Mixing
Symmetric group action on polynomial ring
0
$begingroup$
Let the symmetric group $S_4$ act on $mathbb R[x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4]$ by permuting the 1st $4$ variables and again permuting the last $4$ variables. We can restrict the action to the alternating group $A_4$. It seems there is a degree $4$ polynomial which is invariant under the $A_4$ action but not invariant under the $S_4$ action. But I can't figure out what is that polynomial. Any help in this direction will be helpful.
polynomials representation-theory symmetric-groups symmetric-polynomials invariant-theory
asked Jan 30 at 16:50
jackjack
1
$endgroup$
|
show 4 more comments
0
$begingroup$
Let the symmetric group $S_4$ act on $mathbb R[x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4]$ by permuting the 1st $4$ variables and again permuting the last $4$ variables. We can restrict the action to the alternating group $A_4$. It seems there is a degree $4$ polynomial which is invariant under the $A_4$ action but not invariant under the $S_4$ action. But I can't figure out what is that polynomial. Any help in this direction will be helpful.
polynomials representation-theory symmetric-groups symmetric-polynomials invariant-theory
asked Jan 30 at 16:50
jackjack
1
$endgroup$
$begingroup$
Are you familiar with the proof that the parity of a permutation (defined as the parity of the number of transpositions in any expression of the permutation as a product of transpositions) is well defined, by acting on a discriminant?
$endgroup$
– Arturo Magidin
Jan 30 at 17:02
$begingroup$
yes, I am familiar with even and odd permutations. But here the discriminant is of degree $36$.
$endgroup$
– jack
Jan 30 at 17:06
$begingroup$
You only need to do it with some of the variables, since the action is independent. You only need to figure out how to drop it from 6 to 4, not from 36 to 4.
$endgroup$
– Arturo Magidin
Jan 30 at 17:08
$begingroup$
Yes, I tried all possible degree 4 combinations like (x_i-y_j).... but it doesn't work.
$endgroup$
– jack
Jan 30 at 17:11
2
$begingroup$
No, the point is that it acts on the $x$s separately from how it acts on the $y$s. You should look at polynomials that contain only $x$s, only $y$s, or have them in separate monomials that mirror each other. For example, if you just take the discriminant of the $x$s, $prod_{ilt j}(x_j-x_i)$, then this is invariant under the action of $A_4$ but not under the action of $S_4$. (And you could of course take the sum of the discriminant of the $x$s and the discriminant of the $y$s). I know this is degree 6, not 4, but you shouldn't be mixing your $x$s and $y$s, I think...
$endgroup$
– Arturo Magidin
Jan 30 at 17:25
|
show 4 more comments
0
0
0
$begingroup$
Let the symmetric group $S_4$ act on $mathbb R[x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4]$ by permuting the 1st $4$ variables and again permuting the last $4$ variables. We can restrict the action to the alternating group $A_4$. It seems there is a degree $4$ polynomial which is invariant under the $A_4$ action but not invariant under the $S_4$ action. But I can't figure out what is that polynomial. Any help in this direction will be helpful.
polynomials representation-theory symmetric-groups symmetric-polynomials invariant-theory
asked Jan 30 at 16:50
jackjack
1
$endgroup$
Let the symmetric group $S_4$ act on $mathbb R[x_1,x_2,x_3,x_4,y_1,y_2,y_3,y_4]$ by permuting the 1st $4$ variables and again permuting the last $4$ variables. We can restrict the action to the alternating group $A_4$. It seems there is a degree $4$ polynomial which is invariant under the $A_4$ action but not invariant under the $S_4$ action. But I can't figure out what is that polynomial. Any help in this direction will be helpful.
polynomials representation-theory symmetric-groups symmetric-polynomials invariant-theory
polynomials representation-theory symmetric-groups symmetric-polynomials invariant-theory
asked Jan 30 at 16:50
jackjack
1
asked Jan 30 at 16:50
jackjack
1
asked Jan 30 at 16:50
jackjack
1
asked Jan 30 at 16:50
jackjack
1
asked Jan 30 at 16:50
jackjack
1
1
$begingroup$
Are you familiar with the proof that the parity of a permutation (defined as the parity of the number of transpositions in any expression of the permutation as a product of transpositions) is well defined, by acting on a discriminant?
$endgroup$
– Arturo Magidin
Jan 30 at 17:02
$begingroup$
yes, I am familiar with even and odd permutations. But here the discriminant is of degree $36$.
$endgroup$
– jack
Jan 30 at 17:06
$begingroup$
You only need to do it with some of the variables, since the action is independent. You only need to figure out how to drop it from 6 to 4, not from 36 to 4.
$endgroup$
– Arturo Magidin
Jan 30 at 17:08
$begingroup$
Yes, I tried all possible degree 4 combinations like (x_i-y_j).... but it doesn't work.
$endgroup$
– jack
Jan 30 at 17:11
2
$begingroup$
No, the point is that it acts on the $x$s separately from how it acts on the $y$s. You should look at polynomials that contain only $x$s, only $y$s, or have them in separate monomials that mirror each other. For example, if you just take the discriminant of the $x$s, $prod_{ilt j}(x_j-x_i)$, then this is invariant under the action of $A_4$ but not under the action of $S_4$. (And you could of course take the sum of the discriminant of the $x$s and the discriminant of the $y$s). I know this is degree 6, not 4, but you shouldn't be mixing your $x$s and $y$s, I think...
$endgroup$
– Arturo Magidin
Jan 30 at 17:25
|
show 4 more comments
$begingroup$
Are you familiar with the proof that the parity of a permutation (defined as the parity of the number of transpositions in any expression of the permutation as a product of transpositions) is well defined, by acting on a discriminant?
