Homogeneous products of variables

Question

Prove that the sum of homogeneous products of $n$ dimensions which can be formed of the letters $a, b, c$ and their powers is

$$\frac {{a^{n+2}(b-c)}+{b^{n+2}(c-a)}+{c^{n+2}(a-b)}} {{a^2(b-c)}+{b^2(c-a)}+{c^2(a-b)}}.$$

Answer 1

A Proof without Knowing Schur Polynomials (Tedious and Inelegant Work)

Let $P_n$ be the required expression. For each partition $\mu=(\mu_1,\mu_2,\mu_3)$ of $n$ with at most three parts (i.e., $\mu_1,\mu_2,\mu_3$ are integers s.t. $\mu_1\ge \mu_2\ge \mu_3\ge0$ and $\mu_1+\mu_2+\mu_3=n$), let $m_\mu$ be the sum of all monomials of the form $a^{\mu_i}b^{\mu_j}c^{\mu_k}$ with $\{i,j,k\}=\{1,2,3\}$. Here are some examples: $m_{(3,3,3)}=a^3b^3c^3$, $m_{(2,1,1)}=a^2bc+ab^2c+abc^2$, $m_{(2,1,0)}=a^2(b+c)+b^2(c+a)+c^2(a+b)$.

We want to show by induction on $n$ that $P_n=\sum_{\mu\in Y^3_n} m_\mu$ where $Y^3_n$ is the set of partitions of $n$ with at most three parts. The base cases $n=0,1,2$ are trivial. Suppose now that $n\ge 3$. Since $P_n$ is symmetric in $a,b,c$ and is homogeneous of degree $n$, we see that $$P_n=\sum_{\mu\in Y^3_n}t_\mu m_\mu$$ for some constants $t_\mu$.

For $\nu=(\nu_1,\nu_2,\nu_3) \in Y_{n-1}^3$, we observe that $$(a+b+c) m_\nu=\left\{\begin{array}{ll}m_{\nu+(1,0,0)}&\text{if }\nu_1=\nu_2=\nu_3,\\ m_{\nu+(1,0,0)}+3m_{\nu+(0,0,1)}&\text{if }\nu_1=\nu_2=\nu_3+1\\ m_{\nu+(1,0,0)}+m_{\nu+(0,0,1)}&\text{if }\nu_1=\nu_2>\nu_3+1,\\ m_{\nu+(1,0,0)}+2m_{\nu+(0,1,0)}&\text{if }\nu_1=\nu_2+1=\nu_3+1,\\ m_{\nu+(1,0,0)}+m_{\nu+(0,1,0)}&\text{if }\nu_1>\nu_2+1=\nu_3+1,\\ m_{\nu+(1,0,0)}+2m_{\nu+(0,1,0)}+2m_{\nu+(0,0,1)}&\text{if }\nu_1=\nu_2+1=\nu_3+2\\ m_{\nu+(1,0,0)}+2m_{\nu+(0,1,0)}+m_{\nu+(0,0,1)}&\text{if }\nu_1=\nu_2+1>\nu_3+2\\ m_{\nu+(1,0,0)}+m_{\nu+(0,1,0)}+2m_{\nu+(0,0,1)}&\text{if }\nu_1>\nu_2+1=\nu_3+2\\ m_{\nu+(1,0,0)}+m_{\nu+(0,1,0)}+m_{\nu+(0,0,1)}&\text{if }\nu_1>\nu_2+1>\nu_3+2. \end{array}\right.\tag{1}$$ Likewise for $\kappa=(\kappa_1,\kappa_2,\kappa_3)\in Y_{n-2}^3$, we observe that $$(ab+bc+ca)m_\kappa=\left\{\begin{array}{ll}m_{\kappa+(1,0,0)}&\text{if }\kappa_1=\kappa_2=\kappa_3,\\ m_{\kappa+(1,1,0)}+2m_{\kappa+(1,0,1)}&\text{if }\kappa_1=\kappa_2=\kappa_3+1\\ m_{\kappa+(1,1,0)}+m_{\kappa+(1,0,1)}&\text{if }\kappa_1=\kappa_2>\kappa_3+1,\\ m_{\kappa+(1,1,0)}+3m_{\kappa+(0,1,1)}&\text{if }\kappa_1=\kappa_2+1=\kappa_3+1,\\ m_{\kappa+(1,1,0)}+m_{\kappa+(0,1,1)}&\text{if }\kappa_1>\kappa_2+1=\kappa_3+1,\\ m_{\kappa+(1,1,0)}+2m_{\kappa+(1,0,1)}+2m_{\kappa+(0,1,1)}&\text{if }\kappa_1=\kappa_2+1=\kappa_3+2\\ m_{\kappa+(1,1,0)}+m_{\kappa+(1,0,1)}+2m_{\kappa+(0,1,1)}&\text{if }\kappa_1=\kappa_2+1>\kappa_3+2\\ m_{\kappa+(1,1,0)}+2m_{\kappa+(1,0,1)}+m_{\kappa+(0,1,1)}&\text{if }\kappa_1>\kappa_2+1=\kappa_3+2\\ m_{\kappa+(1,1,0)}+m_{\kappa+(1,0,1)}+m_{\kappa+(0,1,1)}&\text{if }\kappa_1>\kappa_2+1>\kappa_3+2. \end{array}\right.\tag{2}$$ Finally for $\xi=(\xi_1,\xi_2,\xi_3)\in Y_{n-3}^3$, we have $$(abc)m_\xi=m_{\xi+(1,1,1)}.\tag{3}$$

Note that $$P_n=(a+b+c)P_{n-1}-(ab+bc+ca)P_{n-2}+(abc)P_{n-3}.\tag{4}$$ For a fixed $\mu=(\mu_1,\mu_2,\mu_3)\in Y_n^3$, we consider the following cases.

