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Let $m_\lambda(X_1,X_2,...X_N)$ be a monomial symmetric function with partition $\lambda$.

For example:

$$ m_{(3,1,1)}(X_1,X_2,X_3) =X_1^3X_2X_3 + X_1X_2^3X_3 + X_1X_2X_3^3 $$

Is there a general formula for roots of $m_\lambda$ if $X_j$ is restricted to elements of $\mathbb C$ with $\| X_j \|=1$?

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Could you define what is a symmetric monomial with partition $\lambda$ ? – user18119 Nov 12 '11 at 22:39
Ah sorry, that's not very good explained. See – draks ... Nov 12 '11 at 23:03
"monomial" means one term; what you have are polynomials. – Gerry Myerson Nov 13 '11 at 3:25
The usual term is not a "symmetrical monomial", it is a "monomial symmetric function". – Phira Nov 13 '11 at 12:48
up vote 1 down vote accepted

I doubt it. Let's look at $m_{(1,0,0,0,0)}$. You want the solutions of $$a+b+c+d+e=0$$ with all variables on the unit circle. It will be hard enough to find a formula for that special case, much less for the general case.

Note that $m_{(1,0)}$ is $a+b=0$ which is solved by $a=e^{it}$, $b=e^{i(t+\pi)}$. Then $m_{(1,0,0)}$ is $a+b+c=0$, which forces $a,b,c$ to be vertices of an equilateral triangle. Next, $m_{(1,0,0,0)}$ is $a+b+c+d=0$, and with a bit of work you can show that $a,b,c,d$ must be vertices of a rectangle. But once you get up to 5 unknowns the geometric argument doesn't give you anything that simple.

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staying geometrically this linear example, this means that the barycentre is at the origin, right? – draks ... Nov 17 '11 at 14:47
I guess so. The barycenter of a finite number of points is just their average, right? If the sum is zero, then certainly the average is zero. – Gerry Myerson Nov 17 '11 at 23:38
I agree. What we be an extension to the barycenter in non-linear cases? For quardatic terms the objects we deal with are planes then, right? – draks ... Nov 18 '11 at 12:13
Not sure I follow you. $m_{(2,0)}$ gets us $a^2+b^2=0$ which is still a one-dimensional problem. – Gerry Myerson Nov 19 '11 at 11:05
hm, right, maybe I was confused by the squares... – draks ... Nov 22 '11 at 6:58

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