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Inspired by the kind of argument discussed in this answer, I begin by considering the following quadratic:

$$ax^2 + bx +c=0$$

And I want a functions $ \zeta(a,b,c)$ which maps from the coefficients to the root, there must be two such functions via the fundamental theorem of algebra. Now, I make a few observations.

  1. $\zeta(\lambda a , \lambda b , \lambda c) = \zeta(a,b,c)$
  2. $\zeta(0,b,c) = - \frac{b}{c}$

I'm not very sure how to construct a polynomial after this but after considering the first property and taking derivative with $\lambda$ we get:

$$ \frac{\partial \zeta}{\partial (\lambda a)} a+ \frac{\partial \zeta}{\partial (\lambda b)} b+ \frac{\partial \zeta}{\partial (\lambda c)} c=0$$

Not sure what to do after this..

Attempt 2.0

Consider a quadratic:

$$ x^2 + Bx + C=0$$

This is equivalent to the original one as we can divide through by $a$, now we have the following properties for the root function $ \zeta(B,C)$:

  1. $\zeta(0,0) = 0$

And I can spot no other significant properties, the problem with this way is that we no longer have the $\lambda$ trick in the previous one.

Any help is appreciated!

tryst with freedom
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    Why $ \zeta(0,0,0) = 0$? – Raffaele Feb 14 '21 at 17:40
  • Oh shoot, you're right, yes all real numbers solve it if that is true @Raffaele – tryst with freedom Feb 14 '21 at 17:46
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    Why polynomial? – Ted Shifrin Feb 14 '21 at 17:51
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    It would certainly be fun if there were such a polynomial, but there's no reason, a priori, to expect the function to be polynomial. There's also no reason to expect it to be well-defined, in the sense that 'consistently choosing one of the two roots' in a way that varies continuously with the coefficients may not be possible, even if you eliminate the coefficient triple $(0,0,0)$, which is probably a good idea. – John Hughes Feb 14 '21 at 18:03
  • I was trying to pull of something similar to what I saw in the linked post @TedShifrin – tryst with freedom Feb 14 '21 at 18:15
  • @JohnHughes suppose it was not polynomial, but still some 'function' existed ( which we know does), how would one find searching for one based arguements similar to the linked post? – tryst with freedom Feb 14 '21 at 18:20
  • You probably want $\zeta$ to be a map from $\mathbb P^2$ – Kenta S Feb 14 '21 at 18:25
  • What exactly is a $P^2$? – tryst with freedom Feb 14 '21 at 18:27
  • @Buraian it is the projective space. – Kenta S Feb 14 '21 at 18:43
  • The question you linked is not analogous to this question. Firstly, there is no "the" function; there are two. It's the two functions you need to be looking for, as a pair, and looking for equations they must satisfy. – Ted Shifrin Feb 14 '21 at 19:42
  • Right, my bad for using singular (However, I had used two in the post), the question is not exactly analogous but I'm trying to argue in similar lines to what was shown there @TedShifrin – tryst with freedom Feb 14 '21 at 19:52
  • So you're looking for $z,w$, homogeneous functions of $(a,b,c)$ of degree $0$, satisfying $zw=a/c$ and $z+w=-b/c$. This is really no easier than the original problem of solving a quadratic equation. The fact that these will not be polynomial functions makes a solution parallel to the linked one impossible. – Ted Shifrin Feb 14 '21 at 20:00
  • By the way, your equation with partial derivatives — although I absolutely hate your notation — is a special case of Euler's Theorem for homogeneous functions; as I noted, these are homogeneous functions of degree $0$. – Ted Shifrin Feb 14 '21 at 20:03
  • How would you denote the partials then? @TedShifrin hmm about that function point.. good point, but maybe there is a method to modify that technique there such that it works here – tryst with freedom Feb 15 '21 at 08:36

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The following is probably the best way to formalize the map $\zeta$:

Let $S=\mathbb C^2/\sim$ where $(x,y)\sim(z,w)$ iff $(x,y)=(z,w)$ or $(x,y)=(w,z)$. Also let $\mathbb P_\mathbb C^2$ be the projective space, the quotient $(\mathbb C^3\setminus\{(0,0,0)\})/{\sim}$ where $(a,b,c)\sim(a',b',c')$ iff there exists a $\lambda\in\mathbb C^*$ with $(a',b',c')=(\lambda a,\lambda b,\lambda c)$.

Then we have a well-defined map $\zeta\colon\mathbb P^2_\mathbb C\setminus\{(0,0,1)\}\to S$ sending $(a,b,c)\in\mathbb P^2_\mathbb C\setminus\{(0,0,1)\}$ to the solutions of $ax^2+bx+c=0$. Then the map $\zeta$ becomes continuous (giving $\mathbb P^2_\mathbb C$ and $S$ the quotient topology).

Kenta S
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