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Calculating the Hilbert Series for symmetric polynomials

0

$begingroup$

Let $S = mathbbC[x_1,...,x_n]$ be the polynomial ring in $n$ variables, $S_d subset S$ the subspace of homogeneous polynomials of degree $d$, i.e., the polynomials with the property
beginalign
f(tx) = t^d f(x).
endalign

Let G be the symmetric group consisting of $n times n$ permutation matrices. Then $S^G subset S$ is defined as the subspace of invariant polynomials under G, i.e. the set
beginalign
S^G &= f in S~ \&= ~ f(x_sigma(1),...,x_sigma(n)) = f(x_1,...,x_n) ~for~all~sigma in mathcalS_n
endalign

So all the invariant Polynomials are the symmetric Polynomials, which are generated by the $n$ elementary symmetric polynomials
beginalign
sigma_1 (x) &= sum_ix_i \
sigma_2 (x) &= sum_i<j x_i x_j \
&...... \
sigma_n (x) &= x_1 cdot x_2cdot... cdot x_n
endalign

The Hilbert Series, which I want to find out, is defined as the formal power series
beginalign P(t) = sum_d geq 0 dim(S^G cap S_d)t^d in mathbbZ[[t]].
endalign

I started calculating the $dim(S^G cap S_d)$ for each $d$ and I got:
$dim(S^G cap S_d) = 1 $ for all $d$, since I took the $sigma_i$ as a basis for each summand $S^G cap S_d$.

But this must obviously be wrong, because then I get the Hilbert Series
beginalign
frac11-t.
endalign

In Mukai it says the Hilbert Series for the symmetric Group is

beginalign
frac1(1-t)(1-t^2)...(1-t^n).
endalign
So I am missing a few factors in my calculation but don't know how to get them. Can anyone show me how to calculate the Hilbert Series in the correct way ?

invariant-theory hilbert-polynomial

share|cite|improve this question

asked Mar 19 at 21:51

GilligansGilligans

496

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your graded ring generated by symmetric polynomials will be similar to $R(n) = BbbC[y_1^1,y_2^2,ldots,y_n^n]$ graded by degree. Then $R(n)_d = bigoplus_mn le d y_n^mn R(n-1)_d-mn$ so $dim(R(n)_d) = sum_mnle d dim(R(n-1)_d-mn)$ and $P(n)(t) = sum_d=0^infty dim(R(n)_d)t^d= frac11-t^n P(n-1)(t)=frac1(1-t)(1-t^2)...(1-t^n)$.
  $endgroup$
  – reuns
  Mar 20 at 0:47
  
  

  
  
  
  

add a comment | 

0

$begingroup$

Let $S = mathbbC[x_1,...,x_n]$ be the polynomial ring in $n$ variables, $S_d subset S$ the subspace of homogeneous polynomials of degree $d$, i.e., the polynomials with the property
beginalign
f(tx) = t^d f(x).
endalign

Let G be the symmetric group consisting of $n times n$ permutation matrices. Then $S^G subset S$ is defined as the subspace of invariant polynomials under G, i.e. the set
beginalign
S^G &= f in S~ \&= ~ f(x_sigma(1),...,x_sigma(n)) = f(x_1,...,x_n) ~for~all~sigma in mathcalS_n
endalign

So all the invariant Polynomials are the symmetric Polynomials, which are generated by the $n$ elementary symmetric polynomials
beginalign
sigma_1 (x) &= sum_ix_i \
sigma_2 (x) &= sum_i<j x_i x_j \
&...... \
sigma_n (x) &= x_1 cdot x_2cdot... cdot x_n
endalign

The Hilbert Series, which I want to find out, is defined as the formal power series
beginalign P(t) = sum_d geq 0 dim(S^G cap S_d)t^d in mathbbZ[[t]].
endalign

I started calculating the $dim(S^G cap S_d)$ for each $d$ and I got:
$dim(S^G cap S_d) = 1 $ for all $d$, since I took the $sigma_i$ as a basis for each summand $S^G cap S_d$.

But this must obviously be wrong, because then I get the Hilbert Series
beginalign
frac11-t.
endalign

In Mukai it says the Hilbert Series for the symmetric Group is

beginalign
frac1(1-t)(1-t^2)...(1-t^n).
endalign
So I am missing a few factors in my calculation but don't know how to get them. Can anyone show me how to calculate the Hilbert Series in the correct way ?

invariant-theory hilbert-polynomial

share|cite|improve this question

asked Mar 19 at 21:51

GilligansGilligans

496

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Your graded ring generated by symmetric polynomials will be similar to $R(n) = BbbC[y_1^1,y_2^2,ldots,y_n^n]$ graded by degree. Then $R(n)_d = bigoplus_mn le d y_n^mn R(n-1)_d-mn$ so $dim(R(n)_d) = sum_mnle d dim(R(n-1)_d-mn)$ and $P(n)(t) = sum_d=0^infty dim(R(n)_d)t^d= frac11-t^n P(n-1)(t)=frac1(1-t)(1-t^2)...(1-t^n)$.
  $endgroup$
  – reuns
  Mar 20 at 0:47
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $S = mathbbC[x_1,...,x_n]$ be the polynomial ring in $n$ variables, $S_d subset S$ the subspace of homogeneous polynomials of degree $d$, i.e., the polynomials with the property
beginalign
f(tx) = t^d f(x).
endalign

Let G be the symmetric group consisting of $n times n$ permutation matrices. Then $S^G subset S$ is defined as the subspace of invariant polynomials under G, i.e. the set
beginalign
S^G &= f in S~ \&= ~ f(x_sigma(1),...,x_sigma(n)) = f(x_1,...,x_n) ~for~all~sigma in mathcalS_n
endalign

So all the invariant Polynomials are the symmetric Polynomials, which are generated by the $n$ elementary symmetric polynomials
beginalign
sigma_1 (x) &= sum_ix_i \
sigma_2 (x) &= sum_i<j x_i x_j \
&...... \
sigma_n (x) &= x_1 cdot x_2cdot... cdot x_n
endalign

The Hilbert Series, which I want to find out, is defined as the formal power series
beginalign P(t) = sum_d geq 0 dim(S^G cap S_d)t^d in mathbbZ[[t]].
endalign

I started calculating the $dim(S^G cap S_d)$ for each $d$ and I got:
$dim(S^G cap S_d) = 1 $ for all $d$, since I took the $sigma_i$ as a basis for each summand $S^G cap S_d$.

But this must obviously be wrong, because then I get the Hilbert Series
beginalign
frac11-t.
endalign

In Mukai it says the Hilbert Series for the symmetric Group is

beginalign
frac1(1-t)(1-t^2)...(1-t^n).
endalign
So I am missing a few factors in my calculation but don't know how to get them. Can anyone show me how to calculate the Hilbert Series in the correct way ?

invariant-theory hilbert-polynomial

share|cite|improve this question

asked Mar 19 at 21:51

GilligansGilligans

496

$endgroup$

share|cite|improve this question

asked Mar 19 at 21:51

GilligansGilligans

496

share|cite|improve this question

asked Mar 19 at 21:51

GilligansGilligans

496

asked Mar 19 at 21:51

GilligansGilligans

496

asked Mar 19 at 21:51

GilligansGilligans

496