direct sum of representation of product groups












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Given two finite groups $G_1$ and $G_2$, and some representations $rho_1: G_1 to V_1$ and $rho_2: G_2 to V_2$, it seems the standard way to create a representation for $G_1 times G_2$ is to use the tensor product
$$rho_1(g_1) otimes rho_2(g_2) quad g_1,g_2 in G_1,G_2.$$
It seems to me that one could also use the direct sum
$rho_1(g_1) oplus rho_2(g_2)$,
because the blocks in the matrix form of the representation do not interact and one gets the desired effect. Given that this representation could have a lower dimension than using tensor product, why is it not used?






group-theory finite-groups representation-theory







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    $begingroup$


    Given two finite groups $G_1$ and $G_2$, and some representations $rho_1: G_1 to V_1$ and $rho_2: G_2 to V_2$, it seems the standard way to create a representation for $G_1 times G_2$ is to use the tensor product
    $$rho_1(g_1) otimes rho_2(g_2) quad g_1,g_2 in G_1,G_2.$$
    It seems to me that one could also use the direct sum
    $rho_1(g_1) oplus rho_2(g_2)$,
    because the blocks in the matrix form of the representation do not interact and one gets the desired effect. Given that this representation could have a lower dimension than using tensor product, why is it not used?






    group-theory finite-groups representation-theory







    share|cite|improve this question









    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Given two finite groups $G_1$ and $G_2$, and some representations $rho_1: G_1 to V_1$ and $rho_2: G_2 to V_2$, it seems the standard way to create a representation for $G_1 times G_2$ is to use the tensor product
      $$rho_1(g_1) otimes rho_2(g_2) quad g_1,g_2 in G_1,G_2.$$
      It seems to me that one could also use the direct sum
      $rho_1(g_1) oplus rho_2(g_2)$,
      because the blocks in the matrix form of the representation do not interact and one gets the desired effect. Given that this representation could have a lower dimension than using tensor product, why is it not used?






      group-theory finite-groups representation-theory







      share|cite|improve this question









      $endgroup$




      Given two finite groups $G_1$ and $G_2$, and some representations $rho_1: G_1 to V_1$ and $rho_2: G_2 to V_2$, it seems the standard way to create a representation for $G_1 times G_2$ is to use the tensor product
      $$rho_1(g_1) otimes rho_2(g_2) quad g_1,g_2 in G_1,G_2.$$
      It seems to me that one could also use the direct sum
      $rho_1(g_1) oplus rho_2(g_2)$,
      because the blocks in the matrix form of the representation do not interact and one gets the desired effect. Given that this representation could have a lower dimension than using tensor product, why is it not used?






      group-theory finite-groups representation-theory




      group-theory finite-groups representation-theory






      share|cite|improve this question













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      asked 4 hours ago









      self-educatorself-educator

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      4611






















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          When $V_1$ and $V_2$ are representations of $G_1$ and $G_2$ respectively, I'll use $V_1 boxtimes V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 otimes_{mathbb{C}} V_2$, and $V_1 boxplus V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 oplus V_2$.



          If $V$ is an irreducible representation of $G_1 times G_2$, then $V$ is isomorphic to $V_1 boxtimes V_2$ for some irreducible representations $V_1$ and $V_2$ of $G_1$ and $G_2$ respectively. This means that if we know the representations of $G_1$ and $G_2$, then using the $boxtimes$ construction we can get to all the (irreducible) representations of $G_1 times G_2$. Conversely, the $boxtimes$ product of two irreducible representations always produces an irreducible representation of $G_1 times G_2$.



          On the other hand, $V_1 boxplus V_2$ is always reducible as a $G_1 times G_2$ representation, since both vector subspaces $V_1$ and $V_2$ are stable under the $G_1 times G_2$ action. On the $V_1$ subspace, really only the $G_1$ part of the group acts, and the $G_2$ part acts trivially, and similarly for the $V_2$ subspace. We cannot produce all irreducible representations of $G_1 times G_2$ using this construction, which can already be seen in the example $G_1 = G_2 = mathbb{Z} / 2 mathbb{Z}$.






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            $begingroup$

            When $V_1$ and $V_2$ are representations of $G_1$ and $G_2$ respectively, I'll use $V_1 boxtimes V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 otimes_{mathbb{C}} V_2$, and $V_1 boxplus V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 oplus V_2$.



            If $V$ is an irreducible representation of $G_1 times G_2$, then $V$ is isomorphic to $V_1 boxtimes V_2$ for some irreducible representations $V_1$ and $V_2$ of $G_1$ and $G_2$ respectively. This means that if we know the representations of $G_1$ and $G_2$, then using the $boxtimes$ construction we can get to all the (irreducible) representations of $G_1 times G_2$. Conversely, the $boxtimes$ product of two irreducible representations always produces an irreducible representation of $G_1 times G_2$.



