Homogeneous heterotic supergravity solutions with linear dilaton

Given a coset G/K, we need to determine whether the spin representation over $\mathfrak m=\mathfrak g/\mathfrak k$ contains the trivial $\mathfrak k$-representation as an irreducible component. Suppose for the moment that $\mathfrak k$ is simple, and denote the set of weights of $\mathfrak m$ by $\Omega (\mathfrak m)$, whereas $\Omega ^+(\mathfrak m)$ contains only the positive ones. Then, the weights that appear in $S(\mathfrak m)$ are of the form

$$\begin{equation} \Omega (S(\mathfrak m)) = \bigg \lbrace \frac{1}{2}\sum _{\alpha \in \Omega ^+(\mathfrak m)} \pm \alpha \bigg \rbrace , \end{equation} \tag{ 4.1 }$$

where all combinations of signs appear (this can be understood from lemma 4.1 below). Now only the dominant weights can be highest weights of an irreducible representation of $S(\mathfrak m)$, so it is often enough to determine all the dominant weights in $\Omega (S(\mathfrak m))$. Then, a couple of situations can occur. If the zero weight is not in $\Omega (S(\mathfrak m))$, then the trivial representation is not contained. If it is, one can sometimes conclude by dimensional reasoning that the trivial representation must or must not occur as a component. Sometimes the situation is even simpler.

Consider the coset SO(n + 1)/SO(n) = Sⁿ. In this case, $\mathfrak m$ is simply the fundamental representation of $\mathfrak {so}(n)$, and it follows that $S(\mathfrak m)$ is the (Dirac) spinor representation, which is often reducible, but does not have invariant elements. Thus, Sⁿ with its standard round metric does not admit a homogeneous solution, in accordance with our general result for symmetric spaces.

Despite its elegance, we will not employ the representation-theoretic method in the following, but use the more down-to-earth approach explained in the next paragraph. The reason for that is that the latter method allows us to determine how the three-form H acts on invariant spinors, and thereby how to choose the dilaton appropriately to also solve the dilatino equation $\gamma ({\rm d}\phi -\frac{1}{12} H)\varepsilon =0$. A drawback is that the method is not always applicable, as explained below.

Recall that the action of $\mathfrak k$ on $\mathfrak m$ defines an embedding $\mathfrak k\subset \mathfrak {so}(\mathfrak m)$. Here we give an explicit construction of the spinor space for the case that $\mathfrak k\subset \mathfrak {su}(\mathfrak m)$ for a well-chosen complex structure on $\mathfrak m$. We assume for the time being that rk$(\mathfrak k)<$ rk$(\mathfrak g)$, although for certain maximal rank subgroups we will be able to generalize our construction. A particular example where this is possible is the case where G/K is a six-dimensional nearly Kähler manifold.

A further assumption we want to make is that $\mathfrak k$ and $\mathfrak g$ admit a common root space decomposition, i.e. they are of the form

$$\begin{eqnarray} \mathfrak g\otimes \mathbb C &=& \mathfrak h \bigoplus _{\alpha \in R^+} (\mathfrak g_\alpha \oplus \mathfrak g_{-\alpha }), \\ \ms \mathfrak k \otimes \mathbb C &=& {\rm span}\lbrace H_{p+1},\dots , H_r \rbrace \bigoplus _{\alpha \in S^+} ( \mathfrak g_\alpha \oplus \mathfrak g_{-\alpha }), \end{eqnarray} \tag{ 4.2 }$$

where the Cartan algebra of $\mathfrak g$ is

$$\begin{equation} \mathfrak h = {\rm span}\lbrace H_1,\dots , H_r \rbrace \end{equation} \tag{ 4.3 }$$

with r = rk$(\mathfrak g)>$ rk$(\mathfrak k)=r-p$, and the set of positive roots of $\mathfrak g$ is denoted by R⁺, whereas those of $\mathfrak k$ are contained in S⁺⊂R⁺. The complement $\mathfrak m$ of $\mathfrak k$ in $\mathfrak g$ is then given by

$$\begin{equation} \mathfrak m \otimes \mathbb C= {\rm span}\lbrace H_{1},\dots ,H_p\rbrace \bigoplus _{\alpha \in R^+\setminus S^+} ( \mathfrak g_\alpha \oplus \mathfrak g_{-\alpha }), \end{equation} \tag{ 4.4 }$$

which is usually not a Lie algebra. Our final assumption is that the roots in R⁺∖S⁺ are higher than those in S⁺ (which roots are highest, i.e. more positive than others, is a convention, but it is not always possible to choose the roots in this way; typically if rk$(\mathfrak k)=$ rk$(\mathfrak g)$ then this ordering is not possible). By restriction, we will consider roots of $\mathfrak g$ as roots of $\mathfrak k$ as well. Further we assume a basis of root vectors E_α of the $\mathfrak g_\alpha$ and E_−α of the $\mathfrak g_{-\alpha }$ to be chosen, obeying the commutation relation

$$\begin{equation} [E_\alpha ,E_\beta ] =\left\lbrace \begin{array}{@{}ll@{}}2\sum _j\alpha (H_j)H_j & {\rm if }\ \alpha =-\beta \\ N_{\alpha \beta } E_{\alpha +\beta } & {\rm if }\ \alpha +\beta \ {\rm is\ a\ root}\ \\ 0 & {\rm otherwise}, \end{array}\right. \end{equation} \tag{ 4.5 }$$

where N_αβ are constants, and the normalization can be chosen as [19]

$$\begin{equation} g(H_i,H_j)=\delta _{ij},\qquad g(E_\alpha ,E_\beta ) = 2\delta _{\alpha ,-\beta },\qquad g(H_i,E_\alpha )=0. \end{equation} \tag{ 4.6 }$$

Now extend the metric of $\mathfrak m$ to one on $\mathfrak m^{\prime }:=\mathfrak m \oplus \mathbb R^p$ by using the standard metric on $\mathbb R^p$. Furthermore, we extend the action of $\mathfrak k$ on $\mathfrak m$ trivially to one on $\mathfrak m\oplus \mathbb R^p$, so that we still have

$$\begin{equation} \mathfrak k \subset \mathfrak {so}(\mathfrak m^{\prime }), \end{equation} \tag{ 4.7 }$$

and the spin representation ${\rm d}S: \mathfrak {so}(\mathfrak m^{\prime }) \rightarrow \mathfrak {spin}(\mathfrak m^{\prime })$ embeds $\mathfrak k$ into the spin algebra. A complex structure J on $\mathfrak m^{\prime }$ is obtained by defining holomorphic vectors to be given by the positive root vectors E_α, α ∈ R⁺∖S⁺, as well as the following combinations of Cartan vectors H_j and standard basis vectors ∂_j of $\mathbb R^p$:

$$\begin{equation} E_j = H_j - i \partial _j,\qquad j= 1,\dots ,p, \end{equation} \tag{ 4.8 }$$

whereas anti-holomorphic ones are given by the negative root vectors E_−α and the $\overline{E}_j = H_j + {\rm i} \partial _j$. The H_j are assumed to be elements of the real Lie algebra $\mathfrak g$, so that roots wrt them are purely imaginary. From the commutation relation (4.5) and our ordering of the roots, we deduce the following relations:

$$\begin{eqnarray} {[}\mathfrak k,\mathfrak m^+]&\subset& \mathfrak m^+,\qquad \quad [\mathfrak k,\mathfrak m^-]\subset \mathfrak m^-, \nonumber\\ {[}\mathfrak m^+,\mathfrak m^+] &\subset& \mathfrak m^+,\qquad [\mathfrak m^-,\mathfrak m^-]\subset \mathfrak m^-, \end{eqnarray} \tag{ 4.9 }$$

where $\mathfrak m^+$ consists of holomorphic vectors in $\mathfrak m^{\prime }\otimes \mathbb C$, and $\mathfrak m^-$ of anti-holomorphic ones. Thus, our subalgebra $\mathfrak k$ commutes with the complex structure, and is contained in the unitary subalgebra of $\mathfrak {so}(\mathfrak m^{\prime })$:

$$\begin{equation} \mathfrak k \subset \mathfrak u(\mathfrak m^{\prime },J). \end{equation} \tag{ 4.10 }$$

It follows from this that the complex structure J extends to an almost-complex structure on the manifold $\mathbb R^p\times G/K$, which is just a different way of saying that its structure group is contained in $U(\mathfrak m^{\prime })$. Let e^α be the holomorphic one-forms dual to the E_α, and e^j be dual to E_j, and denote by W the space of all holomorphic one-forms on $\mathfrak m^{\prime }$. Then, the spinor space is the exterior algebra

$$\begin{equation} S= \Lambda W. \end{equation} \tag{ 4.11 }$$

The Clifford algebra Cl$(\mathfrak {m^{\prime }}^*)$ is the quotient of the tensor algebra over ${\mathfrak m^{\prime }}^*$ by the relation

$$\begin{equation} v\otimes w - w\otimes v = 2g^{-1}(v,w), \qquad \forall v,w \in {\mathfrak m^{\prime }}^*, \end{equation} \tag{ 4.12 }$$

where g⁻¹ is the metric induced by g on the dual ${\mathfrak m^{\prime }}^*$. As a vector space, the Clifford algebra is isomorphic to the exterior algebra $\Lambda \mathfrak {m^{\prime }}^*$, but not as an algebra of course. The vector space isomorphism is just the quantization map γ. Cl$(\mathfrak {m^{\prime }}^*)$ acts on S as follows; associate with every holomorphic one-form e^α an operator ζ^α := γ(e^α), an operator $\overline{\zeta }^\alpha =\gamma ({\rm e}^{-\alpha })$ to every anti-holomorphic one-form e^−α, as well as ζ^j = γ(e^j) to e^j and $\overline{\zeta }^j=\gamma (\overline{e}^j)$ to $\overline{e}^j$. They act on S = ΛW as creation and annihilation operators:

