The invariance of the Auslander-Reiten Formula for hereditary algebras

Let Λ be a finite dimensional hereditary algebra over a field k, with Auslander-Reiten translate τ and inverse translate τ -. The Auslander-Reiten Formula says that there is a natural isomorphism of bifunctors Ext 1 (X, Y ) ∼ = D Hom(τ -Y, X) for all finite dimsional modules X, Y ∈ mod Λ. Alternatively, we can express this as a bifunctorial perfect pairing {-, -} ′ : Ext 1 (X, Y ) × Hom(τ -Y, X) → k.

If now X has no non-zero injective direct summands, then τ -ζ is again exact for all ζ ∈ Ext 1 (X, Y ), so given f ∈ Hom(τ -Y, X), we can compute both {ζ, f } ′ and {τ -ζ, τ -f } ′ . Our main result is the these two expressions always agree. Beyond settling this natural question, this result plays a key role in showing that, over the associated preprojective algebra Π, we have a natural isomorphism of bifunctors Ext 2 Π (X, Y ) ∼ = D Hom(Y, X) [5].

We remark that our result holds for all finite dimensional hereditary algebras, and not just for those which are tensor algebras. This phenomenon can occur whenever the semisimple algebra Λ/J(Λ) is not separable over k, which happens even for some tame hereditary algebras. Of course, if one restricts to tensor algebras, or even further to path algebras of quivers, then certain constructions admit canonical splittings which can be exploited to simplify the proof.

Acknowledgements. This work was supported by Deutsche Forschungsgemeinschaft (Project-ID 491392403 -TRR 358).

We fix a base field k. Let Λ be a finite dimensional hereditary algebra. It is known that we can always write Λ = A ⊕ J such that A is a semisimple algebra and J is the Jacobson radical [1,9]. It follows that the epimorphism J → J/J 2 splits as left A-modules, and also as right A-modules, but in general it will not be split as A-bimodules. In fact, this happens if and only if Λ is isomorphic to the tensor algebra T A (J/J 2 ).

Remark 2.1. The natural map J → J/J 2 splits as A-bimodules in the following situations.

(1) The ext-quiver of Λ is a tree. This includes all hereditary algebras of finite representation type. See [2, Proposition 10.2]. (2) There are no indecomposable projective modules P, P ′ with rad n (P, P ′ ) ̸ = 0

for n = 1, r for some r > 1. See [3, Section 5].

(3) The k-algebra A is separable. See [7].

On the other hand, there are finite dimensional hereditary algebras which are not tensor algebras. See [8,3], where it is shown that this fails even for tame hereditary algebras.

Associated to Λ is the fundamental short exact sequence

where p is the usual multiplication map. Note that this splits as left Λ-modules, and as right Λ-modules. The kernel Ω is sometimes called the bimodule of noncommutative 1-forms on Λ over A. Just as in the construction of the bar resolution we can identify Ω with the image of the bimodule homomorphism

We write dµ for the image of 1 Section 10], though the relative case is not considered there).

In the special situation that Λ = T A (M ) is a tensor algebra, then Ω ∼ = Λ ⊗ A M ⊗ A Λ, and Der A (Λ, X) ∼ = Hom A-A (M, X). Proposition 2.2. Every right Λ-module X admits a standard projective resolution

as well as a standard injective coresolution

Here, and elsewhere except in Section 3, unadorned tensor products, and homomorphism and extension spaces, will be over Λ.

Proof. We have P X = X ⊗ P, and this is exact as P splits as a sequence of left Λ-modules. As A is semisimple the middle term X ⊗ A Λ is a projective right Λ-module, and hence so too is X ⊗ Ω since Λ is hereditary.

Similarly, we have the short exact sequence Hom(P, X), and up to sign this equals I X . Again, the middle term is an injective right Λ-module, and hence so too is Hom(Ω, X). □ Remark 2.3. We note that, regarding P as a chain complex in degrees {-2, -1, 0}, and the module X as a stalk complex in degree 0, the sign conventions from [6, Section 0FNG] tell us that Hom(P, X) is a chain complex in degrees {0, 1, 2} having the differentials -Hom(p, X) and Hom(i, X). We have chosen instead to take j X = Hom(p, X) and q X = -Hom(i, X). The important point, though, is that there is some sign involved, so we should not take both Hom(p, X) and Hom(i, X). This viewpoint is further justified Lemma 4.8.

Remark 2.4. The dual result for left Λ-modules also holds.

Remark 2.5. For a projective module P we know that both P and (P/P J) ⊗ A Λ are projective covers of P/P J, so are isomorphic. In particular, we can apply this to

For, considering the standard projective resolution P ⊗ A gives Ω ⊗ A ∼ = J as Λ-A-bimodules, so X ⊗ Ω ⊗ A ∼ = X ⊗ J as right A-modules. Also, as Λ is hereditary, J i is projective, and hence X ⊗ J i ∼ = XJ i . Thus, as right A-modules, we have

Similarly, we have

proving the claim. In general, however, the isomorphism X ⊗ Ω ∼ = X ⊗ A M ⊗ A Λ is not natural in X. In fact, if it is natural in X, then taking X = A yields an isomorphism of A-Λ-bimodules J ∼ = A ⊗ Ω ∼ = M ⊗ A Λ. The canonical map J → J/J 2 thus becomes M ⊗ A Λ → M , which is clearly split as A-bimodules,

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