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direct sum of representation of product groups

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direct sum of representation of product groups

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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

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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 $V_1 otimes_mathbbC V_2$, 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 $G_1 = G_2 = mathbbZ / 2 mathbbZ$.

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answered 4 hours ago

JoppyJoppy

5,818421

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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 $V_1 otimes_mathbbC V_2$, 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 $G_1 = G_2 = mathbbZ / 2 mathbbZ$.

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 $V_1 otimes_mathbbC V_2$, 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 $G_1 = G_2 = mathbbZ / 2 mathbbZ$.

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 $V_1 otimes_mathbbC V_2$, 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 $G_1 = G_2 = mathbbZ / 2 mathbbZ$.

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