Timeline for Frame-bundle reduction from spinor-bundle reduction
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15 events
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| Apr 28, 2015 at 17:16 | comment | added | Bilateral | @RobertBryant: Thanks. The way I understood this, and now I am afraid it might not be correct, is that from the complex representations of $Spin(d)$ one can obtain all the others. For example, if $d=7$, the complex representation of $Spin(7)$ must have an invariant real structure which defines a unique real representation inside the complex one, and which coincides with the real representation induced by the real Clifford group $Cl(7)$. In particular, when the real representation of $Spin(d)$ is of complex type, then it is the same as the complex representation of $Spin(d)$. | |
| Apr 28, 2015 at 16:26 | comment | added | Robert Bryant | @Bilateral: Yes, those are what I would call the complex spin representations. However, when you say 'very simple', you should be aware that this really only applies to the dimension formulae. The actual orbit structure of the action defined by the representation can be quite complicated (and is, when $d$ gets large). | |
| Apr 28, 2015 at 16:25 | comment | added | Bilateral | @RobertBryant: Thanks Robert. Yes, it answers my question. If I am not mistaken, you are obtaining such real representations by restriction of the real representations of the real Clifford algebra $Cl(d)$, which are of the complex or quaternionic type depending also on the dimension. However, one can consider the complexified Clifford algebra $\mathbb{C}l(d)$, whose complex representations are very simple. These induce complex representations on $Spin(d)\subset \mathbb{C}l(d)$,irreducible in odd dimension a reducible in even. These are what you would call complex representations of $Spin(d)$? | |
| Apr 28, 2015 at 15:20 | comment | added | Robert Bryant | @Bilateral: This is a question of terminology. In my terminology, an irreducible representation is always real; it is complex if and only if the commuting ring contains a copy of $\mathbb{C}$; and it is quaternionic if and only if the commuting ring contains a copy of $\mathbb{H}$ (in which case the commuting ring has to be equal to $\mathbb{H}$). In general, the spinor representation is complex if and only if $d\ \mathrm{mod}\ 8$ is in $\{2,3,4,5,6\}$ and is quaternionic if and only if $d\ \mathrm{mod}\ 8$ is in $\{3,4,5\}$. This answer your question about $\mathrm{Spin}(3)$, I think. | |
| Apr 28, 2015 at 15:17 | comment | added | Bilateral | @RobertBryant: I think I understand now what you mean. It looks like that what I call (perhaps incorrectly) complex representations of $Spin(d)$ are real representations of $Spin(d)$ of complex type, namely they are real representations equipped with an invariant complex structure. Is this correct? | |
| Apr 28, 2015 at 15:14 | comment | added | Robert Bryant | @Bilateral: The spinor representation is the lowest dimensional faithful (real) representation of $\mathrm{Spin}(d)$ on which the nontrivial element in the kernel of $\pi: \mathrm{Spin}(d)\to\mathrm{SO}(d)$ acts as $-1$. When $d\not\equiv0\ \mathrm{mod}\ 4$, this is irreducible, but when $d\equiv0\ \mathrm{mod}\ 4$ it is the direct sum of two representations, neither of which is faithful. These representations are not always complex; this depends on $d\ \mathrm{mod}\ 8$. Complex representations are a different story. If you want the complex representations, you should specify this. | |
| Apr 28, 2015 at 14:20 | comment | added | Bilateral | @RobertBryant: Thanks for the clarifications, I just wanted to understand the terminology. However now I am confused by something: you say that the spinor representation is the lowest faithful real representation. However, I was under the impression that the $Spin(d)$ group does not always admit real irreps. From 3.6.1 of Joyce's book on holonomy and calibrations I got the impression that real irreps for $Spin(d)$ only exist when $d=8k-1,8k,8k+1$. For example, what would be the real spin representation of $Spin(3)\simeq SU(2)$? (thanks for the patience). | |
| Apr 28, 2015 at 12:42 | comment | added | Bilateral | 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 $\rho\colon Spin(d)\to U(\mathbb{S})$. | |
| Apr 28, 2015 at 12:38 | comment | added | Bilateral | @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. | |
| Apr 28, 2015 at 12:18 | vote | accept | Bilateral | ||
| Apr 28, 2015 at 10:03 | comment | added | Robert Bryant | @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. | |
| Apr 28, 2015 at 6:22 | comment | added | Chris Gerig | Why does it fail to be transitive for $d\ge 10$? | |
| Apr 26, 2015 at 15:03 | history | edited | Robert Bryant | CC BY-SA 3.0 |
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| Apr 26, 2015 at 13:04 | comment | added | Bilateral | 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. | |
| Apr 26, 2015 at 12:31 | history | answered | Robert Bryant | CC BY-SA 3.0 |