Harmonic flow on the Young lattice

Question

Let me begin with some preliminary concepts: A positive real-valued function $\varphi: P \rightarrow \Bbb{R}_{>0}$ on a locally finite, ranked poset $(P, \trianglelefteq)$ is harmonic if $\varphi(\emptyset)=1$ and

\begin{equation} \varphi(u)=\sum_{\stackrel{\scriptstyle u \, \triangleleft \, v}{|v| \, = \, |u| + 1}} \varphi(v) \end{equation}

where $\emptyset$ is the unique bottom element of $P$ (which we require to exist) and where $|u|$ denotes the rank of an element $u \in P$. The poset $P$ is 1-differential if in addition

$\bullet$ The number of elements covered by both $u$ and $v$ equals the number of elements in $P$ covering both $u$ and $v$ whenever $u \ne v$.

$\bullet$ If $u \in P$ covers exactly $k$ elements then $u$ is covered by exactly $k+1$ elements.

In (https://arxiv.org/abs/math/9712266) Goodman and Kerov introduced a semi-group flow on the space of harmonic functions ${\frak{H}}(P)$ when $P$ is 1-differential. Specifically, for $\tau \in [0,1]$ and $\varphi \in {\frak{H}}(P)$

\begin{equation} C_\tau(\varphi)(v) \, := \, \sum_{k=0}^{|v|} {\tau^k (1-\tau)^{|v|-k} \over {(|v| -k )! }} \sum_{|u| = k} \varphi(u) \dim(u,v) \end{equation}

where $\dim(u,v)$ is the number of saturated chains $p_{|u|} \! \lhd \cdots \lhd p_{|v|}$ in $P$ starting at $p_{|u|} = u$ and ending at $p_{|v|}= v$. A fairly easy calculation reveals that $C_\tau(\varphi)$ is harmonic whenever $\varphi$ is and that $C_\tau (C_\sigma(\varphi)) = C_{\tau \sigma}(\varphi)$. Furthermore we recover the original function $\varphi$ when $\tau = 1$ while we obtain the function

\begin{equation} v \mapsto {1 \over {|v|!}} \dim(\emptyset, v) \end{equation}

when $\tau = 0$, which is known to be harmonic whenever $P$ is 1-differential.

Now consider the Young lattice $(\Bbb{Y}, \subseteq)$ of all integer partitions, ordered by inclusion of their respective Young diagrams. In virtue of the Pieri rule we know that the function

\begin{equation} \varphi(\lambda) \, := \, {s_\lambda({\bf x}) \over {s^n_{\Box}({\bf x})}} \quad \text{where $\lambda \vdash n$} \end{equation}

is a harmonic function on $\Bbb{Y}$ where $s_\lambda({\bf x})$ is the Schur function associated to $\lambda$ and $s_\Box({\bf x}) = x_1 + x_2 + x_3 + \cdots < \infty$ is the Schur function associated to the partition $(1)$. Let's apply the Goodman-Kerov flow to this function: For $\lambda \vdash n$ we get

\begin{equation} C_\tau (\varphi)(\lambda) \, = \, {1 \over {s^n_\Box({\bf x})}} \, \underbrace{\sum_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} s_\Box^{n-k}({\bf x}) \sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda)}_{\text{call this $s_\lambda({\bf x};\tau)$}} \end{equation}

We may now expand $s_\lambda({\bf x}; \tau)$ as $\sum_{\rho \vdash n} a_{\lambda, \rho}(\tau) s_\rho({\bf x})$.

Question: What can be said about the polynomials $a_{\lambda, \rho}(\tau)$? Have they already been identified/considered in the literature?

Sub-question: If the coefficient polynomials $a_{\lambda, \rho}(\tau)$ are (in general) messy, does anything nice happen with $s_\lambda({\bf x}; \tau)$ when we perform either the principal or content specializations, i.e.

\begin{equation} \begin{array}{ll} x_i \mapsto \ \ q^{i-1} \ \ \text{for all $i \geq 1$} & \\ x_i \mapsto \left\{ \begin{array}{ll} 1 &\text{for all $i \leq d$} \\ 0 &\text{for all $i > d$} \end{array} \right. &\text{for some fixed but far out $d \geq 1$} \end{array} \end{equation}

thanks, ines.

Answer 1

Let us identify $s_\lambda$ with the character of the irreducible representation $S^\lambda$ of the symmetric group $S_n$ indexed by the partition $\lambda$. Then $$ s_\Box^{n-k}({\bf x}) \sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda) $$ defines the character of a certain representation of $S_n$. We can explicitly describe it as follows. Since the Young lattice is the branching graph of representations of $S_n$, $\dim(\mu,\lambda)$ is exactly the multiplicity of $S^\mu$ in the restriction of $S^\lambda$ from $S^n$ to $S^k$ (where $n = |\lambda|$ and $k = |\mu|$), so $\sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda)$ is nothing but the character of the restriction of $S^\lambda$ to $S_k$. Similarly, it is a well known fact that multiplication by $s_\Box$ corresponds to induction from $S_m$ to $S_{m+1}$. So the quantity displayed above is the character of $$ \mathrm{Ind}_{S_k}^{S_n}\left( \mathrm{Res}_{S_k}^{S_n}\left(S^\lambda \right)\right) = \mathrm{Ind}_{S_k}^{S_n}(1) \otimes S^\lambda. $$ So this is the same thing as tensoring $S^\lambda$ with the representation obtained by inducing the trivial representation from $S_k$ to $S_n$. The induced representation $\mathrm{Ind}_{S_k}^{S_n}(1)$ that we are tensoring with might be more familiar under the notation $M^{(k, 1^{n-k})}$ (it is a permutation module). Its character expresses as $h_k s_\Box^{n-k}$, where $h_k$ is the complete symmetric function of degree $k$.

