Octonion-Valued Forms and the Canonical 8-Form on Riemannian Manifolds with a Spin(9)-Structure

Abstract

It is well known that there is a unique Spin(9)-invariant 8-form on the octonionic plane that naturally yields a canonical differential 8-form on any Riemannian manifold with a weak Spin(9)-structure. Over the decades, this invariant has been studied extensively and described in several equivalent ways. In the present article, a new explicit algebraic formula for the Spin(9)-invariant 8-form is given. The approach we use generalises the standard expression of the Kähler 2-form. Namely, the invariant 8-form is constructed only from the two octonion-valued coordinate 1-forms on the octonionic plane. For completeness, analogous expressions for the Kraines form, the Cayley calibration and the associative calibration are also presented.

Appendix

We now express the form $\Psi _8$ explicitly in terms of the dual basis $\{\mathrm{{d}}x^0,\dots ,\mathrm{{d}}x^7,\mathrm{{d}}y^0,\dots ,\mathrm{{d}}y^7\}$ of $\bigwedge ^1(\mathbb {O}^2)^*$ corresponding to the standard basis $\{e_0,\dots ,e_7\}$ of $\mathbb {O}$. Although the computations we perform to this end are slightly more technical, they are based on very elementary algebraic properties of the octonions. Basically, we just use the formula (38) together with the rule $R_{\overline{e_i}} R_{e_j}=-R_{\overline{e_j}} R_{e_i}$, whenever $i\ne j$, following easily from (11). We shall also keep the notation from the proof of the main theorem and omit the wedge product symbol for the sake of brevity.

1.1 The Parts $\Psi _{80}$ and $\Psi _{08}$

As already shown in Lemma 5.2, both these parts consist of one element each. Namely, $\Psi _{80}=8! \,\mathrm{{d}}x^0\cdots \mathrm{{d}}x^7$ and $\Psi _{08}=8! \,\mathrm{{d}}y^0\cdots \mathrm{{d}}y^7$.

1.2 The Parts $\Psi _{62}$ and $\Psi _{26}$

According to (23) we have

$$\begin{aligned} 4\,\Psi _{62}&=4\sum \langle R_{e_{i_4}} R_{\overline{e_{i_5}}}R_{e_{{i_6}}}R_{\overline{e_ {i_7}}} R_{e_{{i_3}}} R_{\overline{e_{i_2}}} R_{e_{{i_1}}} R_{\overline{ e_{i_0}}}(1), 1\rangle \\ {}&\quad \,\mathrm{{d}}y^{i_0}\mathrm{{d}}x^{i_1}\mathrm{{d}}x^{i_2}\mathrm{{d}}x^{i_3}\mathrm{{d}}y^{i_4}\mathrm{{d}}x^{i_5}\mathrm{{d}}x^{i_6}\mathrm{{d}}x^{i_7}, \end{aligned}$$

or, after reordering the canonical 1-forms,

$$\begin{aligned} 4\,\Psi _{62}&=-4\sum \langle R_{e_{i_7}} R_{\overline{e_{i_3}}}R_{e_{{i_4}}}R_{\overline{e_ {i_5}}} R_{e_{{i_2}}} R_{\overline{e_{i_1}}} R_{e_{{i_0}}} R_{\overline{ e_{i_6}}}(1), 1\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_5}\mathrm{{d}}y^{i_6}\mathrm{{d}}y^{i_7}. \end{aligned}$$

Clearly, a general term

$$\begin{aligned} -4 \langle R_{e_{i_7}} R_{\overline{e_{i_3}}}R_{e_{{i_4}}}R_{\overline{e_ {i_5}}} R_{e_{{i_2}}} R_{\overline{e_{i_1}}} R_{e_{{i_0}}} R_{\overline{ e_{i_6}}}(1), 1\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_5}\mathrm{{d}}y^{i_6}\mathrm{{d}}y^{i_7} \end{aligned}$$

(41)

of this sum is possibly non-trivial only if $\#\{i_0,\dots ,i_5\}=6$ and $\#\{i_6,i_7\}=2$. Hence, there are just three eventualities for $\#\{i_0,\dots ,i_7\}$: 6, 7 or 8. Further necessary condition on non-triviality of (41) is obviously, in a sense of Remark 5.3,

$$\begin{aligned} \prod _{k=0}^7e_{i_k}=\pm 1. \end{aligned}$$

(42)

First, suppose $\#\{i_0,\dots ,i_7\}=8$. This means all indices in (41) are distinct and the inner product there is thus totally skew-symmetric. Therefore, there are ${8\atopwithdelims ()2}=28$ distinct terms of this kind, each corresponding to a different set $\{i_6,i_7\}$, all with coefficients $\pm 4\cdot 2!\cdot 6!=\pm 8\cdot 6!$.

