Plethystic Notation

We have seen that the symmetric functions $p_{\lambda}$ with $\lambda$ a partition of $n$ form a basis for $\Lambda_n$ Recall that each $p_{\lambda}$ is a product of the power functions $p_n$ where

$$p_n = \sum_i x^n$$ We can say that the ring of symmetric functions $\Lambda$ is freely generated as an algebra by $p_1,p_2,\cdots$ in other words

$$\Lambda = \mathbb{K} [p_1,p_2,\cdots]$$ Technically, we don't need variables to represent each symmetric function: each one can be expressed as a polynomial combination of the power symmetric functions and we can regard them as abstract symbols without further meaning. The plethystic notation is a very flexible way of handling different sets of variables to evaluate.

To specify an algebra homomorphism from $\Lambda$ it suffices to specify it on each $p_n$ Consider a Laurent series $A$ in some indeterminates $a_1,a_2, a_3\cdots$ Given a symmetric function $f$ the plethystic substitution $f[A]$ is the image of $f$ under the homomorphism sending $p_n$ to $p_n[A]$ where $p_n[A]$ is obtained by changing every indeterminate $a_i$ by $a_i^n$ Lets make sense of this with some examples

• If $A=a_1+a_2+...+a_k$ then for each $n$ $p_n[A] = a_1^n+a_2^n+\cdots a_k^n$ so that $f[A]$ is simply $f(a_1,a_2,\cdots,a_k)$ Recovering the notion of variables that we had in mind. In general we denote the sum of indeterminates by their capital letter. For example $Z=z_1+z_2+z_3+\cdots$ so that $f[Z]$ is the usual evaluation $f(z_1,z_2,\cdots)$
• The same arguments work for any expression that has a series expansion as a sum of monomials. If $A=1/(1-t)=1+t+t^2+t^3+\cdots$ then $f[1/(1-t)]=f(1,t,t^2,t^3,\cdots)$ In this notation we have the following extension of the hook length formula ​ $$s_{\lambda}[1/(1-t)] = \dfrac{t^{n(\lambda)}}{\prod_{x\in \lambda}(1-t^{h(x)})}$$ where $h(x)$ is the hook length of the cell $x$ in the diagram of $\lambda$
• The substitution $f[Z+a]$ means simply that we add a new variable $a$ to the set of variables. Sometimes this new $a$ is special and we want to keep track of it. For instance we have the following formula, the dual Pieri rule $$s_{\lambda}[Z+a]=\sum_k a^k \sum_{\lambda/\upsilon}s_{\upsilon}$$ For an explicit proof using jeu du taquin, consider each semi standard filling of the tableau using the variable $a$ $k$ times, and forward slide all the $a$ entries. The fact that it was a semi standard filling ensures that the positions of the $a$ 's after forward sliding them to the border is a horizontal strip of length $k$
• In the same train of thought the substitution $f[Z-z_i]$ means we are kicking out the variable $z_i$ Also note that $f[Z-(1-t)z_i]$ first kick out $z_i$ but then replace it by $tz_i$
• One of the simplest substitution is $A=1$ This sends $p_n\to 1$ for every $n$ Now recall the expansion​ $$s_\lambda = \dfrac{1}{n!}\sum_{w\in S_n} \chi^\lambda (w)p_{\tau(w)}$$ Which means $s_\lambda\left[1\right]=\dfrac{1}{n!} \sum_{w\in S_n}\chi^{\lambda}(w)1$ and that's precisely the inner product of the character $\chi^{\lambda}$ with the trivial character. Then the above expression is zero in all cases except when $\lambda=(n)$ where it is equal to 1. This substitution is simply giving the coefficient of $s_{(n)}$ in the ​expansion, in other words​ $$f[1] = \langle f,s_{(n)}\rangle$$
• The substitution $f[-Z]$ is very interesting. It sends $p_n\to -p_n$ for every $n$ so it sends $p_\lambda \to (-1)^{l(\lambda)}p_\lambda$ where $l(\lambda)$ is the length, or number of parts. Applying it to the power symmetric function of a partition of $m$ we get $$p_\lambda = (-1)^{l(\lambda)}p_\lambda = (-1)^{m}(-1)^{m+l(\lambda)}p_\lambda = (-1)^m \omega p_\lambda$$ so in general, for $f$ of degree $d$ $$f[-Z] = (-1)^d \omega(f)$$ where $\omega$ is the involution

An important thing to keep in mind is the difference between indeterminates and actual values, because the plethystic just affects the indeterdeminates. For example, it is easy to see that if $f$ is a symmetric function of degree $d$ then

$$f[tZ] = t^df[Z]$$ but it is not true that $f[-Z] = (-1)^df[Z]$ as the last example showed.

There is another important player in this theory. Define $$\Omega = \textrm{exp}\left( \displaystyle\sum_{k=1}^{\infty} p_k/k \right)$$ the obvious identities $p_k[A+B]=p_k[A]+p_k[B]$ and $p_k[-A]=-p_k[A]$ translates into $$\Omega[A+B]=\Omega[A]\Omega[B], \qquad \Omega[-A]=1/\Omega[A]$$ plus the single variable evaluation $$\Omega[x]=\textrm{exp}\left( \displaystyle\sum_{k=1}^{\infty} x^k/k \right)=1/(1-x)$$ we have $$\Omega[Z] = \prod \dfrac{1}{1-x_i}$$

The cauchy identity can be reformulated as: two bases $u$ and $v$ are dual with respect to the hall inner product if and only if

$$\displaystyle \sum_{\lambda} u_\lambda[X]v_{\lambda}[Y] = \Omega[XY]$$

so the bases $u$ and $v[AY]$ are dual for any Laurent series $A$ if and only if $\sum_{\lambda} u_\lambda[X]v_{\lambda}[AY] = \Omega[XY]$, and doing $Y\to AY$ gives $\sum_{\lambda} u_\lambda[X]v_{\lambda}[Y] = \Omega[XY/A]$ a condition which is symmetric in $u$ and $v$ So the following identity holds for dual bases

$$\langle u[AZ],v \rangle = \langle u,v[AZ]\rangle$$

and it implies that the following holds for any $f,g$

$$\langle f[AZ],g \rangle = \langle f,g[AZ]\rangle$$

The substitution $Z\to Z/(1-t)$ which is one of the most used, has the explicit inverse $Z\to Z(1-t)$ This substitutions means that the variables changing are the $z_i$ but not $t$ In other words, the operator $\Pi_{(1-t)}$ which sends $Z\to Z(1-t)$ has the inverse $\Pi_{(1-t)}^{-1}$ which is $Z\to Z(1-t)$ Moreover, by the above discussion $\Pi_{(1-t)}$ is self adjoint in the hall inner product.