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

THE STANDARD RESOLUTION.
¡ ¥ ¡ ¥ ¡ ¡
t(¢ s(¢
Let A=kQ. If X is a left A-module, there is an exact sequence
0
f
Ae e X
) k )
n
g
Ae e X X 0
i k i
¡ ¢
x¡ a¡
¦¢ a¡ x¡ t(¢ s(¢
a¢¡
t(¢ s(¢
Q1 i=1
where g(a x) = ax for Ae , e X, and
i i
f(a x) = x - a x for Ae
)
and e X
)
in ) ) component.
PROOF. Clearly g© f=0 and g is onto. If ¢
of the sequence, we can write it uniquely in the form
¢
n
is an element of the middle term
= ¤ ¤ a x (x ¡ e X almost all zero)
a a s(a)
i=1 paths a with s(a)=i
and define degree( ¢
) = length of the longest path a with x £ 0.
a
If a is a non-trivial path with s(a)=i, then we can express it as a product
a=a¥ ¢ with ¢ an arrow with s(¢ )=i, and a¥ another path. Viewing a¥£ x as an
a
element in the ¢ ’th component of left hand term, we have
x ) = a x - a¥£ ¦¢ x .
f(a¥£
a a a
We claim that ¢
¢
has degree d>0, then
¢
n
+ Im(f) always contains an element of degree 0. Namely, if
- f( ¤ ¤ a¥£ x )
a
i=1 paths a with s(a)=i and length d
has degree < d, so the claim follows by induction.
Im(f)=Ker(g): If ¢
¢
+Im(f) have degree zero. Thus
Ker(g), let ¡ ¢ ¡ ¥
0 = g( ¢
) = g( ¢ ) = g(¤ e x¥ ) = ¤ x¥
i i e e
¥
i i
n
Now this belongs ¥ to X , so each term in the final sum must be zero.
i=1 i
Thus ¢ =0, and the assertion follows.
¥
Ker(f)=0: we can write an element ¢
¡
¢
¢ s(¢ ¡ ¢ s(a)=t(¢ ¢
£ ¢ ¢
Ker(f) in the form
Q
1
paths a with ) ,a ,a )
Let a be a path of maximal length such that x 0 (some ). Now
,a
f( ¢
= ¤ ¤ a x (x ¡ e X almost all 0).
) = ¤ ¤ a¢¡ x - ¤ ¤ a ¦¢ x
¢ a ¢ ,a ¢ a ¢ ,a
so the coefficient of a¢ in f( ¢
) is x . A contradiction.
¢ ,a
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