Proof. The assertion (a) was proved in [Reference Azad3]. We review the proof here. Let $\pi :{{\mathrm G}}\to {{\mathrm G}}/{{\mathrm K}}$ be the projection and consider a closed $3$-form $\phi$ on ${{\mathrm M}}$. As ${{\mathrm G}}$ is compact, we can suppose that $\phi$ is ${{\mathrm G}}$-invariant. If $\hat \phi \in \Omega ^{3}({{\mathrm G}})$ is the closed $3$-form $\pi ^{*}(\phi )$, then $\hat \phi$ is invariant under left ${{\mathrm G}}$-translations and right ${{\mathrm K}}$-translations. Since ${{\mathrm G}}$ is simple, the third cohomology group $H^{3}({{\mathrm G}},\,\mathbb {R})$ is generated by the standard $3$-form $\omega$ defined in (2.1). Thus, $\hat \phi = c\,\omega + d\xi$, for some $c\in {{\mathbb {R}}}$ and some $2$-form $\xi \in \Omega ^{2}({{\mathrm G}})$. Again by compactness, we can suppose that $\xi$ is invariant under left ${{\mathrm G}}$-translations and right ${{\mathrm K}}$-translations. This implies that, given $X,\,Y\in \frak g$ and $Z\in \mathfrak {k}$, we have

\[ \xi([Z,X],Y) + \xi(X,[Z,Y]) = 0, \]

where we see $\xi$ as an element of $\Lambda ^{2}(\frak g^{*})$. Moreover, $\imath _Z\hat \phi =0$, so that

\[ c\,\omega(Z,X,Y) ={-} d\xi(Z,X,Y) = \xi([X,Y],Z). \]

Hence

\[ c B([X,Y],Z) = \xi([X,Y],Z). \]

As ${{\mathrm G}}$ is simple, we have $\frak g= [\frak g,\,\frak g]$ and therefore for every $U\in \frak g$

(2.2)\begin{equation} \xi(U,Z) = cB(U,Z). \end{equation}

Suppose now $\frak k\neq \{0\}$ and choose a non-zero element $Z\in \frak k$. By (2.2) we have that $0=c B(Z,\,Z)$, hence $c=0$ and $\imath _{\frak k}\xi =0$. This means that $\xi$ descends to a $2$-form on ${{\mathrm M}}$ with $d\xi =\phi$. Consequently, every element in $H^{3}({{\mathrm M}},\,\mathbb {R})$ must be trivial, a contradiction.

We now prove (b). Using the same notation as in (a), suppose $0\neq [\phi ]\in H^{3}({{\mathrm M}},\,\mathbb {R})$ and consider the corresponding $3$-form $\hat \phi \in \Omega ^{3}({{\mathrm G}})$. The form $\hat \phi$ can be written as $\hat \phi = c_1\,\omega _1 + c_2\,\omega _2 + d\xi$, where $\omega _i$ is the standard $3$-form on ${{\mathrm G}}_i$, for $i=1,\,2$. If $Z\in \mathfrak {k}$, then for every $X,\,Y\in \mathfrak {g}$

\[ c_1\,\omega_1(Z,X,Y)+c_2\,\omega_2(Z,X,Y) ={-} d\xi(Z,X,Y) = \xi([X,Y],Z). \]

Hence

\[ \xi([X,Y],Z) = c_1 B_1([X,Y]_1,Z_1) + c_2 B_2([X,Y]_2,Z_2), \]

where $B_i$ is the Cartan–Killing forms on $\frak g_i$, and the subscript ${-}_i$ denotes the projection along the $i^{th}$-component $\frak g_i$. Now, suppose that, say, $p_1$ has a non-trivial kernel containing some $Z\neq 0$ such that $Z_1=0$ and $Z_2\neq 0$. As ${{\mathrm G}}$ is semisimple, we can express $Z = \sum _k[X_k,\,Y_k]$ for some vectors $X_k,\,Y_k\in \frak g$. Hence,

\[ 0 = \xi(Z,Z) = c_2B_2(Z_2,Z_2), \]

forcing $c_2=0$. If $c_1=0$, then $\hat \phi = d\xi$ and $\imath _{\mathfrak {k}}\xi = 0$, so that $\xi$ descends to an invariant $2$-form on ${{\mathrm M}}$ and $\phi$ is exact, a contradiction. Therefore $c_1\neq 0$, and for every $Z'\in \mathfrak {k}$ we have $0=\xi (Z',\,Z') = c_1B_1(Z'_1,\,Z'_1)$ so that $Z'_1=0$, implying $\mathfrak {k}\subseteq \frak g_2$.

Proof. As the $3$-form $H$ is harmonic, the $1$-form $\theta := *_gH$ is also harmonic. Therefore, by Bochner formula, we have

\[ 0=\int_{{\mathrm M}} \langle \Delta \theta,\theta\rangle\, dV_g = \int_{{\mathrm M}} \left({{{\rm Ric}}}(\theta^\#,\theta^\#) + ||\nabla^{g}\theta||^{2}\right)\, dV_g, \]

so that $\theta$ is parallel and $0={{{\rm Ric}}}(\theta ^\#,\,\theta ^\#)=\frac 14 H^{2}(\theta ^\#,\,\theta ^\#) = \frac 14 ||\imath _{\theta ^{\sharp }}H||^{2}$. This implies that the universal cover of ${{\mathrm M}}^{4}$ splits isometrically as a product $\mathrm {N}^{3}\times \mathbb {R}$, for some $3$-dimensional space $\mathrm {N}^{3}$, and that $\imath _vH=0$ whenever $v$ is tangent to the flat factor. The claim now follows from the $3$-dimensional case.

Proof. We consider the compact Lie group ${{\mathrm G}}$ given by the connected full isometry group of $({{\mathrm M}},\,g)$, and we recall that any harmonic form is invariant under the ${{\mathrm G}}$-action. We start noting that we can assume $H\neq 0$, as otherwise by [Reference Alekseevsky and Kimelfeld4] the Ricci flat metric $g$ would be flat and ${{\mathrm M}}$ would be covered by a torus.

