# Understanding Umemura's Theorem for roots of algebraic equations

I am trying to understand Umemura's Theorem for expressing the roots of any algebraic equation by higher genus theta functions. The original paper can be found here: Umemura, H.: Resolution of algebraic equations by theta constants. In: Mumford, D. (Ed.): Tata Lectures on Theta II, Birkhäuser, Boston/Basel/Stuttgart 1984. I apologize if the answers are quite straightforward for those familiar with the theorem.

R. Bruce King states the theorem as follows in his book "Beyond the Quartic Equation":

Let $f(x) = a_{0}x^{n} + a_{1}x^{n-1} + ... + a_{n} = 0$ with $a_{0} \neq 0$ be an algebraic equation irreducible over a certain subfield of the complex numbers. Then a root of this equation, $x_{k}$ can be expressed by the following equation involving theta

functions of zero argument:

$x_{k} = \left( \theta \left( \begin{array}{cccc} 1 & 0 & ... & 0 \\ 0 & ... & 0 & 0 \end{array} \right)(\Omega) \right)^{4} \left( \theta \left( \begin{array}{ccccc} 1 & 1 & 0 & ... & 0 \\ 0 & ... & 0 & 0 & 0 \end{array} \right)(\Omega) \right)^{4} + \left( \theta \left( \begin{array}{ccc} 0 & ... & 0 \\ 0 & ... & 0 \end{array} \right)(\Omega) \right)^{4} \left( \theta \left( \begin{array}{ccccc} 0 & 1 & 0 & ... & 0 \\ 0 & 0 & 0 & ... & 0\end{array} \right)(\Omega) \right)^{4} - \frac{\left( \theta \left( \begin{array}{cccc} 0 & 0 & ... & 0 \\ 1 & 0 & ... & 0 \end{array} \right)(\Omega) \right)^{4} \left( \theta \left( \begin{array}{ccccc} 0 & 1 & 0 & ... & 0 \\ 1 & 0 & ... & 0 & 0 \end{array} \right)(\Omega) \right)^{4} }{2 \left( \theta \left( \begin{array}{cccc} 1 & 0 & ... & 0 \\ 0 & ... & 0 & 0 \end{array} \right)(\Omega) \right)^{4} \left( \theta \left( \begin{array}{ccccc} 1 & 1 & 0 & ... & 0 \\ 0 & 0 & 0 & ... & 0 \end{array} \right)(\Omega) \right)^{4}}$\

where $\Omega$ is the period matrix derived from one of the following hyperelliptic integragls:

$u(a) = \int_1^a \frac{dx}{\sqrt{x(x-1)f(x)}}$ for odd degree of $f(x)$ and

$u(a) = \int_1^a \frac{dx}{\sqrt{x(x-1)(x-2)f(x)}}$ for even degree of $f(x)$.

The theta function is defined as follows:

For row vectors $m_{1}, m_{2} \in \mathbb{R}^g, z \in \mathbb{C}^g$ and a symmetric $g \times g$ matrix $\tau$ with positive definite imaginary part, we define the theta function $\theta \left( \begin{array}{c} m_{1} \\ m_{2} \end{array} \right)(z, \tau) = \Sigma_{\xi \in \mathbb{Z}^g} e(\frac{1}{2}(\xi + m_{1})\tau^{T}(\xi + m_{1}) + (\xi + m_{1})^{T}(z + m_{2}))$ where $e(x) = e^{2 \pi i x}$. We write $\theta \left( \begin{array}{c} m_{1} \\ m_{2} \end{array} \right)(\tau)$ for $\theta \left( \begin{array}{c} m_{1} \\ m_{2} \end{array} \right)(0, \tau)$.

My question about the definition are as follows: the equation gives a matrix, how can this be equal to a scalar root of an algebraic equation? The answer appears to be from Theorem 8.1 in the paper, but I can't understand it and would be grateful if someone could.

Finally, my question that may or may not be of mathematical interest is: can every root of an algebraic equation be found in this manner and can we then conclude that a number algebraic iff it satisfies Umemura's equation?

• I found the original paper of Umemura online so have edited the question to take this into account. I am still struggling to understand, but I think everything is there, if someone would be kind enough to explain. Dec 12 '13 at 15:01
• I apologise, no one seems interested in this question. But for completeness sake, in answer to the first question, it seems that for Riemann Theta functions we define $n \cdot \Omega \cdot n = \Sigma_{i, j = 1}^{g} \Omega_{ij}n_{i}n_{j}$ giving a scalar, where $n \in \mathbb{Z}^{g}$ and $\Omega \in \mathbb{C}^{g \times g}$. I assume $(\xi + m_{1})^{T}(z + m_{2})$ is just the dot product also giving a scalar. Dec 13 '13 at 12:13
• Why does the equation give a matrix? It seems to me to give a scalar. Feb 19 '17 at 1:31
• But I'm confused for two different reasons: 1. How do you get $x_k$ for a different $k$? 2. In all invocation of the $\theta$ function, $m_1$ is always in $\mathbb{Z}^g$, so do we still need $m_1$ in this case? It seems that it can be absorbed by a change of variable $\xi+m_1\mapsto\xi$. Feb 19 '17 at 1:34