Umbral Calculus Derivation of the Bernoulli numbers

$$“(B-1)^n = B^n”$$

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

(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}

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:
X^1 &= \frac{1}{2} \\
B (X_1) &= B(\frac12) \\
B_1 &= \frac12 B(1) = \frac12

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?