TITLE: Classification by Dynkin diagram: Why's there no $E_9$?
QUESTION [6 upvotes]: 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?

REPLY [10 votes]: 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.