Second, let $\#\{i_0,\dots ,i_7\}=7$, i.e. let precisely two indices coincide in (41). Then (42) requires that the product of six distinct basis vectors equals $\pm 1$. According to (38), this would however mean that the product of the two remaining (and distinct) basis elements is also $\pm 1$, which is impossible. We conclude, therefore, that there is no non-trivial term of this kind.

Finally, suppose $\#\{i_0,\dots ,i_7\}=6$, i.e. $\{i_6,i_7\}\subset \{i_0,\dots ,i_5\}$. In particular, $i_6$ agrees with precisely one element in $\{i_0,\dots ,i_5\}$ and thus, after commuting the operator $R_{\overline{e_{i_6}}}$ leftwards, (41) takes the form

$$\begin{aligned} +4\,\langle R_{e_{i_7}} R_{\overline{e_{i_6}}}R_{e_{{i_3}}}R_{\overline{e_ {i_4}}} R_{e_{{i_5}}} R_{\overline{e_{i_2}}} R_{e_{{i_1}}} R_{\overline{ e_{i_0}}}(1), 1\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_5}dy^{i_6}\mathrm{{d}}y^{i_7} \end{aligned}$$

that is totally skew-symmetric in $i_6,i_7$ and in $i_0,\dots ,i_5$, respectively. The inner product is, however, non-zero precisely when the product of the basis elements of indices $\{i_0,\cdots ,i_5\}\backslash \{i_6,i_7\}$ is $\pm 1$. So, as shown during the proof of Lemma 5.4, there are 14 possibilities for the set $\{i_0,\cdots ,i_5\}\backslash \{i_6,i_7\}$ and to each of them there are ${4\atopwithdelims ()2}=6$ choices of $\{i_6,i_7\}$. Therefore, there are $6\cdot 14=84$ terms of this kind, each with prefactor $\pm 4\cdot 2!\cdot 6!=\pm 8\cdot 6!$.

The case of $\Psi _{26}$ is completely analogous.

1.3 The Part $\Psi _{44}$

Now we have

$$\begin{aligned} 6\,\Psi _{44}&=-10\sum \langle ((\overline{e_{i_0}}e_{i_1})\overline{e_{i_2}})e_{i_3}, \overline{((\overline{ e_{i_4}}e_{i_5})\overline{ e_{i_6}})e_{i_7}}\rangle \,\mathrm{{d}}y^{i_0}\mathrm{{d}}x^{i_1}\mathrm{{d}}x^{i_2}dx^{i_3}\mathrm{{d}}x^{i_4}\mathrm{{d}}y^{i_5}\mathrm{{d}}y^{i_6}\mathrm{{d}}y^{i_7}, \end{aligned}$$

so, after reordering, a general term takes the form

$$\begin{aligned} -10 \langle ((\overline{e_{i_4}}e_{i_0})\overline{e_{i_1}})e_{i_2}, \overline{((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}}\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_3}\mathrm{{d}}y^{i_4}\cdots \mathrm{{d}}y^{i_7}, \end{aligned}$$

(43)

that is only non-trivial if $\#\{i_0,\dots ,i_3\}=\#\{i_4,\dots ,i_7\}=4$, i.e. $4\le \#\{i_0,\dots ,i_7\}\le 8$. Due to the higher complexity of this case, we introduce the following product of indices: $(i,j)\mapsto {ij}$, where ij is the (unique) element of $\{0,\dots ,7\}$ such that $e_{ij}=\pm e_ie_j$. Such a product is clearly commutative as well as associative (see Remark 5.3). The condition (42), which of course still applies, translates in this language as

$$\begin{aligned} \prod _{k=0}^7i_k=0. \end{aligned}$$

(44)