We first study the case where ${{\mathrm G}}$ is semisimple. Let ${{\mathrm K}}$ denote the isotropy group of the ${{\mathrm G}}$-action on ${{\mathrm M}}$ at a fixed point $p$. Then, $\dim {{\mathrm K}}$ belongs to the set $\{1,\,2,\,3,\,4,\,5,\,6\}$. Indeed, the isotropy representation embeds ${{\mathrm K}}$ into ${{\mathrm {SO}}}(5)$ and ${{\mathrm K}}$ is a proper subgroup, as the standard representation of ${{\mathrm {SO}}}(5)$ has no non-trivial invariant $3$-forms. Moreover, a proper subgroup of ${{\mathrm {SO}}}(5)$ has dimension at most $6=\dim {{\mathrm {SO}}}(4)$, with equality if and only if ${{\mathrm K}}$ is conjugate to the standard ${{\mathrm {SO}}}(4)$. Again, this case can be ruled out as there are no non-trivial invariant $3$-forms.

The case $\dim {{\mathrm K}}=5$ cannot occur, as there are no $5$-dimensional subgroups of ${{\mathrm {SO}}}(5)$ (the rank is at most two: if the rank is $1$, then ${{\mathrm K}}\cong {{\mathbb {T}}}^{1}$ or ${{\mathrm {SU}}}(2)$, if the rank is $2$, then either ${{\mathrm K}}\cong {{\mathbb {T}}}^{2}$ or ${{\mathrm K}}\cong {{\mathbb {T}}}^{1}\cdot {{\mathrm {SU}}}(2)$, never with dimension $5$).

If $\dim {{\mathrm K}}=4$, then, by dimensional reasons, ${{\mathrm K}}$ and ${{\mathrm G}}$ are locally isomorphic to ${{\mathbb {T}}}^{1}\times {{\mathrm {SU}}}(2)$ and ${{\mathrm {SU}}}(2)^{3}$, respectively. We denote by $\frak z$ the centre of $\mathfrak {k}$ and by $\frak k_s\cong \mathfrak {su}(2)$ the semisimple part of $\frak k$. If $\mathfrak {g}=\oplus _{i=1}^{3}\mathfrak {g}_i$, with $\mathfrak {g}_i\cong \mathfrak {su}(2)$, and $p_i:\mathfrak {g}\to \mathfrak {g}_i$ are the projections for $i=1,\,2,\,3$, then we may suppose that $p_1(\mathfrak {z})\neq \{0\}$, so that $p_1(\mathfrak {k}_s)=\{0\}$. Since $\mathfrak {k}_s$ cannot coincide with $\mathfrak {g}_2$ or $\mathfrak {g}_3$ and since it is simple, we see that $p_i(\mathfrak {k}_s)=\mathfrak {g}_i$ and $p_i(\mathfrak {z})=\{0\}$ for $i=2,\,3$. Therefore, the manifold ${{\mathrm M}}$ is (up to a finite covering) ${{\mathrm G}}$-diffeomorphic to ${{\mathrm {SU}}}(2)/{{\mathbb {T}}}^{1}\times {{\mathrm {SU}}}(2)^{2}/\Delta {{\mathrm {SU}}}(2)\cong S^{2}\times S^{3}$. In this case, the isotropy representation contains two inequivalent modules and therefore the above diffeomorphism maps the metric $g$ to a product metric. As the $3$-form $H$ on $S^{2}\times S^{3}$ is the pull-back of a $3$-form on the $S^{3}$-factor, we see that the Ricci tensor on the $S^{2}$-factor should vanish, a contradiction.

If $\dim {{\mathrm K}} = 3$, then $\dim {{\mathrm G}}= 8$ and, being semisimple, this implies ${{\mathrm G}}\cong {{\mathrm {SU}}}(3)$, which is simple. This contradicts theorem 2.1.

The case $\dim {{\mathrm K}} = 2$ can be ruled out as there are no semisimple groups of dimension $7$. The last case $\dim {{\mathrm K}} = 1$ forces ${{\mathrm G}}\cong {{\mathrm {SU}}}(2)^{2}$ and theorem 2.1 says that ${{\mathrm K}}$ is embedded diagonally into ${{\mathrm G}}$, unless $\mathfrak {k}$ is contained in one of the factors $\mathfrak {su}(2)$. When this occurs, ${{\mathrm M}}$ is ${{\mathrm G}}$-diffeomorphic (up to a covering) to ${{\mathrm {SU}}}(2)\times S^{2}$. As the isotropy $\mathfrak {k}$ acts on the tangent space of $S^{2}$ with no non-zero invariant vector and trivially on the tangent space of ${{\mathrm {SU}}}(2)$, we see that the above splitting is isometric. Moreover, an invariant $3$-form $H$ on ${{\mathrm M}}$ is of the form $H_1+H_2$, where $H_1$ is an invariant form on ${{\mathrm {SU}}}(2)$, while $H_2 = \theta \wedge \sigma$, where $\theta$ is any invariant $1$-form on ${{\mathrm {SU}}}(2)$ and $\sigma$ is the volume form of $S^{2}$. As $dH_2=d\theta \wedge \sigma$ and $dH_1\in \Lambda ^{3}(\mathfrak {su}(2))$, we see that $dH=0$ forces $H=H_1$. This means that $\imath _vH=0$ for every vector $v$ tangent to the $S^{2}$ factor, implying that ${{\rm Ric}}_{S^{2}}\equiv 0$, a contradiction.

We now deal with the non-semisimple case. Let ${{\mathrm Z}}$ be the connected centre of ${{\mathrm G}}$ and let ${{\mathrm L}}$ be the semisimple part of ${{\mathrm G}}$, which is a compact normal subgroup. We first summarize some basic observations:

1. (1) all ${{\mathrm L}}$-orbits are diffeomorphic. Indeed, ${{\mathrm L}}$ is a normal subgroup of ${{\mathrm G}}$, so for every $p\in {{\mathrm M}}$ and $g\in {{\mathrm G}}$ we have ${{\mathrm L}}\cdot gp=g({{\mathrm L}}\cdot p)$. A generic ${{\mathrm L}}$-orbit will be denoted by $\mathcal {O}$;

