Permutation and Symmetric Polynomials
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Given a polynomial $f(x_1, · · · , x_n) ∈ F[x_1, · · · , x_n]$ and a permutation
$Ã ∈ S_n$, I want to show that both
$prod_Ã∈Sn
à · f(x_1, · · · , x_n)$ and $sum_Ã∈Sn
à · f(x_1, · · · , x_n)$ are symmetric polynomials.
I know a symmetric polynomial is a polynomial $F(X_1, …, X_n)$ in $n$ variables, such that if any of the variables are interchanged, one obtains the same polynomial. Or P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ..., n$ one has $F(X_Ã(1),…, X_Ã(n)) = P(X_1,…, X_n)$. But I am not sure how to proceed in the above example.
calculus real-analysis linear-algebra abstract-algebra number-theory
asked Aug 23 at 5:37
Homaniac
48019
add a comment |Â
up vote
1
down vote
favorite
Given a polynomial $f(x_1, · · · , x_n) ∈ F[x_1, · · · , x_n]$ and a permutation
$Ã ∈ S_n$, I want to show that both
$prod_Ã∈Sn
à · f(x_1, · · · , x_n)$ and $sum_Ã∈Sn
à · f(x_1, · · · , x_n)$ are symmetric polynomials.
I know a symmetric polynomial is a polynomial $F(X_1, …, X_n)$ in $n$ variables, such that if any of the variables are interchanged, one obtains the same polynomial. Or P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ..., n$ one has $F(X_Ã(1),…, X_Ã(n)) = P(X_1,…, X_n)$. But I am not sure how to proceed in the above example.
calculus real-analysis linear-algebra abstract-algebra number-theory
asked Aug 23 at 5:37
Homaniac
48019
You can't say "given a permutation $sigmain S_n$" and then sum over $sigmain S_n$.
– joriki
Aug 23 at 5:49
Hint: for each $tau in S_n$, multiplying by $tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group.
– 4-ier
Aug 23 at 6:12
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Given a polynomial $f(x_1, · · · , x_n) ∈ F[x_1, · · · , x_n]$ and a permutation
$Ã ∈ S_n$, I want to show that both
$prod_Ã∈Sn
à · f(x_1, · · · , x_n)$ and $sum_Ã∈Sn
à · f(x_1, · · · , x_n)$ are symmetric polynomials.
I know a symmetric polynomial is a polynomial $F(X_1, …, X_n)$ in $n$ variables, such that if any of the variables are interchanged, one obtains the same polynomial. Or P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ..., n$ one has $F(X_Ã(1),…, X_Ã(n)) = P(X_1,…, X_n)$. But I am not sure how to proceed in the above example.
calculus real-analysis linear-algebra abstract-algebra number-theory
asked Aug 23 at 5:37
Homaniac
48019
Given a polynomial $f(x_1, · · · , x_n) ∈ F[x_1, · · · , x_n]$ and a permutation
$Ã ∈ S_n$, I want to show that both
$prod_Ã∈Sn
à · f(x_1, · · · , x_n)$ and $sum_Ã∈Sn
à · f(x_1, · · · , x_n)$ are symmetric polynomials.
I know a symmetric polynomial is a polynomial $F(X_1, …, X_n)$ in $n$ variables, such that if any of the variables are interchanged, one obtains the same polynomial. Or P is a symmetric polynomial if for any permutation $Ã$ of the subscripts $1, 2, ..., n$ one has $F(X_Ã(1),…, X_Ã(n)) = P(X_1,…, X_n)$. But I am not sure how to proceed in the above example.
calculus real-analysis linear-algebra abstract-algebra number-theory
asked Aug 23 at 5:37
Homaniac
48019
asked Aug 23 at 5:37
Homaniac
48019
asked Aug 23 at 5:37
Homaniac
48019
asked Aug 23 at 5:37
Homaniac
48019
48019
You can't say "given a permutation $sigmain S_n$" and then sum over $sigmain S_n$.
– joriki
Aug 23 at 5:49
Hint: for each $tau in S_n$, multiplying by $tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group.
– 4-ier
Aug 23 at 6:12
add a comment |Â
You can't say "given a permutation $sigmain S_n$" and then sum over $sigmain S_n$.
– joriki
Aug 23 at 5:49
Hint: for each $tau in S_n$, multiplying by $tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group.
– 4-ier
Aug 23 at 6:12
You can't say "given a permutation $sigmain S_n$" and then sum over $sigmain S_n$.