$$\begin{eqnarray} \zeta ^a \cdot (w^{b_1}\wedge \cdots \wedge w^{b_r}) &=& w^a \wedge w^{b_1} \wedge \cdots \wedge w^{b_r},\nonumber\\ \overline{\zeta }^a \cdot ( w^{b_1}\wedge \cdots \wedge w^{b_r}) &=& \sum _{i=1}^ r (-1)^{i-1} \delta ^{a b_i} w^{b_1} \wedge \cdots \wedge \breve{w}^{b_i} \wedge \cdots \wedge w^{b_r}, \nonumber \end{eqnarray} \tag{ 4.13 }$$

where $\breve{w}^{b_i}$ means leaving out the element, and the w^a are elements of W. This action can be uniquely extended to one of the full Clifford algebra. Inside the Clifford algebra Cl(2n) := Cl($\mathbb R^{2n})$ sits the spin algebra $\mathfrak {spin}(2n)$, which is the image of the map

$$\begin{equation} {\rm d}S : \mathfrak {so}(2n) \rightarrow {\rm Cl}(2n),\quad A\mapsto \case{1}{4} g_{ac}{A^c}_b \gamma ^a \gamma ^b, \end{equation} \tag{ 4.14 }$$

where γ^a = γ(dx^a). One can check that this action of the orthogonal Lie algebra restricts to the usual action of $\mathfrak {su}(n)$ on $\Lambda \mathbb C^{n*}$, in particular, the decomposition

$$\begin{equation*} S =\oplus _l S^l =\mathbb C \oplus \mathbb C^{n*} \oplus \Lambda ^2 \mathbb C^{n*} \oplus \cdots \oplus \Lambda ^n \mathbb C^{n*} \end{equation*}$$

is preserved by this subalgebra, and then even by $\mathfrak u(n)$, although the additional $\mathfrak u(1)$-component acts in a non-standard way. To show that $\mathfrak k$ leaves at least one spinor invariant, it is therefore enough to show that it is contained in $\mathfrak {su}(\mathfrak m^{\prime })$, as this acts trivially on S⁰⊕Sⁿ. We can easily determine the action of the Cartan generators of $\mathfrak k$ on spinors (recall $\widetilde{{\rm ad}}(X)=$ dS ○ ad(X) for $X\in \mathfrak k$).

Lemma 4.1. We have for the $H_j\in \mathfrak k$, j = p + 1, ..., r

$$\begin{equation} \widetilde{{\rm ad}}(H_j) = \frac{1}{2} \sum _{\alpha \in R^+\setminus S^+} \alpha (H_j)( \overline{\zeta }^\alpha \zeta ^\alpha - \zeta ^\alpha \overline{\zeta }^\alpha ). \end{equation} \tag{ 4.15 }$$

Proof. According to the definition of dS, we get

$$\begin{eqnarray*} \widetilde{{\rm ad}}(H_j) & =\case{1}{4} g_{ac} {(H_j )^c}_b\gamma ^a\gamma ^b \nonumber \\ &= \case{1}{4} g_{\alpha \overline{\gamma }} {(H^j)^{\overline{\gamma }}}_{\overline{\beta }} \zeta ^\alpha \overline{\zeta }^\beta + \case{1}{4} g_{\overline{\beta }\gamma } {(H^j)^{\gamma }}_{\alpha } \overline{\zeta }^\beta \zeta ^\alpha , \end{eqnarray*}$$

where we introduced [H_j, E_α] ≕ (H_j)^β_αE_β. But this just means that (H_j)^β_α = δ^β_αα(H_j) and ${(H_j)^{\overline{\beta }} }_{\overline{\alpha }} = -{\delta ^\beta }_\alpha \alpha (H_j)$. Furthermore, $g_{\alpha ,\overline{\beta }}=g_{\alpha ,-\beta }=2\delta _{\alpha \beta }$.

Proposition 4.2. Under the assumptions of this section suppose that

$$\begin{equation} \sum _{\alpha \in R^+\setminus S^+} \alpha =0 \end{equation} \tag{ 4.16 }$$

as roots of $\mathfrak k$. Then, $\mathfrak k$ acts trivially on the completely empty and completely filled states

$$\begin{equation*} S^0 \subset S\qquad {\rm and} \qquad S^{m} \subset S, \end{equation*}$$

where $m=\dim (\mathfrak m^{\prime })$.

Proof. We know already that $\mathfrak k\subset \mathfrak u(\mathfrak m^{\prime })$ leaves invariant the decomposition S = ⊕_lS^l. The subrepresentations on S⁰ and S^m are one dimensional, and according to lemma 4.1 the Cartan elements of $\mathfrak k$ act by multiplication with

$$\begin{equation*} \widetilde{{\rm ad}}(H_j)\Big |_{S^0} = \frac{1}{2} \sum _{\alpha \in R^+\setminus S^+} \alpha (H_j), \qquad \widetilde{{\rm ad}}(H_j)\Big |_{S^{m}} = -\frac{1}{2} \sum _{\alpha \in R^+\setminus S^+} \alpha (H_j). \end{equation*}$$

If these vanish, the representations are trivial.

A closer inspection shows that the trivial representation occurs with multiplicity at least 2^{p + 2}, as $\mathfrak k$ acts trivially on all of the holomorphic vectors E_j. The requirement that the supersymmetry generator in heterotic supergravity be Majorana–Weyl will reduce the amount of supersymmetry preserved in the space orthogonal to $\mathbb R^p\times G/K$ to $\mathcal N=2^{p}$, whereas the dilatino equation reduces it even further.

A different way to write condition (4.16) is in terms of the Weyl vector $\rho =\frac{1}{2} \sum _{\alpha \in R^+}\alpha$. Then, it reads

$$\begin{equation} \rho _{\mathfrak g} \big |_{\mathfrak k}=\rho _\mathfrak k. \end{equation} \tag{ 4.17 }$$

As we mentioned already, it is equivalent to $\mathfrak k\subset \mathfrak {su}(\mathfrak m)$. The condition cannot be satisfied by a subalgebra of maximal rank, as this would imply $\sum _{\alpha \in R^+\setminus S^+} \alpha =0$ as roots of $\mathfrak g$, contradicting the positivity of all the roots in R⁺∖S⁺. In this case, one can sometimes choose the complex structure in a different way, i.e. such that holomorphic vectors do not coincide with positive root vectors, and apply the procedure presented here analogously. This will be demonstrated for G₂/SU(3) below.

Summarizing the results of this section, we have proven that given a naturally reductive subalgebra $\mathfrak k\subset \mathfrak g$ of lower rank, admitting a common root space decomposition with $\mathfrak g$ of the form (4.2) such that the roots not belonging to $\mathfrak k$ are all higher than those of $\mathfrak k$, and such that their sum acts trivially on the Cartan generators of $\mathfrak k$, it admits invariant spinors over $\mathfrak m\oplus \mathbb R^p$, where p = rk($\mathfrak g)-$ rk($\mathfrak k$).

With lemma 3.3, we conclude that there are invariant, and thereby ∇⁻-parallel spinors over $\mathbb R^p \times G/K$. Furthermore, we have a very explicit construction of these spinors, which will allow us to also solve the dilatino equation in this case.

For the dilatino equation, we also need to know how the three-form H acts on the invariant spinors, i.e. we have to determine γ(H)ε for ε ∈ S⁰⊕S^m. Using the commutation relations (4.5), we can determine H:

$$\begin{eqnarray} H&=& -\frac{1}{6} f_{abc} \,{\rm e}^{abc} \nonumber\\ &=& 2\sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} \alpha (H_j) \,{\rm e}^{-\alpha }\wedge \,{\rm e}^{\alpha } \wedge h^j \\ &&- 2\sum _{\underset{\scriptstyle {\alpha <\beta }}{\alpha ,\beta \in R^+\setminus S^+}} [N_{\alpha \beta } \,{\rm e}^{\alpha }\wedge e^\beta \wedge \,{\rm e}^{-(\alpha +\beta )} +N_{-\alpha ,-\beta } \,{\rm e}^{-\alpha }\wedge \,{\rm e}^{-\beta }\wedge \,{\rm e}^{\alpha +\beta }], \nonumber \end{eqnarray} \tag{ 4.18 }$$

with h^j dual to H_j. Under the quantization map γ the last line eliminates S⁰ and S^m, and therefore we only need to consider

$$\begin{equation} H^{\prime } = 2\sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} \alpha (H_j) \,{\rm e}^{-\alpha }\wedge \,{\rm e}^{\alpha } \wedge h^j . \end{equation} \tag{ 4.19 }$$

By construction the one-form h^j is quantized to

$$\begin{equation} \gamma (h^j) = \zeta ^j + \overline{\zeta }^j, \end{equation} \tag{ 4.20 }$$

and we conclude that

$$\begin{equation} \gamma (H^{\prime }) =6 \sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} \alpha (H_j)(\overline{\zeta }^\alpha \zeta ^\alpha - \zeta ^\alpha \overline{\zeta }^\alpha ) (\zeta ^j+\overline{\zeta }^j), \end{equation} \tag{ 4.21 }$$

so that

$$\begin{eqnarray} &&\gamma (H) |_{\mathcal S^0} \ :\ \ \mathcal S^0 \rightarrow \mathcal S^1,\ 1\mapsto 6\sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} \alpha (H_j) w^j ,\nonumber\\ &&\gamma (H)|_{\mathcal S^m} \ :\ \ \mathcal S^m\rightarrow \mathcal S^{m-1}, \\ &&w^1\wedge \dots \wedge w^m\mapsto 6\sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} (-1)^j\alpha (H_j) w^1\wedge \cdots \wedge \breve{w}^j\wedge \cdots \wedge w^m.\nonumber \end{eqnarray} \tag{ 4.22 }$$

Therefore, the dilatino equation $\gamma ({\rm d}\phi - {\textstyle \frac{1}{12}}H) \varepsilon =0$ is not satisfied if we choose ϕ = 0 and ε ∈ S⁰⊕S^m. To overcome this problem we can introduce a linear dilaton on the additional $\mathbb R^p$-factor. Take the real valued

$$\begin{equation} \phi (x) = \frac{i}{2}\sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} \alpha (H_j) x^j. \end{equation} \tag{ 4.23 }$$

