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would any one explicitly and precisely help me to understand how the map from the set consists of monic complex polynomials with degree n to the set consists of n complex numbers without order by mapping each polynomial to its roots is continuous? if I understand this fact then i will be able to prove this one : Let X be a compact subset of $GL_n(\mathbb{C})$ and Y= set of all eigenvalues of matrices in X, then Y is compact in $\mathbb{C}$

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Ref'ed question here:… – Chris Eagle Apr 20 '12 at 14:49
Dumb question, but what's the topology on the set of $n$ complex numbers without order? Is it just the quotient topology on $\mathbb{C}^n/S_n$? – Jason DeVito Apr 20 '12 at 14:55
I think you can use Rouche's theorem to prove a small disturbation of a polynomial $f$ change the roots of $f$ slightly. – Yuchen Liu Apr 20 '12 at 14:56
@JasonDeVito: Yes, I'm talking about the quotient topology. – Yuchen Liu Apr 20 '12 at 14:58
@jerrysciencemath: Are you Makuasi? If so, please get those accounts merged to avoid confusion. If not, it seems to make little sense to answer a question about the question directed at the OP. – joriki Apr 20 '12 at 16:21
up vote 4 down vote accepted

Here's an approach that you may not like, ’cause it’s much more algebraic than analytic, and it works only for the open set in the domain corresponding to $n$-tuples of distinct complex numbers.

Consider the reverse map $(\rho_1,\rho_2,\cdots,\rho_n)\mapsto (S_1(\rho),S_2(\rho),\cdots,S_n(\rho))$, where the polynomials $S_i$ are the elementary symmetric functions, $S_j$ is the sum of all the possible products of $j$ of the roots $\rho_i$. In particular, $S_1$ is the sum of all the roots, and $S_n$ is the product of them all. So you know that the polynomial whose roots are the $\lbrace\rho_i\rbrace$ is $X^n - S_1X^{n-1}+\cdots+(-1)^nS_n$.

Now look at this as a map from $n$-space to $n$-space, defined by polynomials, and thus as differentiable as you could like. What is the Jacobian determinant of this map? This requires a proof, but up to sign, it’s the square root of the discriminant, namely $$ \prod_{i<j}(\rho_j-\rho_i)\>. $$ From this formula, you see that the Jacobian determinant is nonzero exactly when the roots are distinct, and so the Jacobian matrix is invertible under exactly the same circumstances. Now, in the very last step, you apply the Inverse Function Theorem, and get not only continuity but infinite differentiability of your map.

This has, in fact, nothing to do with real or complex analysis: I first made my own use of this in the context of $p$-adic analysis.

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Dear Sir, first I would like to share my great pleasure to have the answer from you. Sir, would you please tell why is the reverse map is like that? my original map was say $f: $P_n(z)\longrightarrow $\mathbb{C}^n$ defined by $f(p(z))=(z_1,\dots,z_n)$, isn't it? I am not getting the reverse map you said. – Un Chien Andalou Apr 21 '12 at 18:39
@Makuasi, the pleasure is mine. I hope that you are familiar with the connection between the roots of a polynomial and the elementary symmetric polynomials. This should be described in any basic Algebra text. If you haven’t seen this relationship, try it out for a quadratic polynomial (I know you've seen this!) and for a cubic. That will likely persuade you of the truth of the general fact, before you work out a proof for yourself. I should also say that the set of unordered $n$-tuples of complex numbers is definitely not ${\mathbb{C}}^n$: this latter is the set of ordered $n$-tuples. – Lubin Apr 22 '12 at 21:26

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