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This is a bijection $\pi$ between $\mathbb C^n$ and the set of monic polynomials of degree $n$ defined by \begin{align*} \pi(\alpha_{n-1}, \alpha_{n-2}, \dots, \alpha_0) = t^n + \alpha_{n-1} t^{n-1} + \dots + \alpha_0. \end{align*} Let \begin{align*} E = \{ \zeta \in \mathbb C^n: \text{All roots of }\pi(\zeta) \text{ have negative real parts}\}. \end{align*} By Vieta's formulas, we know $E$ is connected. Indeed, if we let $\Delta = \{z \in \mathbb C: \text{Re}(z) < 0\}$. Then $E$ is the image of $\underbrace{\Delta \times \Delta \times \dots \times \Delta}_{n \text{ times}}$ under a continuous function. I am wondering whether it is easy to construct some explicit path. That is, suppose we have $a = (a_{n-1}, \dots, a_0), b= (b_{n-1}, \dots, b_0) \in E$, can we construct a continuous path $\gamma: [0,1] \to E$ with $\gamma(0) = a$ and $\gamma(1)=b$ explicitly?

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  • $\begingroup$ It is irritating to begin your question with "Suppose we have a monic polynomial ...". I suggest to do it like that: There is bijection $\pi$ between $\mathbb{C}^n$ and the set of monic polynomials of degree $n$ defined by $\pi(\alpha_{n-1},...,\alpha_0) = ...$. Then define $E = \{ \zeta \in \mathbb{C}^n \mid \text{All roots of } \pi(\zeta) \text{ have a negative real part} \}$. Can you make explicit which continuous function has image $E$? $\endgroup$ – Paul Frost Jul 16 '18 at 23:13
  • $\begingroup$ @PaulFrost: Thanks for your editing suggestion. The function is elementary symmetric functions of the roots given by Vieta's formulas. $\endgroup$ – user1101010 Jul 17 '18 at 4:33
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If $e_0, e_1, ... ,e_n$ denote the elementary symmetric polynomials in $n$ variables, we get a continuous map $$\sigma : \mathbb{C}^n \to \mathbb{C}^n, \sigma(\zeta) = (-e_1(\zeta), e_2(\zeta),...,(-1)^n e_n(\zeta)) .$$

You may interpret $\sigma(\zeta)$ as the coefficients of a monic polynomial $p_\zeta$ of degree $n$ having as roots the coordinates of $\zeta$. In fact, $p_\zeta = \pi(\sigma(\zeta))$ where $\pi$ was introduced in the question.

We have $E = \sigma(\Delta^n)$. This means that the set $E$ could alternatively be defined without reference to monic polynomials and their roots. Let $s : \Delta^n \to E$ denote the restriction.

Now it depends on what you understand by "explicitly construct some path from $a$ to $b$". You find $a', b' \in \Delta^n$ such that $s(a') = a, s(b') = b$. Define $u : [0,1] \to \Delta^n, u(t) = (1-t)a' + tb'$. This is a path connecting $a'$ and $b'$, therefore $s \circ u$ is a path connecting $a$ and $b$. However, it is not obvious how to find $a',b'$. There is no general method to express $a',b'$ as a function of $a,b$ and in that sense $a',b'$ cannot be made explicit (if you could, you would have a solution formula for polynomials of arbitrary degree $n$ which in fact only exists for $n \le 4$).

Let us finally consider the map $\sigma$. For each $\eta \in \mathbb{C}^n$ the inverse image $\sigma^{-1}(\eta)$ consists of all $\zeta = (z_1,....z_n)$ such that the $\{ z_1,....z_n \}$ is the set of all roots of $\pi(\eta)$. Hence $\sigma$ is surjective. Moreover, the symmetric group $S_n$ of $n$ elements operates by permutation of coordinates on $\mathbb{C}^n$; the fibres $\sigma^{-1}(\eta)$ agree with the orbits of this operation. Therefore $\sigma$ induces a bijection $\sigma': \mathbb{C}^n/S_n \to \mathbb{C}^n$.

We show that $\sigma$ is a closed map. It is known that $\max\{1, \lvert a_{n-1} \rvert, ... , \lvert a_0 \rvert \}$ is an upper bound for the abolute values of the roots of $z^n + a_{n-1}z^{n+-1} + ... + a_1z + a_0$. This implies that $\sigma^{-1}(B)$ is bounded if $B$ is bounded. Now let $A \subset \mathbb{C}^n$ be closed and $(\eta_m)$ be a sequence in $\sigma(A)$ converging to some $\eta \in \mathbb{C}^n$. Then $Y = \{ \eta_m \mid m \in \mathbb{N} \}$ is bounded, hence $Z = \sigma^{-1}(Y)$ is bounded. Choose $\zeta_m \in A$ such that $\sigma(\zeta_m) = \eta_n$. Since $\zeta_m \in Z$, we see that $(\zeta_m)$ is a bounded sequence and has a convergent subsequence with limit $\zeta$. Since $A$ is closed, $\zeta \in A$. By continuity of $\sigma$ we conclude $\eta \in \sigma(A)$.

This implies that $\sigma$ is an identification map. Therefore, if we give $\mathbb{C}^n/S_n$ the quotient topology induced by the canonical quotient function $\mathbb{C}^n \to \mathbb{C}^n/S_n$, we see that $\sigma'$ is a homeomorphism.

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