## Models of Formal Group Laws of Every Height

This is not yet air-tight, but it is better to put this out there in the hopes people will find it helpful. I’ll continue to iron it out as time goes. This paper is with an application in topology in mind.

everyheight

## 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 $$\text{der}(A) = \text{Lie}(\text{Aut}(A))$$

## What does an algebraic integer have to be?

What does an integer have to be?

• No matter how you extend $\mathcal{Q}$, the integers which lie in $\mathcal{Q}$ must lie in $\mathcal{Z}$.
• If $\alpha$ is an integer, then so are its conjugates.
• The sums and products of integers are also integers.

From this we may describe what an algebraic integer must be.

Start with a root $\alpha$.

Look at all of it’s conjugates. Continue reading What does an algebraic integer have to be?