Now we will use the internal product of symmetric functions defined by $p_\mu * p_\nu = \delta_{\mu, \nu} z_\mu p_\mu$. So in particular, homogeneous symmetric functions of different degrees multiply to zero. It is convenient for us because it describes the tensor product of representations of symmetric groups: $s_\mu * s_\nu = \sum_{\lambda} k_{\mu, \nu}^\lambda s_\lambda$, where $k_{\mu, \nu}^\lambda$ is a Kronecker coefficient.

So now we may express $$ s_\lambda({\bf x}; \tau) = \sum_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} s_\Box^{n-k}({\bf x}) \sum_{|\mu| = k} s_\mu({\bf x}) \dim(\mu,\lambda) = \left( \sum_{k=0}^{n} {\tau^k (1-\tau)^{n-k} \over {(n -k )! }} h_k s_\Box^{n-k}\right) * s_\lambda. $$ To tidy up this equation, we sum over all possible values of $n$ and $k$ (there is no harm in doing this because the introduced terms will become zero upon taking the internal product with $s_\lambda$), which gives us a generating function $$ \left(\left( \sum_{k} \tau^k h_k \right)\left( \sum_l \frac{(1-\tau)^l s_\Box^{l}}{l!}\right)\right) * s_\lambda = \left(H(\tau) \exp((1-\tau)s_\Box) \right) * s_\lambda, $$ where $l = n-k$, and $H(\tau) = \sum_{k \geq 0} \tau^k h_k$ is the generating function of complete symmetric functions. Now, we can express $H(\tau)$ in terms of power-sum symmetric functions as $$ H(\tau) = \exp\left( \sum_{i \geq 1} \frac{p_i \tau^i}{i} \right). $$ Using the fact that $s_\Box = p_1$, we are left with $$ s_\lambda({\bf x}; \tau) = \exp\left( p_1 + \sum_{i \geq 2} \frac{p_i \tau^i}{i} \right) * s_\lambda = \sum_{|\mu| = n} \frac{\chi_\mu^\lambda p_\mu}{z_\mu} \tau^{n-m_1(\mu)}, $$ where $m_1(\mu)$ is the number of parts of size 1 in $\mu$ and we used the equation $s_\lambda = \sum_\mu \frac{\chi_\mu^\lambda p_\mu}{z_\mu}$.

But the quantity that was asked about, $a_{\lambda, \rho}(\tau)$, is the coefficient of $s_\rho$ in this expression: $$ a_{\lambda, \rho}(\tau) = \langle s_\rho, \exp\left( p_1 + \sum_{i \geq 2} \frac{p_i \tau^i}{i} \right) * s_\lambda \rangle = \sum_{|\mu| = n} \frac{\chi_\mu^\rho \chi_\mu^\lambda}{z_\mu} \tau^{n - m_1(\mu)}, $$ So this is like the inner product between two Schur functions with respect to a deformed inner product where power sums $p_i$ are weighted by a factor of $\tau$ if $i \geq 2$.

As a function of $\tau$, this interpolates between the case $\tau = 1$, where we just get $\langle s_\rho, s_\lambda \rangle = \delta_{\rho, \lambda}$, and the case where $\tau = 0$ (where only $\mu = (1^n)$ has a nonzero contribution to the sum) which gives $\frac{\dim(\rho)\dim(\lambda)}{n!}$. I do not recall seeing this particular interpolation before.

Regarding the evaluations you mention, they can be described using homomorphisms $\varphi: \Lambda \to \mathbb{C}$ defined by $$ \varphi(p_n) = 1^n + q^n + q^{2n} + \cdots = (1-q^n)^{-1} $$ (or $\varphi(p_n) = d$ in the latter case). The upshot here is that we can describe the result of applying $\varphi$ to $s_\lambda({\bf x}; \tau)$ as the result of applying a different homomorphism to the Schur function $s_\lambda$. Specifically, $\varphi(s_\lambda({\bf x}; \tau)) = \psi(s_\lambda)$, where $$ \psi(p_n) = \frac{\tau^n}{1-q^n} $$ for $n \geq 2$ and $\psi(p_1) = (1-q)^{-1}$. (In the latter case we take $\psi(p_n) = d\tau^n$ for $n\geq 2$ and $\psi(p_1) = d$.) The fact that these descriptions are not uniform (have a different case for $n=1$) makes me suspect there might not be nice formulas for the evaluations.