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Consider a polynomial in one variable with complex coefficient

$$f(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0$$

we are interested in its roots. Babylonian solved for $n = 2$, and Cardano did it for $n = 3, 4$. Later group theorists (Galois and Abel et al) proved that radical solutions (i.e. expressions in $+-\times\div \sqrt[k]{\cdot}$) do not exist in general for $n \geq 5$. However, exact solutions are available in general if we allow theta functions [1][2].

In [1], the author very slightly modified $f(x)$ to $F(x) = x(x-1)f(x)$, and create a compact Riemann surface $$C_F := (y^2 = F(x))$$ which doubly branch-covered $\mathbb{C}P^1$, whose branch points are exactly the roots. Then, by specifying an order on the roots (assumed to be seperated), one can write down the period matrix $\Omega_F$ and the zeta function $\Theta(\vec{z}; \Omega_F)$. A root of $f$ can then be written explicitly in $\Theta$.

Questions

  1. Instead of an explicit expressions of roots, I'd like to understand it from a higher point of view. We know that $\Omega_F$ is closely related to the function theory of the curve $C_F$. Why, in principle, is it capable of extracting the locations of the branch points over $\mathbb{C}P^1$?

  2. The introduction of general $\sqrt[k]{\cdot}$ (in particular, imaginary numbers) was crucial for solving equations. What's remarkable in Cramer's formulae is that we essentially only have to blackbox the solutions to $x^k = c$. In this sense, what is the "typical" equation that $\Theta$ solves? (For $n=5$, see Bing radical.)

References

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  • $\begingroup$ 1. Theta functions are functions on the jacobian of the curve; one way to understand these formulas is to understand what happens to the branch points of $C$ via the Abel-Jacobi map $C\to Jac(C)$. 2) $\Theta$ is not really designed to solve any specific equation - note also that it depends on the curve $C$. $\endgroup$
    – Ennio Mori cone
    Commented Oct 19, 2022 at 8:01
  • $\begingroup$ 2) Exactly. That's why I had this question. It seems like a nice coincidence. 1) Interestingly, a priori $p: C \to \mathbb{C}P^1$ doesn't say too much about $C$ itself. SO I wonder why the Abel-Jacobi map "sees" $p$. $\endgroup$
    – Student
    Commented Oct 19, 2022 at 13:49
  • $\begingroup$ Well, the map $p:C\to P^1$ determines $C$ uniquely up to isomorphism, because you can recover the branch points from where the map ramifies. In fact, this is why Umemura works with the curve $x(x-1)p(x)$ ranther than $y^2=p(x)$ -- he is forcing three of the branch points to lie at $0,1\infty$, so the remaining roots are determined by the cross-ratios (which appear in his formulas!). $\endgroup$
    – Ennio Mori cone
    Commented Oct 19, 2022 at 15:51
  • $\begingroup$ Note also that the Abel-Jacobi map $C\to Jac(C)$ also allows us to recover $C$ (and thus the roots). In fact the image of $C$ under the composition of Abel-Jacobi with $C\to Jac(C) \to Jac(C)/\pm$ is a rational curve inside the Kummer variety $Jac(C)/\pm$. The Theta constants allow us to write down a projective model for the map $Jac(C)\to Jac(C)/\pm \to P^N$ into some big projective space. Thus looking at the partial derivatives, one recovers the branch points. $\endgroup$
    – Ennio Mori cone
    Commented Oct 19, 2022 at 15:55
  • $\begingroup$ @EnnioMoricone Conversely, is $C$ as an abstract Riemann surface (assume hyperbolic) enough to determine $p: C \to P^1$ (in particular, its set of roots)? I suspect this should be true, and due to this, we can write down $\Omega$ (up to permutation) from the Riemann surface $C$ itself. If both of these are true, these facts will be the answer to my question. $\endgroup$
    – Student
    Commented Oct 19, 2022 at 17:08

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