2. (2) let ${{\mathrm U}} := \{ z\in {{\mathrm Z}} {\ |\ }z({{\mathrm L}}\cdot p) = {{\mathrm L}}\cdot p,\,~\forall p\in {{\mathrm M}}\}$. Then, ${{\mathrm U}}$ is a closed subgroup of the torus ${{\mathrm Z}}$ and we can find a closed subgroup ${{\mathrm Z}}_1\subseteq {{\mathrm Z}}$ with ${{\mathrm Z}}_1\cap {{\mathrm U}}=\{e\}$, ${{\mathrm Z}} = {{\mathrm U}}\cdot {{\mathrm Z}}_1$ and ${{\mathrm L}}\cdot {{\mathrm Z}}_1$ acting transitively on ${{\mathrm M}}$. Indeed, at each point $q\in {{\mathrm M}}$ we have that $T_q{{\mathrm M}} = T_q({{\mathrm Z}}_1\cdot q) \oplus T_q({{\mathrm L}}\cdot q)$. In detail, if $X\in \mathfrak {z}_1$ with $\hat X_q\in T_q({{\mathrm L}}\cdot q)$, then for every $g\in {{\mathrm G}}$ we have $\hat X_{gq}= g_*\hat X_q\in T_{gq}({{\mathrm L}}\cdot gq)$, meaning that $X\in \mathfrak {u}$, whence $X=0$;

3. (3) the manifold ${{\mathrm M}}$ is ${{\mathrm G}}$-diffeomorphic (up to a finite covering) to the product ${{\mathrm Z}}_1\times \mathcal {O}$. Indeed, the map $F : {{\mathrm Z}}_1\times \mathcal {O}\to {{\mathrm M}}$ given by $F(z,\,p)=z\cdot p$ is a local diffeomorphism, hence a covering thanks to the compactness of the involved spaces.

Since ${{\mathrm L}}$ is non-trivial, its orbits have dimension at least $2$, whence $\dim {{\mathrm Z}}_1 = {\mathrm {chm}}({{\mathrm M}},\,{{\mathrm L}})\leq 3$. We now proceed by looking at the possible dimensions of ${{\mathrm Z}}_1$:

1. (a) $\dim {{\mathrm Z}}_1 = 1$. The generic ${{\mathrm L}}$-orbit $\mathcal {O}$ has dimension $4$ and, up to a finite covering, it is ${{\mathrm L}}$-diffeomorphic to ${{\mathrm {SO}}}(5)/{{\mathrm {SO}}}(4)$, ${{\mathrm {SU}}}(3)/\mathrm {S}({{\mathrm U}}(1){{\mathrm U}}(2))$ or ${{\mathrm {SO}}}(4)/{{\mathbb {T}}}^{2}$ (cf. [Reference Bérard-Bergery5]). As the isotropy representation has no non-trivial invariant vector in the tangent space of $\mathcal {O}$, we see that the splitting in (3) is isometric. As the space of invariant $3$-forms on $\mathcal {O}$ is trivial, the form $H$ is given by $H=dt\wedge \sigma$, where $dt\in \Lambda ^{1}({{\mathbb {T}}}^{1})^{{{\mathbb {T}}}^{1}}$ and $\sigma$ is an ${{\mathrm L}}$-invariant $2$-form. Then, ${{\rm Ric}}_g(\partial _t,\,\partial _t)=0$, while $||\imath _{\partial _t}H||^{2} \neq 0$, a contradiction;

2. (b) $\dim {{\mathrm Z}}_1=2$. In this case, $\dim \mathcal {O}=3$, and ${{\mathrm L}}$ must be either ${{\mathrm {SU}}}(2)$ or ${{\mathrm {SO}}}(3)$, up to covering. In the former case, ${{\mathrm M}}$ is a Lie group up to covering; in the latter, $\mathcal {O}$ is covered by $S^{3}$ and the same argument as above shows that the splitting ${{\mathrm M}}={{\mathrm Z}}_1\times S^{3}$ is isometric. In this case, $H$ is a multiple of the volume form on the $S^{3}$ factor, as this has no non-trivial invariant $1$- or $2$-forms. Moreover, the metric splits as the product of a flat metric on ${{\mathrm Z}}_1$ and a multiple of the standard metric on $S^{3}$. Consequently, the associated Bismut connection is flat;

3. (c) $\dim {{\mathrm Z}}_1=3$. Here $\mathcal {O} \cong S^{2} = {{\mathrm {SU}}}(2)/{{\mathbb {T}}}^{1}$, so that the splitting (3) is isometric. The form $H$ can be expressed as $H=\phi \wedge \nu$, where $\phi$ is an invariant $1$-form on ${{\mathrm Z}}_1$ and $\nu$ is the volume form on $S^{2}$. Again, $||\imath _vH||^{2}\neq 0$, for every $v$ in $T{{\mathrm Z}}_1$, while ${{\rm Ric}}_g(v,\,v)=0$, a contradiction.

Proof. We begin observing that ${{\mathrm M}}_{p,q}$ does not admit any Bismut flat connection. Indeed, by [Reference Garcia-Fernández and Streets15, Thm. 3.54], a simply connected compact Riemannian manifold admitting a Bismut flat connection is isometric to a product of compact simple Lie groups with bi-invariant metrics. We obtain our claim by observing that there are no $5$-dimensional compact semisimple Lie groups.

In the Lie algebra $\mathfrak {su}(2)$ we select the elements

\[ H= \begin{pmatrix} \frac{i}{2} & 0\\ 0 & -\frac{i}{2}\end{pmatrix},\quad E = \begin{pmatrix} 0 & \frac 1{2\sqrt{2}}\\-\frac 1{2\sqrt{2}} & 0\end{pmatrix},\quad V= \begin{pmatrix} 0 & \frac i{2\sqrt{2}}\\\frac i{2\sqrt{2}} & 0\end{pmatrix}, \]

so that $[H,\,E]=V$, $[H,\,V]=-E$, $[H,\,V]= -E$ and $[E,\,V]=\frac 12 H$. If $B$ denotes the Cartan–Killing form of $\mathfrak {su}(2)$, then $B(E,\,E)=B(V,\,V)=-1$, $B(H,\,H)=-2$. Then, we can choose the following basis of $\mathfrak {g}$

\begin{align*} & e_1 = (qH,-pH), \quad e_2 = (E,0),\quad e_3 = (V,0),\\ & e_4 = (0,E),\quad e_5 = (0,V),\quad e_6 = (pH,qH), \end{align*}

so that $\mathfrak {k}_{p,q}=\mathbb {R} e_6$, while $\mathfrak {m}:= {\mathrm {span}_{{\mathbb {R}}}}(e_1,\,\ldots,\,e_5)$ is an $\textrm{ad} (\mathfrak {k}_{p,q})$-invariant subspace and $\frak g= \mathfrak {k}_{p,q}+\mathfrak {m}$ is an $\textrm{ad} (\mathfrak {k}_{p,q})$-invariant $B$-orthogonal decomposition. An $\textrm{ad} (\mathfrak {k}_{p,q})$-irreducible decomposition of $\mathfrak {m}$ is given by

