18

In algebraic combinatorics, conjectures that certain numbers are integers or are positive or are unimodal are really implicit challenges to find the hidden structure. And finding hidden structure is a large part of what mathematics is all about.
As lambda said in the comments, in the specific case of combinatorially defined symmetric functions whose Schur-...

answered Nov 10 '17 at 3:12

Timothy Chow

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14

Let $t$ be an indeterminate. Let
$\vartheta:\Lambda
\rightarrow \Lambda[t]$
be the specialization (homomorphism) defined by
$$ \vartheta(p_k)=t+\sum_{i=1}^k {k\choose i}p_i, $$
where $p_i$ is a power sum summetric function. According to item 75 of
http://math.mit.edu/~rstan/ec/...

answered Jan 22 '16 at 1:00

Richard Stanley

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14

In the special case mentioned in the problem, I'll show the bound
$$
|e_{n/2}(x_1,\ldots, x_n)| \le 2^{n/2}.
$$
Let
$$
F(z) = \prod_{j=1}^{n} (1+zx_j) = C \prod_{j=1}^{n} (z + \overline{x_j}),
$$
where $C= \prod_{j} x_j$ has magnitude $1$. (The roots of $F$ are $-\overline{x_j}$.)
Now by Cauchy's theorem
$$
e_{n/2}(x_1,\ldots,x_n) = \frac{1}{2\pi ...

12

I would like to suggest an interpretation using super symmetric functions. These are symmetric functions that are symmetric in two sets of variables $\{x_i\}$ and $\{y_j\}$ separately. They satisfy the property that setting a single $x_i$ variable to equal $z$ and a single $y_j$ variable equal to $z$ gives a polynomial independent of $z$, in addition to ...

12

This was recently proven by Suvrit Sra, on the arXiv. (Make sure to read version 3.)

10

We can use the fact that $N(\lambda)=\left|\text{SSYT}(\lambda)\right|$, the number of semistandard Young tableaux of shape $\lambda$, and that $d!\cdot\left(\frac{\prod_{1\le i < j\le n}(\lambda_i - \lambda_j + j - i)}{\prod_{1 \le i \le n} (\lambda_i + n - i)!}\right)=\left|\text{SYT}(\lambda)\right|$, the number of standard Young tableaux of shape $\...

answered Jul 25 '16 at 22:49

Gjergji Zaimi

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9

This is discussed in Stanley's paper Some combinatorial aspects of the Schubert calculus. Corollary 3.7 says that under the natural isomorphism given by the Borel presentation of $H^*(G/P)$ which sends an ordinary Schur function $s_{\lambda}$ to the class of the Schubert variety $X_{\lambda}$, a skew Schur function $s_{\lambda / \mu}$ is sent to the class of ...

8

Yes, your prediction is correct. The determinant identity in this case is theorem 3.5 in Division and the Giambelli Identity, by Wu and Yang (also published at Linear Algebra Appl. 406 (2005), 301-309).

answered Nov 26 '18 at 18:00

Gjergji Zaimi

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7

It follows from the Cauchy identity and Exercise 7.43 of Enumerative
Combinatorics, vol. 2, that
$$ 1+ (u+t)\sum_{1\leq l\leq d} s_{l,1^{d-l}}(x)u^{l-1} t^{d-l} =
\prod_i\frac{1+tx_i}{1-ux_i} $$
$$ \qquad = \exp \sum_{n\geq 1}\frac 1n p_n(x)(u^n-(-t)^n). $$
If the decomposition you want actually exists then it should follow
from the above formula, ...

answered Dec 20 '17 at 16:03

Richard Stanley

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7

To expand a bit on the answer of Matt Samuel, let
$x$ and $y$ be two sets of variables (of any
length) and $\lambda$ any partition. Then
\begin{eqnarray*} s_\lambda(x,y) & = & \sum_{\mu\subseteq\lambda}
s_\mu(x)s_{\lambda/\mu}(y)\\ & = &
\sum_{\mu,\nu} c_{\mu\nu}^\lambda s_\mu(x)s_\nu(y).
\end{eqnarray*}
See for instance equation (7....

answered Oct 10 '15 at 14:49

Richard Stanley

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6

The sum (in any number of variables) is equal to the determinant
$$\det(B_{j-i})_{1\le i,j\le n},$$
where
$$B_i=\sum_{l=0}^\infty e_{l+i}(x)e_l(y),$$
and $e_l$ is the elementary symmetric function, with $e_l=0$ for $l<0$.
This follows by applying the involution $\omega$ in $x$ and $y$ to Theorem 16 of my paper Symmetric functions and P-recursiveness, ...

