10.8 Symmetric Polynomials

Some Symmetric Polynomials: p, e and h

Definition

A polynomial in $m$ variables is called symmetric if it is stable under all permutations $\sigma \in S_m$, that is, $\sigma (p(x_1,\cdots ,x_n))=p(x_{\sigma (1)},\cdots ,x_{\sigma (n)})$.

The algebra of symmetric polynomials is denoted by ${\Lambda }_m$.

Examples. They are symmetric polynomials.

• $p_n=x_1^n+\cdots +x_m^n$ for all $n\in {\mathbb{N}}_{+}$, where the corresponding generated function is $P(t)={\sum }_{n=1}^{\infty }{p}_n{t}^{n-1}={\sum }_{i=1}^m\frac{x_i}{1-x_{i}t}$.
• The elementary symmetric polynomials $e_n={\sum }_{i_1<\cdots <i_n}{x}_{i_1}\cdots x_{i_n}$ for $n⩾1$. Furthermore, define $e_0=1$. Note that $e_n=0$ when $n⩾m$, and the corresponding generated function is $E(t)={\sum }_{n=0}^t{e}_n{t}^n={\prod }_{i=1}^m(1+x_{i}t)$.
• The complete symmetric polynomials $h_n={\sum }_{i_1⩽\cdots ⩽i_n}{x}_{i_1}\cdots x_{i_n}$ for $n⩾1$. We also define $h_0=1$, and the corresponding generated function is $H(t)={\sum }_{n=0}^{\infty }{h}_n{t}^n={\prod }_{i=1}^{\infty }\frac{1}{1-x_{i}t}$.

Since $H(t)E(-t)=1$, we have ${\sum }_{n=0}^{\infty }(-1{)}^r{e}_r{h}_{n-r}=0$. Next, notice that $p(t)=d/dt\ln {\prod }_{i=1}^{\infty }\frac{1}{1-x_{i}t}$, then

P(t)=d/dt\ln(H(t))=\frac{H'(t)}{H(t)}. $$ ^skztji Similarly $P(-t)=E'(t)/E(t)$. It deduces the Newton identity. > [!lemma] the Newton identity > > With the definition above, we have $nh_n=\sum_{r=1}^\infty p_rh_{n-r}$ and $ne_n=\sum_{r=1}^n(-1)^{r-1}p_re_{n-r}$. ^sbpikb > [!proposition] > > If $A$ is an $m \times m$ matrix, then the coefficients of the characteristic polynomial $\operatorname{det}(A-t I)$ are given by the elementary symmetric functions of the eigenvalues $\left\{\lambda_1, \lambda_2, \ldots, \lambda_m\right\}$, with alternating signs $\pm 1$ depending on the degree of each term. The power sums of the eigenvalues coincide with $\operatorname{tr}A^n=\lambda_1^n+\cdots+\lambda_m^n$. **Remark.** There is some connection between symmetric polynomials and the rooted solution of polynomial equations. See [here](https://math.stackexchange.com/a/96329/1445401). When the degree of polynomial is less than or equal to $4$, the roots can be obtained by operating the coefficients by combination of adding, subtracting, multiplication, division, and taking roots. # Schur Polynomial > [!definition] > > Suppose $\lambda=(\lambda_1,\cdots,\lambda_m)$ is a partition of $n$ of length at most $m$. The Schur polynomial $\mathscr s_\lambda$ is defined by > > $$\mathscr s_\lambda=\frac{\left|\begin{matrix} x_1^{\lambda_1+m-1} & \cdots & x_m^{\lambda_1+m-1} \\ x_1^{\lambda_2+m-2} & \cdots & x_m^{\lambda_2+m-2} \\ \vdots & & \vdots \\ x_1^{\lambda_m} & \cdots & x_m^{\lambda_m} \end{matrix}\right|}{\left|\begin{matrix} x_1^{m-1} & \cdots & x_m^{m-1} \\ x_1^{m-2} & \cdots & x_m^{m-2} \\ \vdots & & \vdots \\ 1 & \cdots & 1 \end{matrix}\right|}.$$ > > ^i6y60w For example, when $\lambda=(n)$, $\mathscr s_\lambda=h_n$. When $\lambda=1^n$ and $m=n$, $\mathscr s_\lambda=x_1\cdots x_n$. We can prove them by computing directly, or by the following proposition. > [!proposition] Jacobi-Trudi formula > > We have > > $$ > \mathscr s_\lambda=\left|\begin{matrix}h_{\lambda_1} & h_{\lambda_1+1} & \cdots & h_{\lambda_1+m-1} \\ h_{\lambda_2-1} & h_{\lambda_2} & \cdots & h_{\lambda_2+m-2}\\ \vdots & \vdots & & \vdots \\ h_{\lambda_m-m+1} & h_{\lambda_m-m+2} & \cdots & h_{\lambda_m}\end{matrix}\right|. > $$ ^jrsjoc `\begin{proof}` Let $\alpha=(\alpha_1,\cdots,\alpha_m)$ be a composition. Put $$A_\alpha=\begin{bmatrix} x_1^{\alpha_1} & \cdots & x_m^{\alpha_1} \\ \vdots & & \vdots \\ x_1^{\alpha_m} & \cdots & x_m^{\alpha_m} \end{bmatrix}\mbox{ and }H_\alpha =\begin{bmatrix} h_{\alpha_1-m+1} & \cdots & h_{\alpha_1} \\ \vdots & & \vdots \\ h_{\alpha_m-m+1} & \cdots & h_{\alpha_m}\end{bmatrix}.$$ Consider the elementary symmetric polynomial $e_n^{(k)}$ in variables $x_1,\cdots,\hat x_k,\cdots,x_m$ with $1\leqslant k\leqslant m$ and write them to $m\times m$ matrix $$M=\begin{bmatrix} (-1)^{m-1}e_{m-1}^{(1)} & \cdots & (-1)^{m-1}e_{m-1}^{(m)} \\ \vdots & & \vdots \\ (-1)e_1^{(1)} & \cdots & (-1)e_1^{(m)} \\ 1 & \cdots & 1\end{bmatrix}.$$ We claim that $A_\alpha=H_\alpha M$. Consider generating function for $e_n^{(k)}$