\[ \mathfrak{m} = \mathfrak{m}_0 \oplus \mathfrak{m}_1 \oplus \mathfrak{m}_2, \]

where $\mathfrak {m}_0={{\mathbb {R}}}e_1$, $\mathfrak {m}_1= \mathrm {span}_{{\mathbb {R}}}(e_2,\,e_3)$ and $\mathfrak {m}_2=\mathrm {span}_{{\mathbb {R}}}(e_4,\,e_5)$. Moreover, we have

\begin{align*} & \textrm{ad}(e_6)e_1 = 0,\quad \textrm{ad}(e_6)e_2 = p\,e_3,\quad \textrm{ad}(e_6)e_3 ={-}p\,e_2,\\ & \textrm{ad}(e_6)e_4 = q\,e_5,\quad \textrm{ad}(e_6)e_5 ={-}q\,e_4, \end{align*}

thus the module $\mathfrak {m}_0$ is trivial, while the modules $\mathfrak {m}_1$ and $\mathfrak {m}_2$ are non-trivial and inequivalent if $p\neq q$, and non-trivial and equivalent if $p=q$. We will discuss the cases $p\neq q$ and $p=q$ separately.

In the following, $\mathcal {B}^{*} = (e^{1},\,e^{2},\,e^{3},\,e^{4},\,e^{5})$ denotes the dual basis of $\mathcal {B}$, and the shortening $e^{ijk\cdots }$ is used to denote the wedge product of covectors $e^{i}\wedge e^{j} \wedge e^{k} \wedge \cdots$. Moreover, we fix the orientation on $\mathfrak {m}$ for which $\mathcal {B}$ is positively oriented.

Case $\boldsymbol {p\neq q}$. We have

\[ (\Lambda^{3}\mathfrak{m}^{*})^{\mathfrak{k}_{p,q}} = \mathfrak{m}_0^{*}\otimes \Lambda^{2}\mathfrak{m}_1^{*} \oplus \mathfrak{m}_0^{*}\otimes \Lambda^{2}\mathfrak{m}_2^{*}, \]

and a generic invariant 3-form is given by

\[ H = h_1 e^{123} + h_2 e^{145}, \]

for some $h_1,\,h_2\in \mathbb {R}$. Using the Koszul formula for the differential of invariant forms on $\mathfrak {m}$

(3.1)\begin{equation} dH(X_0,X_1,X_2, X_3) = \sum_{i< j} ({-}1)^{i+j}\ H \left([X_i,X_j]_{\mathfrak{m}},X_k,X_l\right),\quad X_0,\ldots, X_3\in \mathfrak{m}, \end{equation}

where $[X_i,\,X_j]_{\mathfrak {m}}$ denotes the projection of $[X_i,\,X_j]$ onto $\mathfrak {m}$, we see that $H$ is closed if and only if $(h_1,\,h_2) = (\lambda q,\,\lambda p)$, for some $\lambda \in {{\mathbb {R}}}$. Now, the generic invariant metric on $\mathfrak {m}$ has the following expression

\[ g = \mu^{2} e^{1}\odot e^{1} + a^{2} \left(e^{2}\odot e^{2} + e^{3} \odot e^{3} \right) + b^{2} \left(e^{4}\odot e^{4} + e^{5} \odot e^{5} \right), \]

for some positive real numbers $\mu,\,a,\,b$, where $e^{i}\odot e^{j} := \frac 12(e^{i}\otimes e^{j} + e^{j} \otimes e^{i})$. We can then easily compute the Hodge dual of $H$ obtaining

\[ *_gH = \frac{a^{2}}{\mu\,b^{2}}\,h_2\,e^{23} + \frac{b^{2}}{\mu\,a^{2}}\,h_1\,e^{45}, \]

and we see that it is always closed. Thus, up to a constant multiple, the generic invariant harmonic 3-form on $\mathfrak {m}$ is given by

\[ H = q\,e^{123} + p\, e^{145}. \]

Now, we compute the symmetric $2$-tensor $H^{2}$ and we see that its non-zero components are the following

\begin{align*} H^{2}(e_1,e_1) & = 2\, \frac{a^{4}p^{2}+b^{4}q^{2}}{a^{4}b^{4}},\\ H^{2}(e_2,e_2) & = 2\, \frac{q^{2}}{a^{2}\mu^{2}} = H^{2}(e_3,e_3), \\ H^{2}(e_4,e_4) & = 2\, \frac{p^{2}}{b^{2}\mu^{2}} = H^{2}(e_5,e_5). \end{align*}

As for the Ricci tensor ${{\rm Ric}}_g$, we choose a $g$-orthonormal basis $(E_1,\,E_2,\,E_3,\,E_4,\,E_5)$ of $\mathfrak {m}$, and we compute its components with respect to the basis $\mathcal {B}$ using the formula [Reference Besse6, 7.38]. Since $\mathfrak {g}$ is unimodular, this formula reads

(3.2)\begin{equation} {{\rm Ric}}_g(X,X) ={-}\frac 12 \sum_{i=1}^{5} ||[X,E_i]_{\mathfrak{m}}||^{2} -\frac 12 B(X,X) +\frac 12 \sum_{1\leq i < j \leq 5} g([E_i,E_j]_{\mathfrak{m}},X)^{2}, \end{equation}

for every $X\in \mathfrak {m}$. Since

\[ [e_2,e_3]_{\mathfrak{m}} = \frac{q}{2\left(p^{2}+q^{2}\right)}\,e_1,\quad [e_4,e_5]_{\mathfrak{m}} ={-}\frac{p}{2\left(p^{2}+q^{2}\right)}\,e_1, \]

and $[e_i,\,e_j]_{\mathfrak {m}} = [e_i,\,e_j]$ otherwise, we obtain that the only non-zero components of ${{\rm Ric}}_g$ are the following

\begin{align*} {{\rm Ric}}_g(e_1,e_1) & = \mu^{4}\frac{a^{4}p^{2} + b^{4}q^{2}}{8 a^{4} b^{4} \left(p^{2} + q^{2}\right)^{2}}, \\ {{\rm Ric}}_g(e_2,e_2) & = \frac{4a^{2}\left(p^{2} + q^{2}\right)^{2} - \mu^{2} q^{2}}{8a^{2} \left(p^{2} + q^{2}\right)^{2}} = {{\rm Ric}}_g(e_3,e_3), \\ {{\rm Ric}}_g(e_4,e_4) & = \frac{4b^{2}\left(p^{2} + q^{2}\right)^{2} - \mu^{2} p^{2}}{8b^{2} \left(p^{2} + q^{2}\right)^{2}} = {{\rm Ric}}_g(e_5,e_5). \end{align*}

