Lectures on Representations of Quivers by William Crawley-Boevey ...
Lectures on Representations of Quivers by William Crawley-Boevey ...
Lectures on Representations of Quivers by William Crawley-Boevey ...
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THE STANDARD RESOLUTION.<br />
¡ ¥ ¡ ¥ ¡ ¡<br />
t(¢ s(¢<br />
Let A=kQ. If X is a left A-module, there is an exact sequence<br />
0<br />
f<br />
Ae e X<br />
) k )<br />
n<br />
g<br />
Ae e X X 0<br />
i k i<br />
¡ ¢<br />
x¡ a¡<br />
¦¢ a¡ x¡ t(¢ s(¢<br />
a¢¡<br />
t(¢ s(¢<br />
Q1 i=1<br />
where g(a x) = ax for Ae , e X, and<br />
i i<br />
f(a x) = x - a x for Ae<br />
)<br />
and e X<br />
)<br />
in ) ) comp<strong>on</strong>ent.<br />
PROOF. Clearly g© f=0 and g is <strong>on</strong>to. If ¢<br />
<strong>of</strong> the sequence, we can write it uniquely in the form<br />
¢<br />
n<br />
is an element <strong>of</strong> the middle term<br />
= ¤ ¤ a x (x ¡ e X almost all zero)<br />
a a s(a)<br />
i=1 paths a with s(a)=i<br />
and define degree( ¢<br />
) = length <strong>of</strong> the l<strong>on</strong>gest path a with x £ 0.<br />
a<br />
If a is a n<strong>on</strong>-trivial path with s(a)=i, then we can express it as a product<br />
a=a¥ ¢ with ¢ an arrow with s(¢ )=i, and a¥ another path. Viewing a¥£ x as an<br />
a<br />
element in the ¢ ’th comp<strong>on</strong>ent <strong>of</strong> left hand term, we have<br />
x ) = a x - a¥£ ¦¢ x .<br />
f(a¥£<br />
a a a<br />
We claim that ¢<br />
¢<br />
has degree d>0, then<br />
¢<br />
n<br />
+ Im(f) always c<strong>on</strong>tains an element <strong>of</strong> degree 0. Namely, if<br />
- f( ¤ ¤ a¥£ x )<br />
a<br />
i=1 paths a with s(a)=i and length d<br />
has degree < d, so the claim follows <strong>by</strong> inducti<strong>on</strong>.<br />
Im(f)=Ker(g): If ¢<br />
¢<br />
+Im(f) have degree zero. Thus<br />
Ker(g), let ¡ ¢ ¡ ¥<br />
0 = g( ¢<br />
) = g( ¢ ) = g(¤ e x¥ ) = ¤ x¥<br />
i i e e<br />
¥<br />
i i<br />
n<br />
Now this bel<strong>on</strong>gs ¥ to X , so each term in the final sum must be zero.<br />
i=1 i<br />
Thus ¢ =0, and the asserti<strong>on</strong> follows.<br />
¥<br />
Ker(f)=0: we can write an element ¢<br />
¡<br />
¢<br />
¢ s(¢ ¡ ¢ s(a)=t(¢ ¢<br />
£ ¢ ¢<br />
Ker(f) in the form<br />
Q<br />
1<br />
paths a with ) ,a ,a )<br />
Let a be a path <strong>of</strong> maximal length such that x 0 (some ). Now<br />
,a<br />
f( ¢<br />
= ¤ ¤ a x (x ¡ e X almost all 0).<br />
) = ¤ ¤ a¢¡ x - ¤ ¤ a ¦¢ x<br />
¢ a ¢ ,a ¢ a ¢ ,a<br />
so the coefficient <strong>of</strong> a¢ in f( ¢<br />
) is x . A c<strong>on</strong>tradicti<strong>on</strong>.<br />
¢ ,a<br />
7