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Let $n \geq 1$ be an integer and consider the symmetric function

$$D_n = \sum_{d|n} p_d^{n/d},$$

where $p_{d}$ are the power-sum symmetric functions.

It can be checked up to $n=35$ that the symmetric function $D_n$ is Schur-positive.

The multiplicity of the Schur function $s_n$ in $D_n$ is given by the number of divisors of $n$ (sequence A000005 in the OEIS).

One can also easily compute the multiplicity of the Schur function $s_{1^n}$ in $D_n$:

$$\sum_{d|n} (-1)^{n+d},$$

which is sequence A112329 in the OEIS.

Q1: Is the Schur-positivity of $D_n$ for all $n$ known ?

Q2: Has this been considered somewhere in the literature ?

I have alas not much more context for this.

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  • $\begingroup$ I believe the answer might be hidden here: arxiv.org/pdf/2107.06389.pdf See definition 4.2, where a more general series is defined... $\endgroup$ Commented Aug 4, 2021 at 17:47
  • $\begingroup$ Thanks. But Def. 4.2 does not seem to include the case here, and has a division by $n$. $\endgroup$
    – F. C.
    Commented Aug 4, 2021 at 18:07
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    $\begingroup$ But $D_n$ is not divisible by n. And I cannot see how Def 4.2 could specialise to $D_n$. $\endgroup$
    – F. C.
    Commented Aug 4, 2021 at 19:22
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    $\begingroup$ Ah, yes, Eq. (2.6) is the more general expression, but your case does not seem to be treated. Well, S. Sundaram is the person to ask about this types of questions, she has some cool conjectures about Schur-positive sums of power-sum symmetric functions. $\endgroup$ Commented Aug 4, 2021 at 20:16
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    $\begingroup$ In fact, I already alerted Sheila about this question. $\endgroup$ Commented Aug 4, 2021 at 20:43

2 Answers 2

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Talking to Sheila Sundram, as suggested in the comments, was a good idea. After some conversation, the following proof became apparent. I don't know of any proof already in the literature.

Let $g$ be an $n$-cycle in $S_n$. Since the inverse Frobenius characteristic of the given symmetric function is supported only on classes that intersect $\langle g \rangle$ nontrivially, one might hope that it is induced from an actual character of $\langle g \rangle$, and this turns out to be the case.

For each divisor $d$ of $n$, write $\rho_d$ for the linear character of $\langle g \rangle$ sending $g$ to $e^{2 \pi i/d}$, and consider the induced character $\theta_d:=\rho_d\uparrow_{\langle g \rangle}^{S_n}$.

For positive integers $a$,$b$, we write $c_a(b)$ for the sum of the $b^{th}$ powers of all of the primitive complex $a^{th}$ roots of $1$ (a Ramanujan sum).

If $w \in S_n$ does not have cycle type $m^{n/m}$ for some divisor $m$ of $n$, then $\theta_d(w)=0$. If $w$ has shape $m^{n/m}$, a direct computation shows that $$\theta_d(w)=\frac{1}{n}|C_{S_n}(w)|c_{n/d}(m).$$ It follows that the Frobenius characteristic of $\theta_d$ is $$ \frac{1}{n}\sum_{m|n}c_{n/d}(m)p_m^{n/m}.$$ So, it suffices to show that there are nonnegative integers $\alpha_d$ ($d|n$) such that $$ \sum_{d} \alpha_dc_{n/d}(m)=n $$ for all divisors $m$ of $n$.

We write $S(n)$ for the matrix with rows and columns indexed by divisors of $n$ with entry $c_{n/d}(m)$ in the position indexed by $n$ and $d$, and $\alpha$ for the column vector with rows indexed as are the rows of $S(n)$ and entry $\alpha_d$ in the position indexed by $d$.

It suffices now to show that $S(n)$ is invertible and that the row sums of $nS(n)^{-1}$ are nonnegative integers, as these row sums are the entries of $\alpha$.

We can appeal to results found in the paper ``Ramanujan sums as supercharacters", by C. F. Fowler, S. R. Garcia and G. Karaali. According to Theorem 3.5 of that paper, if one orders the row and column indices appropriately, one gets $S(n)^2=nI$. So, the row sums of $nS(n)^{-1}$ are the row sums of $S$. A special case of Theorem 4.5 of the same paper implies that the row sums of $S$ are non-negative.

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Just adding that the expression for the Frobenius characteristic of $\theta_d$ is classically attributed to H. O. Foulkes, see e.g. Richard Stanley's EC2, Problem 7.88.

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