Let $\#\{i_0,\dots ,i_7\}=8$, i.e. $\{i_0,\dots ,i_3\}\cap \{i_4,\dots ,i_7\}=\emptyset$. We shall distinguish two cases here. First, suppose $i_0i_1i_2i_3=0$. Then (44) is only fulfilled if $i_4i_5i_6i_7=0$ too, i.e. if $i_5i_6i_7=i_4$. Since $i_3\ne i_4$, one has $i_5i_6i_7\ne i_3$ and thus $i_3i_5i_6i_7\ne 0$. Therefore $\overline{((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}}=-((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}$ and so (43) takes the form

$$\begin{aligned}&+10\,\langle ((\overline{e_{i_4}}e_{i_0})\overline{e_{i_1}})e_{i_2}, ((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_3}\mathrm{{d}}y^{i_4}\cdots \mathrm{{d}}y^{i_7}\\&\quad =+10\langle R_{e_{i_3}} R_{\overline{e_{i_5}}}R_{e_{{i_6}}}R_{\overline{e_ {i_7}}} R_{e_{{i_2}}} R_{\overline{e_{i_1}}} R_{e_{{i_0}}} R_{\overline{ e_{i_4}}}(1), 1\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_3}\mathrm{{d}}y^{i_4}\cdots \mathrm{{d}}y^{i_7}, \end{aligned}$$

that is again totally skew-symmetric and thus the coefficient is $\pm 10\cdot 4!\cdot 4!=\pm 8\cdot 6!$. We have already shown above that there exist 14 distinct sets $\{i_0,\dots ,i_3\}$, such that $i_0i_1i_2i_3=0$, and there are therefore 14 terms of this kind. Second, if $i_0i_1i_2i_3\ne 0$ then, by (44), also $i_4i_5i_6i_7\ne 0$ and thus $i_5i_6i_7\ne i_4$. If, for instance, $i_5i_6i_7= i_5$, then $i_6=i_7$, which is impossible. Similarly one shows that $i_5i_6i_7\ne i_6$ and $i_5i_6i_7\ne i_7$. It is therefore necessary that $i_5i_6i_7\in \{i_0,i_1,i_2,i_3\}$. If $i_5i_6i_7=i_3$, we have $\overline{((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}}=((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}$ and (43) reads

$$\begin{aligned} -10\,\langle ((\overline{e_{i_4}}e_{i_0})\overline{e_{i_1}})e_{i_2}, ((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}\rangle \,\mathrm{{d}}x^{i_0}\cdots dx^{i_3}\mathrm{{d}}y^{i_4}\cdots \mathrm{{d}}y^{i_7}. \end{aligned}$$

In the three other cases $i_5i_6i_7\in \{i_0,i_1,i_2\}$, $\overline{((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}}=-((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}$ and (43) equals

$$\begin{aligned} +10\,\langle ((\overline{e_{i_4}}e_{i_0})\overline{e_{i_1}})e_{i_2}, ((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}\rangle \,\mathrm{{d}}x^{i_0}\cdots \mathrm{{d}}x^{i_3}\mathrm{{d}}y^{i_4}\cdots \mathrm{{d}}y^{i_7}. \end{aligned}$$

Hence, the coefficient in front of such a term is $\pm \left( \frac{-1+3}{4}\right) \cdot 10\cdot 4!\cdot 4!=\pm 4\cdot 6!$. As discussed in the proof of Lemma 5.4, there are 56 sets $\{i_0,\dots ,i_3\}$ with $i_0i_1i_2i_3\ne 0$ and so is the number of the corresponding terms.

If $\#\{i_0,\dots ,i_7\}=7$, then (44) could never be fulfil from exactly the same reason as in the case of $\Psi _{62}$. There is, hence, no such term again.

