Lectures on Representations of Quivers by William Crawley-Boevey ...

PROOF.
(2) By inspection the given vector § is radical, eg if there are no loops
or multiple edges, we need to check that
2 §
= ¤ § .
i neighbours j of i j
Now q is positive semi-definite by the lemma. Finally, since some § ¦ §¡ = ¡ ¦ §
.
=1,
i
rad(q) =
n
~
(1) Embed the Dynkin diagram in the corresponding Euclidean diagram , and
~
note that the quadratic form for is strictly positive on non-zero,
non-sincere vectors.
(3) It is not hard to show that has a Euclidean subgraph ¥ , say with
radical vector § . If all vertices of are in ¥ take ¤ = § . If i is a vertex
not in ¥ , connected to ¥ by an edge, take ¤ =2 § + i
EXTENDING VERTICES.
If is Euclidean, a vertex e is called an extending vertex if § =1. Note
e
(1) There always is an extending vertex.
(2) The graph obtained by deleting e is the corresponding Dynkin diagram.
NOW SUPPOSE that is Dynkin or Euclidean, so q is positive semi-definite.
ROOTS.
We define
n ¢ ¡ ¦ ¢ ¤ £ q(¤ {¤ )¤ = 0, 1}, the set of roots.
A root ¤ is real if q(¤ )=1 and imaginary if q(¤ )=0.
REMARK.
One can define roots for any graph , and more generally for valued
graphs (in which situation the Dynkin diagrams B ,C ,F ,G also arise). In
n n 4 2
case the graph has no loops, this can be found in Kac’s book on infinite
dimensional Lie algebras. In case there are loops, the definition can be
found in V.G.Kac, Some remarks on representations of quivers and infinite
root systems, in Springer Lec. Notes 832.
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