Now, it is easy to see that ${{\rm Ric}}_g = \frac 14 H^{2}$ has a unique solution under the constraints $\mu >0,\,a>0,\,b>0$ and $p^{2}+q^{2}\neq 0$, namely

\[ \mu = \sqrt{2\left(p^{2}+q^{2}\right)},\quad a = \sqrt{\frac{q^{2}}{p^{2}+q^{2}}},\quad b = \sqrt{\frac{p^{2}}{p^{2}+q^{2}}}. \]

Thus, when $p\neq q$, the homogeneous space ${{\mathrm M}}_{p,q}={{\mathrm G}}/{{\mathrm K}}_{p,q}$ admits an invariant metric $g_o$ and an invariant harmonic form $H_o$ giving rise to a Bismut Ricci flat connection. Moreover, the pair $(g_o,\,H_o)$ is unique up to scaling.

Case $\boldsymbol {p= q}$. We have $p=q=1$ and the modules $\mathfrak {m}_1$ and $\mathfrak {m}_2$ are equivalent. The space of $\textrm{ad} (\mathfrak {k}_{1,1})$-invariant 3-forms is given by

\[ (\Lambda^{3}\mathfrak{m}^{*})^{\mathfrak{k}_{1,1}} = \mathfrak{m}_0^{*}\otimes \Lambda^{2}\mathfrak{m}_1^{*} \oplus \mathfrak{m}_0^{*}\otimes \Lambda^{2}\mathfrak{m}_2^{*} \oplus \left(\mathfrak{m}_0^{*} \otimes \mathfrak{m}_1^{*}\otimes \mathfrak{m}_2^{*}\right)^{\mathfrak{k}_{1,1}}, \]

with $\dim (\mathfrak {m}_0^{*} \otimes \mathfrak {m}_1^{*}\otimes \mathfrak {m}_2^{*})^{\mathfrak {k}_{1,1}} = 2$.

The generic invariant 3-form $H\in (\Lambda ^{3}\mathfrak {m}^{*})^{\mathfrak {k}_{1,1}}$ has the following expression

(3.3)\begin{equation} H = h_1\,e^{123} + h_2\,e^{145} + h_3 \left(e^{125}-e^{134}\right) + h_4\left(e^{124} + e^{135} \right), \end{equation}

where $h_i\in {{\mathbb {R}}}$, for $i=1,\,2,\,3,\,4$. Using the Koszul formula (3.1), we see that $H$ is closed if and only if $h_2=h_1$.

The generic invariant metric on $\mathfrak {m}$ is given by

(3.4)\begin{equation} \begin{aligned} g & = \mu^{2} e^{1}\odot e^{1} + a^{2} \left(e^{2}\odot e^{2} + e^{3} \odot e^{3} \right) + b^{2} \left(e^{4}\odot e^{4} + e^{5} \odot e^{5} \right) \\ & \quad + 2c\left(e^{2}\odot e^{4} + e^{3} \odot e^{5}\right) +2s\left(e^{2}\odot e^{5} - e^{3}\odot e^{4}\right), \end{aligned} \end{equation}

where $\mu,\, a,\, b,\, c,\, s$ are real constants such that

\[ \mu>0,\quad a>0,\quad b>0,\quad a^{2}b^{2}-c^{2} + s^{2}>0. \]

We remark here that it is always possible to assume $s=0$. Indeed, the one-dimensional torus ${{\mathrm U}}:= \exp (\mathbb {R} e_1)$ centralizes ${{\mathrm K}}_{1,1}$ and therefore for every $u\in {{\mathrm U}}$ we can consider the ${{\mathrm G}}$-equivariant diffeomorphism $\tau _u\in {\rm {Dif{}f}}({{\mathrm M}})^{{\mathrm G}}$ given by $\tau _u(x{{\mathrm K}}_{1,1}) = xu{{\mathrm K}}_{1,1}$, for $x\in {{\mathrm G}}$. Then, for every $g\in S^{2}(\mathfrak {m})^{\mathfrak {k}_{1,1}}$, which is given by the data $(\mu,\,a,\,b,\,c,\,s)$ as in (3.4), the ${{\mathrm G}}$-invariant symmetric tensor $\tau _u^{*}g$ corresponds to the data $(\mu,\,a,\,b,\,c'=c\,\cos t + s\, \sin t,\,s'= -c\, \sin t + s\, \cos t)$, for $u=\exp (te_1)\in {{\mathrm U}}$. Therefore, we immediately see that we have $s'=0$ for an appropriate choice of $u\in {{\mathrm U}}$.

Now, we have

\begin{align*} *_gH & = \frac{h_1\left(a^{4}+ c^{2}\right) - 2a^{2}c h_3 }{\mu\left(a^{2}b^{2} - c^{2}\right)}\, e^{23} + \frac{ h_1\left(b^{4} + c^{2}\right) - 2b^{2}ch_3}{\mu\left(a^{2}b^{2} - c^{2}\right)}\, e^{45}\\ & \quad -\frac{h_4}{\mu}\, \left(e^{24}+e^{35}\right) - \frac{h_3\left( a^{2} b^{2} + c^{2}\right) -c h_1\left( a^{2} + b^{2}\right) }{\mu\left(a^{2}b^{2} - c^{2}\right)} \left(e^{25}-e^{34}\right), \end{align*}

and using again the Koszul formula, we obtain

\[ d*_gH = 2\,\frac{h_4}{\mu}\left(e^{125} - e^{134}\right) -2\,\frac{h_3\left(a^{2} b^{2} +c^{2}\right) -c h_1\left( a^{2} + b^{2}\right) }{\mu\left(a^{2}b^{2} - c^{2}\right)}\left(e^{124} + e^{135} \right). \]

