Here is an extract of the doctoral thesis of C. Lewis under the supervision of D. Joyce (https://people.maths.ox.ac.uk/joyce/theses/LewisDPhil.pdf, 1998):

2.6 Spin Bundles and the Dirac Operator

To consider spin bundles over a $Spin(7)$ manifold $M$, it is usually best to first consider Clifford algebras.

Let $V$ be a finite dimensional vector space with an inner product defined upon it. Let $e_1$, $e_2$, ... , $e_n$ be an orthonormal basis for $V$.

Then the Clifford algebra, $C_n$, of $V$ is defined to be the algebra generated by the elements $e_1$, $e_2$, ... , $e_n$ subject to the relations

$$e_i^2 =-1$$, $$e_ie_j + e_je_i = 0\text{ for }i \neq j$$

Considered as a vector space $C_n$ is of dimension $2n$, spanned by elements of the form

$$e_1^{\delta_1}e_2^{\delta_2}\cdots e_n^{\delta_n}$$

where $\delta_i = 0$ or 1.

Now consider the case $n = 8$. In this case it can be shown that

$$C_8 = \mathbb{R} (16),$$

the algebra of $16\times 16$ matrices with values in $\mathbb{R}$. [Sal, p.171]

Look also here

Thus we may consider $\mathbb{R}$ (16) as a $C_8$ module.

We may define the group $Spin(8)$ as the subset of $C_8$ consisting of all even products $x_1x_2\cdots x_{2r-1}x_{2r}$ of elements of $V$, with each $\|x_i\| = 1$. (Similarly we might have defined $Spin(7)$ as the subset of $C_7$ consisting of all even products $x_1x_2\cdots x_{2r-1}x_{2r}$ of elements of $\mathbb{R}^8$, with each $\|x_i\| = 1$.)

Now let us consider the element $v = e_1e_2\cdots e_8$ of $C_8$. Then $v$ is a involution of $C_8$, and commutes with every element of $Spin(8)$, and hence $\mathbb{R}^{16}$ splits as a $Spin(8)$ module into the eigenspaces of $v$.

Thus $\mathbb{R}^{16}=\Delta_+\oplus\Delta_-$, where $\Delta_+$ is the $+1$ eigenspace of $v$, and $\Delta_-$ is the $-1$ eigenspace of $v$. We call them the positive and negative spin representations of $Spin(8)$.

Now suppose that $M$ is a $Spin(7)$ manifold. Then we have that $M$ is a spin manifold i.e. there exists a spin structure of $M$, a principal $Spin(8)$ bundle $\tilde{E}$ covering the $SO(8)$ bundle of frames for the tangent bundle.

Now since we have a principal $Spin(8)$ bundle, and the two $Spin(8)$ modules (namely $\Delta_+$ and $\Delta_-$), we may form two vector bundles associated to the principal spin bundles by means of the two spin representations.

We call these bundles $S_+$ and $S_-$, the positive and negative spinor bundles, and their sections are known as positive and negative spinors. It is perhaps worth noting at this point that the group $Spin(7)$ is the subgroup of $SO(8)$ preserving a spinor, and hence the manifold $M$ will possess a constant spinor. Thus we will have isomorphisms $S_+ \equiv\Lambda^0\otimes \Lambda^2_7$ and $S_-\equiv \Lambda^1$.

The question is: Why is it true that $S_+ \equiv\Lambda^0\otimes \Lambda^2_7$ and $S_-\equiv \Lambda^1$?

Any suggestion is welcome.


1 Answer 1


You're really asking an algebra question about how the various representations of $\mathrm{Spin}(8)$ interact. There are lots of places where you can read about this, but here is a set of notes that I wrote, Remarks on Spinors in Low Dimensions, that explains this and that some people have found useful.

Once you have read the first few pages, you'll see why there is a canonical 'nondegenerate' bilinear pairing $S_-\otimes S_+ \longrightarrow T$, where $T$ is the tangent bundle of $M$. Then, once you have a unit-size positive spinor $s$, i.e., a section of $S_+$ of unit size, then $S_+$ splits as the line bundle $\mathbb{R}s$ and its orthogonal $V$, which is a $7$-plane bundle. The map above restricted to $S_-\simeq S_-\otimes \mathbb{R}s\longrightarrow T$ then becomes an isomorphism, so $S_-$ is isomorphic to the tangent bundle $T$. Then, considering the map $S_-\otimes V\to T$ and the fact that all three of these bundles are irreducible under $\mathrm{Spin}(7)$, i.e., the stabilizer of $s$ (which also is proved in the above reference), you'll get, by duality, an embedding of $V$ into $T\otimes S_- = T\otimes T$, and, again using the above notes, you'll see that this goes into $\Lambda^2(T)\subset T\otimes T$ and must be perpendicular to the $\mathrm{Spin}(7)$-irreducible subbundle ${\frak{spin}}(7) = \Lambda^2_{21}\subset \Lambda^2(T)$. By definition $\Lambda^2_7$ is the orthogonal complement to $\Lambda^2_{21} $ in $ \Lambda^2(T)$.

  • $\begingroup$ Thank you very much, really. The answer is concise and exact. $\endgroup$
    – Jjm
    Jan 27, 2015 at 15:13
  • $\begingroup$ I have a doubt about the canonical nondegenerate bilinear pairing $S_-\otimes S_+\longrightarrow T$. This is what I have understood: 1) An explicit isometry $T_x\simeq\mathbb{O}$ induces (according to the bundle construction, some details needed) explicit isometries $S_{\pm x}\simeq\mathbb{O}$ 2) The pairing is related to some operation with octonions (it could be product), provided it is well defined under change of isometry $T_x\simeq\mathbb{O}$. Is this right or have I misunderstood? $\endgroup$
    – Jjm
    Feb 2, 2015 at 15:44
  • 1
    $\begingroup$ Yes. More precisely (following the notes): There is an embedding of the unit octonions into $\mathrm{Spin}(8)$ so that there are three representations $\rho_i:\mathrm{Spin}(8)\to\mathrm{SO}(V_i)=\mathrm{SO}(8)$ on $V_i=\mathbb{R}^8=\mathbb{O}$ such that $\rho_1(u) = L_u$, $\rho_2(u) = L_uR_u$, and $\rho_3(u) = R_u$. These representations are used to construct the bundles $S_1$, $T$, and $S_+$, respectively. The fact that there is a map of representations $V_1\otimes V_3\to V_2$ that is $x\otimes y \mapsto xy$ (octonion multiplication) then gives the desired bundle map $S_1\otimes S_+\to T$. $\endgroup$ Feb 2, 2015 at 16:01
  • $\begingroup$ Why should $S_+$ have a unit-size positive spinor and $S_-$ not? The role they play seem quite similar, so it is rather strange this symmetry breaking. $\endgroup$
    – Jjm
    Mar 1, 2015 at 15:15
  • 1
    $\begingroup$ You perceive this as 'symmetry breaking' because you aren't seeing the full symmetry: A choice of orientation determines which bundle one designates as $S_+$; switching the orientation exchanges the two semi-spinor bundles. People sometimes don't realize at first that there are two distinct $\mathrm{SO}(8)$-conjugacy classes of $\mathrm{Spin}(7)$-subgroups in $\mathrm{SO}(8)$ (of course, the two are conjugate in $\mathrm{O}(8)$). Thus, there are two kinds of $\mathrm{Spin}(7)$-structures on an orientable, spinnable $8$-manifold; each determined by a unit-size section of a semi-spinor bundle. $\endgroup$ Mar 1, 2015 at 16:44

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