# A question about the coefficient of a specific polynomial

Let \begin{aligned} f &= f(x_1,x_2,x_3,y_1,y_2,y_3,z_1,z_2,z_3)\\ &=(x_2+x_3-y_1-z_1)(x_1+x_3-y_2-z_2)(x_1+x_2-y_3-z_3)\\ &\times(y_2+y_3-x_1-z_1)(y_1+y_3-x_2-z_2)(y_1+y_2-x_3-z_3)\\ &\times(z_2+z_3-x_1-y_1)(z_1+z_3-x_2-y_2)(z_1+z_2-x_3-y_3) \end{aligned}

be a real multivariate polynomial.

I want to ask whether the coefficient of $x_1x_2x_3y_1y_2y_3z_1z_2z_3$ in $f$ is nonzero.

Define 6 variables $$\begin{cases} A_1 &= x_1 + x_2 + x_3\\ A_2 &= y_1 + y_2 + y_3\\ A_3 &= z_1 + z_2 + z_3 \end{cases} \quad\text{ and }\quad \begin{cases} B_1 &= x_1 + y_1 + z_1\\ B_2 &= x_2 + y_2 + z_2\\ B_3 &= x_3 + y_3 + z_3 \end{cases}$$ The product at hand can be rewritten as

f = \prod_{i=1}^3\prod_{j=1}^3 (A_i - B_j) = \begin{align} \phantom{ \times } & (A_1 - B_1)(A_1 - B_2)(A_1-B_3)\\ \times & (A_2-B_1)(A_2-B_2)(A_2-B_3)\\ \times & (A_3-B_1)(A_3-B_2)(A_3-B_3) \end{align}

Consider following bijection $\varphi$ on the 9 variables $$(x_1, x_2, x_3, y_1, y_2, y_3, z_1, z_2, z_3) \quad\mapsto\quad (x_1, y_1, z_1, x_2, y_2, z_2, x_3, y_3, z_3)$$ It can be extended to an involution of the polynomial ring formed from these 9 variables.
Under this involution, we have

$$\varphi(A_i) = B_i \quad\text{ and }\quad \varphi(B_i) = A_i\quad\text{ for } i = 1,2,3$$ This leads to

$$\varphi(f) = \prod_{i=1}^3 \prod_{j=1}^3 \varphi(A_i - B_j) = (-1)^9 \prod_{i=1}^3 \prod_{j=1}^3 (A_j - B_i) = - f$$

Since the term we want to extract the coefficient: $x_1 y_1 z_1 x_2 y_2 z_2 x_3 y_3 z_3$ is invariant under $\varphi$. The corresponding coefficient will be equal to negative of itself. This forces that coefficient to be $0$.

• I guess it's easier to see if we re-write the factors using $$x_2 + x_3 - y_1 - z_1 = ( (x_1 + x_2 + x_3) - (x_1+ y_1 + z_1))$$ I saw this too but I thought of using this to expand all the terms and didn't try thinking about a proof. Your convolution has the effect of transposing the $9$ factors displayed above in a $3 \times 3$-matrix fashion, giving a minus sign to each factor. Apparently it works! +1 Jul 5 '15 at 7:20
• @PatrickDaSilva this is actually how I get the answer... Jul 5 '15 at 7:22
• I know, but you didn't write the re-writing explicitly so I wrote it down for the others. I'm aware that you knew! Jul 5 '15 at 7:23
• Thank you very much! Your method also solve the case when $n$ is odd(my problem above is the case when $n=3$.But when $n$ is even, I think the coefficient is nonzero. But I do not know how to prove it. Jul 5 '15 at 7:37
• @user173856, at least when $n = 2$, it is non-zero (equal to $4$). Jul 5 '15 at 8:29

If my computations are correct, the coefficient is $0$.

• You mean you threw this in some software? Jul 5 '15 at 7:01
• That's certainly the easiest thing to do. Jul 5 '15 at 7:07
• Isreal : I agree, a lot of terms seems to actually contribute to evaluating this coefficient. (I found more than three patterns which generate a bunch of non-zero terms if you expand...) Jul 5 '15 at 7:09
• If you expand it out, there are some $13410$ terms. Not too bad for a CAS. Jul 5 '15 at 7:13
• You want me to prove that Maple has no bugs? Jul 5 '15 at 7:17