The tilting module correpsonding to a tilting object of cluster category $mathcal{C}$












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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?






category-theory modules representation-theory homological-algebra







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    $begingroup$


    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?






    category-theory modules representation-theory homological-algebra







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      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?






      category-theory modules representation-theory homological-algebra







      share|cite|improve this question









      $endgroup$




      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?






      category-theory modules representation-theory homological-algebra




      category-theory modules representation-theory homological-algebra






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      asked Sep 28 '18 at 8:30









      Xiaosong PengXiaosong Peng

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          $begingroup$

          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 $B cong End_{mathcal{C}}(tau^ntilde{T})$.



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






          share|cite|improve this answer









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            $begingroup$

            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 $B cong End_{mathcal{C}}(tau^ntilde{T})$.



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






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              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 $B cong End_{mathcal{C}}(tau^ntilde{T})$.



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






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                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 $B cong End_{mathcal{C}}(tau^ntilde{T})$.



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






                share|cite|improve this answer









                $endgroup$



                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 $B cong End_{mathcal{C}}(tau^ntilde{T})$.



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







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 17 '18 at 20:43









                Pierre-Guy PlamondonPierre-Guy Plamondon

                8,88511739




                8,88511739






























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