The multi-moment maps of the nearly Kähler $S^3\times S^3$

Abstract

The homogeneous nearly Kähler structure on $S^3\times S^3$ is the only known complete 6-dimensional strictly nearly Kähler structure which is invariant under the action of a 3-torus. We investigate the multi-moment maps associated to such an action and compare with the usual moment map of a toric manifold.

1 Introduction

An almost Hermitian manifold (M, g, J) is nearly Kähler if $\nabla ^g J$ is skew -symmetric. We say a nearly Kähler manifold is strict if it is not Kähler. The minimum dimension admitting strict nearly Kähler manifolds is 6, where the nearly Kähler condition is equivalent to the existence of an ${{\mathrm{SU}}}_3$ structure $(\omega ,\Omega )$ satisfying

$$\begin{aligned} d\omega = 3{{\mathrm{Re}}}\Omega , \qquad d\,{{\mathrm{Im}}}\Omega = -2\omega \wedge \omega , \end{aligned}$$

where $\omega$ is the Kähler form, and $\Omega$ is a complex volume form [9].

There are only a handful of known examples of compact strictly Kähler 6-manifolds. The homogeneous spaces ${\mathbb {S}}^6, {\mathbb {S}}^3\times {\mathbb {S}}^3,\mathbb {CP}^3$ and ${{\mathrm{SU}}}_3/{\mathbb {T}}^2$ admit strict nearly Kähler structures, and there are no other homogeneous strict nearly Kähler 6-manifolds [2]. There are locally homogeneous structures on quotients of ${\mathbb {S}}^3\times {\mathbb {S}}^3$ [3], and cohomogeneity one structures on ${\mathbb {S}}^6$ and ${\mathbb {S}}^3\times {\mathbb {S}}^3$, which are conjectured to be the only cohomogeneity one examples [5].

If one wants to look for higher cohomogeneity examples, one could look for strict nearly Kähler 6-manifolds admitting the action of a 3-torus ${\mathbb {T}}^3$ by isometries. Of the list of known examples in the previous paragraph, only the homogeneous ${\mathbb {S}}^3\times {\mathbb {S}}^3$ and some of its locally homogeneous quotients admit such a symmetry group. The purpose of this paper is to explore these examples.

We will make analogies to the case of toric Kähler manifolds. On a compact Kähler manifold M, every Killing vector field K is symplectic. Moreover, if M is simply-connected, then K is Hamiltonian. Thus a 2m-dimensional compact simply-connected Kähler manifold $M^{2m}$ admitting an effective isometric ${\mathbb {T}}^m$ action is toric. Such a manifold could be studied with use of the moment map $\mu$, which is a ${\mathbb {T}}^3$-equivariant map from the manifold to the dual Lie algebra of the torus, $\mathfrak t^*$. Each fiber of $\mu$ is a ${\mathbb {T}}^m$ orbit, and the image of $\mu$ is the polytope which is the convex hull of the $\mu$-image of the fixed points of the ${\mathbb {T}}^m$ action.

The Kähler form $\omega$ of a strictly nearly Kähler 6-manifold M is not closed, so there can be no moment map. However, there is a generalization of the moment map called the multi-moment map [6], which can be applied to invariant closed forms of any degree. If M is not the round ${\mathbb {S}}^6$, then every Killing vector field preserves the whole ${{\mathrm{SU}}}_3$ structure [7]. Thus if M admits an effective isometric ${\mathbb {T}}^3$ action, then the exact forms $d\omega$ and $\omega \wedge \omega$ are preserved by the ${\mathbb {T}}^3$ action. Thus we can associate $d\omega$ and $\omega \wedge \omega$ to multi-moment maps, $\nu$ and $\sigma$ respectively, induced from the ${\mathbb {T}}^3$ action.

We compute $\nu$ and $\sigma$ for the homogeneous nearly Kähler ${\mathbb {S}}^3\times {\mathbb {S}}^3$. We find in this case that $\nu$ has many qualitative similarities to the toric moment map $\mu$. In particular, we find that the multi-moment map image $\Delta :=\nu ({\mathbb {S}}^3\times {\mathbb {S}}^3)$ of ${\mathbb {S}}^3\times {\mathbb {S}}^3$ is convex and that its boundary $\partial \Delta$ contains the 1-skeleton of a regular tetrahedron. However, $\Delta$ bulges beyond the faces of the tetrahedron, and $\partial \Delta$ is smooth away from the vertices. In fact, $\Delta$ is the 3-elliptope, which is the compact convex solid body whose boundary is contained in Cayley’s nodal cubic surface. Along $\partial \Delta$, each $\nu$-fiber is a ${\mathbb {T}}^3$ orbit, but in the interior, each fiber contains two orbits. The other multi-moment map $\sigma$ distinguishes between these two orbits in the interior fibers, so the map $\nu \oplus \sigma$ has connected fibers, but its image is only the boundary of a convex region $\hat{\Delta }$. However, $\hat{\Delta }$ can be interpreted using the multi-moment maps of the $G_2$ structure on the cone over M. The following table compares the fiber types for the multi-moment map of ${\mathbb {S}}^3\times {\mathbb {S}}^3$ to the moment map of a toric 3-manifold:

