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I had a few questions about the roots of power sum symmetric polynomials:

  1. Given that $x_1^k+x_2^k+x_3^k= 0$ for all $k \not \equiv 0\mod 3$ and is non-zero otherwise, if we assume none of the $x_i$'s equal 0, does this imply that the $\{x_i\|1\leq i \leq 3 \}=\{a\zeta_3^k|1\leq k\leq 3 \}$ where $\zeta_3$ is a primitive third root of unity and $a$ is an arbitrary constant.

  2. If we had 6 variables instead of 3 (i.e. $x_1^k+x_2^k+x_3^k+x_4^k+x_5^k+x_6^k = 0$ for all $k \not \equiv 0\mod 3$ and is non-zero otherwise), under the same assumption that none of the $x_i$'s is $0$, could we still conclude that the $\{x_i\}=\{a\zeta_3^k|1\leq k\leq 3 \}\cup \{b\zeta_3^k|1\leq k\leq 3 \}$ where $\zeta_3$ is a primitive third root of unity and $a$ and $b$ are arbitrary constants?

I appreciate any insight any of you have. Thanks!

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2 Answers

up vote 2 down vote accepted

Complete rewrite to make it more general.

Let $x_1,...,x_n$ be non-zero complex numbers such that $G_k=\sum_{i=1}^n x_i^k = 0$ iff $k\not\equiv 0 \pmod m$.

Define the generating function $$g(z)=\sum_{k=0}^\infty G_k z^k = \sum_{i=1}^n \frac{1}{1-x_iz}$$

Now since $G_k=0$ when $k\not\equiv 0\pmod m$, let $\zeta_m$ be a primitive $m$th root of unity, we see that $g(\zeta_m z) = g(z)$ for all $z$ where the series converges, and hence for all $z$ in the full analytic continuation.

In particular, then $z\to \zeta_m z$ must send the poles of $g(z)$ to the poles of $g(z)$. The poles of $g(z)$ are the numbers of the form $\frac{1}{x_i}$, so for each $i$, there must be a $j$ such that $\zeta_m/x_i = 1/x_j$ , or$\zeta_m x_j = x_i$. That means that $z\to\zeta_m z$ must permute the $x_i$. In particular, we can partition $x_i$ into orbits under this action. (In particular, it must be the case that $m|n$.)

Part 1 is the case $m=n=3$. You have to deal with the case where some or more of the $x_i$ are zero, but that's easy because that reduces to the problem when $n<3, m=3$, which only has zero solutions.

Part 2 is the case $m=3, n=6$. Again, you have to take into account the cases where an $x_i$ is zero, whih reduces to the case where $m=n=3$ or all the $x_i$ are zero.

Note, we didn't actually use that the $G_{mk}\not = 0$. That was a red herring.

(There's a slight wrinkle when the $x_i$ are not distinct, but then you can just remove one instance each of $x_1,\zeta_m x_1,...\zeta_m^{m-1} x_1$ from the set and apply the result for the subset by induction. Essentially, if $x$ occurs $j$ times in $x_1,...,x_n$ the $\zeta_m x$ occurs $j$ times, too.]

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Hint for 1: Apply Vieta's Formula. What can we say about $\sum x_i$ and $\sum x_i x_j$, and $x_1 x_2 x_3$? Hence the roots are ...


Hint for 2:

The Vandemonde matrix has non-zero determinant (viewed as a polynomial in x_k).

Newton's identities tells us that the values of $\sum x_k, k\geq 6$ are determined by the values for $k=0$ to 5. From the question, the image has 2 degrees of freedom.

Hence, it is sufficient to find a set of 2 linearly independent vectors.

(This works for 1 too, but uses slightly more machinery.)

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Thanks! I didn't think of Vieta's Formula. That's helpful. –  Fortunato Jan 9 '13 at 21:39
    
I'm confused about this statement: "Newton's identities tells us that the values of "$\sum x_k$, k≥6 are determined by the values for k=0 to 5. From the question, the image has 2 degrees of freedom." Did you mean $\sum x^k$? Also, where do the two degrees of freedom come from? When $k=3$ and when $k \neq 3$? –  Fortunato Jan 10 '13 at 15:17
1  
Construct the matrix $A_{i,j} = x_j ^i$ for $i, j = 1$ to $6$. This has det $\prod x_i \prod(x_i-x_j)$. (the initial $\prod x_i$ arises because the true Vandemonde uses all 1's; doesn't make a difference). Then, you're asking for the set of solutions to $A \cdot 1 = (0, 0, a, 0, 0, b)^T$. Since $A$ has full rank, this set of solutions is a deg 2 subspace. (so the 2 degrees of freedom came from the choices of $\sum x_i ^3, \sum x_i ^6$ –  Calvin Lin Jan 10 '13 at 15:28
    
And so I'm then looking for 2 linearly independent vector solutions of the form $(x_1,\ldots , x_6)$ that satisfy both $\sum x_i^3 =a$ and $\sum x_i ^6 =b$. And the only possibilities (up to a permutation of variables) are of the form $(c1 \zeta _3, c1 \zeta _3^2, c1, c2 \zeta _3, c2 \zeta _3^2, c2)$ where $c1$ and $c2$ are constants and $\zeta _3$ is a primitive 3rd root of unity? –  Fortunato Jan 10 '13 at 16:02
    
I'm still having trouble seeing how this leads to the conclusion. Can I have a nudge? –  Fortunato Jan 10 '13 at 18:45
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