Reflection Functor of a Quiver












8












$begingroup$


Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.



Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$



The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
end{cases}$$



Alternatively one can think of the sequence
$$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
in which the first map is the trivial map and the second is inclusion.



My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?



My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.



Is this correct?



If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?



Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?



As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.










share|cite|improve this question











$endgroup$

















    8












    $begingroup$


    Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.



    Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$



    The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
    kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
    end{cases}$$



    Alternatively one can think of the sequence
    $$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
    in which the first map is the trivial map and the second is inclusion.



    My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?



    My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.



    Is this correct?



    If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?



    Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?



    As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.










    share|cite|improve this question











    $endgroup$















      8












      8








      8


      2



      $begingroup$


      Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.



      Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$



      The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
      kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
      end{cases}$$



      Alternatively one can think of the sequence
      $$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
      in which the first map is the trivial map and the second is inclusion.



      My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?



      My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.



      Is this correct?



      If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?



      Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?



      As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.










      share|cite|improve this question











      $endgroup$




      Fix a Quiver $Q=(Q_{0},Q_{1})$, where $Q_{0}$ is the set of vertices and $Q_{1}$ is the set of edges. For $jin Q_{0}$ define $sigma_{j}Q$ to be the quiver in which all that include the vertex $j$ are flipped.



      Suppose $iin Q_{0}$ is a sink (all edges including $i$ are directed towards $i$) then we have a reflection functor $$F_{i}^{+}:Rep(Q)to Rep(sigma_{i}Q).$$



      The functor is defined as follows; given a representation $Xin Rep(Q)$, that is an assignment of a finite-dimensional vector space $X_{j}$ to each $jin Q_{0}$ and a linear map $f_{alpha}:X_{j}to X_{k}$ for each $alpha: jto kin Q_{1}$, define $$(F_{i}^{+}X)_{j}=begin{cases} X_{j} &jneq i,\
      kerleft(bigoplus_{alpha:jto i} f_{alpha}right) & i=j.
      end{cases}$$



      Alternatively one can think of the sequence
      $$ 0longrightarrow (F_{i}^{+}X)_{i} longrightarrow bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}longrightarrow X_{i}$$
      in which the first map is the trivial map and the second is inclusion.



      My question is what is the third map and how do I think about $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$?



      My guess is that you take the direct sum of all the vector spaces $X_{alpha}$ which map into the sink $X_{i}$ and then the map is just given component-wise, i.e. some tuple $(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}$ is mapped to $(f_{1}x_{1},ldots,f_{n}x_{n})$ where $f_{j}:X_{j}to X_{i}$ is the map in the representation of $Q$.



      Is this correct?



      If so does this mean that $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)=bigoplus_{alpha:jto i} ker(f_{alpha})$?



      Finally, do we have any information about the dimension of $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)$ in general?



      As a reference I am reading these notes, in particular the part on reflection functors is in chapter 3 on page 9.







      representation-theory quiver






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 18 '18 at 0:57







      user551642

















      asked Dec 18 '18 at 0:24









      user551642user551642

      2527




      2527






















          2 Answers
          2






          active

          oldest

          votes


















          4












          $begingroup$

          I think I have figured it out;



          let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
          $$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$



          Although I still don't know anything about the dimension in general.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 3:56












          • $begingroup$
            Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 4:03





















          2












          $begingroup$

          If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Nice, this will be very useful. Thank you.
            $endgroup$
            – user551642
            Dec 18 '18 at 23:29











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          2 Answers
          2






          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          4












          $begingroup$

          I think I have figured it out;



          let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
          $$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$



          Although I still don't know anything about the dimension in general.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 3:56












          • $begingroup$
            Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 4:03


















          4












          $begingroup$

          I think I have figured it out;



          let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
          $$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$



          Although I still don't know anything about the dimension in general.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 3:56












          • $begingroup$
            Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 4:03
















          4












          4








          4





          $begingroup$

          I think I have figured it out;



          let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
          $$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$