Fiber of a point in ...	$\mu$ toric 6-manifold	$\nu$ for nearly Kähler $S^3\times {\mathbb {S}}^3$
A vertex	{point}	${\mathbb {T}}^2$
An edge	${\mathbb {S}}^1$	${\mathbb {T}}^3$
A face	${\mathbb {T}}^2$	${\mathbb {T}}^3$
The interior	${\mathbb {T}}^3$	${\mathbb {T}}^3 \sqcup {\mathbb {T}}^3$

Finally, we find that for the action of a 3-torus ${\mathbb {T}}$ on ${\mathbb {S}}^3\times {\mathbb {S}}^3$ to descend to a locally homogeneous quotient $\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3)$, then the finite group $\Gamma$ must be a subgroup of ${\mathbb {T}}$. It follows that the multi-moment maps on $\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3)$ and ${\mathbb {S}}^3\times {\mathbb {S}}^3$ have the same image, and the fibers have the same diffeomorphism type.

2 Homogenous nearly Kähler ${\mathbb {S}}^3\times {\mathbb {S}}^3$

We begin by reviewing the definition of the homogenous nearly Kähler structure on ${\mathbb {S}}^3\times {\mathbb {S}}^3$, following the work in [4].

We identify ${\mathbb {S}}^3$ with the unit sphere in the quaternions $\mathbb H$. For any $p\in {\mathbb {S}}^3$, $T_p{\mathbb {S}}^3\subset T_p\mathbb H$ is the image of $T_1{\mathbb {S}}^3$ by the pushforward of left-multiplication by p. Identifying $T_p{\mathbb {S}}^3\subset T_p\mathbb H$ with $p^\perp \subset \mathbb H$, this pushforward is simply quaternionic multiplication by p. Thus the basis $\{i,j,-k\}$ of ${{\mathrm{Im}}}\,{\mathbb {H}}$ which is identified with $T_1{\mathbb {S}}^3$ gives a frame for $T_{(p,q)}{\mathbb {S}}^3\times {\mathbb {S}}^3=T_p{\mathbb {S}}^3\oplus T_q{\mathbb {S}}^3:$

$$\begin{aligned} E_1(p,q) = (pi,0), \qquad&F_1(p,q) = (0,qi),\\ E_2(p,q) = (pj,0), \qquad&F_2(p,q) = (0,qj),\\ E_3(p,q) = (pk,0), \qquad&F_3(p,q) = (0,qk), \end{aligned}$$

where i, j, k are imaginary quaternions satisfying $ij=k$. Note that this frame differs by the one in [4] by a sign in $E_3$ and $F_3$. Let $\{E^1,E^2,E^3,F^1,F^2,F^3\}$ be the dual frame.

The almost complex structure for the homogenous nearly Kähler ${\mathbb {S}}^3\times {\mathbb {S}}^3$ is given in this frame by

$$\begin{aligned} J = \frac{1}{\sqrt{3}}\sum _{n=1}^3\left( -E_n\otimes E^n+F_n\otimes F^n + 2 F_n\otimes E^n - 2 E_n\otimes F^n\right) . \end{aligned}$$

The metric g is given by the average of $g_{\mathbb H^2}$ and $g_{\mathbb H^2}(J\cdot , J\cdot )$, where

$$\begin{aligned} g_{\mathbb H^2}=\sum _{n=1}^3\left( \left( E^n\right) ^2+\left( F^n\right) ^2\right) \end{aligned}$$

is the flat metric from $\mathbb H^2$ restricted to ${\mathbb {S}}^3\times {\mathbb {S}}^3$. This gives

$$\begin{aligned} g&= \frac{4}{3}\sum _{n=1}^3\left( (E^n)^2-E^nF^n+(F^n)^2\right) ,\\ \omega&=\frac{4}{\sqrt{3}}\sum _{n=1}^3 E^n\wedge F^n. \end{aligned}$$

2.1 Torus actions

For unit quaternions $a,b,c\in {\mathbb {S}}^3$, the map

$$\begin{aligned} F_{a,b,c}:{\mathbb {S}}^3\times {\mathbb {S}}^3\rightarrow {\mathbb {S}}^3\times {\mathbb {S}}^3:(p,q)\mapsto (apc^{-1},bqc^{-1}) \end{aligned}$$

is a holomorphic isometry [8].

Lemma 2.1

The map

$$\begin{aligned} F:{\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3\rightarrow {Aut}({\mathbb {S}}^3\times {\mathbb {S}}^3,J)\cap {Isom}({\mathbb {S}}^3\times {\mathbb {S}}^3,g):(a,b,c)\mapsto F_{a,b,c} \end{aligned}$$

is a homomorphism with kernel $\{(\sigma ,\sigma ,\sigma )\}_{\sigma \in \pm 1}$.