Let $\#\{i_0,\dots ,i_7\}=6$, and denote $\{j_0,\dots ,j_3\}:=\{i_0,\dots ,i_3\}$ and $\{j_2,\dots ,j_5\}:=\{i_4,\dots ,i_7\}$. According to (44), we may assume $j_0j_1j_4j_5=0$, i.e. $j_0j_1=j_4j_5$. Let $\varepsilon _1,\varepsilon _2=\pm 1$ be such that

$$\begin{aligned} \overline{((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}}&=\varepsilon _1((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7},\\ R_{e_{i_3}} R_{\overline{e_{i_5}}}R_{e_{{i_6}}}R_{\overline{e_ {i_7}}} R_{e_{{i_2}}} R_{\overline{e_{i_1}}} R_{e_{{i_0}}} R_{\overline{ e_{i_4}}}&=\varepsilon _2 R_{e_{i_5}} R_{\overline{e_{i_6}}}R_{e_{{i_7}}}R_{\overline{e_ {i_4}}} R_{e_{{i_3}}} R_{\overline{e_{i_2}}} R_{e_{{i_1}}} R_{\overline{ e_{i_0}}}. \end{aligned}$$

Using this notation, a general term (43) takes the form

$$\begin{aligned} 6\,\Psi _{44}=-10\, \varepsilon _1\varepsilon _2 \langle R_{e_{i_5}} R_{\overline{e_{i_6}}}R_{e_{{i_7}}}R_{\overline{e_ {i_4}}} R_{e_{{i_3}}} R_{\overline{e_{i_2}}} R_{e_{{i_1}}} R_{\overline{ e_{i_0}}}(1), 1\rangle \,\mathrm{{d}}x^{i_0}\cdots dx^{i_3}\mathrm{{d}}y^{i_4}\cdots \mathrm{{d}}y^{i_7}. \end{aligned}$$

In what follows, we shall discuss how the sign $\varepsilon _1\varepsilon _2$ alternates for different positions of $i_3$ within $\{j_0,\dots ,j_3\}$ and of $i_4$ within $\{j_2,\dots ,j_5\}$. Let us distinguish two separate cases. First, assume $j_0j_1j_2j_3=0$ or equivalently $j_0j_1=j_2j_3$. Then, if $i_3=j_0$ and $i_4=j_2$, for instance, one has $\{i_5,i_6,i_7\}=\{j_3,j_4,j_5\}$ and $\{i_0,i_1,i_2\}=\{j_1,j_2,j_3\}$. Therefore,

$$\begin{aligned} i_3i_5i_6i_7=j_0j_3j_4j_5=j_0j_3j_2j_3=j_0j_2\ne 0, \end{aligned}$$

meaning $((\overline{ e_{i_3}}e_{i_5})\overline{ e_{i_6}})e_{i_7}\ne \pm 1$ and thus $\varepsilon _1=-1$. Further, since $i_3\notin \{i_5,i_6,i_7\}$, $i_4\notin \{i_0,i_1,i_2\}$ and $i_3\ne i_4$, respectively, we can write

$$\begin{aligned} R_{e_{i_3}} R_{\overline{e_{i_5}}}R_{e_{{i_6}}}R_{\overline{e_ {i_7}}} R_{e_{{i_2}}} R_{\overline{e_{i_1}}} R_{e_{{i_0}}} R_{\overline{ e_{i_4}}}&= -R_{e_{i_5}} R_{\overline{e_{i_6}}}R_{e_{{i_7}}}R_{\overline{e_ {i_3}}} R_{e_{{i_2}}} R_{\overline{e_{i_1}}} R_{e_{{i_0}}} R_{\overline{ e_{i_4}}} \\&= -R_{e_{i_5}} R_{\overline{e_{i_6}}}R_{e_{{i_7}}}R_{\overline{e_ {i_3}}} R_{e_{{i_4}}} R_{\overline{e_{i_2}}} R_{e_{{i_1}}} R_{\overline{ e_{i_0}}} \\&= R_{e_{i_5}} R_{\overline{e_{i_6}}}R_{e_{{i_7}}}R_{\overline{e_ {i_4}}} R_{e_{{i_3}}} R_{\overline{e_{i_2}}} R_{e_{{i_1}}} R_{\overline{ e_{i_0}}}, \end{aligned}$$

