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Let $s_\lambda(x_1,\ldots,x_n)$ be a Schur polynomial in $\mathbb{C}[x_1,\ldots,x_n]$ with $\lambda=(\lambda_1\geq\lambda_2\geq\ldots\geq\lambda_n=0)$ and $\gcd(\lambda_1+n-1,\lambda_2+n-2,\dots,\lambda_n)=1$. Set of its roots on the unit $n$-circle is specified with $$ V=\left\{(\theta_1,\ldots,\theta_n)\in[0,2\pi]^n\big|s_{\lambda}\left(e^{j\ \theta_1},\ldots,e^{j\ \theta_n}\right)=0\right\}. $$ Considering $I=\{i\ |\lambda_i\geq i\}$, $$ V_i=\left\{(\theta_1,\ldots,\theta_n)\Bigg|\theta_k=\cfrac{2\pi k}{\lambda_i+n-i}\text{ for }k=1,\ldots,n\right\},\quad i\in I $$ are subsets of $V$, because the last and $i$th rows (by Wikipedia notation) of Generalized Vandermonde Determinant are identical to $[1,\ldots,1]$. Also it is obvious: $\dim \cup_{i\in I}V_i=1$.

Question:

  1. Is it true that $\cup_{i\in I}V_i=V$?
  2. If the first statement is not true, what is the dimension of $V$?
  3. If specifying the $\dim V$ is not possible, does there exist an informative upper bound for it?
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    $\begingroup$ I find your formula for $V_i$ confusing but, in any case, there are some other $1$-dimensional subsets of $V$ to consider. For any $1 \leq i < j \leq n$ and any $(k_1, \ldots, k_n) \in \mathbb{Z}^n$, we can consider the set of all vectors of the form $(\theta, \theta, \ldots, \theta) + \tfrac{2 \pi}{(\lambda_i+n-i) - (\lambda_j+n-i))}(k_1, \ldots, k_n)$, and the $i$-th and $j$-th rows of the generalized Vandermonde determinant will be proportional. $\endgroup$
    – David E Speyer
    Commented Sep 8, 2017 at 0:57
  • $\begingroup$ Also, $s_{(n-1)(n-2) \cdots 210}(z_1, \ldots,z_n) = \prod_{1 \leq i < j \leq n} (z_i+z_j)$, so in that case $V$ is codimension $1$. $\endgroup$
    – David E Speyer
    Commented Sep 8, 2017 at 0:57
  • $\begingroup$ $s_{(n-1)(n-2) \cdots 210}(z_1, \ldots,z_n) =1$. I'd like to know the dimension of distinct roots i.e. $\theta_1\neq\ldots\neq\theta_n$ and $\in [0,2\pi)^n$. $\endgroup$
    – M. Hosseini
    Commented Sep 8, 2017 at 6:47
  • $\begingroup$ I don't understand your last remark, but I also see that I missed the hypothesis that $\mathrm{GCD}(\lambda_i+n-i)=1$, so this rules out taking $\lambda_i=n-i$. $\endgroup$
    – David E Speyer
    Commented Sep 8, 2017 at 15:18

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