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In a paper I was reading about Kac-Moody algebras there was remark that interested me, but I couldn't really make sense of. When discussing extended Dynkin diagrams the authors dicussed that there is a "gain in symmetry" due to the one extra root compared to the finite Dynkin diagrams.

The gain in symmetry can be related in a precise way to the center of the simply connected group whose Lie algebra is $\mathfrak{g}$: this center is isomorphic to a normal subgroup of the symmetry group of the extended Dynkin diagram, and the corresoponding quotient map is isomorphic to the symmetry group of the original Dynkin diagram.$^1$

The $1$ refers to a paper by Olive & Turok. The authors of the original paper I was reading make an example, but I'm struggeling to understand this example. We consider the Lie group $SU(\ell+1)$ and notice that it has center $\mathbb{Z}_{\ell+1}$, which is the cyclic symmetry of the diagram for $\hat A_\ell$ (where the $\hat{}$ is just to say we consider the extended Dynkin diagram)

enter image description here

whose full symmetry is the $2(\ell+1)$ dihedral group. They then state that the quotient is $\mathbb{Z}_2$, the symmetry of the $A_r$ (finite) diagram.

I see that the cycly group of order $\ell+1$ is a normal subgroup of the dihedral group of order $2(\ell +1)$, which I think is the statement made in the first sentence of the quote. But I fail to understand the "quotient map" part of the second sentence. What quotient is exactly isomorphic to the symmetry of the finite Dynkin diagram? And how exactly does this work for a diagram whose finite version doens't have a really clear symmetry (for example $D_\ell$)?

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If the last part of the quote read

and the corresponding quotient is isomorphic to the symmetry group of the original Dynkin diagram

then, even though I have no idea how to prove that, at least it matches known results.

Namely, in your example of Dynkin diagrams of type $A_\ell$, the quotient of the dihedral group of order $2(\ell+1)$ ($\simeq$ symmetry group of extended Dynkin diagram $\hat{A_\ell}$) by the cyclic group of order $\ell+1$ ($\simeq$ centre of $SU(\ell+1)$, "the" simply connected (compact?) group belonging to that Dynkin diagram) is $\mathbb Z/2$ ($\simeq$ symmetry group of Dynkin diagram $A_\ell$).

As for Dynkin diagrams of type $D_\ell$, I was sloppy in an earlier comment (sorry about that), now I think the following holds:

For $\ell \ge 5$, the symmetry group of the extended Dynkin diagram $\hat{D_\ell}$ is the dihedral group of order $8$ (I needed a while to convince myself of this, but I'm pretty sure now).

[Added: To see this, let's use these notations: enter image description here Let $a$ be the symmetry sending $K_0 \mapsto K_{n-1}, K_1 \mapsto K_n, K_n \mapsto K_0, K_{n-1} \mapsto K_1, K_i \mapsto K_{n-i}$ for all $i \ge 2$. Let $x$ be the symmetry flipping $K_0 \leftrightarrow K_1$ and leaving all other $K_i$ fixed. Then we have $a^4 = x^2 =id$ and $xax^{-1}=a^{-1}$, which presents the dihedral group of order $8$.]

That group has both a normal subgroup $\simeq \mathbb Z/4$ and a normal subgroup $\simeq \mathbb Z/2 \oplus \mathbb Z/2$ which, according to the parity of $\ell$, are $\simeq$ the centre of the simply connected group $Spin(2\ell)$ belonging to that diagram. In both cases, the quotient is $\simeq \mathbb Z/2$, the symmetry group of the usual Dynkin diagram $D_\ell$. (And here I would really like to see from a proof how the correct normal subgroup gets picked as centre of the Lie group according to the parity of $\ell$).

For $\ell=4$, the extended Dynkin diagram $\hat{D}_\ell$ is a central vertex with edges to four indistinguishable vertices, hence its symmetry group is the full symmetric group $S_4$ of order $24$. According to the above link about Spin groups, the centre of $Spin(8)$ is isomorphic to the Klein four-group $\mathbb Z/2 \oplus \mathbb Z/2$,and indeed $S_4$ does have one such normal subgroup; and the corresponding quotient is $\simeq S_3$, the symmetric group of order $6$ which we all know and love as the symmetry group of the Dynkin diagram $D_4$.

By the way, since for the types $B_n$ and $C_n$ ($n \ge 2$) the extended diagram has symmetry group of order $2$, the normal diagram has no symmetries, this also matches the fact that $Spin(2n+1)$ resp. the (compact) symplectic group $Sp(n)$ have centre $\simeq \mathbb Z/2$. Further, for the exceptional types $E_8, F_4, G_2$, where neither the extended nor the usual Dynkin diagram have any symmetries, indeed the adjoint (centreless) group is simply connected. Finally, for type $E_6$ this predicts $\mathbb Z/3$ as centre for the compact simply connected group (symmetries of extended diagram: symmetric group $S_3$, quotient i.e. symmetries of usual $E_6$ diagram: $\mathbb Z/2$), and for $E_7$ it predicts $\mathbb Z/2$ (symmetries of extended diagram, normal diagram has no symmetries). I could not find a confirmation for that quickly, but it sounds right.

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  • $\begingroup$ Thank you for your answer! Please give me some time to look through it in more detail. If there are unclear things, I'll comment again. $\endgroup$
    – Sito
    Mar 20, 2020 at 15:21
  • $\begingroup$ I finally had some time to look into this, sorry for the dealy! I can follow most of the things you say, but I have a hard time seeing how the dihedral group of order 8 is the symmetry group of $\hat{D}_\ell$ for $\ell>4$. Could you maybe elaborate on your reasoning? $\endgroup$
    – Sito
    Apr 10, 2020 at 13:30
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    $\begingroup$ @Sito: See addition. $\endgroup$ Apr 10, 2020 at 22:56

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