Its differential acts on $\mathcal S$ as

$$\begin{equation} \gamma ({\rm d}\phi ) = {\frac{1}{2}}\sum _{\underset{j=1,\dots ,p}{\alpha \in R^+\setminus S^+}} \alpha (H_j) ( \zeta ^j-\overline{\zeta }^j), \end{equation} \tag{ 4.24 }$$

which implies

$$\begin{equation} \gamma ({\rm d}\phi )\Big |_{\mathcal S^0\oplus \mathcal S^m} = \case{1}{12} \gamma (H)\Big |_{\mathcal S^0\oplus \mathcal S^m}. \end{equation} \tag{ 4.25 }$$

How do we then construct the supersymmetry generator ε? For the whole construction to be useful in heterotic string theory, the dimension of $\mathbb R^p \times G/K$ must be less than 10, i.e. we need

$$\begin{equation} l:= \dim G-\dim K + {\rm rk}\,G - {\rm rk}\,K <10; \end{equation} \tag{ 4.26 }$$

then, we complete this to a ten-dimensional Lorentzian manifold $M= \mathbb R^{9-l+p ,1} \times G/K$. The spinor bundle over M is the tensor product of the one over $\mathbb R^{l-1,1}$ and the one over $\mathbb R^p \times G/K$, and there is an (anti-linear) charge conjugation C: S → S mapping S⁰ to S^m times a complement which does not affect the action of forms over $\mathbb R^p \times G/K$. Let η be any constant, positive chirality spinor on $\mathbb R^{l-1,1}$. Then, ε is chosen proportional to 1⊗η + C(1⊗η) with $1 \in S^0\simeq \mathbb C$, and this is parallel wrt ∇⁻, annihilated by $\gamma ({\rm d}\phi -\frac{1}{12} H)$, and Majorana–Weyl, as required by heterotic string theory.

There is a little subtlety related to the amount of supersymmetry preserved. Suppose that the dimension of $\mathbb R^p\times G/K$ is such that it admits Majorana spinors, and that $\mathcal S^0$ is invariant. Then, there are at least two invariant Majorana spinors, given by

$$\begin{equation} 1 + C\cdot 1 \qquad {\rm and} \qquad {\rm i}(1-C\cdot 1). \end{equation} \tag{ 4.27 }$$

These generate just one supersymmetry however, as two Majorana spinors η + Cη and i(η − Cη) over $\mathbb R^{l-1,1}$ tensored with our invariant spinors lead to the same set of spinors over the total ten-dimensional space.

What is then the amount of supersymmetry preserved in $\mathbb R^{9-l,1}$? Consider all those spinors that are obtained from the completely empty one by acting with an arbitrary number of creation operators ζ^j, j = 1, ..., p. These generate a complex 2^p-dimensional space $\tilde{S}=\tilde{S}^0\oplus \cdots \oplus \tilde{S}^p$, which is invariant under γ(H) and γ(dϕ), and we chose ϕ such that $\tilde{S}^0=S^0$ is annihilated by $\gamma ({\rm d}\phi - \frac{1}{12} H)$. On the higher subspaces $\tilde{S}^k$ the condition $\gamma ({\rm d}\phi - \frac{1}{12} H)\psi =0$ becomes one linear equation, so that generically there should be 2^p − p invariant elements. Indeed one can check that $\tilde{S}^p$ is not invariant, whereas $\tilde{S}^2$ has one invariant spinor, and so on. We conclude that our solutions have

$$\begin{equation} \mathcal N\ge 2^p-p \end{equation} \tag{ 4.28 }$$

supersymmetry, with equality if there are no additional invariant spinors over $\mathbb R^p\times G/K$ to those we constructed explicitly. In particular for p = 0 and p = 1 this leads to $\mathcal N\ge 1$, and these are the cases we will consider in the examples below. For equal rank subgroups, we will have to add a factor $\mathbb R$ by hand on which the linear dilaton lives, and the orthogonal space will be $\mathbb R^{8-\dim G+\dim K,1}$.

The simplest examples are the ones with K the trivial group. In this case, the existence of invariant spinors is trivial and need not be checked. The manifold on which our fields live is $\mathbb R^p \times G$ with p = rk(G), and the dilaton is a linear function on $\mathbb R^p$. What is further special about this case is that $H^3(G,\mathbb Z) = \mathbb Z$, so in order to satisfy the condition $[H]\in H^3(M,4\pi ^2 \alpha ^{\prime }\mathbb Z)$ we have to adjust the scale of G, whereas the Bianchi identity is trivially satisfied, with dH = 0 and R⁺ = R⁻ = 0. For cosets G/K on the other hand, the third cohomology may well be trivial, and it is the Bianchi identity which fixes the scale.

The models we obtain here are the low-energy description of Wess–Zumino–Witten models and well known. SU(2) is the simplest of the group manifolds, with rank 1. Thus, we have to consider $\mathbb R\times$SU(2). Choose a basis I₁, I₂, I₃ of the Lie algebra $\mathfrak {su}(2)$ satisfying [I_i, I_j] = _ij^kI_k. Then, there is a corresponding basis of left-invariant vector fields on SU(2), which we denote by I_{1, 2, 3} again, and a dual basis of left-invariant one-forms e¹, e², e³. The metric on $\mathbb R\times$SU(2) is given by

$$\begin{equation} g = {\rm d}x\otimes {\rm d}x + \sum _j e^j \otimes e^j, \end{equation} \tag{ 6.2 }$$

the H-field becomes proportional to the volume form of SU(2):

$$\begin{equation} H = -e^1\wedge e^2 \wedge e^3, \end{equation} \tag{ 6.3 }$$

and the dilaton is

$$\begin{equation} \phi (x) = -\case{1}{2} x. \end{equation} \tag{ 6.4 }$$

Contrary to the Lévi-Cività connection, both connections ∇⁺ and ∇⁻ are flat, i.e. R⁺ = R⁻ = 0, and the Bianchi identity dH = 0 holds. So far all of this remains true for higher-dimensional groups as well. In this particular case, one can check that besides the supersymmetry equations also the heterotic equations of motion are satisfied (and the higher order corrections in α' vanish). This is not true for simple groups in general, but should not be expected either for groups of dimension ⩾8. The amount of supersymmetry preserved in $\mathbb R^{5,1}$ is $\mathcal N=1$.

The only condition we did not check so far was the quantization condition on H. In order to calculate the cohomology class of H, we pull it back to S³ along the map

$$\begin{equation} \iota :S^3\rightarrow SU(2),\ (\eta ,\xi _1,\xi _2)\mapsto \left( \begin{array}{@{}cc@{}}\rm e^{{\rm i}\xi _1}\sin (\eta ) & \,{\rm e}^{{\rm i}\xi _2}\cos (\eta ) \\ -{\rm e}^{-{\rm i}\xi _2}\cos (\eta ) & {\rm e}^{-{\rm i}\xi _1}\sin (\eta )\\ \end{array}\right) \end{equation} \tag{ 6.5 }$$

(where $\eta \in [0,\frac{\pi }{2}),\ \xi _1,\xi _2\in [0,2\pi )$) to calculate its integral

$$\begin{equation*} \int _{SU(2)} H =\int _{S^3}\iota ^\ast H=-16 \pi ^2. \end{equation*}$$

Thus, the cohomology class of H is

$$\begin{equation} [H] = -16\pi ^2\in H^3\big ({\rm SU(2)};\mathbb R\big )\simeq \mathbb R. \end{equation} \tag{ 6.6 }$$

To satisfy the integrality constraint, we have to rescale the metric as g' = μg, with

$$\begin{equation} \mu = \frac{\alpha ^{\prime } n}{4},\qquad n\in \mathbb N, \end{equation} \tag{ 6.7 }$$

which automatically rescales H' = μH (as H is defined in terms of the metric), and leaves invariant the supersymmetry equations. There is thus no continuous volume modulus in the game, as is well known for WZW models. For higher-dimensional simple groups, the analysis is very similar.

Here the difference of the ranks is always 1, so that we get a model based on $\mathbb R\times S^{2n+1}$, where the metric on the sphere is different from the round one [9]. A Cartan basis of $\mathfrak {su}(n+1)$ is given by

$$\begin{eqnarray} \fl \lambda _1&=&\left({{\begin{array}{@{}cccccccc@{}}i & && && &&\\ & -i && && &&\\ & && 0&& &&\\ & && && \ddots && \\ & && && && 0 \end{array}}}\right), \\ \ms \lambda _2&=&\left({{\begin{array}{@{}cccccccccc@{}}0 && & && && && \\ && i & && && &&\\ && & -i && && &&\\ && & && 0&& &&\\ && & && && \ddots && \\ && & && && && 0 \end{array}}}\right), \dots \ \lambda _n=\left({{\begin{array}{@{}cccccccc@{}}0 && && && &\\ && \ddots && && &\\ && && 0&& &\\ && && && i & \\ && && && & -i \end{array}}}\right) \end{eqnarray} \tag{ 6.8 }$$

with positive root vectors E_ij for j > i and (E_ij)_ab = δ_aiδ_bj. The corresponding negative root vectors are −E_ji. We have

$$\begin{equation} {[}\lambda _a,E_{ij}] = i[E_{a,a}-E_{a+1,a+1},E_{ij}] = i(\delta _{ai}-\delta _{a+1,i}-\delta _{a,j} + \delta _{a+1,j} )E_{ij}, \end{equation} \tag{ 6.9 }$$

so that the positive roots with respect to λ₁, ..., λ_n are

$$\begin{equation} \alpha _{ab} = i(0,\dots ,0,\underset{(a-1)}{-1},\underset{(a)}{+1},0,\dots ,0,\underset{(b-1)}{+1},\underset{(b)}{-1},0,\dots ,0)_\lambda , \end{equation} \tag{ 6.10 }$$

with 1 ⩽ a < b ⩽ n + 1 (entries beyond the n components of the root are dropped). The Cartan generators λ_a are not orthogonal though a better basis is given by

$$\begin{equation} H_l = \frac{1}{\sqrt{l(l+1)}} \sum _{a=1}^l a\lambda _a, \end{equation} \tag{ 6.11 }$$

or more explicitly

$$\begin{eqnarray} \fl H_1 &=& \frac{1}{\sqrt{2}}\left(\begin{array}{@{}cccc@{}}{\rm i} & & & \\ & -{\rm i} & &\\ & & & \\ & & & \end{array} \right),\qquad \\ \ms H_2 &=& \frac{1}{\sqrt{6}}\left(\begin{array}{@{}ccccc@{}}{\rm i} & & & & \\ & {\rm i} & & &\\ & & -2{\rm i} & & \\ & & & &\\ & & & & \end{array} \right),\ \dots , H_n = \frac{1}{\sqrt{n(n+1)}}\left(\begin{array}{@{}cccc@{}}{\rm i} & & & \\ & \ddots & &\\ & & {\rm i} & \\ & & & -n{\rm i} \end{array} \right). \end{eqnarray} \tag{ 6.12 }$$

