Representation stability formulated as rationality in generating function of symmetric functions?

Question

We know that for a sequence of real numbers $(a_n)$, the fact that $\lim_n a_n=a\neq 0$ is "more or less" equivalent to saying that the generating function $\sum_n a_n t^n$ has radius of convergence $1$ and a pole at $t=1$ with residue $a$. (Thank Richard to point out that they are not exactly equivalent; but if in addition $(1-t)\sum_n a_n t^n$ has radius of convergence greater than $1$, then it is sufficient to imply $\lim_n a_n=a$.) Often if we expect a limit, we would like to think of the generating function as the "master object" so the limit can be recovered and one can discover finer asymptotic too.

Now consider the $S_n$-equivariant analogue. Let $\Lambda=\mathbb Z[e_1,e_2,\dots]$ be the ring of symmetric functions, and $\Lambda_n$ be the homogeneous degree $n$ piece. Recall $s_\lambda$ (for partitions $\lambda$) are the Schur function, and they are orthonormal under the Hall inner product $\langle \cdot,\cdot \rangle$. For a sequence of symmetric function $(f_n)$ with $f_n\in \Lambda_N$, the following two (equivalent?) notions of (numerical shadows of) representation stability are commonly seen:

1. For each fixed partition $\lambda$, the sequence $\langle f_n, s_{\lambda+(n-|\lambda|)}\rangle$ defined for $n\geq |\lambda|$ stabilizes as $n\to \infty$. Here $\lambda+(n-|\lambda|)$ means the partition obtained from $\lambda$ by enlarging its first part until the size is $n$.

2. For any homogeneous polynomial $P(x_1,x_2,\dots)\in \mathbb Q[x_1,x_2,\dots]$ (where $\deg x_k=k$), for any $n$, consider the class function on $S_n$: $\chi_P(\sigma)=P(c_1(\sigma),\dots)$, where $c_i$ is the number of $i$-cycles of $\sigma$. In symmetric function language (i.e., passing to Frobenius character), $$\chi_P=\sum_{\lambda\vdash n} \frac{p_\lambda}{z_\lambda} P(m_1(\lambda),\dots),$$ where $m_i(\lambda)$ is the multiplicity of $i$ in $\lambda$, and $z_\lambda=\prod_i i^{m_i}m_i!$ is the centralizer count of cycle type $\lambda$. Then $\langle \chi_P, f_n\rangle$ converges as $n\to \infty$.

These two notions are probably related by a change of coordinate, but I haven't thought this through.

Question. Is there an intrinsic framework to talk about $\lim_n f_n$?

Here is a guess of how it could be approached. The motivating example is when $f_n=s_{(n)}$. It satisfies representation stability 1, so we kind of want to say the generating function $\sum_{n\geq 0} s_(n) t^n$ has a "pole" at "$t=1$''. On the other hand, using the infinite set of variables $x_1,x_2,\dots$ allows the infinite product $$\sum_{n\geq 0} s_{(n)}(x_1,\dots) t^n=\frac{1}{(1-x_1t)(1-x_2t)\cdots},$$ so it does have a "pole" in some sense.

Question. Is there a framework to study the "asymptotic" of $f_n$ (and make this notion precise) in terms of a notion of "poles" of $\sum_n f_n t^n$?

Some testing question includes, for example, what is the "denominator" of $$\sum_{n\geq 1} s_{(n-1,1)}(x_1,\dots) t^n,$$ and can it "see" the stability of the sequence $s_{(n-1,1)}$?

Answer 1

A small quibble is that representation stability as originally defined refers to more than the numerical consequences (see Church-Farb's definition). More significantly, there is a stronger consequence than your condition 1 (at least for finite dimensional cases): up to adding rows there are only finitely many $\lambda$ for which $\langle f_n, s_{\lambda + |n|-n} \rangle$ is nonzero. I'll discuss an answer to your question for this stronger version of 1.

As your post suggests, the key case is $$H = \sum_{m} h_m = \sum_{m} s_{(m)}$$ The generating function of any representation stable sequence (in the stronger sense) can always be written as $\tilde a + \tilde b H$ where $\tilde a, \tilde b$ are finite symmetric functions. The "stable limit" of the sequence is determined by $\tilde b$.

Under the analogy with generating functions, representation stable sequences correspond to rational functions with denominator $1-t$. The decomposition above corresponds to how any such function can be written as a polynomial $a(t)$ plus $b/(1-t)$, where $b$ is the constant that can be extracted by multiplying by $(1-t)$ and plugging in $1$.

At least formally, to extract $\tilde b$ we can multiply by $1/H(x) = E(-x) = \prod_{i} (1- x_i)$ and plug in $x_1 = 1$, leaving a symmetric function in infinitely many variables that records $g$. In your test example, we have that $\sum_{n} s_{(n,1)} = s_1 H - H + 1$, so the "residual symmetric function" in this case is $\tilde b = s_1 - 1$.