Proof

It is clear from the definition of F that $F_{a,b,c}\circ F_{a',b',c'}= F_{aa',bb',cc'}$, so that F is a homomorphism. To compute the kernel, let $(a,b,c)\in \ker F$, so that $F_{a,b,c}=Id$. Then

$$\begin{aligned} (1,1)=F_{a,b,c}(1,1)=(ac^{-1},ab^{-1}), \end{aligned}$$

so that $a=b=c$. For any $(p,q)\in {\mathbb {S}}^3\times {\mathbb {S}}^3$,

$$\begin{aligned} (p,q)=F_{a,a,a}(p,q)=(apa^{-1},aqa^{-1}). \end{aligned}$$

Since this is true for all $p,q\in {\mathbb {S}}^3$, we find that a lies in the center $\{\pm 1\}$ of ${\mathbb {S}}^3$. This gives the result, identifying $\sigma$ with a. $\square$

Since the projection of ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ onto any of its factors is a homomorphism, any abelian subgroup of ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ must be a product of abelian subgroups of ${\mathbb {S}}^3$. But every non-trivial abelian subgroup of ${\mathbb {S}}^3$ is of the form $\left\{ e^{At}\right\} _{t\in \mathbb R}$ for some unit imaginary quaternion A. Thus, a maximal torus in ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ is of the form

$$\begin{aligned} \{(e^{At_1},e^{Bt_2},e^{Ct_3})\}_{(t_1,t_2,t_3)\in \mathbb R^3}, \end{aligned}$$

for some $A,B,C\in {\mathbb {S}}^2$, identifying ${\mathbb {S}}^2$ with the unit imaginary quaternions. A routine computation shows that the image of such a torus under F is generated by the Killing vector fields

$$\begin{aligned} K_1&= (Ap,0),\\ K_2&= (0,Bq), \\ K_3&= (-pC,-qC). \end{aligned}$$

3 Multi-moment maps

A multi-moment map associated to a closed r-form $\beta$ invariant under the action of a torus ${\mathbb {T}}$ is a smooth map $\tau :M\rightarrow \Lambda ^{r-1}\mathfrak t^*$ defined by the extending linearly the relation

$$\begin{aligned} \left\langle d\tau ,X^1\wedge \cdots \wedge X^{r-1}\right\rangle = \tau (X^1_M,\dots ,X^{r-1}_M,\cdot ), \end{aligned}$$

where for each j, $X^j_M$ is the vector field on M induced by $X^j\in \mathfrak t$. Note that when the multi-moment map exists, it is unique up to a constant. Also note that when $r=2$, we recover the usual moment map up to a sign convention.

We are interested in computing the multi-moment maps for the ${\mathbb {T}}^3$ action preserving the exact forms $d\omega$ and $\omega \wedge \omega$. In the following result, which has been translated into our simple case, we see that in the exact case, multi-moment maps always exist, and take a convenient form:

Proposition 3.1

([6] Proposition 3.1) Let M be a manifold admitting the action of a torus ${\mathbb {T}}$ that preserves an r-form $\alpha$. Then the map $\tau :M\rightarrow \Lambda ^r\mathfrak t^*$ given by the expression

$$\begin{aligned} \langle \tau ,X^1\wedge X^2\wedge \dots \wedge X^r\rangle = (-1)^r\alpha (X^1_M,X^2_M,\dots ,X^r_M) \end{aligned}$$

is a multi-moment map for $d\alpha$ under the action of ${\mathbb {T}}$.

3.1 The multi-moment map $\nu$ for $d\omega$

In this subsection, we describe the multi-moment map $\nu$ of $d\omega$. We will describe its image and the structure of its fibers.

For $X\in {\mathbb {S}}^2$, let us define a map

$$\begin{aligned} \pi _X:{\mathbb {S}}^3\rightarrow {\mathbb {S}}^2:p\mapsto \bar{p} X p. \end{aligned}$$

When $X=i$, this is the usual Hopf fibration. For general X, $\pi _X$ also identifies ${\mathbb {S}}^3$ as a ${\mathbb {S}}^1$ bundle over ${\mathbb {S}}^2$.

Proposition 3.2

With respect to a suitable basis for $\Lambda ^2\mathfrak t^*$, $\nu =\bar{\nu }\circ (\pi _A\times \pi _B),$ where

$$\begin{aligned} \bar{\nu }:{\mathbb {S}}^2\times {\mathbb {S}}^2\rightarrow \Lambda ^2\mathfrak t^*:(x,y)\mapsto (x\cdot y, x\cdot C, y\cdot C). \end{aligned}$$

Proof

For any $p\in {\mathbb {S}}^3$, $Ap=p\bar{p}Ap=p\pi _A(p)$. Thus the Killing vector fields can be written in terms of the frame $(E_1,E_2,E_3,F_1,F_2,F_3)$ as

$$\begin{aligned} K_1&= \big (\pi _A(p)\cdot i\big ) E_1 + \big (\pi _A(p)\cdot j\big ) E_2 + \big (\pi _A(p)\cdot k\big ) E_3, \\ K_2&= \big (\pi _B(q)\cdot i\big ) F_1 + \big (\pi _B(q)\cdot j\big ) F_2 + \big (\pi _B(q)\cdot k\big ) F_3, \\ -K_3&= (C\cdot i) (E_1+F_1) + (C\cdot j) (E_2+F_2) + (C\cdot k) (E_3+F_3), \end{aligned}$$

where $\cdot$ is the dot product on ${\mathbb {H}}$. This allows us to compute

$$\begin{aligned} \frac{\sqrt{3}}{4}\omega (K_1,K_2)&= \pi _A(p)\cdot \pi _B(q), \\ \frac{\sqrt{3}}{4}\omega (K_1,K_3)&= -\pi _A(p)\cdot C, \\ \frac{\sqrt{3}}{4}\omega (K_2,K_3)&= \pi _B(q)\cdot C. \end{aligned}$$