The roots wrt these generators are

$$\begin{eqnarray} &&\fl \alpha _{ab}={\rm i}\Bigg (0,\dots ,0,\underset{(a-1)}{-\sqrt{\frac{a-1}{a}}},\frac{1}{\sqrt{a(a+1)}},\frac{1}{\sqrt{(a+1)(a+2)}},\dots \nonumber\\ &&\dots , \underset{(b-2)}{\frac{1}{\sqrt{(b-2)(b-1)}}},\underset{(b-1)}{\sqrt{\frac{b}{b-1}}},0,\dots ,0\Bigg )_H. \end{eqnarray} \tag{ 6.13 }$$

The upper-left corner $\mathfrak {su}(n)$ subalgebra of $\mathfrak {su}(n+1)$ has Cartan generators H₁, ..., H_{n − 1}, whereas H_n generates the orthogonal torus. The roots belonging to $\mathfrak m$ are

$$\begin{equation} R^+\setminus S^+ = \lbrace \alpha _{a,n+1}\ |\ a=1,\dots ,n\rbrace . \end{equation} \tag{ 6.14 }$$

Their action on the Cartan generators of $\mathfrak {su}(n)$ is as follows. To every H_a with 1 ⩽ a < n we have one root α_{a + 1, n + 1} and a roots α_{j, n + 1} with 1 ⩽ j ⩽ a such that

$$\begin{equation} \alpha _{a+1,n+1}(H_a)=-{\rm i}\sqrt{\frac{a}{a+1}},\qquad \alpha _{j,n+1}(H_a) = \frac{i}{\sqrt{a(a+1)}}. \end{equation} \tag{ 6.15 }$$

Therefore, the relation

$$\begin{equation} \sum _{\alpha \in R^+\setminus S^+} \alpha \ \Big |_{\mathfrak {su}(n)} =0 \end{equation} \tag{ 6.16 }$$

holds, and the conditions of section 4 are satisfied. Thus, we can introduce a linear dilaton on $\mathbb R$ and find an invariant spinor (in both S⁰ and S^{n + 1}) satisfying all of the supersymmetry conditions, as well as the Bianchi identity. The amount of supersymmetry preserved is $\mathcal N=1$, which follows from the fact that $\mathfrak {su}(n)\subset \mathfrak {spin}(2n)$ has exactly two invariant Majorana spinors (if they exist), and $\mathfrak {so}(\mathfrak m)= \mathfrak {so}(2n+1)$, with $\mathfrak {su}(n)$ embedded in the upper-left $\mathfrak {so}(2n)$ in the standard way. As we have no relations between three roots of $\mathfrak m$, the three-form is simply

$$\begin{equation} H \sim \ \sum _{a=1}^n \,{\rm e}^{\alpha _{a,n+1}} \wedge \,{\rm e}^{-\alpha _{a,n+1}}\wedge h, \end{equation} \tag{ 6.17 }$$

with conventions explained at the beginning of the section.

For the spheres S^{4n + 3} we have this alternative representation as cosets, again with rank-difference 1. Here $\mathfrak {sp}(n)$ denotes the compact real form of the Lie algebra C_n, not the symplectic Lie algebra which is a non-compact real form. The definition is

$$\begin{equation} \mathfrak {sp}(n)=\lbrace X \in \mathbb H^{n\times n}\ |\ X^\dagger +X=0\rbrace , \end{equation} \tag{ 6.18 }$$

where $\mathbb H$ are the quaternions, and † denotes the (quaternion) conjugate transpose. The quaternion conjugate is explicitly given by

$$\begin{equation*} (a+ib+jc+kd)^\ast = a-ib-jc-kd. \end{equation*}$$

A matrix satisfying X† + X = 0 has only imaginary entries on the diagonal and the off-diagonal entries are completely determined by the upper-triangular part, so that the real dimension of $\mathfrak {sp}(n)$ is 3n + 2n(n − 1) = 2n² + n. A set of Cartan generators is given by the diagonal matrices H_a, a = 1, ..., n, with entry i at the (a, a)-th position, and zeros everywhere else. In particular, the rank is n. The root space decomposition takes the following form. Besides the Cartan generators, we have the basis elements

$$\begin{eqnarray} Q_a &=&{\rm diag}(0,\dots ,0,j,0,\dots , 0),\\ \ms P_a &=& {\rm diag}(0,\dots ,0,k,0,\dots ,0) \end{eqnarray} \tag{ 6.19 }$$

for a = 1, ..., n, and

$$\begin{eqnarray} E_{rs} = \left(\begin{array}{@{}cccc@{}}0 & & & \\ & \ddots &1 & \\ & -1 & \ddots & \\ & & & 0 \end{array}\right) ,\qquad I_{rs} = \left(\begin{array}{@{}cccc@{}}0 & & & \\ & \ddots &i & \\ & & \ddots & \\ & i & & 0 \end{array}\right) , \\ \ms J_{rs} = \left(\begin{array}{@{}cccc@{}}0 & & & \\ & \ddots &j & \\ & j & \ddots & \\ & & & 0 \end{array}\right) ,\qquad K_{rs} = \left(\begin{array}{@{}cccc@{}}0 & & & \\ & \ddots & k & \\ & k & \ddots & \\ & & & 0 \end{array}\right) , \end{eqnarray} \tag{ 6.20 }$$

i.e. (E_rs)_ab = δ_arδ_bs − δ_asδ_br, etc for 1 ⩽ r < s ⩽ n. In the following, we will consider the complexification of $\mathfrak {sp}(n)$, and denote the imaginary unit in this space by i, whereas the quaternionic number denoted by i so far will no longer occur explicitly. We have the commutation relations

$$\begin{eqnarray} {[}H_a, Q_b \mp {\rm i} P_{b}]& = (\pm {\rm i}) 2\delta _{ab} (Q_b \mp {\rm i}P_b),\nonumber \\ {[}H_a, E_{rs} \mp {\rm i} I_{rs}] &= (\pm {\rm i})\big (\delta _{ar}-\delta _{as}\big )(E_{rs}\mp {\rm i} I_{rs}),\\ {[}H_a, J_{rs}\mp {\rm i}K_{rs}&= (\pm {\rm i})\big (\delta _{ar} + \delta _{as}\big )(J_{rs}\mp {\rm i}K_{rs})\nonumber \end{eqnarray} \tag{ 6.21 }$$

and choose the positive roots to be

$$\begin{eqnarray} \alpha _b(H_a) &= \sqrt{2}{\rm i} \delta _{ab},\qquad \qquad \quad b=1,\dots ,n,\nonumber \\ \beta _{rs}(H_a) &= \frac{{\rm i}}{\sqrt{2}}\big (\delta _{ar}+\delta _{as}\big ) ,\qquad 1\le r<s\le n, \\ \gamma _{rs}(H_a) &= \frac{{\rm i}}{\sqrt{2}}\big (\delta _{as}-\delta _{ar}\big ),\qquad 1\le r<s\le n,\nonumber \end{eqnarray} \tag{ 6.22 }$$

wrt the properly normalized Cartan generator $H^{\prime }_j= \frac{1}{\sqrt{2}}H_j$.

The coset. Again we choose the embedding of $\mathfrak {sp}(n)$ into the upper-left corner of $\mathfrak {sp}(n+1)$. Then, the orthogonal torus is generated by H_{n + 1}, and the 2n + 1 positive root vectors of $\mathfrak m$ are

$$\begin{eqnarray} &&Q_{n+1} - {\rm i} P_{n+1},\qquad E_{r,n+1} + {\rm i} I_{r,n+1} ,\qquad J_{r,n+1} - {\rm i} K_{r,n+1} , \end{eqnarray} \tag{ 6.23 }$$

with corresponding positive roots

$$\begin{equation} R^+\setminus S^+ = \lbrace \alpha _{n+1}, \gamma _{r,n+1}, \beta _{r,n+1}\ |\ r=1,\dots ,n \rbrace . \end{equation} \tag{ 6.24 }$$

The action of these roots on the Cartan generators H_a (a = 1, ..., n) of $\mathfrak k$ is

$$\begin{equation} \fl \alpha _{n+1}(H_a)=0,\qquad \beta _{r,n+1}(H_a) = -\frac{{\rm i}}{\sqrt{2}}\delta _{ar},\qquad \gamma _{r,n+1}(H_a) = \frac{{\rm i}}{\sqrt{2}}\delta _{ar} \end{equation} \tag{ 6.25 }$$

from which we read off that

$$\begin{equation*} \sum _{\alpha \in R^+\setminus S^+} \alpha \ \Big |_{\mathfrak {sp}(n)} =0 \end{equation*}$$

is again satisfied. There are n + 1 invariant spinors, as $\mathfrak {so}(\mathfrak m) =\mathfrak {so}(4n+1)$, with the standard embedding of $\mathfrak {sp}(n)$ into the upper left $\mathfrak {so}(4n)$. They lead to $\mathcal N=\frac{n}{2} +1$ SUSY for even n, and $\mathcal N=\frac{n+1}{2}$ SUSY for odd n. The three-form is somewhat more complicated now, due to the relations

$$\begin{equation} \beta _{r,n+1} + \gamma _{r,n+1} = \alpha _{n+1} \end{equation} \tag{ 6.26 }$$

between roots in R⁺∖S⁺. We get

$$\begin{eqnarray} &&\fl H \, \sim \ 2 \,{\rm e}^{\alpha _{n+1}} \wedge \,{\rm e}^{-\alpha _{n+1}} \wedge h +\sum _{r=1}^n ({\rm e}^{\beta _{r,n+1}} \wedge \,{\rm e}^{-\beta _{r,n+1}} \wedge h + \,{\rm e}^{\gamma _{r,n+1}} \wedge \,{\rm e}^{-\gamma _{r,n+1}} \wedge h ) \nonumber\\ &&-\sqrt{2}{\rm i} \sum _{r=1}^n ( N_{\beta ,\gamma } \,{\rm e}^{\beta _{r,n+1}} \wedge \,{\rm e}^{\gamma _{r,n+1}} \wedge \,{\rm e}^{-\alpha _{n+1}} + N_{-\beta ,-\gamma } \,{\rm e}^{-\beta _{r,n+1}} \wedge \,{\rm e}^{-\gamma _{r,n+1}} \wedge \,{\rm e}^{\alpha _{n+1}}), \nonumber\\ \end{eqnarray} \tag{ 6.27 }$$

where the constants N_{α, β} are defined by [E_α, E_β] = N_{α, β}E_{α + β} for E_α the root vector corresponding to α.