Answer 2

Let me attempt to answer my own question. After doing some related work, I believe I have found a satisfying framework, which, while not fundamentally new, gives a robust recipe for encoding "rational generating functions" and recover the common operations like inner product, skewing operator, and Kronecker products, "as a whole", without expanding into any basis. The essential ingredients are present in first chapter of Macdonald’s Symmetric Functions and Hall Polynomials [1] and are implicitly present in many papers, but what appears to be never emphasized enough is recognizing that a few identities suffice to handle many rationality statements without invoking any particular basis or deeper combinatorics when not necessary.

See also [1], p. 91, Ex. 25, and p. 95, Ex. 29.

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Notation

• A bold letter (e.g., $\mathbf{x}$) denotes a countably infinite alphabet $(x_1, x_2, \dots)$, while a plain letter (e.g., $t$) denotes $(t,0,0,\dots)$.
• The notation $\mathbf{x}\mathbf{y}$ stands for the set $\{x_i y_j\}_{i,j \geq 1}$, and the notation $\mathbf{x},\mathbf{y}$ stands for $\{x_i, y_i\}_{i\geq 1}$.
• Define the Cauchy kernel: $$ \mathcal{H}(\mathbf{x}) := \prod_{i \geq 1} \frac{1}{1 - x_i} = \sum_{n\geq 0} h_n(\mathbf{x}). $$
• For example, Cauchy identity (in monomial/homogeneous basis) is: $$ \mathcal{H}(\mathbf{x} \mathbf{y}) = \sum_{\lambda} m_\lambda(\mathbf{x}) h_\lambda(\mathbf{y}). $$

Operators:

• $\langle -,- \rangle$ is the Hall inner product on $\Lambda$.
• $f^\perp$ denotes the adjoint of multiplication by $f$ with respect to this inner product.
• $f \ast g$ denotes the Kronecker product (tensor product of $S_n$-representations under Frobenius characteristic).

If multiple alphabets are involved, we clarify the variable involved by subscript, e.g., $\langle f(\mathbf{x}), g(\mathbf{x})h(\mathbf{y}) \rangle_{\mathbf{x}}=\langle f,g\rangle h(\mathbf{y})$.

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Key Formulas (to be proved at the end)

Let $f, g \in \Lambda$:

1. Skewing by $\mathcal{H}$: $$ \mathcal{H}(\mathbf{x}\mathbf{y})^{\perp_{\mathbf{x}}} f(\mathbf{x}) = f(\mathbf{x}, \mathbf{y}). $$
2. General skewing: $$ f(\mathbf{x})^{\perp_{\mathbf{x}}} g(\mathbf{x}) = \langle f(\mathbf{x}), g(\mathbf{x}, \mathbf{y}) \rangle_{\mathbf{y}}. $$
3. Kronecker product with $\mathcal{H}$: $$ \mathcal{H}(\mathbf{x} \mathbf{y}) \ast_{\mathbf{x}} f(\mathbf{x}) = f(\mathbf{x} \mathbf{y}). $$
4. General Kronecker product: $$ f(\mathbf{x}) \ast_{\mathbf{x}} g(\mathbf{x}) = \langle f(\mathbf{y}), g(\mathbf{x} \mathbf{y}) \rangle_{\mathbf{y}}. $$

Easy observations:

5. $\mathcal{H}(\mathbf{x}, \mathbf{y}) = \mathcal{H}(\mathbf{x}) \cdot \mathcal{H}(\mathbf{y})$.
6. $\langle f(\mathbf{x}), 1 \rangle_{\mathbf{x}} = f(0)$ (evaluation at all variables zero).

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Methodology

When computing a Kronecker product, especially for generating functions like $\sum_n a_n \ast b_n$, the strategy is:

1. Express generating functions as something like $\mathcal{H}(\mathbf{x}; \mathbf{y}) \cdot f(\mathbf{x})$.
2. Play with the identities above. They will be robust enough, see the next example.

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Example 1:

Let $f, g \in \Lambda$. Consider: $$ f(\mathbf{x}) \mathcal{H}(\mathbf{x} \mathbf{y}) \ast_{\mathbf{x}} g(\mathbf{x}) \mathcal{H}(\mathbf{x} \mathbf{z}). $$

Apply Formula 4: $$ = \left\langle f(\mathbf{w}) \mathcal{H}(\mathbf{w} \mathbf{y}),\ g(\mathbf{x} \mathbf{w}) \mathcal{H}(\mathbf{x} \mathbf{w} \mathbf{z}) \right\rangle_{\mathbf{w}}. $$

To remove the factors involving the infinite product $\mathcal H$, we move it to the other side using adjointness: $\langle f,gh\rangle = \langle h^\perp f, g\rangle$, and apply Formula 1 to simplify the skewing: $$ = \left\langle f(\mathbf{w}, \mathbf{x} \mathbf{z}) \cdot \mathcal{H}(\mathbf{w} \mathbf{y}, \mathbf{x} \mathbf{y} \mathbf{z}),\ g(\mathbf{x} \mathbf{w}) \right\rangle_{\mathbf{w}}. $$