Choosing the basis $\left\{ \frac{\sqrt{3}}{4}(K_1\wedge K_2)^*,\frac{\sqrt{3}}{4}(K_3\wedge K_1)^*,\frac{\sqrt{3}}{4}(K_2\wedge K_3)^*\right\}$ for $\Lambda ^2\mathfrak t^*$, Proposition 3.1 allows us to write the multi-moment map in the claimed form. $\square$

Note that $\nu ^{-1}(0,0,0)$ is the union of the Lagrangian torus orbits. The example of a Lagrangian torus in [4] can be found with the values $A=B=i$ and $C=j$.

Let $\Delta =\nu ({\mathbb {S}}^3\times {\mathbb {S}}^3)=\bar{\nu }({\mathbb {S}}^2\times {\mathbb {S}}^2)$ with interior $\mathring{\Delta }$. Clearly $\Delta \subset [-1,1]^3$. The following lemma gives more information about $\bar{\nu }$ and $\Delta$:

Lemma 3.3

Let $(X,Y,Z)\in [-1,1]^3$. Then

$$\begin{aligned} \bar{\nu }^{-1}(X,Y,Z)\cong \left\{ \begin{matrix} \emptyset &{} \text {if }F(X,Y,Z)<0 \\ {\mathbb {S}}^1\sqcup {\mathbb {S}}^1 &{} \text {if } F(X,Y,Z)>0 \\ \{\text {point}\} &{} \text {if } (X,Y,Z)\in V \\ {\mathbb {S}}^1 &{} \text {if } (X,Y,Z)\in \ker F\backslash V \end{matrix}\right. , \end{aligned}$$

where

$$\begin{aligned} F(X,Y,Z)&:= 1+2XYZ-X^2-Y^2-Z^2, \\ V&:=\big \{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\big \}. \end{aligned}$$

Proof

Let $(X,Y,Z)=(\cos (\theta ),\cos (\phi ),\cos (\psi ))\in [-1,1]^3$. Without loss of generality, by changing coordinates we may assume that $C=i$.

Let $(x,y)\in \bar{\nu }^{-1}(X,Y,Z)$. The relations

$$\begin{aligned} \cos (\phi ) = Y = x\cdot i, \qquad \cos (\psi ) = Z = y\cdot i \end{aligned}$$

allow us to write

$$\begin{aligned} x = \cos (\phi )i + \sin (\phi )e^{i\xi }j, \qquad y = \cos (\psi )i + \sin (\psi )e^{i\eta }j, \end{aligned}$$

for real numbers $\xi$ and $\eta$. Now

$$\begin{aligned} \cos (\theta )=X=x\cdot y = \cos (\phi )\cos (\psi )+\sin (\phi )\sin (\psi )\cos (\alpha ), \end{aligned}$$

(1)

where $\alpha =\eta -\xi$. So

$$\begin{aligned} F(X,Y,Z)&= (1-Y^2)(1-Z^2)-(X-YZ)^2 \\&= \sin ^2(\phi )\sin ^2(\psi ) -(\sin (\phi )\sin (\psi )\cos (\alpha ))^2 =\sin ^2(\phi )\sin ^2(\psi )\sin ^2(\alpha ). \end{aligned}$$

Note that $F(X,Y,Z)\ge 0$, giving the first case of our claim.

If $F(X,Y,Z)>0$, then none of $\sin (\phi ), \sin (\psi )$ or $\sin (\alpha )$ vanish. We observe that (X, Y, Z) determines $\cos (\alpha )$, thus $\alpha$ up to a sign. Since $\sin (\alpha )\ne 0$, there are two choices for $\alpha$. Since there is a ${\mathbb {S}}^1$ of choices for $\xi +\eta$, we find that $\bar{\nu }^{-1}(X,Y,Z)\cong {\mathbb {S}}^1\sqcup {\mathbb {S}}^1.$

If $F(X,Y,Z) = 0$, then $0=\sin (\phi )\sin (\psi )\sin (\alpha )$. If $\sin (\phi )\ne 0 \ne \sin (\psi )$, then $\sin (\alpha )=0$. $\cos (\alpha )$ is then determined by (1), giving a single choice for $\alpha$. Since there is a ${\mathbb {S}}^1$ of choices for $\xi +\eta$, we find that $\bar{\nu }^{-1}(X,Y,Z)\cong {\mathbb {S}}^1.$ If $\sin (\phi )=0\ne \sin (\psi )$, then $x=i$ and there are no restrictions on $\eta$. Thus $\bar{\nu }^{-1}(X,Y,Z)\cong {\mathbb {S}}^1$. Similarly, if $\sin (\phi )\ne 0 = \sin (\psi )$, then $\bar{\nu }^{-1}(X,Y,Z)\cong {\mathbb {S}}^1$. If $\sin (\phi )=\sin (\phi )=0$, then we find that $\bar{\nu }^{-1}(X,Y,Z)$ is a single point. Note that this is the case when $Y^2=Z^2=1$. The condition $F(X,Y,Z)=0$ then forces $(X,Y,Z)\in V$. $\square$