We use the notation of the last example. Here the roots γ_rs, corresponding to matrices with entries $1,i\in \mathbb H$, belong to $\mathfrak {su}(n)$, whereas the other ones with entries j and k belong to $\mathfrak m$. The Cartan basis of $\mathfrak {su}(n)$ is spanned by the elements λ_i = H_{i + 1} − H_i for i = 1, ..., n − 1, and there is an orthogonal torus generated by ∑_iH_i. Adding up the roots of $\mathfrak m$, we get

$$\begin{equation} \sum _{a=1}^n \alpha _a + \sum _{1\le a<b\le n} \beta _{ab} = {\rm i}\frac{n+1}{\sqrt{2}}\big (1,\dots ,1\big ) \end{equation} \tag{ 6.28 }$$

in the basis H₁, ..., H_n, and this vanishes on the Cartan algebra of $\mathfrak {su}(n)$, so that our condition is satisfied. The amount of supersymmetry could be larger than $\mathcal N=1$; we have $\dim \mathfrak m= n^2+n+1$, and $\mathcal N$ depends on the number of invariant spinors for the corresponding embedding $\mathfrak {su}(n)\subset \mathfrak {spin}(n^2+n)$. There are no relations involving only three roots of $\mathfrak m$, so that the three-form becomes

$$\begin{equation} H\sim \sum _{r<s} \,{\rm e}^{\beta _{rs}}\wedge \,{\rm e}^{-\beta _{rs}}\wedge h + \sum _{r} \,{\rm e}^{\alpha _{r}}\wedge \,{\rm e}^{-\alpha _{r}}\wedge h . \end{equation} \tag{ 6.29 }$$

Again the rank difference one is 1. We define $\mathfrak {su}(n)$ as the subalgebra of $\mathfrak {so}(2n)$ leaving the subspace S⁰⊂S in the spin representation of $\mathfrak {so}(2n)$ invariant. Upon introduction of the canonical complex structure on $\mathbb R^{2n}$ the subalgebra

$$\begin{equation} \mathfrak {su}(n) = \lbrace X\in \mathfrak {so}(2n)\ |\ {\rm d}S(X)\cdot S^0=0\rbrace \end{equation} \tag{ 6.30 }$$

is determined by the equations

$$\begin{equation} X_{12} +X_{34} +\cdots +X_{2n-1,2n}=0 \end{equation} \tag{ 6.31 }$$

(where X_ab = δ_acX^c_b) and

$$\begin{equation} X_{2a-1,2b-1} = X_{2a,2b},\qquad X_{2a-1,2b}= - X_{2a,2b-1} \end{equation} \tag{ 6.32 }$$

for 1 ⩽ a < b ⩽ n. Denote by E_rs (1 ⩽ r < s ⩽ 2n) the standard basis of $\mathfrak {so}(2n)$, with (E_rs)_ab = δ_raδ_sb − δ_saδ_rb. A Cartan basis of $\mathfrak {so}(2n)$ is given by λ_a = E_{2a − 1, 2a} for a = 1, ..., n or

$$\begin{equation} \lambda _1=\left({{\begin{array}{@{}cc@{}}0 & 1\\ -1& 0\\ \end{array}}}\right), \dots ,\quad \lambda _n=\left({{\begin{array}{@{}cc@{}}0 & 1\\ -1& 0 \end{array}}}\right). \end{equation} \tag{ 6.33 }$$

A basis of $\mathfrak {su}(n)$ is given by the Cartan generators

$$\begin{equation} H_a = \lambda _{2a-1} - \lambda _{2a},\qquad a=1,\dots , n-1, \end{equation} \tag{ 6.34 }$$

and combinations of the type (1 ⩽ a < b ⩽ n)

$$\begin{equation} A^\pm _{ab}=E_{2a-1,2b-1} \pm {\rm i} E_{2a,2b-1} \mp {\rm i}(E_{2a-1,2b} \pm {\rm i} E_{2a,2b}), \end{equation} \tag{ 6.35 }$$

where all signs are correlated. They satisfy

$$\begin{equation} [\lambda _c, A_{ab}^\pm ] = \pm {\rm i} (\delta _{ac} -\delta _{bc})A^\pm _{ab}. \end{equation} \tag{ 6.36 }$$

The Cartan basis H₁, ..., H_{n − 1} can be extended to one of $\mathfrak {so}(2n)$ by adding $H_n= \sqrt{\frac{2}{n}}\sum _{a=1}^n \lambda _a$, which gives rise to an orthogonal torus again. A basis of $\mathfrak m$ is given by H_n plus

$$\begin{equation} B^\pm _{ab} = E_{2a-1,2b-1} \pm {\rm i} E_{2a,2b-1} \pm {\rm i}(E_{2a-1,2b} \pm {\rm i} E_{2a,2b}). \end{equation} \tag{ 6.37 }$$

They have the commutation relations

$$\begin{equation} [\lambda _c, B_{ab}^\pm ] = \pm {\rm i} (\delta _{ac} +\delta _{bc})B^\pm _{ab}, \end{equation} \tag{ 6.38 }$$

so that the roots β_ab ∈ R⁺∖S⁺, corresponding to B⁺_ab in the (λ₁, ..., λ_n)-basis, are

$$\begin{equation} \beta _{ab} = {\rm i}\left (0,\dots ,0,\underset{(a)}{1},0,\dots ,0,\underset{(b)}{1},0,\dots ,0\right ). \end{equation} \tag{ 6.39 }$$

In the basis H₁, ..., H_{n − 1}, they satisfy the rules as given in table 1.

We conclude that

$$\begin{equation*} \sum _{\beta \in R^+\setminus S^+} \beta \ \Big |_{\mathfrak {su}(n)} =0. \end{equation*}$$

Here, $\dim \mathfrak m=n^2-n +1$, and again we might have $\mathcal N>1$ SUSY, with $\mathcal N$ equal to the number of invariant spinors for $\mathfrak {su}(n)\subset \mathfrak {spin}(n^2-n)$. But for n > 3, this is certainly not relevant for heterotic string theory, and for n ⩽ 3 we have the exceptional isomorphisms $\mathfrak {so}(6)=\mathfrak {su}(4)$, $\mathfrak {so}(4) = \mathfrak {su}(2)\oplus \mathfrak {su}(2)$, so that we do not get any essentially new models. The three-form is simply

$$\begin{equation} H\sim \sum _{a<b} \,{\rm e}^{\beta _{ab}} \wedge \,{\rm e}^{-\beta _{ab}} \wedge h. \end{equation} \tag{ 6.40 }$$

We use the notation of the previous example. Contrary to $\mathfrak {so}(n)/\mathfrak {so}(n-1),$ this space admits invariant spinors and the common root space decomposition. Suppose first that n is odd, so that we consider $\mathfrak {so}(2k)/\mathfrak {so}(2k-2)$. Denote the common Cartan generators by λ₁, ..., λ_{k − 1}, and the remaining one in $\mathfrak m$ by λ_k. The positive roots of $\mathfrak {so}(2k)$ in the λ-basis are

$$\begin{equation} \alpha _{ab} = i\left(\dots ,\underset{(a)}{-1},\dots ,\underset{(b)}{1},\dots \right),\qquad \beta _{ab} = i\left(\dots ,\underset{(a)}{1},\dots ,\underset{(b)}{1},\dots \right) \end{equation} \tag{ 6.41 }$$

for 1 ⩽ a < b ⩽ k, and the ones belonging to $\mathfrak m$ are

$$\begin{equation} R^+\setminus S^+ = \lbrace \alpha _{ak},\beta _{ak}\ |\ a=1,\dots ,k-1\rbrace . \end{equation} \tag{ 6.42 }$$

