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Let $\Lambda^d_n$ the space of symmetric polynomials in $n$ variables, with maximum 'partial degree' of each variable $d$. A basis for this space is the set of symmetrized monomials $m_\lambda$, where $\lambda$ is a partition with maximally $n$ parts, with each part $\leq d$.

Take $n= m N$ and $d = N$ (with $m>1$, and both $m$ and $N$ finite) and define the following specialization (or plethysm) $\mathcal{C}$, $$ \mathcal{C}: \Lambda^{N}_{m N} \rightarrow \Lambda^{mN}_{N} $$ that conflates the $m N$ variables to $N$ variables, via $x_{m(i-1)+j} \rightarrow y_i$, for $i=1,2,\ldots,N$ and $j=1,2,\ldots,m$. Specifically, for $p$ a symmetric polynomial in $\Lambda^N_{m N}$, we have $$ \mathcal{C} \bigl( p(x_1,\ldots,x_{mN}) \bigr) = p( \underbrace{y_1,\ldots,y_1}_{m}, \underbrace{y_2,\ldots,y_2}_{m},\ldots,\underbrace{y_{N},\ldots,y_{N}}_{m}) \ . $$

We strongly suspect that the map $\mathcal{C}$ as defined above is bijective, but could not find this or a similar result in the literature. Is it known that $\mathcal{C}$ is indeed bijective, and if so, how does one prove this?

Note that the restriction on the `partial degree' is essential for $\mathcal{C}$ to be injective. A simple example shows that $\mathcal{C}$ is not injective on, for instance, $\Lambda_2$, where $\Lambda_n$ is the space of symmetric polynomials in $n$ variables. Namely, take $p_2 (x_1,x_2) - 1/2 (p_1(x_1,x_2))^2 = 1/2 (x_1-x_2)^2$. It is clear that $\mathcal{C} \Bigl( p_2 (x_1,x_2) - 1/2 (p_1(x_1,x_2))^2 \Bigr) = 0$, showing that $\mathcal{C}$ is not injective on $\Lambda_2$.

We tried to find a pair of bases, for which the transition matrix becomes triangular, but did not succeed. Take for instance symmetrized monomials. They can simply be expressed in terms of power sums, for which the action of $\mathcal{C}$ is simple. Expressing the result back in terms of monomials (the power sums are in general not linearly independent), leads to the result that the transition matrix is not triangular, as can be shown by means of a simple example.

Take $m=2$ and $N=3$, and consider the polynomials with total degree 4. The basis for $\Lambda_6$ is $m_\lambda (x)$, with $\lambda \in \{ (4),(3,1),(2,2),(2,1,1),(1,1,1,1) \}$, while the basis for $\Lambda_3$ is given by $m_\lambda (y)$, with $\lambda \in \{ (4),(3,1),(2,2),(2,1,1)\}$. Note that for three variables, $m_{(1,1,1,1)} = 0$. The action of $\mathcal{C}$ is given, in these bases, by $$ \begin{pmatrix} 2 & 2 & 1 & 0 & 0\\ 0 & 4 & 0 & 4 & 0\\ 0 & 0 & 4 & 4 & 1\\ 0 & 0 & 0 & 8 & 4\\ \end{pmatrix} $$ Taking into account the `partial degree' restriction amounts to deleting the first column of this matrix. The resulting matrix has non-zero determinant, but this does not follow from a triangular structure.

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  • $\begingroup$ Have you managed to compute this map for say, powersum, elementary, or monomial basis? If you do this, express the result in the same basis (or different), and look at the transition matrix. With some luck, and some smart ordering of the elements, you might hope for some triangular system. $\endgroup$ – Per Alexandersson Sep 22 '14 at 12:40
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    $\begingroup$ Let $p_r$ be the power sum symmetric function of degree $r$. Since $\mathcal{C}(p_r(x_1,...,x_{mN})) = mp_r(y_1,..y_N)$ we have $\mathcal{C}(p_r(x)) = mp_r(y)$. Since plethystic substitution is a ring homomorphism, $\mathcal{C}(p_{r_1}(x)\ldots p_{r_k}(x)) = m^k p_{r_1}(y) \ldots p_{r_k}(y)$. So $\mathcal{C}$ is diagonal with non-zero eigenvalues in the power-sum basis of $\Lambda^d_n$. $\endgroup$ – Mark Wildon Sep 22 '14 at 12:51
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    $\begingroup$ @per we tried this with several types of polynomials in several combinations, but we didn't find a combination that gives a triangular structure. $\endgroup$ – eddy ardonne Sep 23 '14 at 13:41
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    $\begingroup$ @MarkWildon This method certainly works in the case of an infinite number of variables, but seems more complicated in the case of a finite number of variables. In that case, we need to take the 'partial degree' restriction into account. The power sums $p_\lambda$ with $\lambda$ a partition of at most $n$ parts, each part $\leq d$ do not form a basis of $\Lambda^d_n$, which complicates matters. Do you mean to say that we do not have to worry about this? If so, could you explain why this isn't a problem? $\endgroup$ – eddy ardonne Sep 23 '14 at 13:47
  • $\begingroup$ My comment above has a serious error: as the example in the question shows, the $p_{r_1}(y)\ldots p_{r_k}(y)$ are not necessarily linearly independent. I've deleted an incorrect answer that built on this comment. $\endgroup$ – Mark Wildon Sep 29 '14 at 13:20
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This is not a complete answer, but maybe this will be helpful. Let

