Characterizing 2 x 2 matrices squaring to the identity

Suppose A is an 2 x 2 real matrix such that $A^2 = I$. Then either $A = \pm I$ or \[ A = \begin{pmatrix} a & b \\ c & -a \end{pmatrix}, a^2 + bc = 1. \]

Some number theory questions

1. For all n > 1, n does not divide $2^n - 1$.
   Proof: Suppose not. Then n is odd as $2^n - 1$ is odd. Let $p$ be the smallest prime dividing n. We have $(p - 1, n) = 1$. By division algorithm, $n = q(p - 1) +k$ for some integers q and k with $0 < k < p$. By Fermat’s Little Theorem, $2^{p - 1} \equiv 1$ (mod p), so $2^k \equiv (2^{p - 1})^q2^k \equiv 2^n \equiv 1$ (mod p). This results in a contradiction by infinite descent.

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Universal enveloping algebra

For a complex Lie algebra $\mathfrak{g}$, we can define its universal enveloping algebra $U(\mathfrak{g})$ as the complex associative algebra $T(\mathfrak{g})/I$ where $T(\mathfrak{g})$ is the tensor algebra over $\mathfrak{g}$ and $I$ is the ideal generated by $\{ X \otimes Y - Y \otimes X - [X, Y]: X, Y \in \mathfrak{g} \}$. We have the natural map $\iota: \mathfrak{g} \rightarrow U(\mathfrak{g})$. The universal enveloping algebra satisfies the following universal property:

For any complex associative algebra $A$ and linear map $f: \mathfrak{g} \rightarrow A$ satisfying $[f(X), f(Y)] = f(X)f(Y) - f(Y)f(X)$ for all $X, Y \in \mathfrak{g}$, there is a unique algebra homomorphism $f^*: U(\mathfrak{g}) \rightarrow A$ such that $f = f^* \circ \iota$.

Theorem (Poincare-Birkhoff-Witt). Let $\{X_1, \ldots, X_m\}$ be a basis for $\mathfrak{g}$. Then $\{(\iota X_1)^{n_1}\cdots(\iota X_m)^{n_m} : n_1, \ldots, n_m \geq 0\}$ is a basis for $U(\mathfrak{g})$.

Chatting

Last night I had dinner with Brian. We talked about a lot of things, from social issues, politics to love affairs. One interesting question we discussed is why quite some friends in our circles are still single, given that they are very good persons. We didn’t arrive at a definite conclusion, but I believed that one factor is that our circle consists of people who are independent. Anyway, I wish every one of my friends can find his/her Miss/Mr. Right.

Recently Brian is reading a book about the philosophy of travel. The author is Alain de Butoon. It seems that he had written some interesting books. I hope I can find some time to read one.

Test post 2

廢up

人生有時就係唔知為乜。女？樓？Bullshit!

日常

存在是否客觀事實？:joy:

Test post

The Lie bracket of $\mathfrak{gl}(n, \mathbb{C})$ is given by

$$[X, Y] = XY - YX.$$

The following is due to Euler:

$$\sum_{n = 1}^\infty \frac{1}{n^s} = \prod_{p \textrm{ prime}} \frac{1}{1 - p^{-s}}.$$

測試：

$$\int_{空間} d\omega$$
$$\require{AMScd} \begin{CD}K(你) @>{人生}>> H(你;\mathbb Q);\\@VVV @VVV \\K(我) @>{人生}>> H(我;\mathbb Q);\end{CD}$$

There are three crazy guys in mathematics:

1. Hilbert
2. von Neumann
3. Grothendieck

滾滾長江東逝水 浪花淘盡英雄

#include <iostream>
using namespace std;

int main()
{
    cout << "Good!!!" << endl;
}

good

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