Mixing
Complex Representations of A4
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Im asked to obtain the number of irreducible complex representations of the group $A_4$ and their repesctive dimensions. I know that the number of irreducible representations is going to be the number of conjugacy classes , so there are 4, but how do i obtain they re dimension? I also know that $|G|= sum_{i=1}^{n} n_i^2$ where the $n_i$ are the dimension but i dont see how that helps me. Thanks.
abstract-algebra representation-theory
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
$endgroup$
add a comment |
$begingroup$
Im asked to obtain the number of irreducible complex representations of the group $A_4$ and their repesctive dimensions. I know that the number of irreducible representations is going to be the number of conjugacy classes , so there are 4, but how do i obtain they re dimension? I also know that $|G|= sum_{i=1}^{n} n_i^2$ where the $n_i$ are the dimension but i dont see how that helps me. Thanks.
abstract-algebra representation-theory
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
$endgroup$
$begingroup$
I'd start with the degree one representations.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 19:54
$begingroup$
I guess i would use the fact that there is a bijection between the representations of degree 1 of $A_4$ and $A_4/[A_4,A_4]$, now just got determine what that is.
$endgroup$
– Pedro Santos
Jan 22 at 19:57
add a comment |
$begingroup$
Im asked to obtain the number of irreducible complex representations of the group $A_4$ and their repesctive dimensions. I know that the number of irreducible representations is going to be the number of conjugacy classes , so there are 4, but how do i obtain they re dimension? I also know that $|G|= sum_{i=1}^{n} n_i^2$ where the $n_i$ are the dimension but i dont see how that helps me. Thanks.
abstract-algebra representation-theory
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
$endgroup$
Im asked to obtain the number of irreducible complex representations of the group $A_4$ and their repesctive dimensions. I know that the number of irreducible representations is going to be the number of conjugacy classes , so there are 4, but how do i obtain they re dimension? I also know that $|G|= sum_{i=1}^{n} n_i^2$ where the $n_i$ are the dimension but i dont see how that helps me. Thanks.
abstract-algebra representation-theory
abstract-algebra representation-theory
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
asked Jan 22 at 19:52
Pedro SantosPedro Santos
1539
1539
$begingroup$
I'd start with the degree one representations.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 19:54
$begingroup$
I guess i would use the fact that there is a bijection between the representations of degree 1 of $A_4$ and $A_4/[A_4,A_4]$, now just got determine what that is.
$endgroup$
– Pedro Santos
Jan 22 at 19:57
add a comment |
$begingroup$
I'd start with the degree one representations.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 19:54
$begingroup$
I guess i would use the fact that there is a bijection between the representations of degree 1 of $A_4$ and $A_4/[A_4,A_4]$, now just got determine what that is.
$endgroup$
– Pedro Santos
Jan 22 at 19:57
$begingroup$
I'd start with the degree one representations.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 19:54
$begingroup$
I'd start with the degree one representations.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 19:54
$begingroup$
I guess i would use the fact that there is a bijection between the representations of degree 1 of $A_4$ and $A_4/[A_4,A_4]$, now just got determine what that is.
$endgroup$
– Pedro Santos
Jan 22 at 19:57
$begingroup$
I guess i would use the fact that there is a bijection between the representations of degree 1 of $A_4$ and $A_4/[A_4,A_4]$, now just got determine what that is.
$endgroup$
– Pedro Santos
Jan 22 at 19:57
add a comment |
1 Answer
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Hint As OP suggests in the comments, one can start by identifying $$A_4 / [A_4 , A_4] cong A_4 / (Bbb Z_2 times Bbb Z_2) cong Bbb Z_3 ,$$ which gives immediately that there are $3$ representations of dimension $1$, and so by the sum-of-squares formula $1$ representation of dimension $3$.
Alternatively, one can get away with using only the count of the conjugacy classes (which you've already found to be $4$) and the sum-of-squares formula, $$12 = |A_4| = sum_{i = 1}^4 n_i^2 ,$$
where $n_i$ is the dimension of the $i$th irreducible representation. Checking manually shows that the only way to write $12$ as a sum of four positive squares is $1^2 + 1^2 + 1^2 + 3^2$. Of course, the existence of the trivial representation means that we need only look for ways to write $11$ as a sum of three squares.
answered Jan 22 at 21:37
TravisTravis
63k767150
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add a comment |
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$begingroup$
Hint As OP suggests in the comments, one can start by identifying $$A_4 / [A_4 , A_4] cong A_4 / (Bbb Z_2 times Bbb Z_2) cong Bbb Z_3 ,$$ which gives immediately that there are $3$ representations of dimension $1$, and so by the sum-of-squares formula $1$ representation of dimension $3$.