Thus, $H$ is coclosed if and only if

\[ h_3 = \frac{c h_1\left( a^{2} + b^{2}\right)}{a^{2} b^{2} +c^{2} }, \quad h_4 = 0. \]

The generic invariant harmonic 3-form on $\mathfrak {m}$ is then given by

(3.5)\begin{equation} H = h_1\left( e^{123} + e^{145} + c\,\frac{ a^{2} + b^{2} }{a^{2} b^{2} +c^{2} } \left(e^{125}-e^{134}\right) \right). \end{equation}

Consequently, the (possibly) non-zero components of the invariant symmetric $2$-tensor $H^{2}$ are the following

\begin{align*} H^{2}(e_1,e_1) & = 2 h_1^{2}\, \frac{a^{4} + b^{4} - 2c^{2}}{\left(a^{2}b^{2} - c^{2}\right)\left(a^{2}b^{2} + c^{2}\right)},\\ H^{2}(e_2,e_2) & = 2 h_1^{2}\, \frac{a^{2}c^{2} \left(a^{4} + b^{4}\right) -c^{4} \left(2a^{2} + b^{2}\right) + a^{4}b^{6}}{\mu^{2}\left(a^{2}b^{2} - c^{2}\right)\left(a^{2}b^{2} + c^{2}\right)^{2}} = H^{2}(e_3,e_3),\\ H^{2}(e_4,e_4) & = 2 h_1^{2}\, \frac{b^{2} c^{2} \left(a^{4} + b^{4}\right) -c^{4}\left(a^{2} + 2b^{2}\right) + a^{6}b^{4}}{\mu^{2}\left(a^{2}b^{2} - c^{2}\right)\left(a^{2}b^{2} + c^{2}\right)^{2}} = H^{2}(e_5,e_5), \\ H^{2}(e_2,e_4) & = 2 c h_1^{2}\, \frac{ a^{2}b^{2}\left( a^{4} + a^{2} b^{2} + b^{4} - 2 c^{2}\right) - c^{4}}{\mu^{2}\left(a^{2}b^{2} - c^{2}\right)\left(a^{2}b^{2} + c^{2} \right)^{2}} = H^{2}(e_3,e_5). \end{align*}

Let us consider the following $g$-orthonormal basis of $\mathfrak {m}$

\begin{align*} E_1 & = \frac{1}{\mu}\,e_1,\quad E_i = \frac{1}{a}\,e_i,\quad i=2,3,\\ E_j & ={-}\frac{c}{a\sqrt{a^{2}b^{2}-c^{2}}}\,e_{j-2} + \frac{a}{\sqrt{a^{2}b^{2}-c^{2}}}\,e_j,\quad j=4,5. \end{align*}

Then, using formula (3.2) and its polarization, and observing that

\[ [e_2,e_3]_{\mathfrak{m}} = \frac{1}{4}\,e_1,\quad [e_4,e_5]_{\mathfrak{m}} ={-}\frac{1}{4}\,e_1,\quad [e_i,e_j]_{\mathfrak{m}} = [e_i,e_j] \mbox{ otherwise,} \]

we obtain the following expressions for the (possibly) non-zero components of the Ricci tensor

\begin{align*} {{\rm Ric}}_g(e_1,e_1) & = \frac{ 2c^{2}\left( 64 c^{2} - 64 a^{2} b^{2} - \mu^{4}\right) + \mu^{4} (a^{4} + b^{4})}{32 \left(a^{2} b^{2} - c^{2}\right)^{2}}, \\ {{\rm Ric}}_g(e_2,e_2) & = \frac{ 64a^{2}c^{2} + \mu^{2}\left(16a^{2}b^{2} -b^{2}\mu^{2} - 16c^{2}\right) }{32 \mu^{2} \left(a^{2}b^{2} - c^{2}\right)} = {{\rm Ric}}_g(e_3,e_3),\\ {{\rm Ric}}_g(e_4,e_4) & = \frac{64b^{2}c^{2} +\mu^{2} \left( 16a^{2}b^{2} -a^{2}\mu^{2} - 16c^{2}\right)}{32\mu^{2}\left(a^{2}b^{2} - c^{2}\right)} = {{\rm Ric}}_g(e_5,e_5), \\ {{\rm Ric}}_g(e_2,e_4) & = c\, \frac{64a^{2}b^{2} - \mu^{4}}{32\mu^{2}\left(a^{2}b^{2} - c^{2}\right)}= {{\rm Ric}}_g(e_3,e_5). \end{align*}

We now look for invariant metrics $g$ and harmonic 3-forms $H$ for which ${{\rm Ric}}_g = \frac 14 H^{2}$. A computation using the above expressions of the components of ${{\rm Ric}}_g$ and $H^{2}$ shows that we need to find points in the open subset

\[ \mathcal{A} := \left\{(\mu,a,b,c,h_1) \in {{\mathbb{R}}}^{5} {\ |\ } \mu>0, a>0, b>0, a^{2}b^{2}-c^{2}>0, h_1\neq 0 \right\} \subset {{\mathbb{R}}}^{5}, \]

where all of the following polynomials vanish

\begin{align*} p_1 & = \left(a^{4}+b^{4}-2 c^{2}\right) \left[\mu^{4} \left(a^{2} b^{2}+c^{2}\right)-16\,h_{1}^{2} \left(a^{2}b^{2} -c^{2} \right) \right] -128\, c^{2} \left(a^{4}b^{4} -c^{4} \right), \\ p_2 & = \left(a^{2} b^{2}+c^{2}\right)^{2} \left[16\,\mu^{2} \left(a^{2} b^{2}-c^{2}\right) +64 a^{2} c^{2} -b^{2} \mu^{4}\right]\\& \quad -16 h_{1}^{2} \left[a^{4} b^{6}+ c^{2}\left(a^{6} +a^{2} b^{4} -2 a^{2} c^{2}-b^{2} c^{2}\right)\right], \\ p_3 & = \left(a^{2} b^{2}+c^{2}\right)^{2} \left[16\,\mu^{2} \left(a^{2} b^{2}-c^{2}\right) +64 b^{2} c^{2} -a^{2} \mu^{4}\right]\\& \quad-16 h_{1}^{2} \left[a^{6} b^{4}+ c^{2} \left(b^{6} +a^{4} b^{2} -2 b^{2} c^{2} -a^{2} c^{2} \right)\right], \\ p_4 & = c \left[\left(a^{2} b^{2}+c^{2}\right)^{2} \left(64 a^{2} b^{2}-\mu^{4}\right)-16 h_{1}^{2} \left(a^{2}b^{2} \left(a^{2} + b^{2}\right)^{2}-\left(a^{2} b^{2}+c^{2}\right)^{2}\right)\right]. \end{align*}