and thus $\varepsilon _2=+1$. The signs corresponding to the other positions of $i_3$ and $i_4$ are computed analogically and summarised in Table 1. One can observe from the table that $\varepsilon _1\varepsilon _2$ equals +1 in precisely 8 cases and -1 in the 8 others, from which we conclude that the corresponding term is trivial in the end. We may thus assume $j_0j_1j_2j_3\ne 0$. Then it is easily seen that the eight indices $j_0$, $j_1$, $j_2$, $j_3$, $j_0j_1j_2$, $j_0j_1j_3$, $j_0j_2j_3$ and $j_1j_2j_3$ are all distinct. Therefore, $j_4$ and $j_5$ must be among the last four ones. The requirement $j_0j_1j_4j_5=0$ however chooses the last two ones. Without loss of generality, we thus have $j_4=j_0j_2j_3$ and $j_5=j_1j_2j_3$. Now we investigate the behaviour of the sign $\varepsilon _1\varepsilon _2$ again, taking into account that $j_0j_4=j_1j_5=j_2j_3$. The results are captured in Table 2. Clearly, $\varepsilon _2$ stays the same as in the case $j_0j_1j_2j_3=0$ but $\varepsilon _1$ alternates so that $\varepsilon _1\varepsilon _2$ is positive only in 4 cases and negative otherwise. This means that the corresponding term appears with the coefficient $\pm \left( \frac{-12+4}{16}\right) \cdot 10\cdot 4!\cdot 4!=\pm 4\cdot 6!$. Regarding the number of such terms, there are 56 options for $\{j_0,\dots ,j_3\}$, $j_0j_1j_2j_3\ne 0$, and for each of them there are ${4\atopwithdelims ()2}=6$ possible partitions into $\{j_0,j_1\}$ and $\{j_2,j_3\}$. Since $\{j_4,j_5\}$ is then uniquely determined, there are altogether $56\cdot 6=336$ terms of this kind.

Table 1 The signs $\varepsilon _1\varepsilon _2$ in the case $j_0j_1j_2j_3=0$

Table 2 The signs $\varepsilon _1\varepsilon _2$ in the case $j_0j_1j_2j_3\ne 0$

Table 3 Explicit expression of the form $-\frac{1}{4\cdot 6!}\Psi _8$ in the standard basis

Further, suppose that $\#\{i_0,\dots ,,i_7\}=5$, i.e. that $\{i_0,\dots ,i_3\}\cap \{i_4,\dots ,i_7\}$ contains precisely three indices. (44) requires that the product of the two (distinct) elements of $\{i_0,\dots ,i_7\}$ that do not belong to this intersection is 0. This is again impossible and thus there are no terms here either.

Finally, let $\#\{i_0,\dots ,,i_7\}=4$, i.e. $\{i_0,\dots ,i_3\}=\{i_4,\dots ,i_7\}$. First, assume $i_0i_1i_2i_3=0$. If $i_3=i_4$, then it is easily seen that $\varepsilon _1=\varepsilon _2=1$. If $i_3\ne i_4$, then $\varepsilon _1=\varepsilon _2=-1$. In any case $\varepsilon _1\varepsilon _2=1$ and so 14 of these terms have all coefficients $\pm 10\cdot 4!\cdot 4!=\pm 8\cdot 6!$. Second, if $i_0i_1i_2i_3\ne 0$, then $\varepsilon _1=-1$ regardless the relation between $i_3$ and $i_4$. Since $\varepsilon _2$ does not change from the previous case, for any $i_3$ we have $\varepsilon _1\varepsilon _2=-1$ if $i_4=i_3$ and $\varepsilon _1\varepsilon _2=1$ in the three other cases of $i_4\ne i_3$. Altogether, the prefactors of these 56 terms are $\pm \left( \frac{-1+3}{4}\right) \cdot 10\cdot 4!\cdot 4!=\pm 4\cdot 6!$.

1.4 Summary

All in all, the expression of $\Psi _8$ in the standard basis possesses 702 non-trivial terms. They are summarised in Table 3. Each block of the table corresponds to one summand in (40). Each row of the table stands for a particular class of terms of $-\frac{1}{4\cdot 6!}\Psi _8$. A general term of the class is stated in the second column and the class is further specified in the third column. In the first column, the coefficient standing in front of the terms from the respective class is given. Let us remark that we scaled the form $\Psi _8$ by $-\frac{1}{4\cdot 6!}$ in order to adhere to the conventions of [57]. Notice that the signs of the coefficients can be explicitly determined directly from the aforedescribed construction. Finally, the number of non-trivial terms within each class is given in the fourth column. Throughout the table, we assume $i_k\ne i_l$ if $k\ne l$. Recall also that the product of indices is taken in the following sense: $e_{ij}=\pm e_ie_j$.