Alternatively, one can get away with using only the count of the conjugacy classes (which you've already found to be $4$) and the sum-of-squares formula, $$12 = |A_4| = sum_{i = 1}^4 n_i^2 ,$$
where $n_i$ is the dimension of the $i$th irreducible representation. Checking manually shows that the only way to write $12$ as a sum of four positive squares is $1^2 + 1^2 + 1^2 + 3^2$. Of course, the existence of the trivial representation means that we need only look for ways to write $11$ as a sum of three squares.
answered Jan 22 at 21:37
TravisTravis
63k767150
$endgroup$
add a comment |
$begingroup$
Hint As OP suggests in the comments, one can start by identifying $$A_4 / [A_4 , A_4] cong A_4 / (Bbb Z_2 times Bbb Z_2) cong Bbb Z_3 ,$$ which gives immediately that there are $3$ representations of dimension $1$, and so by the sum-of-squares formula $1$ representation of dimension $3$.
Alternatively, one can get away with using only the count of the conjugacy classes (which you've already found to be $4$) and the sum-of-squares formula, $$12 = |A_4| = sum_{i = 1}^4 n_i^2 ,$$
where $n_i$ is the dimension of the $i$th irreducible representation. Checking manually shows that the only way to write $12$ as a sum of four positive squares is $1^2 + 1^2 + 1^2 + 3^2$. Of course, the existence of the trivial representation means that we need only look for ways to write $11$ as a sum of three squares.
answered Jan 22 at 21:37
TravisTravis
63k767150
$endgroup$
add a comment |
$begingroup$
Hint As OP suggests in the comments, one can start by identifying $$A_4 / [A_4 , A_4] cong A_4 / (Bbb Z_2 times Bbb Z_2) cong Bbb Z_3 ,$$ which gives immediately that there are $3$ representations of dimension $1$, and so by the sum-of-squares formula $1$ representation of dimension $3$.
Alternatively, one can get away with using only the count of the conjugacy classes (which you've already found to be $4$) and the sum-of-squares formula, $$12 = |A_4| = sum_{i = 1}^4 n_i^2 ,$$
where $n_i$ is the dimension of the $i$th irreducible representation. Checking manually shows that the only way to write $12$ as a sum of four positive squares is $1^2 + 1^2 + 1^2 + 3^2$. Of course, the existence of the trivial representation means that we need only look for ways to write $11$ as a sum of three squares.
answered Jan 22 at 21:37
TravisTravis
63k767150
$endgroup$
Hint As OP suggests in the comments, one can start by identifying $$A_4 / [A_4 , A_4] cong A_4 / (Bbb Z_2 times Bbb Z_2) cong Bbb Z_3 ,$$ which gives immediately that there are $3$ representations of dimension $1$, and so by the sum-of-squares formula $1$ representation of dimension $3$.
Alternatively, one can get away with using only the count of the conjugacy classes (which you've already found to be $4$) and the sum-of-squares formula, $$12 = |A_4| = sum_{i = 1}^4 n_i^2 ,$$
where $n_i$ is the dimension of the $i$th irreducible representation. Checking manually shows that the only way to write $12$ as a sum of four positive squares is $1^2 + 1^2 + 1^2 + 3^2$. Of course, the existence of the trivial representation means that we need only look for ways to write $11$ as a sum of three squares.
answered Jan 22 at 21:37
TravisTravis
63k767150
answered Jan 22 at 21:37
TravisTravis
63k767150
answered Jan 22 at 21:37
TravisTravis
63k767150
answered Jan 22 at 21:37
TravisTravis
63k767150
63k767150
add a comment |
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$begingroup$
I'd start with the degree one representations.
$endgroup$
– Lord Shark the Unknown
Jan 22 at 19:54
$begingroup$
I guess i would use the fact that there is a bijection between the representations of degree 1 of $A_4$ and $A_4/[A_4,A_4]$, now just got determine what that is.
$endgroup$
– Pedro Santos
Jan 22 at 19:57