Skip to main content
MathOverflow

Return to Question

replaced http://mathoverflow.net/ with https://mathoverflow.net/
Source Link
Community Bot
Community Bot
  • 1
  • 2
  • 3

For a better understanding of this question, please see the question and answer herehere.

In $Spin(8)$ there are plenty of copies of $Spin(7)$; consider, for instance, the antiimage of $SO(7)<SO(8)$ by the double cover for any $SO(7)<SO(8)$ obtained by fixing some direction in $\mathbb{R}^8$. So it is true, there are plenty of $Spin(7)<Spin(8)$ (let's call this subgroups primitives).

By the triality automorphism of $Spin(8)$, the $Spin(7)$ subgroups go to other $Spin(7)$ subgroups. But the hot point is that it is not a permutation between primtives subgroups: that is, if we perform triality on a $Spin(7)<Spin(8)$ obtained by fixing $e_1\in\mathbb{R}^8$, then we get a copy $Spin(7)<Spin(8)$ whose proyection onto $SO(8)$ is not a double cover of some subgroup $SO(7)<SO(8)$, but an isomorphism over some $Spin(7)<SO(8)$. And it happens that this $Spin(7)<SO(8)$ may be described as the stabilizer of a 4-form, as described in the link above.

So the corner question is: Why this $Spin(7)$ obtained by performing triality to a primitive $Spin(7)$ should in fact be so strangely described, as the stabilizer of some 4-form? Is there a nice way to understand this fact?

Any idea or suggestion is welcome.

For a better understanding of this question, please see the question and answer here.

In $Spin(8)$ there are plenty of copies of $Spin(7)$; consider, for instance, the antiimage of $SO(7)<SO(8)$ by the double cover for any $SO(7)<SO(8)$ obtained by fixing some direction in $\mathbb{R}^8$. So it is true, there are plenty of $Spin(7)<Spin(8)$ (let's call this subgroups primitives).

By the triality automorphism of $Spin(8)$, the $Spin(7)$ subgroups go to other $Spin(7)$ subgroups. But the hot point is that it is not a permutation between primtives subgroups: that is, if we perform triality on a $Spin(7)<Spin(8)$ obtained by fixing $e_1\in\mathbb{R}^8$, then we get a copy $Spin(7)<Spin(8)$ whose proyection onto $SO(8)$ is not a double cover of some subgroup $SO(7)<SO(8)$, but an isomorphism over some $Spin(7)<SO(8)$. And it happens that this $Spin(7)<SO(8)$ may be described as the stabilizer of a 4-form, as described in the link above.

So the corner question is: Why this $Spin(7)$ obtained by performing triality to a primitive $Spin(7)$ should in fact be so strangely described, as the stabilizer of some 4-form? Is there a nice way to understand this fact?

Any idea or suggestion is welcome.

For a better understanding of this question, please see the question and answer here.

In $Spin(8)$ there are plenty of copies of $Spin(7)$; consider, for instance, the antiimage of $SO(7)<SO(8)$ by the double cover for any $SO(7)<SO(8)$ obtained by fixing some direction in $\mathbb{R}^8$. So it is true, there are plenty of $Spin(7)<Spin(8)$ (let's call this subgroups primitives).

By the triality automorphism of $Spin(8)$, the $Spin(7)$ subgroups go to other $Spin(7)$ subgroups. But the hot point is that it is not a permutation between primtives subgroups: that is, if we perform triality on a $Spin(7)<Spin(8)$ obtained by fixing $e_1\in\mathbb{R}^8$, then we get a copy $Spin(7)<Spin(8)$ whose proyection onto $SO(8)$ is not a double cover of some subgroup $SO(7)<SO(8)$, but an isomorphism over some $Spin(7)<SO(8)$. And it happens that this $Spin(7)<SO(8)$ may be described as the stabilizer of a 4-form, as described in the link above.

So the corner question is: Why this $Spin(7)$ obtained by performing triality to a primitive $Spin(7)$ should in fact be so strangely described, as the stabilizer of some 4-form? Is there a nice way to understand this fact?

Any idea or suggestion is welcome.

Fixed a typo in the title.
Link
Stefan Kohl ♦
Stefan Kohl ♦
  • 19.6k
  • 21
  • 75
  • 137

$Spin(7)$ as stabilizerrstabilizer of a $4$-form revisited

Source Link
Jjm
Jjm
  • 2.1k
  • 14
  • 27

$Spin(7)$ as stabilizerr of a $4$-form revisited

For a better understanding of this question, please see the question and answer here.

In $Spin(8)$ there are plenty of copies of $Spin(7)$; consider, for instance, the antiimage of $SO(7)<SO(8)$ by the double cover for any $SO(7)<SO(8)$ obtained by fixing some direction in $\mathbb{R}^8$. So it is true, there are plenty of $Spin(7)<Spin(8)$ (let's call this subgroups primitives).

By the triality automorphism of $Spin(8)$, the $Spin(7)$ subgroups go to other $Spin(7)$ subgroups. But the hot point is that it is not a permutation between primtives subgroups: that is, if we perform triality on a $Spin(7)<Spin(8)$ obtained by fixing $e_1\in\mathbb{R}^8$, then we get a copy $Spin(7)<Spin(8)$ whose proyection onto $SO(8)$ is not a double cover of some subgroup $SO(7)<SO(8)$, but an isomorphism over some $Spin(7)<SO(8)$. And it happens that this $Spin(7)<SO(8)$ may be described as the stabilizer of a 4-form, as described in the link above.

So the corner question is: Why this $Spin(7)$ obtained by performing triality to a primitive $Spin(7)$ should in fact be so strangely described, as the stabilizer of some 4-form? Is there a nice way to understand this fact?

Any idea or suggestion is welcome.