4
$\begingroup$

I am reading the paper "representation dimension of cluster-concealed algebras", the link is here: https://arxiv.org/pdf/1102.1048v1.pdf

Let $H$ be a finite dimensional hereditary algebra. $\mathcal{C}$ is the cluster category associated to $H$.

In section 2.2 of this paper, there is a theorem: each basic tilting module over $H$ induces a basic tilting object for $\mathcal{C}$ and each basic tilting object in $\mathcal{C}$ is induced by a basic tilting module over a hereditary algebra $H'$, derived equivalent to $H$

At the start of section 3, there are the following words: Let $\widetilde{T}$ be a tilting object in a cluster category $\mathcal{C}$ and let $B=End_{\mathcal{C}}(\widetilde{T})$ be the associated cluster-tilted algebra. To simplify some proofs, we choose without lose of generality $T$ and $\tau T$ without projective summands.

I want to know that why we could choose $T$ and $\tau T$ without projective summands without lose of generality? Could the theorem in section 2.2 make sure we choose $T$ such that $T$ and $\tau T$ without projective summands?

$\endgroup$

1 Answer 1

2
$\begingroup$

The authors of the paper are only interested in the endomorphism algebra $B$ of $\tilde{T}$. If one applies any automorphism $\Phi$ to $\tilde{T}$, one gets an isomorphism between $End_{\mathcal{C}}(\tilde{T})$ and $End_{\mathcal{C}}(\Phi(\tilde{T}))$.

In the last sentence of the first paragraph of Section 3, the authors also assume that $H$ is of infinite representation type. Thus there exists an integer $n$ such that $\tau^n \tilde{T}$ and its Auslander-Reiten translation do not have any projective direct summand. Then $B \cong End_{\mathcal{C}}(\tau^n\tilde{T})$.

Thus the authors can assume, without loss of generality, that $\tilde{T}$ and $\tau\tilde{T}$ do not have any projective direct summands.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.