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As it is shown in the post Subgroups of $E_8$ by using extended Dynkin diagrams it is possible to find subgroups of a group from the extended Dynkin diagram by simply cutting nodes.

However, the relation $G_2 \subset SO(7)$ obviously cannot be obtained, since the Dynkin diagram of the $SO(7)$ doesn't contain a triple edge.

Is there an additional rule besides cutting nodes that allows the derivation of the relation on the level of extended Dynkin diagrams?

p6majo
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    Why is it not possible to follow the minimal rule of providing any kind of context? Why not reading first https://math.meta.stackexchange.com/questions/9959/how-to-ask-a-good-question? – dan_fulea Jul 29 '23 at 22:17
  • Sorry, I hoped that there would be a fast answer that I just didn't find anywhere. – p6majo Jul 29 '23 at 22:27
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    Sometimes this operation is called "folding". Similarly, we can fold $D_4 \rightsquigarrow B_3$, which corresponds to the exceptional inclusion $\operatorname{Spin}(7) \hookrightarrow \operatorname{SO}(8, \Bbb R)$. See, e.g., Section 5.3.1 of https://arxiv.org/abs/math/0510216, but n.b. the notion of folding there by definition begins with a simply laced diagram, which $SO(7)$ does not have. – Travis Willse Jul 29 '23 at 22:59
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    I think you can see this inclusion by considering the root system $B_3$ which comprises the midpoints of the edges and faces of a cube and projecting onto a plane. The key is to ensure that each of the short roots is being matched to a long root (orthogonal to each other). Indeed I think you can choose the plane to go through the other 6 long roots and they will form the long roots in $G_2$. – Callum Jul 31 '23 at 10:30
  • I would begin by using the 7-dimensional fundamental (irreducible) representation of $\mathfrak{g}_2$ (the short roots and $0$ as the weights). That gives us an embedding $\mathfrak{g}_2\to\mathfrak{gl}_7$. With a bit of luck (and checking several known facts) the images actually lies in... – Jyrki Lahtonen Aug 03 '23 at 08:30

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As mentioned in comments, one has to think not of sub-root systems, but rather of "quotients" of root systems here. As also mentioned in a comment, there is some activity on this topic under the name of "folding" in recent years, but I think it is not explored in the generality that it could be. Cf. https://math.stackexchange.com/a/3972084/96384, https://math.stackexchange.com/a/3300945/96384 and links from there.

In fact, as Callum suggests: Choose a set of simple roots $\gamma_1, \gamma_2, \gamma_3$ for $B_3$, where $\gamma_3$ is the short root. Now "mod out" the relation $\gamma_1=\gamma_3$, i.e. project the ambient, three-dimensional space down to the plane orthogonal to $\gamma_1-\gamma_3$.

If we call $\alpha$ the image of $\gamma_1$ (and $\gamma_3$) under this projection, and $\beta$ the image of $\gamma_2$, one easily checks that all positive roots of $B_3$ get projected to $\alpha$, $\beta$, $\alpha+\beta$, $2\alpha+\beta$, $3\alpha+\beta$, $3\alpha+2\beta$, i.e. the projection/quotient/folding is a root system of type $G_2$ with $\alpha$ (short) and $\beta$ (long) forming a set of simple roots. (If one chooses Cartesian coordinates for better visualization, say $\gamma_1 = (0,-1,1), \gamma_2= (-1,1,0), \gamma_3 = (1,0,0)$, then we project down to the plane $z=x+y$ (which already contains $\gamma_2$ but is perpendicular to $(1,1,-1) = \gamma_3-\gamma_1$.)

To bring things back up to the Lie algebra or group level, cf. the near-duplicate Understanding $G_2$ inside Spin(7)? (EDIT: problem solved).

What makes this specific "folding" a bit strange compared to the standard first examples is that it is not induced by any kind of automorphism of the roots; in fact, we identified the long root $\gamma_1$ with the short root $\gamma_3$! Maybe it becomes more plausible if one sees $B_3$ already as a folding of $D_4$ (as mentioned in Travis Willse's comment). This $D_4$, its Dynkin diagram being "three branches from a central node" seems to play the role of kind of a "simply laced covering" for both $B_3$ and $G_2$: For $D_4 \twoheadrightarrow B_3$, one identifies two of the outer nodes of $D_4$ (and similarly one always gets $D_{n+1} \twoheadrightarrow B_n$); whereas for $D_4 \twoheadrightarrow G_2$, one identifies all three of the outer nodes of $D_4$. What we are doing here is factoring the second folding through the first: which explains why we had to identify that remaining one-edged branch of the diagram with the two-edged one on the other side, the central node $\gamma_2$ has to stay long, whatever it takes.

Torsten Schoeneberg
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