Is the eigenvalue map open?

Question

The eigenvalue map in question is $\sigma: {\mathfrak gl}(\mathbb{C}, n) \to S_n \backslash \mathbb{C}^n$, from $n$ by $n$ complex matrices to $\mathbb{C}^n$ vectors modulo permutation of entries by $S_n$, and is known to be continuous.

This map can be constructed by first considering the root map $\rho: \mathbb{C}^n \to S_n \backslash \mathbb{C}^n$ of a monomial of degree $n$ $$ z^n + a_{n-1}z^{n-1} + \ldots + a_1 z + a_0 = (z - \lambda_1) \cdots (z - \lambda_n) $$ given by $$ \rho(a_0, a_1, \ldots, a_{n-1}) = [(\lambda_1, \ldots, \lambda_n)] $$ which is known to be an homemomorphism with inverse given by Vieta's formulae (see here, for example). Now given $A \in {\mathfrak gl}(\mathbb{C}, n)$, the coefficients $(a_0, a_1, \ldots, a_{n-1})$ of its characteristic polynomial $p_A(z) = \det(A - zI)$ are continuous (indeed polynomial) functions on the entries of $A$. The eigenvalue map is then given by $$ \sigma(A) = \rho(a_0, a_1, \ldots, a_{n-1}) $$ so that it is continuous on $A$.

Is $\sigma$ an open map on $A$?

Besides being a natural question, a positive answer would have some nice applications. For example, when one proves that certain subset of matrices are open and dense when their eigenvalues satisfy some open and dense conditions (e.g. hyperbolic matrices, matrices with distinct eigenvalues, etc) one can use the continuity of $\sigma$ to prove openness but has to argue denseness in other ways, for example perturbing the matrices involved. If one could prove that $\sigma$ is an open map, one could use $\sigma$ all the way.

A negative answer by means of a counterexample would also be very instructive.

Answer 1

Yes. Write $D(\lambda_1, \ldots, \lambda_n)$ as the diagonal matrix with diagonal entries $(\lambda_1, \ldots, \lambda_n)$.

Let $A$ be any complex $n \times n$ matrix. We can upper-triangularize $A$ as $A = S (D(\lambda_1, \ldots, \lambda_n)+N) S^{-1}$ where $N$ is upper triangular. Let $U$ be an open ball around $A$. We want to show that $\rho(U)$ contains an open ball $V$ around $(\lambda_1, \ldots, \lambda_n)$.

Indeed, we can choose $V$ small enough that $S (D(\mu_1, \ldots, \mu_n)+N) S^{-1}$ is in $U$ for all $(\mu_1, \ldots, \mu_n)$ in $V$. Then $\rho(U)$ contains $\rho(S (D(\mu_1, \ldots, \mu_n)+N) S^{-1}) = (\mu_1, \ldots, \mu_n)$ for all $(\mu_1, \ldots, \mu_n)$ in $V$, and $\rho(U)$ contains $V$, as desired.

Answer 2

It is well known theorem of Kostant (Lie group representations on polynomial rings) that $\sigma$ is flat as a morphism between algebraic varieties. Thus it is open for the Zariski topology. For the Hausdorff topology you can probably argue that then $\sigma$ is also flat in the analytic category. I don't have reference, though. Then as in this MO answer use Banica-Stanasila, Algebraic methods in the global theory of Complex Spaces, Theorem 2.12 p. 180 to conclude that $\sigma$ is open also in the Hausdorff topology.