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direct sum of representation of product groups

faithful representation related to the centerDirect sum and tensor product of two representations of a groupEvery Irreducible Representation of $G times H$ is tensor product of Irreducible Reps of $G$ and $H$?direct sum and tensor product of representation of lie algebraTensor product of representations of a product group?Show $rho_1 oplus rho_2 : G → GL(V oplus W)$ is a homomorphisma decomposition of a representation of Z_2Is the tensor product of irreducible representations of different groups irreducible?What is the intuition behind the direct product and the direct sum of groups being identical for finite groups?Definition of the tensor product of representations

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

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asked 4 hours ago

self-educatorself-educator

4611

$endgroup$

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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

asked 4 hours ago

self-educatorself-educator

4611

$endgroup$

add a comment | 

$begingroup$

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

asked 4 hours ago

self-educatorself-educator

4611

$endgroup$

share|cite|improve this question

asked 4 hours ago

self-educatorself-educator

4611

share|cite|improve this question

asked 4 hours ago

self-educatorself-educator

4611

asked 4 hours ago

self-educatorself-educator

4611

asked 4 hours ago

self-educatorself-educator

4611

1 Answer
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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

answered 4 hours ago

JoppyJoppy

5,818421

$endgroup$

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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

answered 4 hours ago

JoppyJoppy

5,818421

$endgroup$

add a comment | 

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

answered 4 hours ago

JoppyJoppy

5,818421

$endgroup$

add a comment | 

$begingroup$

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

answered 4 hours ago

JoppyJoppy

5,818421

$endgroup$

share|cite|improve this answer

answered 4 hours ago

JoppyJoppy

5,818421

answered 4 hours ago

JoppyJoppy

5,818421

answered 4 hours ago

JoppyJoppy

5,818421