Calculating the Period Matrix of a Shiga Curve, $y^3 = x^4-1$.

Thanks to Dami Lee for patiently walking through how to compute the period matrix of this 12-fold cyclic cover of a thrice punctured sphere, and thanks Matthias Weber for showing me how to write the 3 fold cover of a 5-punctured sphere as a 12-fold cyclic cover of a thrice punctured sphere. Most of the figures are either hand-drawn or made using Geogebra. Note that I will sometimes use $\tau$ to denote $2 \pi$. All errors are mine and mine alone.

Motivational Sidenote:
This is part of my project in attempting to understand the notion of height in terms of the symmetry of the underlying variety in formal group law theory.

Though it may seem disparate, this post is the computation of the Jacobian of a Shiga curve shown to have height 3 properties by Sebastien Thyssen and Hanno van Woerden. I suspect that the Jacobian of this curve is indeed my abelian variety constructed as a variety with complex multiplication in my last post under the guidance of Kato. Continue reading Calculating the Period Matrix of a Shiga Curve, $y^3 = x^4-1$.

Umbral Calculus Derivation of the Bernoulli numbers

$$“(B-1)^n = B^n”$$

\begin{align*}
(B-1)^2 &= B^2\\
B^2 – 2B^1 + 1 &= B^2 \\
-2B^1+1 & = 0\\
B^1 &= \frac{1}{2} \\
B_1 &= \frac12
\end{align*}

\begin{align*}
(B-1)^3 &= B^3\\
B^3 – 3B^2 + 3B^1 – 1 &= B^3 \\
B^3 – 3B^2 + 3B_1 – 1 &= B^3 \\
-3B^2 + 3(\frac{1}{2})-1& = 0\\
-3B_2 + \frac{1}{2} &= 0 \\
B_2 &= \frac{1}{6}
\end{align*}

Thanks to Laurens Gunnarsen for showing me this strange trick.

I’ve finally understood the principle which allows us to lower the index. The step where we move $B^i$ to be $B_i$ is quite simple. As Rota and Roman say, one method of expressing an infinite sequence of numbers is by a transform method. That is, to define a linear transform $B$ such that $$B x^n = B_n$$

So, the above “lowering of the index” is actually using the relation $(X-1)^n = X^n$, and applying $B$ to both sides of it. To get $B(X-1)^n = B(X^n)$. Let’s look for example at the “lowering step” of the first calculation:
\begin{align*}
X^1 &= \frac{1}{2} \\
B (X_1) &= B(\frac12) \\
B_1 &= \frac12 B(1) = \frac12
\end{align*}

How do I construct the Tits-Freudenthal magic square?

Thanks to Mia Hughes and John Huerta for the helpful discussions on this topic.

I am here taking another quick jab at trying to understand the construction of the Tits-Freudenthal Magic square. Let’s see if we can get into Vinberg’s mindset when he wrote down Vinberg’s construction.

Let’s say we knew the following theorem: $$\text{ the derivations of  } \mathcal{J}_3(\mathbb{O}) = f_4$$ We want to write down derivations of other algebras, $\mathbb{O} \otimes_{\mathbb{R}} \mathbb{D}$, where $\mathbb{D}$ is a division algebra.

Let’s see how we might derive the fact that $$\text{der}(\mathcal{J}_3(\mathbb{A})) \simeq a_3(\mathbb{A}) \oplus \text{der}(\mathbb{A})$$

where $a_3$ denotes the 3×3 trace-free antisymmetric matrices, and $$\text{der}(A) = \text{Lie}(\text{Aut}(A))$$

Continue reading How do I construct the Tits-Freudenthal magic square?