The Height of a Formal Group Law in terms of the Symmetry of the Underlying CM Abelian Variety

This proof was made possible by a couple helpful and fabulous conversations with Yifeng Liu. All errors are mine and mine alone.

This is toward my understanding of the phrase “Why is is height so important as an invariant? Because the height of a formal group law comes from the symmetry of the underlying variety.”

One method of getting a lower dimensional formal group laws from a abelian varieties of higher dimension is via using the theory of complex multiplication — splitting the abelian variety by splitting the prime (as I exposited in my paper here).

I show that for abelian varieties with CM, the height of the formal group law pieces are expressible as a formula in terms of the degree of some field extensions of $Q_p$ (one corresponding to each prime living over $p$) and the dimension of the rational endomorphism ring of the variety as a $Q$-vector space. Continue reading The Height of a Formal Group Law in terms of the Symmetry of the Underlying CM Abelian Variety

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 formal group law theory) in terms of the symmetry of the underlying variety.

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.  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*}