Lemma 3.4

$$\begin{aligned} \Delta&= \{(X,Y,Z)\in [-1,1]^3:F(X,Y,Z)\ge 0\}, \\ \partial \Delta&= \{(X,Y,Z)\in [-1,1]^3:F(X,Y,Z)= 0\}. \end{aligned}$$

Proof

The first statement follows directly from the previous lemma. To prove the second statement, we compute

$$\begin{aligned} F(X,Y,Z)|_{Z^2=1}=-(X-YZ)^2\le 0. \end{aligned}$$

Similar computations show that $F|_{\partial [-1,1]^3}\le 0$. Thus

$$\begin{aligned} \partial \Delta = \bigg (\partial [-1,1]^3\cap F^{-1}\big ([0,\infty )\big )\bigg ) \cup \big ([-1,1]^3\cap \ker F\big ) = [-1,1]^3\cap \ker F. \end{aligned}$$

$\square$

Note that $\ker F$ describes Cayley’s nodal cubic surface, pictured in Fig. 1. $\partial \Delta$ is then the intersection of this surface with $[-1,1]^3$, which is the closure of the compact component of the set of regular points of the cubic.

Fig. 1

The real points of Cayley’s nodal cubic surface [1]

The previous lemma gives a description of $\Delta$ which can be recognized as being a definition of a convex body called a 3-elliptope. In the following lemma, we review this correspondence:

Lemma 3.5

$\Delta$ can be identified with the 3-elliptope

$$\begin{aligned} {\mathcal {E}}_3:=\left\{ (X,Y,Z)\in {\mathbb {R}}^3:\begin{pmatrix} 1 &{} X &{} Y \\ X &{} 1 &{} Z \\ Y &{} Z &{} 1 \end{pmatrix}\text { is positive semi-definite}\right\} . \end{aligned}$$

Proof

By Sylvester’s criterion, a Hermitian matrix is positive semi-definite if and only if all of the principal minors are non-negative. Noting that F(X, Y, Z) is the determinant of the matrix in the definition of ${\mathcal {E}}_3$, we find that

$$\begin{aligned} {\mathcal {E}}_3 = \left\{ (X,Y,Z)\in {\mathbb {R}}^3:(1-X^2),(1-Y^2),(1-Z^2),F(X,Y,Z)\ge 0\right\} =\Delta , \end{aligned}$$

where the last equality follows from the previous lemma. $\square$

Since ${\mathcal {E}}_3$ is obviously convex, we immediately obtain:

Corollary 3.6

$\Delta$ is convex.

We finish this section by describing vertices and edges of $\Delta$, which are well known properties of the Cayley cubic.

Proposition 3.7

The set of singular points of $\partial \Delta$ is

$$\begin{aligned} V:=\big \{(1,1,1),(1,-1,-1),(-1,1,-1),(-1,-1,1)\big \}. \end{aligned}$$

Proof

By Lemma 3.4 The singular points of $\partial \Delta$ are the points where F and $\nabla F$ both vanish. The set of points where $\nabla F$ vanishes are $\{(0,0,0)\}\cup V$. The result follows since $F(0,0,0)=1\ne 0$, while F vanishes on V. $\square$

Proposition 3.8

The line segment between any two points in V lies in $\partial \Delta$.

Proof

We will show that the line segment between (1, 1, 1) and $(1,-1,-1)$ lies in $\partial \Delta$, with the other line segments following similarly. This line segment is parametrized by

$$\begin{aligned} \gamma :[-1,1]\rightarrow {\mathbb {R}}:t\mapsto (1,t,t). \end{aligned}$$

The claim follows easily from Lemma 3.4 and $F|_{X=1}=-(Y-Z)^2.$$\square$

By the previous two propositions, we find that $\partial \Delta$ contains the 1-skeleton of the regular tetrahedron with vertices V, but that the edges of this tetrahedron are not singular in $\partial V$. In Fig. 2, we see that $\Delta$ is a regular tetrahedron with convexly bulging sides.

Fig. 2

The multi-moment map image $\Delta =\nu ({\mathbb {S}}^3\times {\mathbb {S}}^3)$ with the tetrahedral 1-skeleton indicated

3.2 The multi-moment map $\sigma$ for $\omega \wedge \omega$

In this section, we want to compute the multimoment map $\sigma$ for $\omega \wedge \omega =-\frac{1}{2} d\,{{\mathrm{Im}}}\,\Omega$. In order to apply Proposition 3.1 again, we will solve directly for a potential 3-form for $\omega \wedge \omega$. We continue to work in the frame $(E_1,E_2,E_3,F_1,F_2,F_3)$, although for convenience we will make the identifications $E_n=E_{n+3}$ and $F_n=F_{n+3}$ for every $n\in {\mathbb {Z}}$.