E^{(k)}(t)=\sum_{n=0}^{m-1}e_n^{(k)}t^n=\prod_{i\neq k}(1+x_it),

then by $H(t)E(-t)=1$ there is

H(t)E^{(k)}(t)=\frac{1}{1-x_kt}=1+x_kt+\cdots+(x_kt)^\ell+\cdots.

Take the coefficient of $t^{\alpha_i}$ and we have

\sum_{j=1}^m h_{\alpha_i-m+j}(-1)^{m-j}e_{m-j}^{(k)}=x_k^{\alpha_i}\tag{*}.

With $(*)$, we can prove the claim. Thus, $\det A_\alpha=\det H_\alpha\det M$ and so $\det H_\alpha=\det A_\alpha/\det M$. For $\delta=(m-1,m-2,\cdots,1,0)$, the matrix $H_\delta=1$ and so $\det A_\delta=\det M$. It follows that

\det H_\alpha=\frac{\det A_\alpha}{\det A_\delta}=\mathscr s_\lambda,

where $\alpha=\lambda+\delta=(\lambda_1+m-1,\lambda_2+m-2,\cdots,\lambda_m)$. Now we finish the proof. `\end{proof}` We have a more direct formula for $\mathscr s_\lambda$ by considering a generalized tableau of shape $\lambda$. > [!definition] > > We place numbers $\{1,\cdots,n\}$ into tableau $\lambda\vdash n$ allowing repetition. Such tableau is called *semistandard* if rows are weakly increasing (not decreasing) while columns are strictly increasing sequences. > > If $T$ is semistandard with $\{1,\cdots,n\}$, set > > $$ > \mathscr x^T=\prod_{i=1}^n(x_i)^{\operatorname{number of occurrence of }i \in T}. > $$ > [!proposition] > > For $\lambda\vdash n$ with length $m$, the Schur polynomial in $m$ variables is $\mathscr s_\lambda=\sum_T \mathscr x^T$. # Basis of $\Lambda_m$ Now we consider algebra of symmetric polynomials in $m$ variables. First define $\Lambda_m^n=\mathrm{span}\{\mathscr s_\lambda:\lambda\vdash n\}$, the subspace of homogeneous symmetric polynomial of degree $n$, and then define $\Lambda_m=\oplus_{n\geqslant 0}\Lambda_m^n$. For example, when $m=3$, we have $\Lambda_m^0=\mathrm{span}\{1\}$, $\Lambda_m^1=\mathrm{span}\{x_1+x_2+x_3\}$,