Clearly, the polynomial $p_4$ vanishes if $c=0$. When this happens, the remaining polynomials simplify considerably, and they vanish simultaneously if and only if

\[ \mu= 2\sqrt{2}\,t,\quad a = b = t,\quad h_1 ={\pm} 2t^{2}, \]

for some $t>0$. Therefore, when $c=0$, we obtain an invariant BRF pair $(g_o,\,H_o) \in \mathcal {B}({{\mathrm M}}_{1,1})^{{\mathrm G}}$ that is unique up to scaling and up to a sign in the definition of $H_o$, namely

\[ g_o = 8\, e^{1}\odot e^{1} + \left(e^{2}\odot e^{2} + e^{3} \odot e^{3} \right) + \left(e^{4}\odot e^{4} + e^{5} \odot e^{5} \right), \quad H_o = 2e^{123} + 2e^{145}. \]

To prove the last assertion, we first consider the set

\[ \mathcal{P} := \left\{(g,H)\in S^{2}(\mathfrak{m})^{\mathfrak{k}_{1,1}}\times \Lambda^{3}(\mathfrak{m})^{\mathfrak{k}_{1,1}} {\ |\ }g >0,~dH=0,~\delta_gH=0\right\}, \]

and we recall that every invariant metric is uniquely determined by the string $(\mu,\,a,\,b,\,c,\,s)$ as in (3.4). We then consider the subset

\[ \mathcal{P}' := \left\{(g,H)\in \mathcal{P}|\ g(e_2,e_5)=0\right\}. \]

Every point $(g,\,H)\in \mathcal {P}'$ is uniquely determined by a string $(\mu,\,a,\,b,\,c,\,h_1)\in \mathcal {A} \subset \mathbb {R}^{5}$, where $(\mu,\,a,\,b,\,c)$ determines $g$ as in (3.4) and $h_1$ determines $H$ as in (3.5). The set $\mathcal {B}({{\mathrm M}}_{1,1})^{{{\mathrm G}}} \cap \mathcal {P}'$ can be identified with the zero set in $\mathcal {A}$ of the map

\[ F : \mathcal{A} \rightarrow {{\mathbb{R}}}^{4},\quad F(\mu,a,b,c,h_1) = (p_1,p_2,p_3,p_4). \]

Notice that $F(x_o)=0$, where $x_o=(2\sqrt {2},\,1,\,1,\,0,\,2)\in \mathcal {A}$ corresponds to the BRF pair $(g_o,\,H_o)$, and that $F(\gamma (t))=0\in \mathbb {R}^{4}$ for every $t\in {{\mathbb {R}}}^{+}$, where $\gamma (t) = (2\sqrt {2}t,\,t,\,t,\,0,\,2t^{2})$ is the curve through $x_o$ corresponding to $(t^{2}g_o,\,t^{2}H_o)$. If we now compute the differential of $F$ at $x_o$, we obtain

\[ F_{*x_o} = \left( \begin{array}{@{}ccccc@{}} 128 \sqrt{2} & 0 & 0 & 0 & -128 \\ 0 & 256 & 0 & 0 & -64 \\ 0 & 0 & 256 & 0 & -64 \\ 0 & 0 & 0 & -192 & 0 \end{array} \right), \]

and since $\mathrm {rank} (F_{*x_o} ) = 4,$ we see that there exists a neighbourhood $\mathcal {U}'$ of $(g_o,\,H_o)$ in $\mathcal {P}'$ so that $\mathcal {U}' \cap \mathcal {B}({{\mathrm M}}_{1,1})^{{{\mathrm G}}}$ coincides with the set $\left \{(t^{2}g_o,\,t^{2}H_o) {\ |\ }t\in (1-\varepsilon,\,1+\varepsilon ')\right \}$, for suitable $\varepsilon,\,\varepsilon ' >0$.

We now use the ${{\mathrm G}}$-equivariant diffeomorphisms $\tau _u$, $u\in {{\mathrm U}}$, to transform every element of $\mathcal {B}({{\mathrm M}}_{1,1})^{{{\mathrm G}}}\cap \mathcal {P}$ into an element of $\mathcal {B}({{\mathrm M}}_{1,1})^{{{\mathrm G}}}\cap \mathcal {P}'$ via $\tau _u^{*}$. Note that for every $u\in {{\mathrm U}}$ we have $(\tau _u^{*}g_o,\,\tau _u^{*}H_o)=(g_o,\,H_o)$, since $c=s=0$. Moreover, we can easily see that there exists a neighbourhood $\mathcal {U}$ of $(g_o,\,H_o)$ in $\mathcal {P}$ so that $\tau _u^{*}(\mathcal {U})\subseteq \mathcal {U}'$. Therefore, if $(g,\,H)\in \mathcal {B}({{\mathrm M}}_{1,1})^{{{\mathrm G}}}\cap \mathcal {U}$, then we can find $u\in {{\mathrm U}}$ so that $\tau _u^{*}(g,\,H)\in \mathcal {B}({{\mathrm M}}_{1,1})^{{{\mathrm G}}}\cap \mathcal {U}'$. Hence, $\tau _u^{*}(g,\,H)= (t^{2}g_o,\,t^{2}H_o)$, for some $t > 0$, and we have $(g,\,H)=\tau _{u^{-1}}^{*}(t^{2}g_o,\,t^{2}H_o) = (t^{2}g_o,\,t^{2}H_o)$.