$$H_k(t)=\prod_{i=1}^k(1-x_it)^{-1}=\sum_{n\geq0}h_n(x_1,\ldots,x_k)t^n.$$

Applying the map $\mathcal{C}$ to this generating series gives

$$\mathcal{C}(H_{mN}(t))=[H_N(t)]^m=\sum_{n\geq0}h_n^{(m)}t^n$$

where

$$h_n^{(m)}=\sum_{\lambda\in P(n)}{m\choose \ell(\lambda)}{\ell(\lambda)\choose m_1(\lambda),m_2(\lambda),\ldots}h_\lambda.$$

That is $\mathcal{C}(h_n)=h_n^{(m)}$, and we set $h_\lambda^{(m)}=\mathcal{C}(h_\lambda)$

Now, expand

$$\Pi_N(x,y)^m = \prod_{i,j=1}^N(1-x_i y_j)^{-m}=\sum_\lambda h_\lambda^{(m)}(x)m_\lambda(y).$$

The sum above is over $\lambda$ such that $\ell(\lambda)\leq N$.

Now, in the infinite variable version, one has

$$\Pi(x,y)^m=\prod_{i,j}(1-x_iy_j)^{-m}=\sum_{\lambda,\mu}\langle u_\lambda^* ,v_\mu^* \rangle_m u_\lambda(x) v_\mu(y)$$

where the bilinear form $\langle\cdot,\cdot\rangle_m$ is defined on power sums by the formula $\langle p_\lambda, p_\mu \rangle_m = \delta_{\lambda,\mu} z_{\lambda} m^{\ell(\lambda)},$ and $u_\lambda^*$ means dual to $u_\lambda$ with respect to $\langle\cdot,\cdot\rangle_1$. Another orthogonal basis for $\langle\cdot,\cdot\rangle_m$ consists of Jack symmetric functions.

Anyway, specializing to the finite variable case says that the $h_\lambda^{(m)}$ are dual to $m_\lambda$ ($\ell(\lambda)\leq N$) with respect to this form. In particular, the set $$\{ h_\lambda^{(m)}\mid \ell(\lambda)\leq N\}$$ is a basis for $\Lambda_N$.

Now, for $\ell(\lambda)>k$, one has $$(*)\;\;\;h_\lambda^{(m)}(x_1,\ldots,x_k)=\sum_{\mu>\lambda}c_{\mu\lambda}h_\mu^{(m)}(x_1,\ldots,x_k).$$ Hence, matrix for $\mathcal{C}$ in the bases $\{h_\lambda\mid \ell(\lambda)<mN\}$ and $\{h_\lambda^{(m)}\mid \ell(\lambda)<N\}$, restricted to the degree $d$ component, has the form $$[\mathcal{C}]=\begin{pmatrix}1&\cdots&0&*&*\\0&\ddots&0&*&*\\0&\cdots&1&*&*\end{pmatrix}$$ Now, the question of whether the restriction of $\mathcal{C}$ to $\lambda_{mN}^N$ is injective amounts to showing that certain coefficients $c_{\mu\lambda}$ in $(*)$ are nonzero.

It would probably be easier to do something with the power sum symmetric functions directly, but I don't see immediately why $\{p_\lambda\mid \ell(\lambda)\leq N\}$ is a basis for $\Lambda_N$ (thought I am probably just being stupid).