\begin{aligned} \Lambda_m^2&=\mathrm{span}{\mathscr s_{(1,1)},\mathscr s_{(2,0)}}=\mathrm{span}{x_1x_2+x_1x_3+x_2x_3,x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3}, \ \Lambda_m^3&=\mathrm{span}{s_{(3)},s_{(2,1)},s_{(1,1,1)}}=\mathrm{span}{x_1x_2x_3,x_1^2x_2+x_1^2x_3+x_1x_2^2+2x_1x_2x_3+x_1x_3^2+x_2^2x_3+x_2x_3^2,h_3}. \end{aligned}

> [!proposition] > > $\mathscr s_\lambda$ is a basis for $\Lambda_m$ for all $\lambda\vdash n$ with $n\geqslant 0$. (For a fixed $n$, $\lambda\vdash n$ running through all partitions $\{\mathscr s_\lambda:\lambda\vdash n\}$ is a basis for $\Lambda_m^k$.) Then we introduce a few families of symmetric polynomials for algebra $\Lambda_m$. - For any partition $\lambda=(\lambda_1,\cdots,\lambda_m)$, define $p_\lambda=p_{\lambda_1}\cdots p_{\lambda_m}$, $e_\lambda=e_{\lambda_1}\cdots e_{\lambda_m}$ and $h_\lambda=h_{\lambda_1}\cdots h_{\lambda_m}$. For example, when $m=3$ and $n=2$, $p_{(2)}=p_2=x_1^2+x_2^2+x_3^2$ and $p_{(1,1)}=p_1p_1=(x_1+x_2+x_3)^2$, $e_{(2)}=e_2=x_1x_2+x_2x_3+x_1x_3$ and $e_{(1,1)}=e_1e_1=(x_1+x_2+x_3)^2$, $h_{(2)}=h_2=x_1^2+x_2^2+x_3^2+x_1x_2+x_1x_3+x_2x_3$ and $h_{(1,1)}=h_1h_1=(x_1+x_2+x_3)^2$. - Let $\lambda=(\lambda_1,\cdots,\lambda_m)$. Another family of symmetric polynomials is defined as the sum of all different monomials obtained from $x_1^{\lambda_1}\cdots x_m^{\lambda_m}$ under permutation. For example, when $m=3$ and $n=2$, we have $m_{(2)}=x_1^2+x_2^2+x_3^2$, $m_{(1,1)}=2(x_1x_2+x_2x_3+x_1x_3)$. > [!theorem] > > Suppose that $\lambda$ runs over all partitions of $n$ of length at most $m$. Then each of the following families is basis of the space $\Lambda_m^n$: > > $$ > m_\lambda,\mathscr s_\lambda,e_{\lambda'},h_{\lambda'},p_{\lambda'},e_\lambda,h_\lambda,p_\lambda > $$ > > In particular, $\dim \Lambda_m^n=p(n)$ with $m\geqslant n$. `\begin{proof}` We first prove $\{m_\lambda:\lambda\vdash n\}$ is a basis. Note that they are linearly independent: If $\lambda\vdash n$ and $\mu\vdash n$ with $\lambda\neq \mu$, then $m_\lambda$ and $m_\mu$ have no common monomials. Hence, if $\sum_i \alpha_i