          Although I still don't know anything about the dimension in general.






          share|cite|improve this answer











          $endgroup$



          I think I have figured it out;



          let $h:bigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j} to X_{i}$ be the third map in the sequence then we define $h(x_{1},ldots,x_{n}):=f_{1}x_{1}+cdots+f_{n}x_{n}$, where $f_{j}:X_{j}to X_{i}$ is the map associated to the edge $alpha:jto iin Q_{1}$. Then we have
          $$ker(h)={(x_{1},ldots,x_{n})inbigoplus_{substack{alphain Q_{1}\ alpha:jto i}} X_{j}: f_{1}x_{1}+cdots+f_{n}x_{n}=0}.$$



          Although I still don't know anything about the dimension in general.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 1:38

























          answered Dec 18 '18 at 1:33









          user551642user551642

          2527




          2527












          • $begingroup$
            Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 3:56












          • $begingroup$
            Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 4:03




















          • $begingroup$
            Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 3:56












          • $begingroup$
            Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
            $endgroup$
            – Mike Pierce
            Dec 18 '18 at 4:03


















          $begingroup$
          Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
          $endgroup$
          – Mike Pierce
          Dec 18 '18 at 3:56






          $begingroup$
          Yep, that's it. I think that looking at the next lemma, 3.3.2, which explain what the composite functors $F_i^{+}F_i^{-}$ and $F_i^{-}F_i^{+}$ do, helps a bit. I don't know what could be said about the dimension of that kernel in general though. The map $bigoplus_{alpha:jto i} X_{j} to X_i$ can be designed to be between any two vector spaces you like, all depending on the representation $X$. You can say that the dimension of that kernel is at least $dimbigoplus_{alpha:jto i} X_{j} - dim X_i$, but I can't see much more.
          $endgroup$
          – Mike Pierce
          Dec 18 '18 at 3:56














          $begingroup$
          Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
          $endgroup$
          – Mike Pierce
          Dec 18 '18 at 4:03






          $begingroup$
          Oh yeah, and $kerleft(bigoplus_{alpha:jto i} f_{alpha}right)neqbigoplus_{alpha:jto i} ker(f_{alpha})$. Consider an $A_3$ quiver $1 rightarrow 2 leftarrow 3$ and representation $X$ where each $X_i$ is $1$-dimensional and all the maps are the identity map. Individually the maps each have no kernel, but the map $X_1oplus X_3 to X_2$ certainly will.
          $endgroup$
          – Mike Pierce
          Dec 18 '18 at 4:03













          2












          $begingroup$

          If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Nice, this will be very useful. Thank you.
            $endgroup$
            – user551642
            Dec 18 '18 at 23:29
















          2












          $begingroup$

          If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Nice, this will be very useful. Thank you.
            $endgroup$
            – user551642
            Dec 18 '18 at 23:29














          2












          2








          2





          $begingroup$

          If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.






          share|cite|improve this answer











          $endgroup$



          If the representation is indecomposable, the map $oplus_{alpha:j to i} f_alpha$ is surjective (otherwise you could split off a complement to its image as a summand) and hence the dimension of its kernel is the sum over the incoming arrows of the dimensions of the vector spaces at their sources minus the dimension of the vector space at $i$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 18 '18 at 17:02

























          answered Dec 18 '18 at 16:28









          StephenStephen

          10.7k12439




          10.7k12439












          • $begingroup$
            Nice, this will be very useful. Thank you.
            $endgroup$
            – user551642
            Dec 18 '18 at 23:29


















          • $begingroup$
            Nice, this will be very useful. Thank you.
            $endgroup$
            – user551642
            Dec 18 '18 at 23:29
















          $begingroup$
          Nice, this will be very useful. Thank you.
          $endgroup$
          – user551642
          Dec 18 '18 at 23:29




          $begingroup$
          Nice, this will be very useful. Thank you.
          $endgroup$
          – user551642
          Dec 18 '18 at 23:29


















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