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Let the real polynomial $$f_{a,b,c}(x_1,x_2,x_3)=(x_1-x_2)^{2a+1}(x_2-x_3)^{2b+1}(x_3-x_1)^{2c+1},$$ where $a,b,c$ are nonnegative integers.

Let $m_{a,b,c}$ be the coefficient of the monomial $x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$ in the expansion of $f_{a,b,c}(x_1,x_2,x_3)$.

It is easy to see $m_{a,b,c}=0$ when two of $a,b,c$ are equal. I want to ask whether $m_{a,b,c}=0$ or not when $a,b,c$ are pairwise unequal.

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This coefficient $L$ is a constant term of the Laurent polynomial $g(x_1,x_2,x_3)=f(x_1,x_2,x_3)/x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$, this guy $g$ satisfies $g(x_1,x_2,x_3)=-g(1/x_1,1/x_2,1/x_3)$, thus $L=-L$.

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  • $\begingroup$ Fedor Petrov: Thank you very much for your answer! Your proof is very slick! $\endgroup$
    – user173856
    Mar 30 '16 at 7:36
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The coefficient is always zero. I use the superb conventions of Concrete Mathematics: Sums are over all integers unless otherwise indicated, and $\binom{n}{k}$ is $0$ if $k<0$ or $>n \geq 0$.

Expanding by the binomial theorem, $$f_{abc} = \sum_{i,j,k} (-1)^{i+j+k} \binom{2a+1}{i} \binom{2b+1}{j} \binom{2c+1}{k} x_1^{2a+1-i+k} x_2^{2b+1-j+i} x_3^{2c+1-k+j}.$$

We want $$\begin{array}{r@{}c@{}lcr@{}c@{}l} a+c+1 &=& 2a+1-i+k &\implies& c-k &=& a-i\\ a+b+1 &=& 2b+1-j+i &\implies& a-i &=& b-j\\ b+c+1 &=& 2c+1-k+j &\implies& b-j &=& c-k \\ \end{array}$$ so $(i,j,k) = (a+r, b+r, c+r)$ for some $r$. We need to evaluate $$\sum_r (-1)^{a+b+c+3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r}.$$

Pair off the $r$ and $1-r$ terms to get $$(-1)^{a+b+c} \sum_{r \geq 0} \left[ (-1)^{3r} \binom{2a+1}{a+r} \binom{2b+1}{b+r} \binom{2c+1}{c+r} + \right.$$ $$\phantom{(-1)^{a+b+c} \sum_{r \geq 1}} \left. (-1)^{3-3r} \binom{2a+1}{a+1-r} \binom{2b+1}{b+1-r} \binom{2c+1}{c+1-r} \right].$$

The two terms in the square brackets cancel, and the sum is zero.

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  • $\begingroup$ David Speyer: Thank you very much for your answer! $\endgroup$
    – user173856
    Mar 29 '16 at 11:41
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I don't have time to write a detailed answer but the following should help.

Blast your sum with Newton's binomial formula. That gives a sum over three indices $i,j,k$. The three equations you get on these indices by fixing the monomial of interest are linearly dependent, so you are left with a sum over just one index say $i$. I believe what you get is a ${}_3F_{2}$ terminating hypergeometric sum. Your quantity of interest is also a Wigner $3jm$-symbol or Clebsch-Gordan coefficient from the literature on quantum angular momentum. In general, determining the zeros of such numbers is an open problem, see my article "The bipartite Brill-Gordan locus and angular momentum" with Jaydeep Chiplakatti for some references. If your case is sufficiently special (e.g., stretched coefficients) then you might be lucky enough to know if your coefficients are zero or not.


Edit: As I said, I did not have time to look at your coefficient closely yesterday but from the other answers it turns out you are dealing with a trivial $3j$ zero. In case you need more complicated analogues of your coefficients, I am giving more details and perspective.

The Wigner $3jm$ symbol $$ \left(\begin{array}{ccc} j_1 & j_2 & J\\ m_1 & m_2 & m \end{array}\right) $$ is (up to trivial nonzero prefactors) the coefficient of $x_1^{j_1-m_1}x_2^{j_2-m_2}x_3^{J-m}$ in $$ (x_1-x_2)^{j_1+j_2-J}(x_2-x_3)^{J+j_2-j_1}(x_3-x_1)^{J+j_1-j_2}\ . $$

Therefore your coefficient is $$ \left(\begin{array}{ccc} a+c+1 & a+b+1 & b+c+1\\ 0 & 0 & 0 \end{array}\right)\ . $$

If you look up page 2611 of the article by Raynal, Van der Jeugt, Srinivasa Rao and Rajeswari, that we cited in my article mentioned above, you will see that your case (with all magnetic moments $m$ begin zero) is a trivial zero because the sum of the numbers on top is odd. If the sum is even the result is nonzero and this is basically Dixon's Theorem for ${}_3F_2$ series.

You can find many more details on $3j$, $6j$, $9j$ symbols with precise signs, prefactors and all that in my other article with Chipalkatti "The higher transvectants are redundant".

BTW, I just looked up citations to the Raynal et al. article, and there does not seem to have been much progress on understanding the zeros of $3jm$ coefficients.

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  • $\begingroup$ Abdelmalek Abdesselam: Thank you very much for your answer. $\endgroup$
    – user173856
    Mar 29 '16 at 11:40
  • $\begingroup$ no pbm. BTW may I ask what are you working on, how did the question above arise? $\endgroup$ Mar 29 '16 at 13:32
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This answer is really just a convoluted edition of Fedor's answer. By Cauchy's theorem, $$f_{abc} = \frac{1}{(2\pi)^3}\int_{-\pi}^\pi \int_{-\pi}^\pi \int_{-\pi}^\pi \frac{(e^{i\theta_1}-e^{i\theta_2})^{2a+1}(e^{i\theta_2}-e^{i\theta_3})^{2b+1}(e^{i\theta_3}-e^{i\theta_1})^{2c+1}}{e^{i(a+c)\theta_1+i(a+b)\theta_2+i(b+c)\theta_3}} \,d\theta_1\,d\theta_2\,d\theta_3.$$ Advancing each variable by $\pi$ changes the sign of the integrand but not the value of the integral, so its value must be 0.

To obtain the above integral start with the contour integral $$\frac{1}{(2\pi i)^3} \oint\oint\oint \frac{(x_1-x_2)^{2a+1} (x_2-x_3)^{2b+1} (x_3-x_1)^{2c+1}}{x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}} \,dx_1dx_2dx_3 $$ that is just Cauchy's formula applied three times. Now take the contours to be unit circles $x_j=e^{i\theta_j}$ for $j=1,2,3$.

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  • $\begingroup$ Brendan McKay: Thank you very much for your answer. Would you please explain how you got the equality above? Why is the denominator of the integrand $e^{i(2a+2b+2c)}$? $\endgroup$
    – user173856
    Mar 28 '16 at 8:05
  • $\begingroup$ @user173856: I added an explanation. $\endgroup$ Mar 28 '16 at 11:31
  • $\begingroup$ Brendan McKay: I think the denominator of the integrand should be $e^{i(a+c)\theta_1}e^{i(a+b)\theta_2}e^{i(b+c)\theta_3}$ instead of $e^{i(2a+2b+2c)}$, am I wrong? $\endgroup$
    – user173856
    Mar 28 '16 at 14:29
  • $\begingroup$ @User173856: You are not wrong! Fortunately the correct denominator still has the property that it is unchanged if each variable is advanced by $\pi$. $\endgroup$ Mar 28 '16 at 22:07

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