– joriki
Aug 23 at 5:49
You can't say "given a permutation $sigmain S_n$" and then sum over $sigmain S_n$.
– joriki
Aug 23 at 5:49
Hint: for each $tau in S_n$, multiplying by $tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group.
– 4-ier
Aug 23 at 6:12
Hint: for each $tau in S_n$, multiplying by $tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group.
– 4-ier
Aug 23 at 6:12
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
A polynomial $g$ is a symmetric polynomial if $sigma g=g$ for all $sigma in S_n$.
For your case we have $g=sumlimits_tau tau(f)$. If $tau$ runs over $S_n$ then the set $sigma circ tau$ coincides with $S_n$.Then
$$
sigma(g)=sumlimits_tau sigma(tau(f))=sumlimits_tau (sigma circ tau)(f)=sumlimits_tau tau(f)=g,
$$
so $g$ is an symmetric polynomial.
answered Aug 23 at 6:30
Leox
5,1141323
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
yes, the proof is the same
– Leox
Aug 23 at 9:28
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
A polynomial $g$ is a symmetric polynomial if $sigma g=g$ for all $sigma in S_n$.
For your case we have $g=sumlimits_tau tau(f)$. If $tau$ runs over $S_n$ then the set $sigma circ tau$ coincides with $S_n$.Then
$$
sigma(g)=sumlimits_tau sigma(tau(f))=sumlimits_tau (sigma circ tau)(f)=sumlimits_tau tau(f)=g,
$$
so $g$ is an symmetric polynomial.
answered Aug 23 at 6:30
Leox
5,1141323
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
yes, the proof is the same
– Leox
Aug 23 at 9:28
add a comment |Â
up vote
0
down vote
A polynomial $g$ is a symmetric polynomial if $sigma g=g$ for all $sigma in S_n$.
For your case we have $g=sumlimits_tau tau(f)$. If $tau$ runs over $S_n$ then the set $sigma circ tau$ coincides with $S_n$.Then
$$
sigma(g)=sumlimits_tau sigma(tau(f))=sumlimits_tau (sigma circ tau)(f)=sumlimits_tau tau(f)=g,
$$
so $g$ is an symmetric polynomial.
answered Aug 23 at 6:30
Leox
5,1141323
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
yes, the proof is the same
– Leox
Aug 23 at 9:28
add a comment |Â
up vote
0
down vote
up vote
0
down vote
A polynomial $g$ is a symmetric polynomial if $sigma g=g$ for all $sigma in S_n$.
For your case we have $g=sumlimits_tau tau(f)$. If $tau$ runs over $S_n$ then the set $sigma circ tau$ coincides with $S_n$.Then
$$
sigma(g)=sumlimits_tau sigma(tau(f))=sumlimits_tau (sigma circ tau)(f)=sumlimits_tau tau(f)=g,
$$
so $g$ is an symmetric polynomial.
answered Aug 23 at 6:30
Leox
5,1141323
A polynomial $g$ is a symmetric polynomial if $sigma g=g$ for all $sigma in S_n$.
For your case we have $g=sumlimits_tau tau(f)$. If $tau$ runs over $S_n$ then the set $sigma circ tau$ coincides with $S_n$.Then
$$
sigma(g)=sumlimits_tau sigma(tau(f))=sumlimits_tau (sigma circ tau)(f)=sumlimits_tau tau(f)=g,
$$
so $g$ is an symmetric polynomial.
answered Aug 23 at 6:30
Leox
5,1141323
answered Aug 23 at 6:30
Leox
5,1141323
answered Aug 23 at 6:30
Leox
5,1141323
answered Aug 23 at 6:30
Leox
5,1141323
5,1141323
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
yes, the proof is the same
– Leox
Aug 23 at 9:28
add a comment |Â
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
yes, the proof is the same
– Leox
Aug 23 at 9:28
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
Ah how about the case for product rather than sum? Is the proof the same?
– Homaniac
Aug 23 at 9:10
yes, the proof is the same
– Leox
Aug 23 at 9:28
yes, the proof is the same
– Leox
Aug 23 at 9:28
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2891733%2fpermutation-and-symmetric-polynomials%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
You can't say "given a permutation $sigmain S_n$" and then sum over $sigmain S_n$.
– joriki
Aug 23 at 5:49
Hint: for each $tau in S_n$, multiplying by $tau$ on the left is a permutation of the group $S_n$. (the proof of Cayley's theorem). This is a common trick in group theory, averaging over a group.
– 4-ier
Aug 23 at 6:12