Calculating the Hilbert Series for symmetric polynomials

Let $$S = \mathbb{C}[x_1,...,x_n]$$ 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
\begin{align} f(tx) = t^d f(x). \end{align}

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 \begin{align} S^G &= \{f \in S~ |~ f(Ax) = f(x) ~ for ~all ~ A \in G \} \\&= \{ f \in S ~ | ~ f(x_{\sigma(1)},...,x_{\sigma(n)}) = f(x_1,...,x_n) ~for~all~\sigma \in \mathcal{S}_n \} \end{align}

So all the invariant Polynomials are the symmetric Polynomials, which are generated by the $$n$$ elementary symmetric polynomials \begin{align} \sigma_1 (x) &= \sum_{i}x_i \\ \sigma_2 (x) &= \sum_{i

The Hilbert Series, which I want to find out, is defined as the formal power series \begin{align} P(t) = \sum_{d \geq 0} dim(S^{G} \cap S_d)t^d \in \mathbb{Z}[[t]]. \end{align}

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 \begin{align} \frac{1}{1-t}. \end{align}

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

\begin{align} \frac{1}{(1-t)(1-t^2)...(1-t^n)}. \end{align} 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 ?

• Your graded ring generated by symmetric polynomials will be similar to $R(n) = \Bbb{C}[y_1^1,y_2^2,\ldots,y_n^n]$ graded by degree. Then $R(n)_d = \bigoplus_{mn \le d} y_n^{mn} R(n-1)_{d-mn}$ so $\dim(R(n)_d) = \sum_{mn\le d} \dim(R(n-1)_{d-mn})$ and $P(n)(t) = \sum_{d=0}^\infty \dim(R(n)_d)t^d= \frac{1}{1-t^n} P(n-1)(t)=\frac{1}{(1-t)(1-t^2)...(1-t^n)}$. – reuns Mar 20 at 0:47