6

Actually, all of these representations are defined over $\mathbb{Q}$ (i.e., they are absolutely indecomposable). This is stated in the second sentence of the Encyclopedia of Math article on Representation of the symmetric groups. This follows immediately from the fact that the Young symmetrizer is defined over $\mathbb{Q}$ (and thus also $\mathbb{R}$).
If ...

answered Nov 6 '15 at 14:22

6

Here is another argument. Let $\chi^{\mu/k}$ denote the skew character
of the symmetric group $\mathfrak{S}_n$ corresponding to the skew
shape $\mu/k$. Then by Pieri's rule,
$$ \sum_{\mu-\lambda\ \mathrm{is\ a\ horizontal\ strip}}
\sum_{\alpha\vdash|\lambda|}\frac{\chi^\lambda_\alpha}{z_\alpha}
\prod_i p_i^{a_i}
= \sum_k \sum_{\alpha\vdash n-k}
...

answered Nov 15 '20 at 15:19

Richard Stanley

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6

This is a special case of the generating function for Schur polynomials
$$\sum_{\lambda}s_\lambda(x_1,\dots,x_m)s_\lambda(y_1,\dots,y_n)=\prod_{i=1}^m\prod_{j=1}^n\frac 1{1-x_iy_j}.$$
Take $x_1=\dots=x_m=x$ and $y_1=\dots=y_n=1$, gives
$$\sum_{\lambda} x^{|\lambda|}s_\lambda(1^m)s_\lambda(1^n)=\frac 1{(1-x)^{mn}}.$$
It seems you take $m=n=N\kappa=N_f$ and $x=...

5

The answer is yes, simply because they are symmetric functions. However, it can even be done with nonnegative coefficients. The formula I present in Reduction formula for Schubert polynomials combined with the Littlewood-Richardson rule will work, though there's a more intuitive way (still requiring the Littlewood-Richardson rule).
Namely, we can represent ...

5

I found a proof which I don't really like, but I'll share it.
For two (real) diagonal matrices $A,B$, the Harish-Chandra-Itzykson-Zuber (HCIZ) integral is
$$
I(A,B) = \int_{U(n)} e^{\rm{tr}(U^* A U B)} \, \rm{d} U = c_n \frac{\det\left([e^{a_j b_k}]_{j,k=1}^n\right)}{\Delta(a)\Delta(b)},
$$
where $\Delta(a) = \prod_{j<k} (a_k - a_k)$ is the Vandermonde ...

4

Your general determinant has to do with the Wronskian isomorphism for $SL_2$ representations
namely a map
$$
\wedge^m({\rm Sym}^n(\mathbb{C}^2))\rightarrow {\rm Sym}^m({\rm Sym}^{n-m+1}(\mathbb{C}^2)))\ .
$$
If you follow it by the natural map
$$
{\rm Sym}^m({\rm Sym}^{n-m+1}(\mathbb{C}^2))\rightarrow {\rm Sym}^{m(n-m+1)}(\mathbb{C}^2)
$$
which corresponds ...

answered Oct 19 '14 at 20:29

Abdelmalek Abdesselam

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4

There are several approaches.
Use linear algebra. Compute Schur polynomials (using the Jacobi-Trudi identity, say), and then use the fact that the coefficient of $s_\nu$ in the product $s_\lambda s_\mu$ is a Littlewood-Richardson coefficient.
Use a combinatorial interpretation.
One can count so called Littlewood-Richardson tableaux, or lattice points in ...

answered Nov 19 '19 at 11:30

Per Alexandersson

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4

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.

