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.


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 '15 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 '15 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 '15 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 '15 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 '15 at 16:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.