Factor out $\mathcal{H}(\mathbf{x} \mathbf{y} \mathbf{z})$, which is independent of $\mathbf{w}$: $$ = \mathcal{H}(\mathbf{x} \mathbf{y} \mathbf{z}) \cdot \left\langle f(\mathbf{w}, \mathbf{x} \mathbf{z}) \cdot \mathcal{H}(\mathbf{w} \mathbf{y}),\ g(\mathbf{x} \mathbf{w}) \right\rangle_{\mathbf{w}}. $$

Repeat Formula 1 again: $$ = \mathcal{H}(\mathbf{x} \mathbf{y} \mathbf{z}) \cdot \left\langle f(\mathbf{w}, \mathbf{x} \mathbf{z}),\ g(\mathbf{x} \mathbf{w}, \mathbf{x} \mathbf{y}) \right\rangle_{\mathbf{w}}. $$

So the product is a finite symmetric function times a “pole” $\mathcal{H}(\mathbf{x} \mathbf{y} \mathbf{z})$.

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Example 2:

As a toy example to illustrate the method, say we want to know the Frobenius character of $\mathbb{C}^n \otimes \mathbb{C}^n$, where $\mathbb{C}^n$ is the tautological $S_n$ representation. Let's do it in a mechanical way so that we know for sure that this sequence is "stable", and the generating function has Schur positive numerator, without having any prior insights from representation theory or combinatorics.

Let $\{a_n\}$ be the Frobenius character of $\mathbb{C}^n$. Define: $$ A(\mathbf{x}) = \sum_{n \geq 1} a_n(\mathbf{x}) = s_1(\mathbf{x}) \cdot \mathcal{H}(\mathbf{x}). $$ (from Pieri rule).

Now compute the generating function for $a_n \ast a_n$. Note that $a_n\ast a_m=0$ if $m\neq n$ by homogeneity, so: $$ (\sum_{n} a_n \ast a_n)(\mathbf{x})=A(\mathbf{x}) \ast A(\mathbf{x}) = s_1(\mathbf{x}) \mathcal{H}(\mathbf{x}) \ast_{\mathbf{x}} s_1(\mathbf{x}) \mathcal{H}(\mathbf{x}). $$

Apply Example 1 with $\mathbf{y}=\mathbf{z}=1:=(1,0,0,\dots)$: $$ (\sum_{n} a_n \ast a_n)(\mathbf{x})= \mathcal{H}(\mathbf{x}) \cdot \left\langle s_1(\mathbf{w}, \mathbf{x}),\ s_1(\mathbf{x} \mathbf{w}, \mathbf{x}) \right\rangle_{\mathbf{w}}. $$

Explicitly compute:

• $s_1(\mathbf{w}, \mathbf{x}) = s_1(\mathbf{w}) + s_1(\mathbf{x})$,
• $s_1(\mathbf{x} \mathbf{w}, \mathbf{x}) = s_1(\mathbf{x}) s_1(\mathbf{w}) + s_1(\mathbf{x})$.

So: $$ (\sum_{n} a_n \ast a_n)(\mathbf{x})= \mathcal{H}(\mathbf{x}) \cdot \left(s_1(\mathbf{x})^2 + s_1(\mathbf{x})\right) = \mathcal{H}(\mathbf{x}) \cdot \left(s_2(\mathbf{x}) + s_{1,1}(\mathbf{x}) + s_1(\mathbf{x})\right). $$

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The next example shows that it is easy to extract numerical information from the generating function above without unpackaging.

Example 3: The dimension of $(\mathbb{C}^n\otimes \mathbb{C}^n)^{S_n}$

Compute generating function: $$ \sum_n \langle a_n \ast a_n,\ h_n \rangle t^n = \left\langle \sum a_n(\mathbf{x}) \ast a_n(\mathbf{x}),\ \sum h_n(\mathbf{x}) t^n \right\rangle_{\mathbf{x}}. $$

From above: $$ \sum_n \langle a_n \ast a_n,\ h_n \rangle t^n= \left\langle \mathcal{H}(\mathbf{x})(s_1^2 + s_1)(\mathbf{x}),\ \mathcal{H}(\mathbf{x} t) \right\rangle_{\mathbf{x}}. $$

Using $$ \langle \mathcal{H}(\mathbf{x} \cdot \mathbf{y}),\ f(\mathbf{x}) \rangle_{\mathbf{x}} = f(\mathbf{y}), $$ (an easy fact, which can also be extracted from Formula 1 and Observation 6) we get: $$ \sum_n \langle a_n \ast a_n,\ h_n \rangle t^n = \mathcal{H}(t)(s_1^2 + s_1)(t) = \frac{t^2 + t}{1 - t}. $$

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Appendix: Proofs of Formulas

1. Extract the $s_\lambda(\mathbf{z})$ coefficients of [1], p. 71, (a)(b) to prove the case $f=s_\lambda$, and then apply linearity for general $f$.
2. For the case $f=s_\lambda$, extract the $s_\lambda(\mathbf{y})$ coefficient of Formula 1. Then apply linearity.
3. This is [1], p. 116, (7.11).
4. Extract the $s_\lambda(\mathbf{y})$ coefficient of Formula 3.

[1]: Macdonald’s Symmetric Functions and Hall Polynomials