# Classification by Dynkin diagram: Why's there no $E_9$?

from Wiki

According to Dynkin's classification, we have as possibilities these only, where n is the number of nodes:

To me it seems not obvious why there should not be a $E_9$? Further clicking got me to the following:

"Roughly speaking, symmetries of the Dynkin diagram lead to automorphisms of the Bruhat-Tits building associated with the group." from here.

Is there easy answer without going through the details of root systems?

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I don't understand what you are asking? When you say «classify Dynkin diagrams» what do you mean? What are Dynkin diagrams for you? (And that quotation you included in your question seems rather unrelated to everything else, no?) – Mariano Suárez-Alvarez Jan 4 '12 at 23:12
In fact I was interested why there are just a few exceptional Lie Groups. I thought, and that was wrong, that the quote has something to do with it. – draks ... Jan 5 '12 at 7:23
If there were infinitely many exceptional groups (say, infinitely many $E$s) then we would not call them exceptional! (For example, in the Todd-Shephard classification of complex reflection groups, there are 34 "exceptional groups", and 34 is much larger than $5$ :) ) – Mariano Suárez-Alvarez Jan 5 '12 at 7:25

## 3 Answers

I'm not sure what your favorte way of thinking about Dynkin diagrams is. I'm going to go with the presentation where a Dynkin diagram encodes angles between vectors in $\mathbb{R}^n$. An edge means that the vectors meet at $120^{\circ}$, the absence of an edge means that the vectors meet at right angles. So now I need to explain why we can't find $9$ vectors $v_1$, $v_2$, ..., $v_9$ in $\mathbb{R}^9$ which meet as in the $E_9$ diagram.

Normalize the $v_i$ to all have length $1$. Let $V$ be the matrix whose columns are the $v_i$. Then the entries of $V^T V$ are $1$'s on the diagonal, $-1/2$'s where vertices are joined by an edge, and $0$'s elsewhere.

But you can check that this matrix is not positive definite, contradicting the supposition that it is of the form $V^T V$.

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Thanks, this is very simple. If you label the top node #1, the label the other nodes left to right as #2 through #9, then [ 3, 2, 4, 6, 5, 4, 3, 2, 1 ] is an element of the null space of $V^TV$, which cannot be positive definite. – Jack Schmidt Jan 4 '12 at 23:27

As Qiaochu mentioned, there is an $E_9$. In fact, there is a whole infinite E series of diagrams. In general, there is a Kac-Moody algebra associated to any Dynkin Diagram. If a diagram is made up of (finitely) many diagrams from your list, its corresponding Lie algebra is finite dimensional semi-simple. If the diagram is connected (from your list), its algebra will be finite dimensional simple.

Any other diagram necessarily leads to an infinite dimensional Kac-Moody Lie algebra. Connected diagrams lead to (essentially) simple infinite dimensional Lie algebras and non-connected diagrams lead to analogs of semi-simple algebras.

After the finite dimensional simple Lie algebras (which correspond to the diagrams you've listed), the next "nicest" class are the affine Lie algebras. These Kac-Moody Lie algebras are infinite dimensional, but they behave and "look" a lot like the finite dimensional simple algebras. $E_9$ is an affine algebra. The next step beyond affine is hyperbolic. $E_{10}$ is a hyperbolic (Kac-Moody) Lie algebra. A ton is known about $E_9$. Some is known about $E_{10}$. After that little is known.

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Do you happen to know if the affine ones are the ones where David Speyer's matrix (also occasionally attributed to Cartan) has a null space of rank 1? – Jack Schmidt Jan 4 '12 at 23:59
If I read David's post correctly, his matrix is a generalized Cartan matrix rescaled so the diagonals are 1's instead of 2's. The affine algebras are those whose Cartan matrices are positive semi-definite and have nullity 1. I think if you just look at nullity 1, it will include indefinite type examples. – Bill Cook Jan 5 '12 at 0:16
Thanks! The complete graph on 4 points disjoint union an affine dynkin diagram has a generalized Cartan matrix with a 0 and a negative eigenvalue, so we do have to pay attention to both 0s and negative. – Jack Schmidt Jan 5 '12 at 0:26
My matrix is twice the Cartan matrix (notice the off diagonal entries are $-1/2$, not $-1$). I was torn between doing the thing which looks most natural (normalizing vectors to have length $1$) or doing the thing which is most useful further on in the theory (normalizing diagonal entries to be $2$). – David Speyer Jan 5 '12 at 11:20

Actually, there is an $E_9$; it's just not finite-dimensional. That is, you can write down a diagram which deserves to be called $E_9$ and construct the corresponding Lie algebra by generators and relations, but I believe it will be an infinite-dimensional Kac-Moody algebra rather than a finite-dimensional simple Lie algebra.

One way to understand the ADE classification (which gives us the simply laced Dynkin diagrams) is that it gives precisely the simple graphs with greatest eigenvalue less than $2$ in absolute value, and $E_9$ doesn't have this property. This condition is important in the theory of simple Lie algebras for reasons I don't completely understand, but in any case you can extract a straightforward proof of it from the proof of a closely related result in this blog post.

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Ah, cool. So it is also possible to extend all the other diagrams? – draks ... Jan 4 '12 at 23:12