From

$$\begin{equation} \beta _{ak}(\lambda _j) = -\alpha _{ak}(\lambda _j) =i,\qquad \forall j<k, \end{equation} \tag{ 6.43 }$$

it follows immediately that the condition $\sum _{\alpha \in R^+\setminus S^+ }\alpha \big |_{\mathfrak {so}(2k-2)}=0$ is satisfied. What is the amount of supersymmetry preserved? We have that $\mathfrak m$ is twice the fundamental representation of $\mathfrak {so}(2k-2)$, plus the trivial from the additional Cartan generator. The complex spin representation has the nice feature that for two even-dimensional representations V, W we have

$$\begin{equation} S(V\oplus W) = S(V)\otimes S(W), \end{equation} \tag{ 6.44 }$$

so that ignoring the additional Cartan generator which we have taken into account already, $S(\mathfrak m)$ is simply the tensor product of the spin representation with itself. This contains the trivial representation only twice [34]; upon introduction of a $\mathfrak {spin}(2n)$-invariant inner product and an orthonormal basis {ψ_i} of S the invariant elements are given by

$$\begin{equation} \sum _i \psi _i \otimes \psi _i\qquad {\rm and}\qquad \sum _i \psi _i \otimes \Gamma \psi _i, \end{equation} \tag{ 6.45 }$$

with Γ the chirality element. From the embedding $\mathfrak {so}(2k-2)\subset \mathfrak {su}(\mathfrak m),$ we knew there would be at least two spinors invariant under the complexified Lie algebra, and as there are no further ones we get indeed $\mathcal N=1$ SUSY. The situation with $\mathfrak {so}(2k+1)/\mathfrak {so}(2k-1)$ is very similar, the only additional positive roots appearing are the

$$\begin{equation} \gamma _a = (0,\dots ,0,\underset{(a)}{+i},0,\dots ,0 ),\qquad 1\le a\le k, \end{equation} \tag{ 6.46 }$$

with γ_k the only new root in R⁺∖S⁺. But this one vanishes on all of the λ_j for j < k, so we have the condition satisfied here as well. The dimension of SO(n + 1)/SO(n − 1) is 2n − 1, together with the additional $\mathbb R$ factor supporting the linear dilaton this gives dimension 2n, and the only essentially new model with dimension low-enough to be of interest for heterotic string theory is the seven-dimensional SO(5)/SO(3). Below we will consider another embedding of SO(3) into SO(5) which also preserves $\mathcal N=1$. Here, H is

$$\begin{equation} H\sim \sum _{a=1}^{k-1} ({\rm e}^{\alpha _{ak}}\wedge \,{\rm e}^{-\alpha _{ak}}\wedge h + {\rm e}^{\beta _{ak}}\wedge \,{\rm e}^{-\beta _{ak}}\wedge h ) \end{equation} \tag{ 6.47 }$$

for $\mathfrak {so}(2k)/\mathfrak {so}(2k-2)$, whereas on $\mathfrak {so}(2k+1)/\mathfrak {so}(2k-1)$ it has additionally the term ${\rm e}^{\gamma _{k}}\wedge \,{\rm e}^{-\gamma _{k}}\wedge h$.

For this example, we get a solution as well, but our construction is not applicable; the concrete calculations below mainly serve as a preparation for the next example, G₂/SU(3). Here we can use the following argument instead. We have dim$(\mathfrak m)=7$, and $\mathfrak m$ is not a sum of trivial representations of $\mathfrak g_2$. But then it has to be the irreducible seven-dimensional one, as this is the lowest possible dimension for irreducible $\mathfrak g_2$ representations. The spin representation of $\mathfrak g_2$ over $\mathfrak m$ then comes from the standard embedding into $\mathfrak {spin}(7)$, which will be given below, and it is well known that this leaves invariant exactly one Majorana spinor ψ. On the other hand γ(H) commutes with $\mathfrak g_2$ acting on spinors, so that ψ must be an eigenspinor:

$$\begin{equation} \gamma (H)\psi ={\rm i} \lambda \psi ,\qquad \lambda \in \mathbb R. \end{equation} \tag{ 6.48 }$$

For a properly normalized linear function ϕ on $\mathbb R$, we then get γ(dϕ) ∼ id on the one-dimensional spinor space over $\mathbb R$, and from a suitable non-zero 1D spinor κ we can construct an eight-dimensional spinor κ⊗ψ such that

$$\begin{equation} \gamma \big ({\rm d}\phi - \case{1}{12} H\big ) \kappa \otimes \psi =0. \end{equation} \tag{ 6.49 }$$

This solution preserves $\mathcal N=1$ SUSY in the orthogonal two-dimensional space, and the same argument applies to other seven-dimensional spaces with nearly parallel G₂-structure and one Killing spinor. We will work out one example in more detail below for the space SO(5)/SO(3) with a non-standard embedding. Other examples of this type are given by the Aloff–Wallach spaces N(k, l).

In the remainder of the paragraph, we work out the embedding of $\mathfrak g_2$ into $\mathfrak {spin}(7)$, and its root space decomposition. We define $\mathfrak g_2$ as the subalgebra of $\mathfrak {spin}(7)$ fixing a given Majorana spinor [20]. This condition amounts to seven equations, which we write in terms of the antisymmetric X_ab := δ_acX^c_b, $X\in \mathfrak {so}(7)$:

$$\begin{eqnarray} X_{17} &=& - X_{36} - X_{45},\qquad X_{27} =X_{46}- X_{35}, \\ X_{37} &=& X_{25} + X_{16},\qquad \quad X_{47} =X_{15}- X_{26} ,\\ X_{57} &=& - X_{14} - X_{23},\qquad X_{67} =X_{24}- X_{13} , \\ &&X_{12} + X_{34} + X_{56} =0. \end{eqnarray} \tag{ 6.50 }$$

Denote by E_ij the 7× 7-matrix with (E_ij)_ab = δ_aiδ_bj − δ_biδ_aj. A Cartan basis for $\mathfrak {so}(7)$ is given by

$$\begin{eqnarray} \lambda _1 = E_{12} = \left(\begin{array}{@{}ccc@{}}0 & 1 & \\ -1 & 0 & \\ & & 0_{5\times 5} \end{array}\right), \nonumber \\ \lambda _2 = E_{34} = \left(\begin{array}{@{}cccc@{}}0_{2\times 2} & & & \\ & 0 & 1 & \\ & -1& 0 & \\ & & & 0_{3\times 3} \end{array}\right), \\ \lambda _3 = E_{56} = \left(\begin{array}{@{}cccc@{}}0_{4\times 4} & & & \nonumber\\ & 0 & 1 & \\ & -1 & 0 & \\ & & & 0 \end{array}\right). \end{eqnarray} \tag{ 6.51 }$$

and we have

$$\begin{equation*} [\lambda _1 , E_{1j} \pm {\rm i} E_{2_j} ] = \pm {\rm i} (E_{1j} \pm {\rm i}E_{2j}),\qquad j>2, \end{equation*}$$

etc. We can choose a basis of $\mathfrak g_2$ with Cartan generators

$$\begin{equation} \sqrt{6} H_1= 2\lambda _1 -\lambda _2-\lambda _3,\qquad \sqrt{2} H_2 =\lambda _2 -\lambda _3, \end{equation} \tag{ 6.52 }$$

and the other basis elements

$$\begin{eqnarray} &&E_{36}-E_{45} ,\qquad 2E_{17} - E_{36}-E_{45}, \\ \ms &&E_{46}+E_{35}, \qquad 2E_{27} +E_{46}-E_{35}, \end{eqnarray} \tag{ 6.53 }$$

etc (cf (6.50); the elements listed here are deduced from the first line in (6.50), the second and third line give rise to another eight analogous basis elements, so that we end up with 14 generators). Now we introduce the following complex linear combinations of these:

$$\begin{eqnarray} A_1 &=& E_{36}-E_{45} + {\rm i}(E_{46}+E_{35}) \\ \ms A_2 &=& E_{25}-E_{16}-{\rm i}(E_{15}+E_{ 26}) \\ \ms A_3 &=& E_{14}-E_{23} -{\rm i}(E_{24}+E_{13}) \\ \ms B_1 &=& 2(E_{17} + {\rm i}E_{27}) -E_{36}-E_{45} + {\rm i}(E_{46}-E_{35}) \\ \ms B_2 &=& 2(E_{37}+{\rm i}E_{47}) + E_{25}+E_{16} +{\rm i}(E_{15}-E_{26}) \\ \ms B_3 &=& 2(E_{57}-{\rm i}E_{67}) -E_{14}-E_{23}+{\rm i}(E_{13}-E_{24}) . \end{eqnarray} \tag{ 6.54 }$$

The positive roots wrt H₁, H₂ are ${\rm i}/\sqrt{2}$ times:

$$\begin{eqnarray} \alpha _1&=&(0,2),\qquad \quad \alpha _2= (\sqrt{3},1),\qquad \alpha _3= (-\sqrt{3},1), \\ \ms \beta _1&=&(2/\sqrt{3},0),\qquad \beta _2=(-1/\sqrt{3},1),\qquad \beta _3=(1/\sqrt{3},1), \end{eqnarray} \tag{ 6.55 }$$

where the first line has the A-roots, and the second one the B-roots. The ordering of the roots is

$$\begin{equation*} \alpha _1> \alpha _2 > \beta _3>\beta _2 >\alpha _3 >\beta _1. \end{equation*}$$

In the orthogonal complement of $\mathfrak g_2$ in $\mathfrak {so}(7),$ let us introduce the third Cartan generator

$$\begin{equation} H_3 = \sqrt{\case{1}{3}} (\lambda _1+\lambda _2+\lambda _3) . \end{equation} \tag{ 6.56 }$$

Now it turns out that [H₃, A] = 0, but [H₃, B] ≠ 0. Thus, the root space decomposition of $\mathfrak g_2$ cannot be lifted to one of $\mathfrak {spin}(7)$, and our construction does not apply to this coset, but the concrete realization of $\mathfrak g_2$ will still be useful for the following example.