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Here are two small observations on the case when $N=2$. Let $\mathcal{C}[d]$ be the restriction of $\mathcal{C} : \Lambda_{2m}^2 \rightarrow \Lambda_2^{2m}$ to the degree $d$ component of $\Lambda_{2m}^2$.

Writing $\mathrm{mon}$ for monomial symmetric functions (to avoid the clash with $m$), the degree $2m$ component of $\Lambda_{2m}^2$ has basis $$ \{ \mathrm{mon}_{(m+b,m-b)'} : 0 \le b \le m \} $$ Now $$\mathrm{mon}_{(m+b,m-b)'}(x_1,\ldots,x_{2m}) = \sum x_{i_1}^2 \ldots x_{i_{m-b}}^2 x_{j_1}\ldots, x_{j_{2b}} $$ where the sum is over all disjoint subsets $\{i_1,\ldots,i_{m-b} \}, \{j_1,\ldots,j_{2b}\}$ of $\{1,\ldots, 2m\}$. Applying $\mathcal{C}$, we get the lexicographically maximal term when all the variables $x_{i_1},\ldots,x_{i_{m-b}}$ and $b$ of the variables $x_{j_1},\ldots,x_{j_{2b}}$ specialize to $y_1$. So $$\{i_1,\ldots,i_{m-b} \} \subseteq \{1,\ldots,m\} \subseteq \{i_1,\ldots,i_{m-b} \} \cup \{j_1,\ldots,j_{2b}\} \subseteq \{1,\ldots,2m\}$$ and \begin{align*} \mathcal{C}(\mathrm{mon}_{(m+b,m-b)'}(x_1,\ldots,x_{2m}) ) &= \binom{m}{m-b}\binom{m}{b} y_1^{2(m-b)+b} y_2^b + \cdots \\ &= \binom{m}{b}^2 \mathrm{mon}_{(2m-b,b)}(y_1,y_2) + f \end{align*} where $f$ is a linear combination of polynomials $\mathrm{mon}_{(2m-c,c)}(y_1,y_2)$ for $c > b$. Hence $\mathcal{C}[2m]$ is triangular in the monomial bases of the degree $2m$ components of $\Lambda_{2m}^2$ and $ \Lambda_2^{2m}$, with eigenvalues $\binom{m}{b}^2$ for $0 \le b \le m$. In particular $\mathcal{C}[2m]$ is bijective.

The second observation is that, in the monomial bases, the matrix of $\mathcal{C}[d]$ is equal to the matrix of $\mathcal{C}[4m-d]$. Given a partition $\mu = (d_1,d_2)$ with parts $\le 2m$, define $\mu^\star = (2m-d_2,2m-d_1)$. Then $$[\mathrm{mon}_{(d_1,d_2)}(y_1,y_2)]\mathcal{C}(\mathrm{mon}_{(e_1,e_2)'}(x)) = [\mathrm{mon}_{(d_1,d_2)^\star}(y_1,y_2)]\mathcal{C}(\mathrm{mon}_{(e_1,e_2)^{\star\,\prime}}(x)) $$

The idea can be seen in an example: take $d = 4$ and $m=4$ and extend $\Lambda_8$ by allowing negative powers of the variables $x_1,\ldots,x_8$. Extend $\mathcal{C}$ in the obvious way, and note that $\mathcal{C}(x_1^2\ldots x_8^2) = y_1^8y_2^8$. Hence \begin{align*} \mathcal{C}(\mathrm{mon}_{(2,1,1)}(x_1,\ldots,x_8)) &= \mathcal{C}(x_1^2x_2x_3 + \cdots ) \\ &= \mathcal{C}\bigl( x_2^{-1}x_3^{-1}x_4^{-2}\ldots x_8^{-2} + \cdots) (x_1^2\ldots x_8^2) \bigr) \\ &= \mathcal{C}(\mathrm{mon}_{(2^5,1^2)}(x_1^{-1},\ldots,x_8^{-1}) y_1^8y_2^8 . \end{align*} The coefficients of $y_1^3y_2$ in the left-hand side and $y_1^{-5}y_2^{-7}$ in the right-hand side are equal; this gives the claimed result when $(d_1,d_2) = (e_1,e_2) = (3,1)$ and $(d_1,d_2)^\star = (e_1,e_2)^\star = (7,5)$.

The matrix of $\mathcal{C}[2m-1]$ (and so $\mathcal{C}[2m+1]$) is also triangular in the monomial bases. It looks like this fails to hold for all degrees $d$ with $3 \le d \le 2m-2$.

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