Lemma 3.9

$$\begin{aligned} dE^n =2E^{n-1}\wedge E^{n+1}, \qquad dF^n =2F^{n-1}\wedge F^{n+1}. \end{aligned}$$

Proof

The Lie bracket in this frame was computed in [4]. Noting that we treat $E^3$ and $F^3$ as the negatives of their convention, the Lie bracket is given by

$$\begin{aligned}{}[E_{n-1},E_{n+1}]=-2E_n, \qquad [E_n,F_m]=0, \qquad [F_{n-1},F_{n+1}]=-2F_n. \end{aligned}$$

For vector fields X and Y, we have

$$\begin{aligned} dE^n(X,Y)=XE^n(Y)-YE^n(X)-E^n\big ([X,Y]\big ). \end{aligned}$$

Choosing $X,Y\in \{E_1,E_2,E_3,F_1,F_2,F_3\},$ we find that

$$\begin{aligned} dE^n = -E^n\big ([E_{n-1},E_{n+1}]\big )E^{n-1}\wedge E^{n+1}=2E^{n-1}\wedge E^{n+1}. \end{aligned}$$

The computation of $dF^n$ is similar. $\square$

Lemma 3.10

$$\begin{aligned} \omega \wedge \omega = d\left( -\frac{8}{9}\sum _{n=1}^3\left( E^n\wedge F^{n-1}\wedge F^{n+1}+F^n\wedge E^{n-1}\wedge E^{n+1}\right) \right) . \end{aligned}$$

Proof

$$\begin{aligned} \omega \wedge \omega&= \left( \frac{4}{\sqrt{3}}\sum _{n=1}^3E^n\wedge F^n\right) ^2 =-\frac{16}{9}\sum _{n,m=1}^3E^n\wedge E^m\wedge F^n\wedge F^m \\&=-\frac{32}{9}\sum _{n=1}^3E^{n-1}\wedge E^{n+1}\wedge F^{n-1}\wedge F^{n+1} \\&=-\frac{8}{9}d\sum _{n=1}^3\left( E^n\wedge F^{n-1}\wedge F^{n+1}+F^n\wedge E^{n-1}\wedge E^{n+1}\right) . \end{aligned}$$

$\square$

Lemma 3.11

The multi-moment map $\sigma$ associated to the form $\omega \wedge \omega$ is given by $\det \big (\pi _A(p),\pi _B(q),C\big ).$ with respect to some convenient basis.

Proof

Choose the basis $\frac{3}{16}(K_1\wedge K_2\wedge K_3)^*$ for $\Lambda ^3\mathfrak t^*$. With respect to this basis, the previous lemma and Proposition 3.1 tells us that $\sigma$ is given by

$$\begin{aligned}&-\frac{3}{16}\left( -\frac{8}{9}\sum _{n=1}^3\left( E^n\wedge F^{n-1}\wedge F^{n+1}+F^n\wedge E^{n-1}\wedge E^{n+1} \right) \right) (K_1, K_2, K_3) \\&\quad = \frac{1}{6} \sum _{n=1}^3\left( E^n\wedge F^{n-1}\wedge F^{n+1}+F^n\wedge E^{n-1}\wedge E^{n+1} \right) (K_1, K_2, K_3). \end{aligned}$$

We will compute the first term first. Since $F^n(K_1)=0$, we get

$$\begin{aligned} \frac{1}{6}\sum _{n=1}^3\left( E^n\wedge F^{n-1}\wedge F^{n+1}\right) (K_1, K_2, K_3)&= \frac{1}{6}\sum _{n=1}^3 E^n(K_1)(F^{n-1}\wedge F^{n+1})(K_2, K_3) \\&= -\frac{1}{2}(i^*\wedge j^*\wedge k^*) \big (\pi _A(p),\pi _B(q),-C\big ) \\&= \frac{1}{2}\det \big (\pi _A(p),\pi _B(q),C\big ). \end{aligned}$$

Similarly

$$\begin{aligned} \frac{1}{6}\sum _{n=1}^3\left( F^n\wedge E^{n-1}\wedge E^{n+1}\right) (K_1, K_2, K_3) = \frac{\det \big (\pi _A(p),\pi _B(q),C\big )}{2}. \end{aligned}$$

Combining these, we get the result. $\square$

Lemma 3.12

$|\sigma | = \sqrt{F}\circ \nu$.

