The vanishing of sum of coefficients: symmetric polynomials












7












$begingroup$


Denote $pmb{X}_n=(x_1,x_2,dots,x_n)$. Consider the symmetric polynomial
$$f_n(pmb X_n)=prod_{1leq i<jleq n}(x_i+x_j).$$
Expand these in terms of elementary symmetric polynomials, say
$$f_n(pmb{X}_n)=sum_{mu}c_{mu,n}cdot e_{mu}(pmb{X}_n).$$



For example,
begin{align*} f_3&=-e_{(3)}+e_{(2,1)} \
f_5&=-e_{(5,5)}+2e_{(5,4,1)}+e_{(5,3,2)}-e_{(5,2,2,1)}-e_{(4,4,1,1)}-e_{(4,3,3)}+e_{(4,3,2,1)}.
end{align*}




QUESTION 1. Is it true that, for integers $n geq 1$, we have
$$sum_{mu}c_{mu,2n+1}=0?$$




POSTSCRIPT. Fedor's reply (to Question 1) shown below suggests to me to ask:




QUESTION 2. Is it true that, for integers $n geq 1$, we have
$$sum_{mu}c_{mu,2n}=(-1)^{binom{n}2}?$$







reference-request co.combinatorics rt.representation-theory symmetric-functions







share|cite|improve this question











$endgroup$

















    7












    $begingroup$


    Denote $pmb{X}_n=(x_1,x_2,dots,x_n)$. Consider the symmetric polynomial
    $$f_n(pmb X_n)=prod_{1leq i<jleq n}(x_i+x_j).$$
    Expand these in terms of elementary symmetric polynomials, say
    $$f_n(pmb{X}_n)=sum_{mu}c_{mu,n}cdot e_{mu}(pmb{X}_n).$$



    For example,
    begin{align*} f_3&=-e_{(3)}+e_{(2,1)} \
    f_5&=-e_{(5,5)}+2e_{(5,4,1)}+e_{(5,3,2)}-e_{(5,2,2,1)}-e_{(4,4,1,1)}-e_{(4,3,3)}+e_{(4,3,2,1)}.
    end{align*}




    QUESTION 1. Is it true that, for integers $n geq 1$, we have
    $$sum_{mu}c_{mu,2n+1}=0?$$




    POSTSCRIPT. Fedor's reply (to Question 1) shown below suggests to me to ask:




    QUESTION 2. Is it true that, for integers $n geq 1$, we have
    $$sum_{mu}c_{mu,2n}=(-1)^{binom{n}2}?$$







    reference-request co.combinatorics rt.representation-theory symmetric-functions







    share|cite|improve this question











    $endgroup$















      7












      7








      7


      1



      $begingroup$


      Denote $pmb{X}_n=(x_1,x_2,dots,x_n)$. Consider the symmetric polynomial
      $$f_n(pmb X_n)=prod_{1leq i<jleq n}(x_i+x_j).$$
      Expand these in terms of elementary symmetric polynomials, say
      $$f_n(pmb{X}_n)=sum_{mu}c_{mu,n}cdot e_{mu}(pmb{X}_n).$$



      For example,
      begin{align*} f_3&=-e_{(3)}+e_{(2,1)} \
      f_5&=-e_{(5,5)}+2e_{(5,4,1)}+e_{(5,3,2)}-e_{(5,2,2,1)}-e_{(4,4,1,1)}-e_{(4,3,3)}+e_{(4,3,2,1)}.
      end{align*}




      QUESTION 1. Is it true that, for integers $n geq 1$, we have
      $$sum_{mu}c_{mu,2n+1}=0?$$




      POSTSCRIPT. Fedor's reply (to Question 1) shown below suggests to me to ask:




      QUESTION 2. Is it true that, for integers $n geq 1$, we have
      $$sum_{mu}c_{mu,2n}=(-1)^{binom{n}2}?$$







      reference-request co.combinatorics rt.representation-theory symmetric-functions







      share|cite|improve this question











      $endgroup$




      Denote $pmb{X}_n=(x_1,x_2,dots,x_n)$. Consider the symmetric polynomial
      $$f_n(pmb X_n)=prod_{1leq i<jleq n}(x_i+x_j).$$
      Expand these in terms of elementary symmetric polynomials, say
      $$f_n(pmb{X}_n)=sum_{mu}c_{mu,n}cdot e_{mu}(pmb{X}_n).$$



      For example,
      begin{align*} f_3&=-e_{(3)}+e_{(2,1)} \
      f_5&=-e_{(5,5)}+2e_{(5,4,1)}+e_{(5,3,2)}-e_{(5,2,2,1)}-e_{(4,4,1,1)}-e_{(4,3,3)}+e_{(4,3,2,1)}.
      end{align*}




