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I have recently been working on a problem to prove that a particular polynomial is in fact homogeneous. Although I have found out that this is true, I am curious to see whether there might be a deeper meaning to this, perhaps in algebra theory. Therefore, I will pose the problem here:

Let $\{x_1,x_2,\cdots,x_N\}$ be a set of complex variables. We can write for a polynomial $S$ the following expansion $$ S(\vec{x})=\sum_{\vec{m}_\in D^N}d_{m_1,m_2,\cdots,m_N}(\vec{p})\prod_{j=1}^Nx_j^{m_j}, $$ where $\vec{p}=(p_1,p_2,\cdots,p_N)^T$ is complex and the hypercube $D^N$ is defined as $$ D^N=\{0,1,\cdots,N-1\}^N. $$ The allowed powers of each variable are thus $0,1,\cdots,N-1$. Now we impose the condition that $(x_i \partial_i -x_j \partial_j +i/2\kappa(p_i-p_j))S(\vec{x})$ should be divisible by $(x_i-x_j)$ for every $i,j$ satisfying $i\neq j$. By trying to divide out this term we get the following condition on the coefficients: for each $\beta,\rho\in \{1,2,\cdots,N\}$ where $\beta \neq \rho$, $$ \sum_{l\in \mathbb{Z}}^*d_{m_1,\cdots,m_{\beta+l},\cdots,m_{\rho-l},\cdots,m_N}(\vec{p})\left(m_\beta-m_\rho +\frac{i}{2\kappa}(p_\beta-p_\rho) \right)=0, $$ where the star $*$ in the summation indicates that the summation runs over all $l\in \mathbb{Z}$ such that $0\leq m_\beta+l \leq N-1$ and $0\leq m_\rho-l \leq N-1$.

Using this relation, prove that the polynomial $S$ is homogeneous of degree $\frac{N(N-1)}{2}$. This is equivalent to proving that if $\sum_{i=1}^Nm_i\neq \frac{N(N-1)}{2}$, then $d_{m_1,m_2,\cdots,m_N}(\vec{p})=0$.

My own approach consisted in formulating an algorithm which rewrites the coefficient under consideration as a linear combination of coefficients which we are $0$. Consequently, I only showed that this algorithm terminated in a finite number of steps. However, for me this algorithm has no deep interpretation, while it would be very nice to understand the reason for this homogeneity. I would therefore very much like to know other approaches.

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  • $\begingroup$ is this problem from quantum mechanics? $\endgroup$ – Norbert Mar 19 '14 at 13:17
  • $\begingroup$ @Norbert Yes. I encountered it while studying a spin 1/2 chain, in which the particles have only spin degrees of freedom. $\endgroup$ – Hrodelbert Mar 19 '14 at 13:40
  • $\begingroup$ May be you should look for physical explanation, and in this case you need to ask this question on physics.stackexchange.com $\endgroup$ – Norbert Mar 19 '14 at 14:13
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    $\begingroup$ @Norbert In my opinion, if physics was important here, it would not have been possible to pose the problem without referring to the physical context of the problem. In this case, the problem is fully set in completely abstract terms and thus a physical explanation would surprise me. But I will keep it in mind in case no deeper mathematical explanation emerges! $\endgroup$ – Hrodelbert Mar 19 '14 at 14:23
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Here's how I would solve it, if I had to use your relations on $d_m$; it may turn out, though, to be equivalent to your algorithm.

Define the energy $E(m)$ of the multi-index $m$ by $E(m_1, \ldots, m_n) = \sum m_i^2.$

Claim 1: Suppose $m$ has a pair of repeated entries $m_a = m_b$. Then $d_m$ can be written as a linear combination of other coefficients $\sum_k d_{n_k}$ with $E(n_k) > E(m)$ for all $k$. Each $n_k$ has the same entry sum as $m$.

Proof: If $m_a=m_b = 0$ or $m_a = m_b = N-1$, then by your relation, $d_m = 0$ and the above holds vacuously. Otherwise, the claim follows from the observation that for any $l>0$, $$2m_a^2 < (m_a+l)^2 + (m_a-l)^2.$$

Claim 2: $d_m$ can be written as a linear combination of other coefficients $\sum_k \alpha_k d_{n_k}$ where none of the $d_{n_k}$ have a repeated entry, and all $n_k$ have the same entry sum as $m$.

Proof: Repeat the process in Claim 1 above every time a coefficient has a repeated entry. This is guaranteed to increase the energy of the coefficient with minimum energy (if it is unique) or decrease the number of coefficients tied for having the minimum energy (if there are several). Since the energy is integer-valued, bounded above, and strictly increases at every step, this process must terminate in a finite number of steps.

Finally, the only way $d_n$ has no repeated entries is if the entries of $n$ sum to $\frac{N(N-1)}{2}$. Therefore if $\sum m_i\neq \frac{N(N-1)}{2}$, the sum $d_m = \sum_k \alpha_k d_{n_k}$ must be empty and $d_m = 0$.


There's another way to get your result, without looking at the coefficients $d_m$ at all.

Claim 1: If $f(x)$ is divisible by $(x_i-x_j)$ for all $i\neq j$, then $$f(x) = g(x)\prod_{i < j}(x_i-x_j) = g(x)\pi(x)$$ for some polynomial $g(x)$. This is because the ring of polynomials is a unique factorization domain and the linear polynomials are primes. Alternatively, treat all variables except $x_0$ as distinct constants. Every $x_j$, with $j > 0$, is a distinct root of $f(x_0)$ and so $f(x)$ factors as

$$f(x) = g_0(x_1,\ldots, x_n)\prod_{0 < j}(x_0-x_j).$$ Now repeat on $g_0(x)$, pulling out all remaining factors depending on $x_1$, etc.

The coefficient of $x_i^{N-1}$ in $\pi(x)$ is $(-1)^i$, so in order for $f(x)$ to have powers of no higher than $N-1$ in $x_i$, $g(x)$ cannot depend on $x_i$ and therefore must be constant. Thus $f(x) = \alpha \pi(x)$ for some constant $\alpha$, which is clearly homogeneous of degree $\frac{N(N-1)}{2}$.

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  • $\begingroup$ Tnx for your answer!The algorithm you propose is exactly the same as the one I came up with.Your second method is a correct solution to the edited problem I wrote last week, but unfortunately, this was an oversimplification in an effort to make it more general. The polynomial becomes way too simple in that case. I have reconstructed the post to mimic the original problem, in which I think your second solution does not work any more.Sorry for the misunderstanding. Could you perhaps generalize your second solution to this case, because this does look like the answer I was looking for? $\endgroup$ – Hrodelbert Apr 7 '14 at 8:07

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