Proof. Let

\[ \overline M := 2\lambda(p^{2}+q^{2}), \]

and let $I=[t_o,\,T)$ be the maximal existence interval for the solution $(M_t,\,A_t,\,B_t)$. We divide the discussion into various steps. step 1. If $M_t$ assumes the value $\overline M$ for some $\overline {t} \in I$, then $M_t=\overline M$ for all $t\in I$, $T=+\infty$ and $\lim _{t\to +\infty }A_t= q^{2}F(\overline M)$, $\lim _{t\to +\infty }B_t= p^{2}F(\overline M)$. Indeed, let $(\overline A_t,\,\overline B_t)$ be the solution of the system

(4.2)\begin{equation} \left\{ \begin{aligned} \frac{d}{dt} A_t & = \frac{q^{2}}{A_t} F(\overline M) - 1,\\ \frac{d}{dt} B_t & = \frac{p^{2}}{B_t} F(\overline M) - 1. \end{aligned} \right. \end{equation}

with initial conditions $(\overline A_{t_o}= A_{t_o},\,\overline B_{t_o}= B_{t_o})$. Then, $(\overline A_t,\,\overline B_t,\, M_t=\overline M)$ satisfies the system (4.1) and therefore we have $M_t=\overline M$ for every $t\in I$ by uniqueness. Now, it is clear that $q^{2}F(\overline M)$ and $p^{2}F(\overline M)$ are particular solutions of the first and second equation of (4.2), respectively. Therefore, either $A_t$, $B_t$ are constant, or their derivatives never vanish. This implies that $A_t$ and $B_t$ remain bounded from above and from below by two positive constants. Then, $|A_t'|$ and $|B_t'|$ are uniformly bounded, and this implies the long time existence and the convergence by standard arguments.

step 2. $M_t$ is monotone and for every $t\in I$ we have either $M_t\in [M_o,\,\overline M]$ or $M_t\in [\overline M,\,M_o]$. In fact, $M_t'$ never vanishes unless $M_t$ is constant. Therefore, $M_t$ is increasing (resp. decreasing) if $M_o<\overline M$ (resp. $M_o>\overline M$). In particular, the limit $\lim _{t\to T}M_t$ exists.

step 3. We have $T=+\infty$. Suppose $T<+\infty$. We consider the first equation in (4.1). First of all, we note that $A_t$ is bounded on $I$. Indeed, by step 2, $C_1 \leq F(M_t) \leq C_2$ on $I$ for some positive constants $C_1,\,C_2$, hence $A_t' \leq \frac {q^{2}{C_2}}{A_t}$, and thus $(A_t^{2})'\leq q^{2}C_2$ and $A_t\leq \sqrt {q^{2}C_2t+A_o^{2}}\leq C_3:= \sqrt {q^{2}C_2T+A_o^{2}}$ for all $t\in I$. In turn this implies that $A_t'\geq \frac {q^{2}{C_1}}{C_3}-1$, so that $|A_t'|$ is uniformly bounded. Consequently, the limit $\lim _{t\to T}A_t$ exists, call it $\alpha \geq 0$. We need to prove that $\alpha \neq 0$. Let us put $y_t=A_t^{2}$, so that $y' = -2\sqrt {y} + 2q^{2}F(M_t)$. If $\alpha =0$, then $\lim _{t\to T}y'(t) = \lim _{t\to T}q^{2}F(M_t) >0$. This is impossible as it implies that $y(t)<0$ in a suitable left neighbourhood of $T$. The same argument applies to $B_t$ and our claim follows.

step 4. If there exists a constant $K>0$ so that $A_t'\leq -1 + \frac {K}{A_t}$, then for every $\epsilon >0$ there exists some $T_\epsilon$ such that

\[ A_t\leq K+\epsilon,\quad\forall\, t\geq T_\epsilon. \]

The same statement holds for $B_t$. As a consequence, the functions $A_t$ and $B_t$ are bounded on $[t_o,\,+\infty )$. If we put $y(t)=A_t^{2}$ as before, then we have $y'\leq -2\sqrt {y} + 2\,K$. First, we note that the inequality $A_t> K+\epsilon$ cannot hold on a half-line $[\tilde {t},\,+\infty )$. Indeed, in that case we would have $y'(t)\leq -2\epsilon$, hence for some constant $C$ we would obtain $y(t)\leq -2\epsilon t+C<0$ for sufficiently large $t$, a contradiction. Thus, there exists a point $\overline {t}$ where $A_{\overline {t}} \leq K+\epsilon$. We claim that $A_t\leq K+\epsilon$ for every $t\geq \overline {t}$. Indeed, suppose there exists $t_1> \overline {t}$ with $A_{t_1}>K+\epsilon$ and consider $t_2\in [\overline {t},\,t_1]$ so that $A_{t_2} = \max _{[\overline {t},\,t_1]}A_t$. If $t_2\in (\overline {t},\,t_1)$, then $0 = A_{t_2}'\leq -1 + \frac {K}{A_{t_2}}$, so that $A_{t_2}\leq K$, a contradiction. Otherwise, $t_2=t_1$ and therefore $A_{t_1}'\geq 0$ implying again $A_{t_1}\leq K$. This proves our first claim. By step 2, $M_t$ takes values into an interval $[C_1,\,C_2]$ with $C_1>0$ and therefore $F(M_t)$ is bounded above by a constant $K$. By the previous arguments, $A_t$ (and similarly $B_t$) is bounded on some positive half-line, hence on $[t_o,\,+\infty )$.

step 5. We have $\lim _{t\to +\infty } M_t = \overline M$. Indeed, $M_t$ is monotone and bounded by step 2, hence the limit $\lim _{t\to +\infty } M_t =m$ exists. Suppose that $m \neq \overline M$. By step 4, the functions $A_t,\,B_t$ are bounded, hence by the first equation in (4.1) there is a constant $C$ so that for sufficiently large $t$

\[ |M_t'| \geq C |\overline M^{2} - M_t^{2}| \geq \frac C2 |\overline M^{2}-m^{2}| >0, \]

contradicting the boundedness of $M_t$.

step 6. We have $\lim _{t\to +\infty }A_t = q^{2} F(\overline M)$ and $\lim _{t\to +\infty }B_t = p^{2} F(\overline M)$. We know that $\lim _{t\to +\infty } M_t = \overline M$, hence $q^{2}F(M_t) \to q^{2}F(\overline M) =: \phi >0$. This means that for every $\epsilon >0$ there exists $S_\epsilon >0$ so that $\phi -\epsilon \leq q^{2}F(M_t)\leq \phi +\epsilon$ for all $t\geq S_\epsilon$. By step 4, there exists $T_\epsilon$ so that for every $t\geq T_\epsilon$ we have $A_t\leq \phi +2\epsilon$. Using similar arguments as in the proof of step 4, we have another point (say again $T_\epsilon$), so that $A_t\geq \phi -2\epsilon$ for all $t\geq T_\epsilon$. Our claim for $A_t$ is then proved, and the same arguments apply for $B_t$.