If we put X_a7 = 0 in (6.50), then we obtain the defining equations of $\mathfrak {su}(3)$ inside $\mathfrak {so}(6)$, and thereby realize $\mathfrak {su}(3)$ as a subalgebra of $\mathfrak g_2$. A basis of $\mathfrak {su}(3)$ is given by the Cartan generators H₁, H₂ and A₁, A₂, A₃ plus their complex conjugates. Therefore, the roots α₁, α₂, α₃ are in S⁺, whereas β_{1, 2, 3,} are in R⁺∖S⁺. The roots of $\mathfrak g_2$ satisfy the following relations:

$$\begin{eqnarray} &&\alpha _1 = \alpha _2 + \alpha _3,\qquad \alpha _1=\beta _2+\beta _3 ,\qquad \alpha _2 = \beta _1+\beta _3 , \nonumber\\ &&\beta _2=\alpha _3+\beta _1 ,\qquad \beta _3=\beta _1+\beta _2, \end{eqnarray} \tag{ 6.57 }$$

contradicting our assumption that roots in R⁺∖S⁺ are higher than those of S⁺. In this case, this is not a matter of bad choice of positive roots, it is simply not possible to satisfy the condition. However, if we now choose the complex structure on $\mathfrak m$ as follows, by giving its +i eigenspace ((1,0)-vectors) and −i eigenspace ((0,1)-vectors):

$$\begin{equation} +i:\quad B_1,B_2,\overline{B}_3, \qquad -i: \quad \overline{B}_1,\overline{B}_2,B_3, \end{equation} \tag{ 6.58 }$$

then (6.57) shows that we have again $\mathfrak k=\mathfrak {su}(3)\subset \mathfrak u(\mathfrak m)$. J is also compatible with the metric, but leads to a three-form H ∈ Ω^{(3, 0)}⊕Ω^{(0, 3)}, instead of the usual H ∈ Ω^{(2, 1)}⊕Ω^{(1, 2)}. Now it is not the positive roots in R∖S corresponding to creation and annihilation operators, but the roots β₁, β₂ and −β₃. Therefore, also the spin representation of the Cartan generators $H_1,H_2\in \mathfrak {su}(3)$ changes slightly (the ζ³ part changes sign) and is now given by

$$\begin{eqnarray} \widetilde{{\rm ad}}(H_1) &=& \frac{{\rm i}}{2\sqrt{6}} (2(\overline{\zeta }^1 \zeta ^1 -\zeta ^1 \overline{\zeta }^1 )- \overline{\zeta }^2 \zeta ^2 +\zeta ^2 \overline{\zeta }^2 -\overline{\zeta }^3 \zeta ^3 +\zeta ^3 \overline{\zeta }^3 ),\\ \ms \widetilde{{\rm ad}}(H_2)&=& \frac{{\rm i}}{2\sqrt{2}} (\overline{\zeta }^2 \zeta ^2 -\zeta ^2 \overline{\zeta }^2 - \overline{\zeta }^3 \zeta ^3 +\zeta ^3 \overline{\zeta }^3). \end{eqnarray} \tag{ 6.59 }$$

This leaves invariant S⁰⊕S^m, being equivalent to β₁ + β₂ − β₃ = 0. The three-form is (independently of the complex structure)

$$\begin{equation} H =-8{\rm i}( {\rm e}^{12\overline{3}} - {\rm e}^{\overline{12}3}) \ \in \ \Omega ^{(3,0)} \oplus \Omega ^{(0,3)}. \end{equation} \tag{ 6.60 }$$

It acts on the spinors through

$$\begin{equation} \gamma (H) = -24{\rm i} \big (\zeta ^1\zeta ^2\zeta ^3 - \overline{\zeta }^1\overline{\zeta }^2\overline{\zeta }^3\big ), \end{equation} \tag{ 6.61 }$$

and in particular on $\mathcal S^0\oplus S^3$ as

$$\begin{equation} \gamma (H)\cdot 1 = -24 {\rm i} w^{123} ,\qquad \gamma (H) w^{123} = -24{\rm i}\cdot 1. \end{equation} \tag{ 6.62 }$$

Here we do not have additional flat directions at our disposal to define the linear dilaton on, as the rank of $\mathfrak k$ is equal to rk$(\mathfrak g_2)$. Now we could try to introduce one by hand and consider $\mathbb R \oplus \mathfrak g_2/\mathfrak {su}(3)$ with ϕ linear on $\mathbb R$. It turns out that this does not lead to any solutions, which is probably due to the fact that the tensor product of the spinor spaces in dimensions 3 and 7 does not give the ten-dimensional spinor space, which in turn follows from dimensional reasons. Let us then try adding $\mathbb R^2$. In this case we can put ϕ(x₁, x₂) = 2x₂, introduce a fourth complex spinor variable $w^4\simeq \frac{1}{2}({\rm d}x^1 +{\rm i}{\rm d}x^2)$ and obtain $\gamma ({\rm d}\phi ) = 2{\rm i}(\overline{\zeta }^4-\zeta ^4)$. Of the four invariant Majorana–Weyl spinors

$$\begin{eqnarray} \eta & = 1+w^{1234},\qquad \mu = w^4 - w^{123} \nonumber\\ \kappa &= {\rm i}(1-w^{1234}) ,\qquad \lambda ={\rm i}(w^4 +w^{123}), \end{eqnarray} \tag{ 6.63 }$$

the first two are annihilated by $\gamma ({\rm d}\phi -\frac{1}{12}H)$, the other two are not. Thus, we get two solutions on $\mathbb R^2 \times G_2/$SU(3). As the dilaton depends only on one of the two additional directions, we should view the other $\mathbb R$ as belonging to the orthogonal space $\mathbb R^{2,1}$, with $\mathcal N=1$ supersymmetry preserved, noting that a three-dimensional Majorana spinor can be combined with η and μ in a unique way into a ten-dimensional Majorana–Weyl spinor.

The almost complex structure on G₂/SU(3) we defined here makes it into a so-called nearly Kähler manifold. In six dimensions, only four manifolds of this type are known, and they are all cosets [15]:

$$\begin{eqnarray*} &&SU(3)/U(1)\times U(1),\qquad Sp(2)/Sp(1)\times U(1),\nonumber\\ &&G_2/SU(3)=S^6, \qquad SU(2)^3/SU(2)_{{\rm diag}}=S^3\times S^3. \end{eqnarray*}$$

The same construction we presented here for G₂/SU(3) can be applied to any of them, but for SU(2)³/SU(2)_diag it has to be modified slightly, with the two Cartan generators of $\mathfrak m$ combined into one complex vector, effectively treating SU(2)³ as a rank 1 group. Except for G₂/SU(3), the nearly Kähler cosets have higher-dimensional generalizations, and one may wonder whether the construction applies to these.

From the well known fact that Sp(n) ×U(1) as a subset of Spin(4n) has no invariant spinors (which is equivalent to the fact that quaternion–Kähler manifolds have no parallel spinors), it can be deduced that the equal rank coset Sp(n)/Sp(n − 1) ×U(1) has no invariant spinors for n > 2. Let us then consider SU(2)ⁿ/SU(2)_diag, where we take n odd for simplicity. $\mathfrak m$ consists of n − 1 copies of the adjoint representation, with weights n − 1 times (2),(0),(−2), in the Dynkin label notation. From this one can determine the weights of the spin representation, and deduce that the trivial representation occurs with multiplicity

$$\begin{equation} \frac{1}{2}(n-1)\left [\left({n-1\atop\frac{n-1}{2}}\right)-\left({n-1\atop\frac{n-3}{2} }\right)\right ] = \frac{n-1}{2n}\left({{n}\atop\frac{n-1}{2}}\right). \end{equation} \tag{ 6.64 }$$

This case thus generalizes, but with a lot of invariant spinors. We leave it open whether or not the dilatino equation can be solved in general, but see the discussion at the end of the following paragraph.

is quite interesting in itself, as the problem of finding invariant spinors becomes a simple combinatorial problem on the roots, so we treat it in detail. Furthermore for odd n, there are no invariant spinors, contrary to all the other non-symmetric examples considered so far. The subalgebra is a Cartan algebra here, and the decomposition of $\mathfrak m\otimes \mathbb C$ into irreducibles is given by the root space decomposition. The positive roots are

$$\begin{equation} \alpha _{ab} = {\rm i}(0,\dots ,0,\underset{(a-1)}{-1},\underset{(a)}{+1},0,\dots ,0,\underset{(b-1)}{+1},\underset{(b)}{-1},0,\dots ,0), \end{equation} \tag{ 6.65 }$$

with 1 ⩽ a < b ⩽ n + 1, and terms that run out of the n entries of the root vector are dropped. They satisfy the relations

$$\begin{equation} \alpha _{ab} + \alpha _{bc} = \alpha _{ac}. \end{equation} \tag{ 6.66 }$$

Similarly, the weights of the spin representation determine the irreducible representations in $S(\mathfrak m)$, and there are invariant spinors if and only if the zero weight occurs. As usual, the weights of $S(\mathfrak m)$ are of the form $\frac{1}{2}\sum _{\alpha \in R^+} \pm \alpha$, with arbitrary combinations of signs.

Proposition 6.1. For even n invariant spinors exist, but not for odd n.

Proof. n even: we have to show that there exists a certain combination of signs such that ∑_ab ± α_ab = 0. Group the roots into pairs of three or four elements according to the following triangle.

Inside each box there is a relation of the type

$$\begin{equation} \alpha _{12} + \alpha _{23} -\alpha _{13} =0,\qquad \alpha _{14}-\alpha _{24} -\alpha _{15} + \alpha _{25} =0,\quad \dots . \end{equation} \tag{ 6.67 }$$

Fix the signs of the roots α_ab accordingly and denote them by sign(a, b); then, we certainly have

$$\begin{eqnarray*} &&\sum _{1\le a<b\le n+1} {\rm sign}(a,b) \alpha _{a,b} =0,\\ \end{eqnarray*}$$

and the corresponding spinor is invariant. On the other hand, this shows clearly that there will be many more invariant spinors for large n, but we will not try to determine their number.

n odd: the reason that we could sum all the roots to zero by an appropriate choice of signs in the even n case was that in every box the indices that appeared, appeared exactly twice, as in relations (6.66). Adding another row to the diagram of the form

$$\begin{equation} \alpha _{1,n+1} \qquad \alpha _{2 ,n+1} \qquad \dots \qquad \alpha _{n,n+1} \end{equation} \tag{ 6.68 }$$

for odd n, we get every index appearing with odd multiplicity. It follows that the roots can no longer add up to zero.