1. $\mu_1=\mu_2=\mu_3$: Then \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=3-3+1=1.$$

2. $\mu_1=\mu_2=\mu_3+1$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=2-1=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+2)-(2+1)+1=1.$$

3. $\mu_1=\mu_2>\mu_3+1$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=2-1=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+2)-(2+1)+1=1.$$

4. $\mu_1=\mu_2+1=\mu_3+1$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2,\mu_3-1)+(1,0,1)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(2+1)-(1+2)+1=1.$$

5. $\mu_1>\mu_2+1=\mu_3+1$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2,\mu_3-1)+(1,0,1)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(2+1)-(1+2)+1=1.$$

6. $\mu_1=\mu_2+1=\mu_3+2$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1)-1=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2,\mu_3-1)+(1,0,1)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1+1)-(1+1+1)+1=1.$$

7. $\mu_1=\mu_2+1>\mu_3+2$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1)-1=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2,\mu_3-1)+(1,0,1)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1+1)-(1+1+1)+1=1.$$

8. $\mu_1>\mu_2+1=\mu_3+2$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1)-1=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2,\mu_3-1)+(1,0,1)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1+1)-(1+1+1)+1=1.$$

9. $\mu_1>\mu_2+1>\mu_3+2$: We have two subcases.

   • $\mu_3=0$: \begin{align}\mu&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1)-1=1.$$
   • $\mu_3>0$: \begin{align}\mu&=(\mu_1,\mu_2,\mu_3-1)+(0,0,1)\\&=(\mu_1,\mu_2-1,\mu_3)+(0,1,0)\\&=(\mu_1-1,\mu_2,\mu_3)+(1,0,0)\\&=(\mu_1,\mu_2-1,\mu_3-1)+(0,1,1)\\&=(\mu_1-1,\mu_2,\mu_3-1)+(1,0,1)\\&=(\mu_1-1,\mu_2-1,\mu_3)+(1,1,0)\\&=(\mu_1-1,\mu_2-1,\mu_3-1)+(1,1,1),\end{align} by $(1)-(4)$ and the induction hypothesis, we get $$t_\mu=(1+1+1)-(1+1+1)+1=1.$$

Therefore $t_\mu=1$ for every $\mu\in Y^3_n$. The proof is now complete.

If you know anything about Schur polynomials $s_\lambda$, the task is much simpler. By definition, $$s_\lambda(x_1,x_2,\ldots,x_k)=\frac{\det\begin{pmatrix}x_1^{\lambda_1+k-1}&x_2^{\lambda_1+k-1}&\cdots &x_k^{\lambda_1+k-1}\\ x_1^{\lambda_2+k-2}&x_2^{\lambda_2+k-2}&\cdots &x_k^{\lambda_2+k-2}\\\vdots&\vdots&\ddots&\vdots\\ x_1^{\lambda_{k-1}+1}&x_2^{\lambda_{k-1}+1}&\cdots&x_k^{\lambda_{k-1}+1}\\x_1^{\lambda_k}&x_2^{\lambda_k}&\cdots&x_k^{\lambda_k}\end{pmatrix}}{\det\begin{pmatrix}x_1^{k-1}&x_2^{k-1}&\cdots &x_k^{k-1}\\ x_1^{k-2}&x_2^{k-2}&\cdots &x_k^{k-2}\\\vdots&\vdots&\ddots&\vdots\\ x_1&x_2&\cdots&x_k\\1&1&\cdots&1\end{pmatrix}}$$ where $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$ is a Young diagram (a partition of $n$) of length at most $k$. Note that $$s_\lambda(x_1,x_2,\ldots,x_k)=\sum_{T\in \operatorname{SSYT}_\lambda^k}x^T,$$ where $\operatorname{SSYT}_n^k$ is the set of all semi-standard Young tableaux $T$ of shape $\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$ of length at most $k$ such that all entries of $T$ are at most $k$, $x=(x_1,x_2,\ldots,x_k)$, and $x^T$ is the monomial of the form $x_1^{t_1}x_2^{t_2}\ldots x_k^{t_k}$ where $t_i$ is the number of times $i$ appears in $T$. However, $$P_n=s_{\lambda}(a,b,c)$$ for $\lambda=(n,0,0)$. It can be seen that each partition $\mu=(\mu_1,\mu_2,\mu_3)$ of $n$, $$m_\mu=\sum_{\substack{T\in \operatorname{SSYT}_\lambda^k\\ \langle t_1,t_2,t_3\rangle =\langle \mu_1,\mu_2,\mu_3\rangle}} x^T,$$ where $\langle\_\rangle$ denotes a multiset and $x=(a,b,c)$. (For example, $\langle 1,1,2\rangle=\langle 2,1,1\rangle$, but $\langle 1,1,2\rangle\neq \langle 2,1,2\rangle$.)