      QUESTION 1. Is it true that, for integers $n geq 1$, we have
      $$sum_{mu}c_{mu,2n+1}=0?$$




      POSTSCRIPT. Fedor's reply (to Question 1) shown below suggests to me to ask:




      QUESTION 2. Is it true that, for integers $n geq 1$, we have
      $$sum_{mu}c_{mu,2n}=(-1)^{binom{n}2}?$$







      reference-request co.combinatorics rt.representation-theory symmetric-functions




      reference-request co.combinatorics rt.representation-theory symmetric-functions






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 7 hours ago







      T. Amdeberhan

















      asked 10 hours ago









      T. AmdeberhanT. Amdeberhan

      17.7k229131




      17.7k229131






















          1 Answer
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          9












          $begingroup$

          Choose $n$ numbers $x_1,dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.



          If $n=2k$ is even, we substitute the roots $w_1,dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)dots (x-w_n)$. Then your claim reads as $$A:=prod_{1leqslant i<jleqslant n} (w_i+w_j)=(-1)^{binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first,
          $$
          |A|^2=prod_{i=1}^n prod_{jne i,1leqslant jleqslant n}|w_i+w_j|=2^{-n}prod_{i=1}^n prod_{j=1}^n|(-w_i)-w_j|=2^{-n}prod_{i=1}^n |f(w_i)|=\=2^{-n}prod_{i=1}^nleft|frac{(-w_i)^{2k+1}+1}{-w_i+1}right|=1,
          $$
          since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,dots,n$ and $prod_{i=1}^n (1-w_i)=f(1)=1$.



          At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.






          share|cite|improve this answer











          $endgroup$









          • 4




            $begingroup$
            An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
            $endgroup$
            – darij grinberg
            7 hours ago








          • 1




            $begingroup$
            Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
            $endgroup$
            – Fedor Petrov
            7 hours ago











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          9












          $begingroup$

          Choose $n$ numbers $x_1,dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.



          If $n=2k$ is even, we substitute the roots $w_1,dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)dots (x-w_n)$. Then your claim reads as $$A:=prod_{1leqslant i<jleqslant n} (w_i+w_j)=(-1)^{binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first,
          $$
          |A|^2=prod_{i=1}^n prod_{jne i,1leqslant jleqslant n}|w_i+w_j|=2^{-n}prod_{i=1}^n prod_{j=1}^n|(-w_i)-w_j|=2^{-n}prod_{i=1}^n |f(w_i)|=\=2^{-n}prod_{i=1}^nleft|frac{(-w_i)^{2k+1}+1}{-w_i+1}right|=1,
          $$
          since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,dots,n$ and $prod_{i=1}^n (1-w_i)=f(1)=1$.



          At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.






          share|cite|improve this answer











          $endgroup$









          • 4




            $begingroup$
            An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
            $endgroup$
            – darij grinberg
            7 hours ago








          • 1




            $begingroup$
            Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
            $endgroup$
            – Fedor Petrov
            7 hours ago
















          9












          $begingroup$

          Choose $n$ numbers $x_1,dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.



          If $n=2k$ is even, we substitute the roots $w_1,dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)dots (x-w_n)$. Then your claim reads as $$A:=prod_{1leqslant i<jleqslant n} (w_i+w_j)=(-1)^{binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first,
          $$
          |A|^2=prod_{i=1}^n prod_{jne i,1leqslant jleqslant n}|w_i+w_j|=2^{-n}prod_{i=1}^n prod_{j=1}^n|(-w_i)-w_j|=2^{-n}prod_{i=1}^n |f(w_i)|=\=2^{-n}prod_{i=1}^nleft|frac{(-w_i)^{2k+1}+1}{-w_i+1}right|=1,
          $$
          since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,dots,n$ and $prod_{i=1}^n (1-w_i)=f(1)=1$.



          At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.






          share|cite|improve this answer











          $endgroup$









          • 4




            $begingroup$
            An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
            $endgroup$
            – darij grinberg
            7 hours ago








          • 1




            $begingroup$
            Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
            $endgroup$
            – Fedor Petrov
            7 hours ago














          9












          9








          9





          $begingroup$

          Choose $n$ numbers $x_1,dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.