m_{\lambda_i}=0$, then $\alpha_i=0$. Suppose $f\in \Lambda_m^n$, and we aim to show $f\in\mathrm{span}\{m_\lambda:\lambda\vdash n\}$. We do induction on the order of the element. Take the greatest element $x_1^{n_1}\cdots x_m^{n_m}$ in [[10.3 Ordering of Partitions#^0ujtr4|lexicographic]] order with $\mu=(n_1,\cdots,n_m)$. Then $f-m_\mu\in\Lambda_m^n$ and $f-m_\mu<f$. By induction, $f$ can be written as linear combination of $\{m_\lambda:\lambda\vdash n\}$. To show $\mathscr s_\lambda$ form a basis, define $A_m$ as the space of skew-symmetric polynomials in $x_1,\cdots,x_m$. We say $g$ is skew-symmetric if $(ij)g=-g$. It is easy to see that any skew-symmetric polynomials is divisible by $\prod_{1\leqslant i<j\leqslant m}(x_i-x_j)$. We have that $\varphi:\Lambda_m\to A_m,f\mapsto f\prod_{1\leqslant i<j\leqslant m}(x_i-x_j)$, where $\varphi$ defines an isomorphism between $\Lambda_m$ and $A_m$. One can [check](https://math.uchicago.edu/~may/REU2020/REUPapers/Graham.pdf?utm_source=chatgpt.com) that basis of $A_m$ can be given by $$b_\lambda=\left|\begin{matrix} x_1^{\lambda_1+m-1} & \cdots & x_m^{\lambda_1+m-1} \\ x_1^{\lambda_2+m-2} & \cdots & x_m^{\lambda_2+m-2} \\ \vdots & & \vdots \\ x_1^{\lambda_m} & \cdots & x_m^{\lambda_m} \end{matrix}\right|$$ where $\lambda$ runs over through partitions of $n$ of length $\leqslant m$. Notice that $\mathscr s_\lambda=b_\lambda/\prod_{1\leqslant i<j\leqslant m}(x_i-x_j)$ and $\varphi(\mathscr s_\lambda)=b_\lambda$. Furthermore, as $\dim\Lambda_m^n=\dim A_m^{n+{m\choose 2}}$, it deduces that $\mathscr s_\lambda$ form a basis. Moreover, recall that [[10.4 Complete List of Irreducible Modules#^i3a31d|^i3a31d]] and similarly we have $\mathscr s_\mu=\sum_{\lambda}\kappa_{\mu\lambda}m_\lambda=m_\mu+\sum_{\lambda\neq \mu}\kappa_{\mu\lambda}m_\lambda$, and it yields that $\left\langle \mathscr s_\mu:\mu\vdash n\right\rangle=\left\langle m_\mu:\mu\vdash n\right\rangle=\Lambda_m^n$. To show $e_{\lambda'}$ form a basis, suppose that $\lambda\vdash n$ with $\lambda=(\lambda_1,\cdots,\lambda_k)$ and $\lambda'=(\rho_1,\cdots,\rho_k)$, and then

