2\in E$. This is equivalent to

$$\begin{aligned} \langle \mu v_1,v_2\rangle ^2 <(x^{\prime }(t))^2 \vert v_1 \vert ^2 \vert v_2 \vert ^2 \end{aligned}$$

(6.6)

for all $v_1,v_2$ in $E\setminus \{0\}$. By the Cauchy–Schwartz inequality, this is equivalent to $-\frac{1}{2}\mathrm {tr}(\mu ^2) <(x^{\prime })^2 (t)$ and since $\mathrm {tr}(\mu ^2)=-2r^2=-\,4t$, (6.6) is equivalent to $x^{\prime }(t) >\sqrt{2t}$. However, this was already known and the proof is finished. $\square$

Remark 6.1

The solutions of the ODE (6.3) of the form $x=kt^l$ with $k,l\in {\mathbb {R}}$ are $x_{1,2}=\pm \frac{2\sqrt{2}}{9}t^{\frac{3}{2}}$ and $x_3=kt^\frac{1}{2}$, corresponding to

$$\begin{aligned} \varphi _{1,2}=\pm \frac{r^3}{9},\qquad \varphi _3=\frac{k}{\sqrt{2}}r. \end{aligned}$$

However, they do not satisfy the positivity requirements from Proposition 6.2.

Solutions to the Cauchy problem (6.3), admissible in the sense of Proposition 6.2, are obtained by requiring the initial data $(t_0, x(t_0), x^{\prime }(t_0))$ belong to

$$\begin{aligned} {\mathcal {S}}:=\left\{ (t,p,q) \in {\mathbb {R}}^3: t>0, \ p>2tq> 2t\sqrt{2t} \right\} . \end{aligned}$$