$endgroup$
– Arturo Magidin
Jan 30 at 17:02
$begingroup$
yes, I am familiar with even and odd permutations. But here the discriminant is of degree $36$.
$endgroup$
– jack
Jan 30 at 17:06
$begingroup$
You only need to do it with some of the variables, since the action is independent. You only need to figure out how to drop it from 6 to 4, not from 36 to 4.
$endgroup$
– Arturo Magidin
Jan 30 at 17:08
$begingroup$
Yes, I tried all possible degree 4 combinations like (x_i-y_j).... but it doesn't work.
$endgroup$
– jack
Jan 30 at 17:11
2
$begingroup$
No, the point is that it acts on the $x$s separately from how it acts on the $y$s. You should look at polynomials that contain only $x$s, only $y$s, or have them in separate monomials that mirror each other. For example, if you just take the discriminant of the $x$s, $prod_{ilt j}(x_j-x_i)$, then this is invariant under the action of $A_4$ but not under the action of $S_4$. (And you could of course take the sum of the discriminant of the $x$s and the discriminant of the $y$s). I know this is degree 6, not 4, but you shouldn't be mixing your $x$s and $y$s, I think...
$endgroup$
– Arturo Magidin
Jan 30 at 17:25
$begingroup$
Are you familiar with the proof that the parity of a permutation (defined as the parity of the number of transpositions in any expression of the permutation as a product of transpositions) is well defined, by acting on a discriminant?
$endgroup$
– Arturo Magidin
Jan 30 at 17:02
$begingroup$
Are you familiar with the proof that the parity of a permutation (defined as the parity of the number of transpositions in any expression of the permutation as a product of transpositions) is well defined, by acting on a discriminant?
$endgroup$
– Arturo Magidin
Jan 30 at 17:02
$begingroup$
yes, I am familiar with even and odd permutations. But here the discriminant is of degree $36$.
$endgroup$
– jack
Jan 30 at 17:06
$begingroup$
yes, I am familiar with even and odd permutations. But here the discriminant is of degree $36$.
$endgroup$
– jack
Jan 30 at 17:06
$begingroup$
You only need to do it with some of the variables, since the action is independent. You only need to figure out how to drop it from 6 to 4, not from 36 to 4.
$endgroup$
– Arturo Magidin
Jan 30 at 17:08
$begingroup$
You only need to do it with some of the variables, since the action is independent. You only need to figure out how to drop it from 6 to 4, not from 36 to 4.
$endgroup$
– Arturo Magidin
Jan 30 at 17:08
$begingroup$
Yes, I tried all possible degree 4 combinations like (x_i-y_j).... but it doesn't work.
$endgroup$
– jack
Jan 30 at 17:11
$begingroup$
Yes, I tried all possible degree 4 combinations like (x_i-y_j).... but it doesn't work.
$endgroup$
– jack
Jan 30 at 17:11
2
2
$begingroup$
No, the point is that it acts on the $x$s separately from how it acts on the $y$s. You should look at polynomials that contain only $x$s, only $y$s, or have them in separate monomials that mirror each other. For example, if you just take the discriminant of the $x$s, $prod_{ilt j}(x_j-x_i)$, then this is invariant under the action of $A_4$ but not under the action of $S_4$. (And you could of course take the sum of the discriminant of the $x$s and the discriminant of the $y$s). I know this is degree 6, not 4, but you shouldn't be mixing your $x$s and $y$s, I think...
$endgroup$
– Arturo Magidin
Jan 30 at 17:25
$begingroup$
No, the point is that it acts on the $x$s separately from how it acts on the $y$s. You should look at polynomials that contain only $x$s, only $y$s, or have them in separate monomials that mirror each other. For example, if you just take the discriminant of the $x$s, $prod_{ilt j}(x_j-x_i)$, then this is invariant under the action of $A_4$ but not under the action of $S_4$. (And you could of course take the sum of the discriminant of the $x$s and the discriminant of the $y$s). I know this is degree 6, not 4, but you shouldn't be mixing your $x$s and $y$s, I think...
$endgroup$
– Arturo Magidin
Jan 30 at 17:25
|
show 4 more comments
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$begingroup$
Are you familiar with the proof that the parity of a permutation (defined as the parity of the number of transpositions in any expression of the permutation as a product of transpositions) is well defined, by acting on a discriminant?
$endgroup$
– Arturo Magidin
Jan 30 at 17:02
$begingroup$
yes, I am familiar with even and odd permutations. But here the discriminant is of degree $36$.
$endgroup$
– jack
Jan 30 at 17:06
$begingroup$
You only need to do it with some of the variables, since the action is independent. You only need to figure out how to drop it from 6 to 4, not from 36 to 4.
$endgroup$
– Arturo Magidin
Jan 30 at 17:08
$begingroup$
Yes, I tried all possible degree 4 combinations like (x_i-y_j).... but it doesn't work.
$endgroup$
– jack
Jan 30 at 17:11
2
$begingroup$
No, the point is that it acts on the $x$s separately from how it acts on the $y$s. You should look at polynomials that contain only $x$s, only $y$s, or have them in separate monomials that mirror each other. For example, if you just take the discriminant of the $x$s, $prod_{ilt j}(x_j-x_i)$, then this is invariant under the action of $A_4$ but not under the action of $S_4$. (And you could of course take the sum of the discriminant of the $x$s and the discriminant of the $y$s). I know this is degree 6, not 4, but you shouldn't be mixing your $x$s and $y$s, I think...
$endgroup$
– Arturo Magidin
Jan 30 at 17:25