e_{\lambda’}=e_{\rho_1}\cdots e_{\rho_k}=(x_1\cdots x_{\rho_1}+\cdots)(x_1\cdots x_{\rho_2}+\cdots)\cdots(x_1\cdots x_{\rho_k}+\cdots)=x_1^{\lambda_1}\cdots x_m^{\lambda_m}+\cdots+\mbox{smaller terms}=m_\lambda+\sum_{\mu\lhd\lambda}m_\mu.

For $h_\lambda$, we can do it by [[#^jrsjoc|^jrsjoc]], as it deduces that $\left\langle \mathscr s_\lambda:\lambda\vdash n\right\rangle\subseteq\left\langle h_\lambda:\lambda\vdash n\right\rangle$. `\end{proof}` > [!definition] > > For each $m$ we have > > $$ > \Lambda_{m+1}\to \Lambda_m,f(x_1,\cdots,x_m,x_{m+1})\mapsto f(x_1,\cdots,x_m,0) > $$ > > Define a [projective limit](https://en.wikipedia.org/wiki/Inverse_limit) algebra $\Lambda=\oplus_{n\geqslant 0}\Lambda^n$ where $f=(f_m:m\geqslant 0)$ is a sequence of $f_m\in\Lambda_m^n$ and $f_{m+1}\mapsto f_m$ for all $m\geqslant 0$. Then $f$ is a formal series in infinitely many variables, and there are several examples: - $p_n=\sum_{i=1}^\infty x_i^n$ - $e_n=\sum_{1\leqslant i_1<\cdots< i_n}x_{i_1}\cdots x_{i_n}$ - $h_n=\sum_{1\leqslant i_1\leqslant\cdots\leqslant i_n}x_{i_1}\cdots x_{i_n}$ For more details of projective limit, see page 489 of [[Algebra chapter 0 - 2009 - Aluffi.pdf]]. The sequence of Schur polynomials $\mathscr s_\lambda$ defines Schur functions, and the sequence of $m_\lambda$ defines the monomial symmetric functions. > [!lemma] > > We have > > $$ > h_n=\sum_{\lambda\vdash n}\frac{p_\lambda}{z_\lambda}\mbox{ and }e_n=\sum_{\lambda\vdash n}\frac{\epsilon_\lambda p_\lambda}{z_\lambda} > $$ > > where $\lambda=(1^{m_1},\cdots,n^{m_n})$, $z_\lambda=1^{m_1}m_1! \cdots n^{m_n}m_n!$ and $\epsilon_\lambda=(-1)^{n-\ell(\lambda)}$. ^dbe3dc `\begin{proof}` Recall that ![[10.8 Symmetric Polynomials#^skztji|^skztji]] It follows

H(t)=\exp(\sum_{k=1}^\infty \frac{p_kt^k}{k})=\prod_{k=1}^\infty\exp(\frac{p_kt^k}{k})=\left( 1+p_1t+\frac{(p_1t)^2}{2!}+\cdots\right)\left( 1+\frac{p_2t}{2}+\frac{(p_1t)^2}{2^2\cdot2!}+\cdots\right)\left( 1+\frac{p_3t}{3}+\frac{(p_3t)^2}{3^2\cdot 2!}+\cdots\right)

and it deduces what we desire. Similarly we can prove $e_n=\sum_{\lambda\vdash n}{\epsilon_\lambda p_\lambda}/{z_\lambda}$. `\end{proof}` > [!proposition] orthogonal properties > > For two families of variables $x=(x_1,\cdots,x_k,\cdots)$ and $y=(y_1,\cdots,y_k,\cdots)$, we have the following properties > > $$ > \begin{aligned} > > \prod_{i,j\geqslant 1}(1-x_iy_j)^{-1}&=\sum_{\lambda}z_\lambda^{-1}p_\lambda(x)p_\lambda(y),\\ > \prod_{i,j\geqslant 1}(1-x_iy_j)^{-1}&=\sum_\lambda h_\lambda(x)m_\lambda(y)=\sum_\lambda m_\lambda(x)h_\lambda(y)=\sum_\lambda \mathscr s_\lambda(x)\mathscr s_\lambda(y). > > \end{aligned} > $$ > `\begin{proof}` By [[#^dbe3dc|^dbe3dc]], we have

\prod_{i\geqslant 1}(1-x_it)^{-1}=\sum_\lambda z_\lambda^{-1}p_\lambda t^{|\lambda|}\tag{*}

where $|\lambda|$ is the length of $\lambda$. Note that $\sum_{i,j}(x_iy_j)^k=p_k(x)p_k(y)$. Use $(*)$ for $x_iy_j$ and set $t=1$, we have

\prod_{i,j\geqslant 1}(1-x_i y_j)^{-1}=\sum_\lambda z_\lambda^{-1} p_\lambda(x_iy_j)=\sum_\lambda z_\lambda^{-1}p_\lambda(x_i)p_\lambda(y_j).

Equip $\Lambda$ with the following form $\left\langle h_\lambda, m_\mu\right\rangle=\delta_{\lambda\mu}$, and it deduces that $\left\langle p_\lambda,p_\mu\right\rangle=z_\lambda \delta_{\lambda\mu}$ and $\left\langle \mathscr s_\lambda,\mathscr s_\mu\right\rangle=\delta_{\lambda\mu}$. With respect to this inner product, $\{\mathscr s_\lambda:\lambda\vdash n\}$ is orthonormal basis of $\Lambda^n$. `\end{proof}`