          If $n=2k$ is even, we substitute the roots $w_1,dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)dots (x-w_n)$. Then your claim reads as $$A:=prod_{1leqslant i<jleqslant n} (w_i+w_j)=(-1)^{binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first,
          $$
          |A|^2=prod_{i=1}^n prod_{jne i,1leqslant jleqslant n}|w_i+w_j|=2^{-n}prod_{i=1}^n prod_{j=1}^n|(-w_i)-w_j|=2^{-n}prod_{i=1}^n |f(w_i)|=\=2^{-n}prod_{i=1}^nleft|frac{(-w_i)^{2k+1}+1}{-w_i+1}right|=1,
          $$
          since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,dots,n$ and $prod_{i=1}^n (1-w_i)=f(1)=1$.



          At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.






          share|cite|improve this answer











          $endgroup$



          Choose $n$ numbers $x_1,dots,x_n$ for which all elementary symmetric polynomials are equal to 1 and substitute them to our $f_n$. We should get zero value for odd $n$. Well, what are these numbers? The roots of $x^{n}-x^{n-1}+x^{n-2}-ldots-1=(x^{n+1}-1)/(x+1)$. This polynomial indeed has two roots with sum equal to 0 when $n$ is odd.



          If $n=2k$ is even, we substitute the roots $w_1,dots,w_n$ of the polynomial $f(x)=x^{2k}-x^{2k-1}+ldots+1=(x^{2k+1}+1)/(x+1)=(x-w_1)dots (x-w_n)$. Then your claim reads as $$A:=prod_{1leqslant i<jleqslant n} (w_i+w_j)=(-1)^{binom{k}2}.$$ This is done by the standard trick (and is well known itself). At first,
          $$
          |A|^2=prod_{i=1}^n prod_{jne i,1leqslant jleqslant n}|w_i+w_j|=2^{-n}prod_{i=1}^n prod_{j=1}^n|(-w_i)-w_j|=2^{-n}prod_{i=1}^n |f(w_i)|=\=2^{-n}prod_{i=1}^nleft|frac{(-w_i)^{2k+1}+1}{-w_i+1}right|=1,
          $$
          since $1+(-w_i)^{2k+1}=2$ for all $i=1,2,dots,n$ and $prod_{i=1}^n (1-w_i)=f(1)=1$.



          At second, we need to find the argument of the complex number $A$. This may be done for example as follows: all pairs $w_i+w_j$ for which $w_i$ and $w_j$ are not complex conjugate are partitioned onto complex conjugate pairs. In each pair the product is positive reals. If $w_i$ and $w_j$ are complex conjugate, the sum $w_i+w_j$ is a real number whose sign is the sign of the real part of $w_i$. Therefore $A$ is the real number whose sign equals $(1)^{m/2}$, where $m$ is the number of $w$'s in the left half-plane. It is easy to see that $m/2=[k/2]$ and that $(-1)^{[k/2]}=(-1)^{k(k-1)/2}$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 8 hours ago

























          answered 9 hours ago









          Fedor PetrovFedor Petrov

          50.3k6115230




          50.3k6115230








          • 4




            $begingroup$
            An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
            $endgroup$
            – darij grinberg
            7 hours ago








          • 1




            $begingroup$
            Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
            $endgroup$
            – Fedor Petrov
            7 hours ago














          • 4




            $begingroup$
            An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
            $endgroup$
            – darij grinberg
            7 hours ago








          • 1




            $begingroup$
            Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
            $endgroup$
            – Fedor Petrov
            7 hours ago








          4




          4




          $begingroup$
          An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
          $endgroup$
          – darij grinberg
          7 hours ago






          $begingroup$
          An alternative to your "standard trick" is to observe that $w_i + w_j = dfrac{w_i^2 - w_j^2}{w_i - w_j}$. This yields $prodlimits_{i<j} left(w_i+w_jright) = dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$. But the $n$ numbers $-w_1^2, -w_2^2, ldots, -w_n^2$ are just a permutation of the $n$ numbers $w_1, w_2, ldots, w_n$, and thus $dfrac{prodlimits_{i<j}left(w_i^2 - w_j^2right)}{prodlimits_{i<j}left(w_i-w_jright)}$ equals a power of $-1$ times the sign of this permutation. Both are easy to compute.
          $endgroup$
          – darij grinberg
          7 hours ago






          1




          1




          $begingroup$
          Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
          $endgroup$
          – Fedor Petrov
          7 hours ago




          $begingroup$
          Yes, this is another standard trick:) Actually possibly the shortest proof is to combine them: the absolute values equals 1 since the differences $w_i^2-w_j^2$ and $w_i-w_j$ are the same up to sign, and the sign may be obtained by looking at the argument.
          $endgroup$
          – Fedor Petrov
          7 hours ago


















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