The cohomology structure of string algebras

THE COHOMOLOGY STRUCTURE OF STRING ALGEBRAS 7
Proof : By construction we have φ ≤ ∈ B n 0
∐ B
n
+ , and φ ≥ ∈ B n −
∐ B
n
0 . □
2.3. Lemma. Let φ ∈ Kerδ n , and α 1 · · · α n ∈ Γ n such that φ(α 1 · · · α n ) ≠ 0.
a) If there exists α n+1 such that α 1 · · · α n+1 ∈ Γ n+1 then φ ≤ (α 1 · · · α n )α n+1 =
0,
b) If there exists α 0 such that α 0 · · · α n ∈ Γ n+1 then α 0 φ ≥ (α 1 · · · α n ) = 0
Proof : We only prove a). Let φ ∈ Kerδ n . Then, since φ − φ ≤ ∈ Imδ n−1 , we have
φ ≤ ∈ Kerδ n , thus
0 = δ n φ ≤ (α 1 · · · α n α n+1 )
= α 1 φ ≤ (α 2 · · · α n+1 ) + (−1) n+1 φ ≤ (α 1 · · · α n )α n+1 .
The statement follows from the fact that paths are linearly independent, and from
the preceding lemma.
□
3. The cohomology structure
The morphisms µ • : K min • (A) K
rad • (A), and ω • : K
rad • (A) K min • (A) allow
us to carry the products defined in C
rad • (A, A) to C• min (A, A). In this way we
obtain a cup product and a bracket, which we still denote ∪ and [−, −]. More
precisely, applying the functor Hom A e(−, A) and making the identifications of section
1.5, we obtain morphisms of complexes µ • : C min • (A, A) C
rad • (A, A), and
ω • : C
rad • (A, A) C min • (A, A). Given f ∈ Cn min (A, A), g ∈ Cm min (A, A), define
f ∪ g ∈ C n+m
min (A, A) as
f ∪ g = µ n+m (ω n f ∪ ω m f).
Thus, given an element α 1 · · · α n β 1 · · · β m ∈ Γ n+m , we have
f ∪ g(α 1 · · · α n β 1 · · · β m ) = f(α 1 · · · α n )g(β 1 · · · β m ).
As usual, we have δ n+m (f ∪ g) = δ n f ∪ g + (−1) m f ∪ δ m g, so the product ∪
defined in C min • (A, A) induces a product at the cohomology level. In fact, the latter
coincides with the cup-product of section 1.5 and the Yoneda product. In what
follows, we work with the product ∪ defined in C min • (A, A).
3.1. Theorem. Let A = kQ/I be a triangular quadratic string algebra, n > 0
and m > 0. Then HH n (A) ∪ HH m (A) = 0.
Proof : Let φ ∈ HH n (A), and ψ ∈ HH m (A). We will show that φ ≤ ∪ ψ ≥ = 0
in HH n+m (A). Assume the contrary is true. In particular there exist an element
α 1 · · · α n β 1 · · · β m ∈ Γ n+m such that
φ ≤ (α 1 · · · α n )ψ ≥ (β 1 · · · β m ) ≠ 0.
With the notations of Lemma 2.2, this is equivalent to
(1) (W + Qα n )(W ′ + β 1 P ) ∉ I.
Now, since α n β 1 ∈ I, Lemma 2.3 gives
(W + Qα n )β 1 ∈ I, and α 1 (W ′ + β 1 P ) ∈ I.
Thus, equation (1) gives W W ′ ∉ I. Let w and w ′ be paths appearing in W and
W ′ respectively, such that ww ′ ∉ I. Moreover, write w = uγ, w ′ = γ ′ u ′ with γ, γ ′
arrows of Q. Since ww ′ ∉ I, we have that γγ ′ ∉ I. But then, since (Q, I) satisfies
(S3), this implies α n γ ′ ∈ I, so that α 1 · · · α n γ ′ ∈ Γ n+1 and, again, Lemma 2.3 gives
(W + Qα n )γ ′ ∈ I so that W γ ′ ∈ I, and, in particular wγ ′ ∈ I, a contradiction. □