Lie Groups and Lie Algebras Lecture 1 notes

Definition: Lie Group

A Lie Group is a space that possesses two structures:
1) structure of a group
2) structure of a smooth manifold

Examples (main are matrix groups):
- General linear group $GL(n, \mathbb{R})$
- Special linear group $SL(n, \mathbb{R})$
- Orthogonal groups $O(n, \mathbb{R})$ , $SO(n, \mathbb{R})$
etc.

Simplest examples:

$SO(2)$ = $\{2\times 2$ matrix $\mathbf{A}$ s.t. $\mathbf{AA}^T = \text{id}$ , det $\mathbf{A} = 1\}$
Then, it's not hard to verify that $SO(2)$ is the set of all rotation matrices.
As a smooth manifold, $SO(2)$ is a circle (trivial to show)

Similarly, the orthogonal group:
$O(2) = \{2\times 2$ matrix $\mathbf{A}$ s.t. $\mathbf{AA}^T = \text{id}\}$
This ends up to be similar to $SO(2)$ , except the polarity of the right side of the matrix is flipped (determinant can be either 1 or -1)
This is a disjoint union of two circles

$SL(2, R) = \{2\times 2$ matrix $\mathbf{A}$ s.t. det $A = 1\}$
Because there are 4 dimensions ( $2\times 2$ ), this defines a second order hypersurface in $\mathbb{R}^4$ #-> Second order hypersurface? Defined by second order polynomial

Theorem: Proposition 1

$SL(2, \mathbb{R})$ is three-dimensional and diffeomorphic to the direct product $S^1 \times \mathbb{R}^2$ .
Sketch of proof:
If we write $a = x + y, d = x - y, b = u + v, c = u - v$ , we get
$ad - bc = 1 \to x^2 - y^2 - u^2 + v^2 = 1$
then, $x^2 + v^2 = 1 + y^2 + u^2$
If $(y, u) \in \mathbb{R}^2$ , then $(x, v)$ lies on a circle $S^1$ (of variable radius). Hence, $((y, u), (x, v)) \in \mathbb{R}^2 \times S^1$ .

Unitary group:

$U(2) = \{2\times 2$ matrix where coeffs in $\mathbb{C}$ , $\mathbf{AA}^* = \text{id}\}$ #-> Implication of conjugate?

Special unitary group:
$SU(2) = \{2\times 2$ matrix where coeffs in $\mathbb{C}$ , $\mathbf{AA}^* = \text{id}$ , and det $\mathbf{A} = 1\}$

Theorem: Proposition 2

$SU(2)$ is three-dimensional and diffeomorphic to 3 dimensional sphere $S^3$ . $U(2)$ is four-dimensional and diffeomorphic to the direct product $S^3 \times S^1$ .
Sketch of proof:
In terms of $a, b, c, d$ , $\mathbf{AA}^* = 1$ can be represented as three equations.
The first, $a\overline{a} + b\overline{b} = 1$ , means that $(a, b)$ in $\mathbb{C}^2 = \mathbb{R}^4$ belongs to the sphere of radius 1 (because $a_1^2 + a_2^2 + b_1^2 + b_2^2 = 1$ )
The second, $c\overline{a} + d\overline{b} = 0$ , means that $(c, d)$ is proportional $(-\overline{b}, \overline{a})$ . This means that $c = -\lambda\overline{b}$ , $d = \lambda\overline{a}$ for some $\lambda$ .
The third, $c\overline{c} + d\overline{d} = 1$ , means that if we plug in our above expressions for $c$ and $d$ , we get $|\lambda| = 1$ , which means that $\lambda = e^{i\phi}$ for some $\phi$ .
Then, matrices in $U(2)$ are $\begin{bmatrix}a & b \\ -\lambda\overline{b} & \overline{a}\end{bmatrix}$ .
Because the determinant of matrices in $U(2)$ are exactly $\lambda$ , setting $\lambda = 1$ gives us matrices in $SU(2)$ : $\begin{bmatrix}a & b \\ -\overline{b} & \overline{a}\end{bmatrix}$ .
Then, there is a natural bijection between the 3-sphere $S^3$ = $\{(a, b, c, d) \in \mathbb{R}^4 \text{ s.t. } a^2 + b^2 + c^2 + d^2 = 1\} = \{(a, b) \in \mathbb{C}^2 \text{ s.t. } |a|^2 + |b|^2 = 1\}$ and $SU(2)$ .
In the case of $U(2)$ , we have an additional parameter $\lambda = e^{i\phi}$ , which defines a circle, so $U(2) = S^3 \times S^1$ .

Definition: Abelian Group

A group is called Abelian if the binary operation is commutative.
Examples:
- Any vector space $V$ over $\mathbb{R}$ can be an abelian lie group, with the binary operation being addition: $(u, v) \mapsto u + v$ .
- A torus $T^n = S^1 \times S^1 \times \dots \text{ n times } \dots \times S^1$ is also an abelian lie group. Representing each point on $T^n$ as an $n$ -tuple $(\phi_1, \phi_2, \dots, phi_n)$ where each $\phi_k$ is an "angle" mod $2\pi$ . The binary operation then is the addition mod $2\pi$ : $(\phi_1, dots, \phi_n) + (\psi_1, \dots, \psi_n) = ((\phi_1 + \psi_1) \mod 2\pi, \dots, (\phi_n + \psi_n) \mod 2\pi)$
Note that $T^n$ can be represented as a matrix lie group, if we assign each $n$ -tuple $(\phi_1, \dots, \phi_n)$ to the diagonal matrix with each component as $e^{i\phi_k}$ .

On a group, we can define many other structures of Lie groups (not all commutative). Ex. on $\mathbb{R}^3$ we can define:
$(x_1, x_2, x_3) \cdot (y_1, y_2, y_3) = (x_1 + y_1, x_2 + y_2, x_3 + y_3 + x_1y_2)$
The smoothness of this operation is evident, even though it's non-linear. #-> Why? Is it because each component is a sum or product?
We have a natural matrix representation, the group of upper triangular matrices: $G = \{\mathbf{A} = \begin{bmatrix}1 & x_1 & x_3 \\ 0 & 1 & x_2 \\ 0 & 0 & 1\end{bmatrix}\}$ . We can see that a product of two of such groups yields the corresponding group.

General Linear Group:
$GL(n, \mathbb{R})$ = all $n\times n$ matrices with non-zero determinants
It's smooth because it's an open subset in the vector space $\mathbb{R}^{n^2}$ .
Because $GL(n, \mathbb{R})$ is not connected, it consists of $\{\mathbf{A}: \text{ det } \mathbf{A} > 0\}$ and $\{\mathbf{A}: \text{ det } \mathbf{A} < 0\}$

Theorem: Proposition 3

In the simplest case $GL(2, \mathbb{R})$ , each of these components (refering to $\{\mathbf{A}: \text{ det } \mathbf{A} > 0\}$ and $\{\mathbf{A}: \text{ det } \mathbf{A} < 0\}$ ) is diffeomorphic to $S^1 \times \mathbb{R}^3$ .
Rough proof sketch:
We can express any matrix in $GL(2, \mathbb{R})$ with positive determinant as a product of some positive constant and a matrix in $SL(2, \mathbb{R})$ .
Then, as $SL(2, \mathbb{R}) \cong S^1 \times \mathbb{R}^2$ and $\mathbb{R}_+ \cong \mathbb{R}$ , the extra scaling coefficient adds a dimension, and thus $GL(2, \mathbb{R}^+) \cong S^1 \times \mathbb{R}^3$ . The same argument can be made for $GL(2, \mathbb{R}_-)$ .