            On the other hand, $V_1 boxplus V_2$ is always reducible as a $G_1 times G_2$ representation, since both vector subspaces $V_1$ and $V_2$ are stable under the $G_1 times G_2$ action. On the $V_1$ subspace, really only the $G_1$ part of the group acts, and the $G_2$ part acts trivially, and similarly for the $V_2$ subspace. We cannot produce all irreducible representations of $G_1 times G_2$ using this construction, which can already be seen in the example $G_1 = G_2 = mathbb{Z} / 2 mathbb{Z}$.






            share|cite|improve this answer









            $endgroup$


















              6












              $begingroup$

              When $V_1$ and $V_2$ are representations of $G_1$ and $G_2$ respectively, I'll use $V_1 boxtimes V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 otimes_{mathbb{C}} V_2$, and $V_1 boxplus V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 oplus V_2$.



              If $V$ is an irreducible representation of $G_1 times G_2$, then $V$ is isomorphic to $V_1 boxtimes V_2$ for some irreducible representations $V_1$ and $V_2$ of $G_1$ and $G_2$ respectively. This means that if we know the representations of $G_1$ and $G_2$, then using the $boxtimes$ construction we can get to all the (irreducible) representations of $G_1 times G_2$. Conversely, the $boxtimes$ product of two irreducible representations always produces an irreducible representation of $G_1 times G_2$.



              On the other hand, $V_1 boxplus V_2$ is always reducible as a $G_1 times G_2$ representation, since both vector subspaces $V_1$ and $V_2$ are stable under the $G_1 times G_2$ action. On the $V_1$ subspace, really only the $G_1$ part of the group acts, and the $G_2$ part acts trivially, and similarly for the $V_2$ subspace. We cannot produce all irreducible representations of $G_1 times G_2$ using this construction, which can already be seen in the example $G_1 = G_2 = mathbb{Z} / 2 mathbb{Z}$.






              share|cite|improve this answer









              $endgroup$
















                6












                6








                6





                $begingroup$

                When $V_1$ and $V_2$ are representations of $G_1$ and $G_2$ respectively, I'll use $V_1 boxtimes V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 otimes_{mathbb{C}} V_2$, and $V_1 boxplus V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 oplus V_2$.



                If $V$ is an irreducible representation of $G_1 times G_2$, then $V$ is isomorphic to $V_1 boxtimes V_2$ for some irreducible representations $V_1$ and $V_2$ of $G_1$ and $G_2$ respectively. This means that if we know the representations of $G_1$ and $G_2$, then using the $boxtimes$ construction we can get to all the (irreducible) representations of $G_1 times G_2$. Conversely, the $boxtimes$ product of two irreducible representations always produces an irreducible representation of $G_1 times G_2$.



                On the other hand, $V_1 boxplus V_2$ is always reducible as a $G_1 times G_2$ representation, since both vector subspaces $V_1$ and $V_2$ are stable under the $G_1 times G_2$ action. On the $V_1$ subspace, really only the $G_1$ part of the group acts, and the $G_2$ part acts trivially, and similarly for the $V_2$ subspace. We cannot produce all irreducible representations of $G_1 times G_2$ using this construction, which can already be seen in the example $G_1 = G_2 = mathbb{Z} / 2 mathbb{Z}$.






                share|cite|improve this answer









                $endgroup$



                When $V_1$ and $V_2$ are representations of $G_1$ and $G_2$ respectively, I'll use $V_1 boxtimes V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 otimes_{mathbb{C}} V_2$, and $V_1 boxplus V_2$ to mean the representation of $G_1 times G_2$ with underlying vector space $V_1 oplus V_2$.



                If $V$ is an irreducible representation of $G_1 times G_2$, then $V$ is isomorphic to $V_1 boxtimes V_2$ for some irreducible representations $V_1$ and $V_2$ of $G_1$ and $G_2$ respectively. This means that if we know the representations of $G_1$ and $G_2$, then using the $boxtimes$ construction we can get to all the (irreducible) representations of $G_1 times G_2$. Conversely, the $boxtimes$ product of two irreducible representations always produces an irreducible representation of $G_1 times G_2$.



                On the other hand, $V_1 boxplus V_2$ is always reducible as a $G_1 times G_2$ representation, since both vector subspaces $V_1$ and $V_2$ are stable under the $G_1 times G_2$ action. On the $V_1$ subspace, really only the $G_1$ part of the group acts, and the $G_2$ part acts trivially, and similarly for the $V_2$ subspace. We cannot produce all irreducible representations of $G_1 times G_2$ using this construction, which can already be seen in the example $G_1 = G_2 = mathbb{Z} / 2 mathbb{Z}$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 4 hours ago









                JoppyJoppy

                5,818421




                5,818421






























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