Let $(M,g)$ be a $d$-dimensional Riemannian oriented, spin manifold, and let us denote by $F(M)$ its frame bundle, by $SP(M)$ its spin bundle and by $S = P(M)\times_{\rho}\Delta$ its spinor bundle, where $\rho\colon Spin(d)\to\Delta$ is the corresponding spinor representation. My question is:

When a reduction on $SP(M)$ implies a reduction on $F(M)$ and vice-versa?

For example: let us suppose that $M$ is seven-dimensional and equipped with a nowhere vanishing real spinor. Then, the spin bundle $SP(M)$ is reduced from $Spin(7)$ to $G_{2}$, the stabilizer of the spin-group action. In that case, since $G_{2}$ is connected an simply connected, its image inside $SO(n)$ by the map of the double-cover $\pi\colon Spin(7)\to SO(7)$ is a unique connected and simply connected subgroup of $SO(7)$ which can be denoted by the same symbol, since $\pi$ gives an isomorphism between them. Therefore in this case, a reduction on the spin bundle implies a reduction on the frame bundle. I would like to understand what are the conditions "if and only if" for this to happen in general, in both directions.


  • $\begingroup$ I don't understand the meaning of $S$ in your post. This seems to be rather standard structure group reduction/extension business, but you should be more specific. Given the covering $\operatorname{Spin}\to SO$, what kind of subgroups in the two do you want to consider? $\endgroup$ Commented Apr 26, 2015 at 12:13

1 Answer 1


Well, in one sense, this is always true. If $Q\subset SP(M)$ is a principal right $H$-bundle, where $H\subset\mathrm{Spin}(n)$ is a subgroup, then $\delta(Q)\subset F(M)$ is a principal right $\pi(H)$-bundle, where $\pi(H)\subset\mathrm{SO}(n)$ is the image subgroup. (Here, I am using $\delta:SP(M)\to F(M)$ to denote the double cover that the OP didn't name or notate.)

The OP may, however, have intended to ask a different question, such as, "When does having a nonvanishing spinor field on $M$ imply a reduction of structure group of $M$?". The answer to this is "When $\mathrm{Spin}(n)$ acts transitively on the unit sphere in its spinor representation $\mathbb{S}$.". (However, one must be precise about that is meant by this. People sometimes say "$\mathrm{Spin}(8)$ acts transitively on the unit sphere in its spin representation." when they mean "$\mathrm{Spin}(8)$ acts transitively on the unit sphere in each of its semi-spin representations.")

Added at the request of the OP: Let $(M,g)$ be an oriented, spinnable Riemannian $d$-manifold, with $\beta: F(M)\to M$ being its oriented orthonormal frame bundle and $\delta:SP(M)\to F(M)$ being a double cover that defines a spin structure on $M$. Let $\pi:\mathrm{Spin}(d)\to\mathrm{SO}(d)$ be the double cover and let $\mathbb{S}$ be the spinor space, with $\rho:\mathrm{Spin}(d)\to\mathrm{SO}(\mathbb{S})$ the spin representation. Then $S = SP(M)\times_\rho\mathbb{S}$ is the spinor bundle of the spin structure, so that a spinor field, i.e., a section of $S\to M$ is represented by a smooth map $s:SP(M)\to \mathbb{S}$ that satisfies $s(u\cdot g) = \rho(g^{-1})s(u)$ for all $u\in SP(M)$ and $g\in \mathrm{Spin}(n)$. The section is nonvanishing if and only if $s$ is nonvanishing, in which case, we can normalize, replacing $s$ by $s/|s|$ to get a unit spinor field. Thus, we can assume $|s| = 1$, i.e., that $s$ maps $SP(M)$ to the unit sphere in $\mathbb{S}$. Now, we can ask, when can we find an element $v\in\mathbb{S}$ with $|v|=1$, such that $s^{-1}(v)\subset SP(M)$ defines a principal subbundle of $SP(M)$ for some subgroup $H\subset \mathrm{Spin}(d)$. Well, the only possibility for $H$ would be the $\rho$-stabilizer of $v$, i.e., the subgroup $H = \{ g\in \mathrm{Spin}(d)\ |\ \rho(g)v = v\}$. But this will only work if $s$ takes values in the orbit $\rho\bigl(\mathrm{Spin}(d)\bigr){\cdot}v$. Now, this will always be true if $\mathrm{Spin}(d)$ acts transitively on the unit sphere in $\mathbb{S}$. Unfortunately, that happens only when $d\in\{2,3,5,6,7,9\}$. In the two cases $d = 4$ and $d=8$, you don't have transitivity because, as a representation of $\mathrm{Spin}(d)$, the space $\mathbb{S}$ is the direct sum of two irreducible sub-representations: $\mathbb{S}=\mathbb{S}_+\oplus\mathbb{S}_-$, corresponding to the fact that a spinor field is the sum of two 'semi-spinors'. It turns out that, in these two cases, $\mathrm{Spin}(d)$ does act transitively on the unit spheres in $\mathbb{S}_\pm$, so you get a structure reduction when there is a nonvanishing semi-spinor.

  • $\begingroup$ Thanks, as usual. Could you please elaborate a little bit this: "When $Spin(n)$ acts transitively on the unit sphere in its spinor representation $\mathbb{S}$." By the way, what about the other direction? If I am not mistaken, in the $G_{2}$ case, the $G_{2}$-reduction of the frame bundle can be "lifted" to a $G_{2}$-reduction on the spin bundle. I guess this question can be reformulated as when the set of tensors defining the reduction on $F(M)$ come from spinors through the isomorphism $S\otimes S = \Omega^{\bullet}(M)$. This is precisely what happens in the $G_{2}$-case. $\endgroup$
    – Bilateral
    Commented Apr 26, 2015 at 13:04
  • $\begingroup$ Why does it fail to be transitive for $d\ge 10$? $\endgroup$ Commented Apr 28, 2015 at 6:22
  • 1
    $\begingroup$ @ChrisGerig: For $d\ge 11$, one has $\mathrm{dim}(\mathbb{S}) > \mathrm{dim}\bigl(\mathrm{Spin}(d)\bigr) + 1$, so transitivity fails for dimension reasons. For $d=10$, it would be possible for dimension reasons, but explicit calculation shows that the action of $\mathrm{Spin}(10)$ on the unit sphere in $\mathbb{S}\simeq \mathbb{R}^{32}$ has cohomogeneity $1$. See www.math.duke.edu/~bryant/Spinors.pdf for details. $\endgroup$ Commented Apr 28, 2015 at 10:03
  • $\begingroup$ @RobertBryant: Thanks Robert. I am confused about something. When you say that $\rho\colon Spin(d)\to SO(\mathbb{S})$ I was assuming that this is already an irrep of $Spin(d)$, and that one constructs the spinor bundle from this irrep. However, from your explanation it looks like you are assuming that this representation is a complex irrep of the Clifford algebra $Spin(d)\subset Cliff(d)$, which splits for even $d$ as two complex irreps of $Spin(d)$ of definite chirality, as you wrote. Not only that, for $d=8k$, there are real chiral irreps. $\endgroup$
    – Bilateral
    Commented Apr 28, 2015 at 12:38
  • $\begingroup$ By the way, shouldn't it be $\rho\colon Spin(d)\to End(\mathbb{S})$? And since (if I am not mistaken) the representation is unitary then one can write $\rho\colon Spin(d)\to U(\mathbb{S})$. $\endgroup$
    – Bilateral
    Commented Apr 28, 2015 at 12:42

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