4

Please excuse that I answer with a link, I only have a phone right now.
http://www.findstat.org/MapsDatabase/Mp00192/

3

If $X = (u_1, u_2, \ldots, u_a, w_1, w_2, \ldots, w_c)$ and $Y = (v_1, v_2, \ldots, v_b, w_1, \ldots, w_c)$ with $u$'s, $v$'s and $w$'s disjoint, then
$$s_{\lambda}(u,w) s_{\mu}(v,w) = \left( \sum_{\alpha,\ \beta} c_{\alpha \beta}^{\lambda} s_{\alpha}(u) s_{\beta}(w) \right) \left( \sum_{\gamma,\ \delta} c_{\gamma \delta}^{\mu} s_{\gamma}(v) s_{\delta}(w) \...

answered Jun 3 '17 at 15:54

David E Speyer

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3

As a first step towards a solution, using the identity
$$\int_{\mathcal{U}(N)} U_{\alpha a}U_{\alpha' a'}\bar{U}_{\beta b}\bar{U}_{\beta' b'}\,dU=\frac{1}{N^{2}-1}\bigl( \delta_{\alpha\beta}\delta_{ab}\delta_{\alpha'\beta'}\delta_{a'b'}+ \delta_{\alpha\beta'}\delta_{ab'}\delta_{\alpha'\beta}\delta_{a'b}\bigr)$$
$$\qquad\qquad\mbox{}-\frac{1}{N(N^{2}-1)}\bigl(...

answered Nov 22 '18 at 11:31

Carlo Beenakker

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3

There is a generalisation of the character formula, although it is usually stated in terms of contents rather than Frobenius coordinates. Also, it applies to conjugacy classes labelled by partitions $(1^{m_1} 2^{m_2} \cdots)$, where $m_2, m_3, \ldots$ are fixed and $m_1$ varies with $n$ (as in the example you gave, where there is one part of size $2$, and $n-...

3

I am skeptical of any "nice" structure while $x$ and $y$ remain so free. It might be somewhat reasonable to check things out under certain specializations.
You may find some interesting quotients, after specializations on $x$ and $y$, in Richard Stanley's Enumerative Combinatorics, Vol. 2, Exercises 7.30 and 7.32.

answered Dec 30 '16 at 17:40

T. Amdeberhan

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3

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 \...

2

For the hook shapes the following bijection becomes especially simple.
Christian Krattenthaler, Another involution principle-free bijective proof of Stanley's hook-content formula, see here.

rt.representation-theory symmetric-groups symmetric-functions schur-functions bijective-combinatorics

2

Couple of quick observations for $\alpha_i(x)=x^{(d_i)}$ ($x^{(d)}=x^d/d!$ as usual).
Note that if $d_i>d_{i+1}$, for all $i$, then $G(x_1,\ldots,x_n)=0$ unless $d_i-d_{i+1}=1$. In particular, if the sequence $\mathbf{d}=(d_i)_{i=1}^n$ is strictly decreasing, then $G\neq 0$ only if $\mathbf{d}=(n-1,\ldots,0)$. In this case $G=1$.
Also, note that if $\...

2

I can't answer your second question, but I have a pretty fast recurrence for computing a single Littlewood-Richardson coefficient for the complete flag variety. It also works for equivariant coefficients, but unlike other equivariant recurrences I've seen you can just set the equivariant part to 0 and get a recurrence for the ordinary coefficients. It does ...

2

$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\varepsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\X}{\mathcal{X}}
\newcommand{\E}{\operatorname{\mathsf E}}
\newcommand{\PP}{\operatorname{\mathsf P}}
\...

answered Apr 17 '18 at 20:57

Iosif Pinelis

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2

Probably not. I don't see an easy proof in general. Similar questions have been asked in Arithmetic Dynamics (look for dynamical analogues of Mordell-Lang). Clearly, the answer is no for $n=2$. For $n=3$, if you had such a situation then you would get infinitely many points on a fixed number field on the curves of the form $P(x_1^{m^k},x_2^{m^k},x_3^{m^k})=0$...

answered Feb 15 '20 at 22:26

Felipe Voloch

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