Proof

We do the computation in the case that $C=i$. Let $(x,y)\in {\mathbb {S}}^2\times {\mathbb {S}}^2$ such that

$$\begin{aligned} \bar{\nu }(x,y) = (X,Y,Z) = \big (\cos (\theta ),\cos (\phi ),\cos (\psi )\big ). \end{aligned}$$

Then as in Lemma 3.3, we have

$$\begin{aligned} x = \cos (\phi )i + \sin (\phi )e^{i\xi }j, \qquad y = \cos (\psi )i + \sin (\psi )e^{i\eta }j, \end{aligned}$$

for real numbers $\xi$ and $\eta$ satisfying

$$\begin{aligned} \cos (\theta )= \cos (\phi )\cos (\psi )+\sin (\phi )\sin (\psi )\cos (\alpha ), \end{aligned}$$

(2)

where $\alpha =\eta -\xi$. Then

$$\begin{aligned} \det (x,y,i) = \sin (\theta )\sin (\phi )\sin (\alpha ). \end{aligned}$$

But in Lemma 3.3, we saw that

$$\begin{aligned} F(X,Y,Z)= \sin ^2(\theta )\sin ^2(\phi )\sin ^2(\alpha ) =\big (\det (x,y,i)\big )^2. \end{aligned}$$

This combined with the previous lemma gives the desired result. $\square$

Note that the the fibers of $\nu \oplus \sigma$ are connected, since when a fiber of $\nu$ has two connected components, the sign of $\sigma$ distinguishes between them.

Lemma 3.13

$\sqrt{F}:\Delta \rightarrow {\mathbb {R}}$ is concave, so that the image of $\nu \oplus \sigma$ is the boundary of a convex region $\hat{\Delta }:=\{(p,t)\in \Delta \times {\mathbb {R}}:t^2\le F(p)\}$.

Proof

By the previous lemma, The image of $\nu \oplus \sigma$ is equal to $\{(p,t)\in \Delta \times {\mathbb {R}}:t^2=F(p)\}=\partial \hat{\Delta }.$ To show that $\hat{\Delta }$ is convex, we will show that ${{\mathrm{Hess}}}\left( \sqrt{F}\right)$ is negative definite.

We first compute

$$\begin{aligned}&\nabla F = 2\begin{pmatrix} -X+YZ \\ -Y+XZ \\ -Z+XY \end{pmatrix}, \qquad {{\mathrm{Hess}}}(F) = 2\begin{pmatrix} -1 &{} Z &{} Y \\ Z &{} -1 &{} X \\ Y &{} X &{} -1 \end{pmatrix}. \end{aligned}$$

Now $(0,0,0)\in \Delta$ with $F(0,0,0)=1$. Thus ${{\mathrm{Hess}}}\left( \sqrt{F}\right) (0,0,0)=-{{\mathrm{Id}}}$ is negative definite. Since $\Delta$ is connected, to show that ${{\mathrm{Hess}}}\left( \sqrt{F}\right)$ is negative definite as required, it suffices to show that $\det \left( \sqrt{F^3}{{\mathrm{Hess}}}\left( \sqrt{F}\right) \right)$ is non-vanishing on the interior of $\Delta$.

$$\begin{aligned}&\sqrt{F^3}{{\mathrm{Hess}}}\left( \sqrt{F}\right) = \tfrac{1}{2} F{{\mathrm{Hess}}}F - \tfrac{1}{4} \nabla F \otimes \nabla F \\&\quad = \begin{pmatrix} -(1-Y^2)(1-Z^2) &{} (Z-XY)(1-Z^2) &{} (Y-XZ)(1-Y^2) \\ (Z-XY)(1-Z^2) &{} -(1-X^2)(1-Z^2) &{} (X-YZ)(1-X^2) \\ (Y-XZ)(1-Y^2) &{} (X-YZ)(1-X^2) &{} -(1-X^2)(1-Y^2) \end{pmatrix}, \\&\qquad \det \big (\sqrt{F^3}{{\mathrm{Hess}}}(\sqrt{F})\big ) =-(1-X^2)(1-Y^2)(1-Z^2)F^2. \end{aligned}$$

Since the functions $\{1-X^2,1-Y^2,1-Z^2,F\}$ are all positive on the interior of $\Delta$, we see that $det\big (\sqrt{F^3}{{\mathrm{Hess}}}(\sqrt{F})\big )$ does not vanish there, completing the proof. $\square$

In the remainder of the section we will interpret $\hat{\Delta }$ in terms of $G_2$ geometry. To any nearly Kähler manifold $(M,g,\omega )$, The metric cone

$$\begin{aligned} \left( M\times (0,\infty ),g_{cone} = dr^2+r^2g\right) \end{aligned}$$

(with r parameterizing $(0,\infty )$) admits a parallel $G_2$ structure $\varphi = \frac{1}{3} d(r^3\omega ).$ The condition that this structure is parallel means that $\varphi$ and its hodge dual $\psi :=*\varphi =-\frac{1}{4} d(r^4{{\mathrm{Im}}}\,\Omega )$ are both closed. In the case when M is nearly toric, then action of the torus ${\mathbb {T}}$ lifts to $M\times (0,\infty )$, and this lifted action preserves $\varphi$ and $\psi$. Applying Proposition 3.1, we find that $\hat{\nu }:=r^3\nu$ and $\hat{\sigma }:=r^4\sigma$ are multi-moment maps for the ${\mathbb {T}}^3$ action preserving the closed forms $\varphi$ and $\psi$ respectively (choosing a convenient basis for $\Lambda ^2{\mathfrak {t}}^*$ and $\Lambda ^3{\mathfrak {t}}^*).$

Since $\nu \oplus \sigma$ has connected fibers, so does $\hat{\nu }\oplus \sigma$. Moreover, the image $(\hat{\nu }\oplus \sigma )(M\times (0,1))$ is the interior of $\hat{\Delta }\backslash \{0\}.$ Thus $\hat{\Delta }$ is the closure of $(\hat{\nu }\oplus \sigma )(M\times (0,1))$.