In the case of even n > 2, there are several invariant Majorana spinors and γ(H) acts non-trivially on them, so that it is far from obvious whether this action can be compensated for by a linear dilaton on an extra $\mathbb R$ factor, and for a well-chosen spinor ε. Analyzing our explicit construction of the spinors, this seems rather unlikely, although a proof that it is not possible probably requires more elaborate techniques. The models with n ⩾ 3 are not relevant for heterotic supergravity anyway, but they would very much break the pattern of the other solutions: they are of even dimension greater than 6, so that their cone cannot support parallel spinors [35], contrary to all the solutions we found so far (cf the discussion in the conclusion). We therefore conjecture that it is not possible to solve the dilatino equation on them. The same applies to SU(2)ⁿ/SU(2) for odd n, and every other naturally reductive coset of even dimension different from 6.

As an example where the common root space decomposition does not exist, and the subalgebra is not in $\mathfrak u(\mathfrak m)$, we consider the following embedding of $\mathfrak {so}(3)$ into $\mathfrak {so}(5)$, whose coset manifold is known to possess a nearly parallel G₂-structure [14, 20]

$$\begin{equation} \left(\begin{array}{@{}ccc@{}}0& \alpha &\beta \\ -\alpha & 0& \gamma \\ -\beta & -\gamma &0 \\ \end{array}\right) \mapsto \left(\begin{array}{@{}ccccc@{}}0& \alpha &-\gamma &\beta & \sqrt{3} \gamma \\ -\alpha & 0& -\beta &-\gamma &-\sqrt{3}\beta \\ \gamma & \beta & 0 & 2\alpha & 0 \\ -\beta & \gamma & -2\alpha & 0&0 \\ -\sqrt{3} \gamma & \sqrt{3}\beta & 0 &0&0 \\ \end{array}\right). \end{equation} \tag{ 6.69 }$$

In terms of our basis vectors

$$\begin{eqnarray} \lambda _1 &=& E_{12},\qquad \lambda _2 = E_{34}, \\ A &=& E_{13} + E_{24} +{\rm i}(E_{23}-E_{14} ),\\ B &=& E_{13} - E_{24} +{\rm i}(E_{23}+E_{14} ),\\ C_1 &=& E_{15} +{\rm i}E_{25}, \\ C_2 &=& E_{35} + {\rm i} E_{45} \end{eqnarray} \tag{ 6.70 }$$

of $\mathfrak {so}(5)$ defined above, the subalgebra and its complement are spanned by

$$\begin{eqnarray} \mathfrak {so}(3) : H_1:&=&\lambda _1 +2\lambda _2,\quad D:= A-\sqrt{3} \overline{C}_1,\quad \overline{D}\\ \ms \mathfrak m : H_2:&=& 2\lambda _1 -\lambda _2 ,\quad B,\overline{B},C_2,\overline{C}_2, \quad E:= \overline{A}+ \case{2}{\sqrt{3}}C_1,\ \ \overline{E}. \end{eqnarray} \tag{ 6.71 }$$

This is not a common root space decomposition, as [H₂, D] ≠ 0 and $[H_2, E]\notin \mathfrak m$. It is a naturally reductive coset though. There is no complex structure on $\mathfrak m$ such that $\mathfrak {so}(3)\subset \mathfrak u(\mathfrak m)$, and it will be important to know precisely the action of $\mathfrak k$ on $\mathfrak m$. After a rescaling

$$\begin{equation*} B\mapsto \sqrt{\frac{5}{2}}B,\quad C_2\mapsto \sqrt{5}C_2,\quad E\mapsto \sqrt{\frac{3}{2}}E, \end{equation*}$$

it is given by

$$\begin{eqnarray} [H_1,B] &=& 3{\rm i}B ,\qquad [H_1,C_2] = 2{\rm i}C_2,\qquad [H_1, E] = {\rm i}E ,\\ [D,B] &=& -\sqrt{6} C_2 ,\qquad [D,C_2] =\sqrt{10} E,\qquad [D,E]= 2\sqrt{6}{\rm i} H_2, \\ [D,\overline{B}] &=& 0, \qquad [D,\overline{C}_2] = \sqrt{6}\overline{B},\qquad [D,\overline{E}]= -\sqrt{10} \overline{C}_2 ,\\ [H_1,H_2] &=&0 ,\qquad [D,H_2] =-\sqrt{6}{\rm i}\overline{E} . \end{eqnarray} \tag{ 6.72 }$$

In particular, the weights of $\mathfrak {so}(3)$ in the Dynkin label notation are (6),(4),(2),(0),(−2),(−4),(−6), which belong to the irreducible seven-dimensional representation with highest weight (6) (in general the highest-weight module to highest weight (n) has dimension n + 1). We write $\mathfrak m = (6)$. For the linear dilaton, we will need to extend this to $\mathfrak m^{\prime }= \mathfrak m \oplus \mathbb R$, which is the $\mathfrak k$-module $\mathfrak m^{\prime }=(6)\oplus (0)$, and the weights of the corresponding spinor module are

$$\begin{equation} \Omega (S(\mathfrak m^{\prime })) = 2\times \lbrace (6),(4),(2),2\times (0),(-2),(-4),(-6) \rbrace , \end{equation} \tag{ 6.73 }$$

showing that $S(\mathfrak m^{\prime }) = 2\times \big ((6)\oplus (0)\big )$. We conclude that there are invariant spinors again, and in order to identify them we need to introduce a complex structure. Take its +i eigenspace to be spanned by

$$\begin{equation} +i :\overline{B},C_2,E , H_2 -{\rm i} \partial _x, \end{equation} \tag{ 6.74 }$$

where ∂_x is the standard basis vector field of $\mathbb R$. Then, the commutation relations above tell us how the basis elements of $\mathfrak k$ are quantized:

$$\begin{eqnarray} \widetilde{{\rm ad}}(H_1) &=& \frac{{\rm i}}{2} ( 3( \zeta ^B\overline{\zeta }^B + \overline{\zeta }^B \zeta ^B) + 2(\overline{\zeta }^C\zeta ^C + \zeta ^C\overline{\zeta }^C) +\overline{\zeta }^E\zeta ^E + \zeta ^E\overline{\zeta }^E),\\ \widetilde{{\rm ad}}(D) &=& \sqrt{6}\overline{\zeta }^C\overline{\zeta }^B + \sqrt{10}\zeta ^C\overline{\zeta }^E +\sqrt{6} {\rm i}\zeta ^E(\zeta ^H + \overline{\zeta }^H), \\ \widetilde{{\rm ad}}(\overline{D}) &=& \sqrt{6}\zeta ^C \zeta ^B + \sqrt{10}\overline{\zeta }^C\zeta ^E -\sqrt{6} {\rm i}\overline{\zeta }^E(\zeta ^H + \overline{\zeta }^H). \end{eqnarray} \tag{ 6.75 }$$

Invariant under all of $\mathfrak {so}(3)$ are

$$\begin{equation} 1 - {\rm i} w^{BCEH},\qquad w^H -{\rm i} w^{BCE},\qquad \end{equation} \tag{ 6.76 }$$

and these multiplied by i, or the Majorana–Weyl spinors

$$\begin{equation} \eta =(1+{\rm i}) 1 +(1-{\rm i}) w^{BCEH} ,\qquad \mu =(1-{\rm i})w^H - (1+{\rm i})w^{BCE}. \end{equation} \tag{ 6.77 }$$

More effort is required to determine the action of γ(H) on these spinors than in the $\mathfrak {su}(n)$ case. We need the commutation relations between elements of $\mathfrak m$. They are as follows:

$$\begin{eqnarray} \fl [H_2, B] &=& {\rm i}B,\qquad [H_2,\overline{B},]= -{\rm i}\overline{B},\qquad [H_2,C_2]= -{\rm i} C_2 ,\qquad [H_2,\overline{C}_2]={\rm i}C_2, \\ \fl [H_2,E] &=& -\sqrt{6}{\rm i} D-{\rm i}E,\qquad \ \ [H_2,\overline{E}] = \sqrt{6}{\rm i} \overline{D}+{\rm i}\overline{E},\\ \fl [B,\overline{B}] &=& 6{\rm i} H_1 + 2{\rm i} H_2 ,\qquad [C_2,\overline{C}_2] = 4{\rm i} H_1 - 2{\rm i}H_2 ,\\ \fl [E,\overline{E}] &=& 2{\rm i}H_1-2{\rm i}H_2, \qquad [B,\overline{C}_2] = -\sqrt{6} \overline{D} +2 E ,\\ \fl [B,C_2] &=& 0, \qquad [B,E] = 0 ,\qquad \ \ [B,\overline{E}] =-2C_2, \\ \fl [C_2,E] &=& 2B,\qquad [C_2,\overline{E}] = \sqrt{10} \overline{D}, \end{eqnarray} \tag{ 6.78 }$$

from which we read off that

$$\begin{equation} H= 4\big ({\rm e}^{B\overline{ C E}}+{\rm e}^{ \overline{B} CE}\big ) -2{\rm i}\big ({\rm e}^{\overline{B}B}+{\rm e}^{C\overline{C}} +{\rm e}^{E \overline{E}}\big )\wedge \big (e^H + {\rm e}^{\overline{H}}\big ), \end{equation} \tag{ 6.79 }$$

leading to the result

$$\begin{equation} \gamma (H)\eta = -42\mu ,\qquad \gamma (H)\mu =42\eta . \end{equation} \tag{ 6.80 }$$

Defining $\phi :\mathbb R\rightarrow \mathbb R$ by $\phi (x)=-\frac{7}{2} x$, we obtain $\gamma ({\rm d}\phi ) = -\frac{7}{2}{\rm i}(\overline{\zeta }^H -\zeta ^H)$, and

$$\begin{eqnarray} \gamma \Big ({\rm d}\phi - \frac{1}{12} H\Big ) \eta &=& 0,\\ \gamma \Big ({\rm d}\phi - \frac{1}{12} H\Big ) \mu &=& -7\eta \ne 0. \end{eqnarray} \tag{ 6.81 }$$

There is thus exactly one Majorana–Weyl spinor on $\mathbb R\times$SO(5)/SO(3) solving the heterotic supersymmetry equations, leading to $\mathcal N=1$ SUSY in the orthogonal two-dimensional space. More abstractly, we could have argued as in the case Spin(7)/G₂, using that SO(5)/SO(3) carries a nearly parallel G₂-structure with one Killing spinor [20].