Proceedings of the 44th Symposium on Ring Theory and ...

1 arrows <strong>of</strong> ˜µ L k (Q Λ, W Λ , C Λ ) and <strong>the</strong> above C ′ is obtained in this way. Thus we identify
degree 1 arrows as a cut.
Because we have P(˜µ L k (Q Λ, W Λ , C Λ )) ∼ = P(µ L k (Q Λ, W Λ , C Λ )), we can rewrite Theorem 7
that we have an algebra isomorphism
End Λ (T k ) ∼ = P(µ L k (Q Λ , W Λ , C Λ )).
3.2. Examples. We explain <strong>the</strong> <strong>the</strong>orem with some examples.
Example 9. We keep <strong>the</strong> assumption <strong>of</strong> Theorem 7.
Λ = KQ and
P(µ L k (Q Λ , W Λ , C Λ )) = P(µ L k (Q, 0, 0)) = K(µ k Q),
If gl.dimΛ = 1, <strong>the</strong>n we have
so that <strong>the</strong> mutation procedure is just reversing arrows having k. Thus <strong>the</strong> above <strong>the</strong>orem
coincides with <strong>the</strong> classical result (Theorem 1).
Example 10. Let Λ = ̂KQ/〈R〉 be a finite dimensional algebra given by <strong>the</strong> following
quiver with a relation.
2 ●
a ●●●● b
✇
1
✇✇✇✇
❊ 4
❊❊❊❊
c 2 d
3
22222
〈R〉 = 〈ab〉.
Then we consider <strong>the</strong> APR tilting module T 1 := τ − P 1 ⊕ Λ/P 1 and calculate Q ′ and R ′
satisfying ̂KQ ′ /〈R ′ 〉 ∼ = End Λ (T 1 ) by <strong>the</strong> following steps.
2 ❈
a ❈❈❈❈❈ b
4
1
44444
❈ 4
❈❈❈❈
c 4 d
3
44444
(Q Λ ,W Λ ,C Λ )
=⇒
2 ❈
a ❈❈❈❈❈ b
4
1
44444 ρ
❈ 4
❈❈❈❈
c 4 d
3
44444
˜µ L 1
=⇒
[ρa]
2
a ∗ ● ●●●●
✇ ✇✇✇✇ b
ρ
1
∗ ●
●●●●
4
d
c ∗ ✇
3
✇✇✇✇
[ρc]
〈R〉 = 〈ab〉. W Λ = ρab. W ′ = [ρa]b+[ρa]a ∗ ρ ∗ +[ρc]c ∗ ρ ∗ .
µ L 1
=⇒ 1
2
2
a ∗
a ∗
5 55555 ρ ∗ P(µ
4
L 1 (Q ✇ ✇✇✇✇ ρ
Λ,W Λ ,C Λ ))
❇ =⇒ Q ′ ❇❇❇❇❇
=
1
∗ ●
●●●●
4
d
✇ ✇✇✇✇
c ∗
3
d
5 55555
[ρc]
W ′ = [ρc]c ∗ ρ ∗ 〈R ′ 〉 = 〈c ∗ ρ ∗ 〉.
Similarly from <strong>the</strong> left-hand side algebra Λ, we obtain <strong>the</strong> quiver and <strong>the</strong> set <strong>of</strong> relations
giving End Λ (T 1 ), which is given by right-hand side picture.
–118–
c ∗
3