4 Torus actions on locally homogeneous quotients of ${\mathbb {S}}^3\times {\mathbb {S}}^3$

Recall that ${\mathbb {S}}^3\times {\mathbb {S}}^3$ is a homogeneous space realized as the quotient

$$\begin{aligned} {\mathbb {S}}^3\times {\mathbb {S}}^3\cong ({\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3)/{\mathbb {S}}^3, \end{aligned}$$

where ${\mathbb {S}}^3$ is a Lie group under its identification as the unit quaternions and ${\mathbb {S}}^3$ is realized as a subgroup of ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ via the diagonal embedding. From [3], a locally homogeneous strict nearly Kähler 6-manifold is of the form

$$\begin{aligned} \Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3):=\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3)/{\mathbb {S}}^3 \end{aligned}$$

for some finite subgroup $\Gamma$ of ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ freely acting on ${\mathbb {S}}^3\times {\mathbb {S}}^3$. We will show that

Theorem 4.1

The action of ${\mathbb {T}}$ on ${\mathbb {S}}^3\times {\mathbb {S}}^3$ descends to $\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3)$ if and only if $\Gamma$ is a subgroup of ${\mathbb {T}}$.

Before we prove this theorem, we begin by studying a simpler situation on ${\mathbb {S}}^3$.

Lemma 4.2

Let $F\subset {\mathbb {S}}^3$ be a finite set. Let $C\le {\mathbb {S}}^3$ be a circle subgroup. If F is preserved under the action by conjugation by C, then $F\subset C$.

Proof

Using the identification of ${\mathbb {S}}^3$ with the unit quaternions, C can be written as $\{e^{\theta x}\}_{\theta \in [0,2\pi )}$ for some unit imaginary quaternion x. $e^{\theta x}$ acts by conjugation on ${\mathbb {S}}^3$ by rotating the imaginary quaternions by an angle of $2\theta$ about the axis generated by x. Since the only finite orbits of this action are fixed points, F must be contained in the set of fixed points of this action. This is the set ${{\mathrm{Span}}}\{1,x\}\cap {\mathbb {S}}^3=C$. $\square$

Proof of Theorem 4.1

First assume that the action of ${\mathbb {T}}$ on ${\mathbb {S}}^3\times {\mathbb {S}}^3$ descends to $\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3)$. Since ${\mathbb {T}}$ and $\Gamma$ are both subgroups of ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ acting on the left of ${\mathbb {S}}^3\times {\mathbb {S}}^3\cong ({\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3)/{\mathbb {S}}^3,$ this implies that the action of ${\mathbb {T}}$ on ${\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3$ to descends to $\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3)$. Thus we have

$$\begin{aligned} t\Gamma g = \Gamma tg,\quad \forall t\in {\mathbb {T}}, g\in {\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3. \end{aligned}$$

Using $g=1$, this relation tells us that $\Gamma$ is fixed by the conjugation action of ${\mathbb {T}}$. For each $i\in \{1,2,3\}$ consider the projection $\pi _i:{\mathbb {S}}^3\times {\mathbb {S}}^3\times {\mathbb {S}}^3\rightarrow {\mathbb {S}}^3$ onto the ith factor. From our discussion in Sect. 2.1, $\pi _i({\mathbb {T}})$ is a circle subgroup of ${\mathbb {S}}^3$. Thus we can apply the previous lemma to find $\pi _i(\Gamma )\subset \pi _i({\mathbb {T}})$. Combining this result for each i implies that $\Gamma \le {\mathbb {T}}$.

Conversely, if $\Gamma \le {\mathbb {T}}$, then

$$\begin{aligned} t\Gamma (p,q) = \Gamma t(p,q), \quad \forall t\in {\mathbb {T}},(p,q)\in {\mathbb {S}}^3\times {\mathbb {S}}^3 \end{aligned}$$

follows from ${\mathbb {T}}$ being abelian. $\square$

In the situation of the previous theorem, we have $\Gamma$ a finite subgroup of the torus ${\mathbb {T}}$. Then ${\mathbb {T}}/\Gamma$ is also a torus. Since the multi-moment map $\nu$ on ${\mathbb {S}}^3\times {\mathbb {S}}^3$ is ${\mathbb {T}}$-equivariant, it naturally descends to the multi-moment map $\nu _\Gamma$ on $\Gamma \backslash ({\mathbb {S}}^3\times {\mathbb {S}}^3)$, so that both multi-moment maps have the same image $\Delta$. For any $P\in \Delta$, the fibers are related by $\nu _\Gamma ^{-1}(P)=\Gamma \backslash \nu ^{-1}(P)$. Since the quotient of a torus by a finite subgroup is still a torus, the fibers have the same diffeomorphism types.