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

LEMMA. The category Rep(Q) is equivalent to kQ-Mod.
PROOF. We only give the construction. If X is a kQ-module, define a
representation X with
X = e X
i i
X (x) = ¢ x = e ¢ x ¡ X for x¡ X .
¢ t(¢ ) t(¢ ) s(¢ )
If X is a representation, define a module X via
n i X=¥ i
X . Let ¡ X ¡ X X be the canonical maps.
i=1 i i i
¢ ...¢ x = X ...X ¡ (x)
1 m t(¢ ) ¢ ¢ s(¢ )
1 1 m m
e x = ¢¡ (x),
i i i
¡
It is straightforward, but tedious, to check that these are inverses and
that morphisms behave, etc. We can now use the same letter for a module and
the corresponding representation, ignoring the distinction.
EXAMPLE. Under this correspondence, the representations S(i) are simple
modules. Moreover, if Q has no oriented cycles, it is easy to see that the
S(i) are the only simple modules.
DEFINITIONS.
(1) The dimension vector of a finite dimensional kQ-module X is the vector
n
dim X ¡ £ , with
¤
(dim X)
i
= dim X
i
= dim e X = dim Hom(Ae ,X).
i i
Thus dim X =
n
i=1
(dim X) .
i
¢ ¡ s(¢ t(¢
n n
i=1 i i Q1 ) )
(2) The Euler form is = ¤ ¤¢¥ - ¤ ¤ ¥ for ¤ ,¥ ¡§¦ .
n
This is a bilinear form on ¦ .
n
(3) The Tits form is q(¤ ) = . This is a quadratic form on ¦ .
(4) The Symmetric bilinear form is (¤ ,¥ ) = + .
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