Definition: Lie Group

A **Lie Group** $G$ is a group which is a smooth manifold such that the group operations are smooth

Standard notation for the identity: $e \in G$

Multiplication in local coordinates: Let $U$ be a neighborhood of $e \in G$ with some local coordinates. Assume that the coordinates of $e$ are $(0, 0, \dots, 0)$.

Take an element $x \in U$ with coords $(x_1, x_2, \dots, x_n)$ and $y$ with coords $(y_1, y_2, \dots, y_n)$. Then, multiplying them via $x, y \mapsto z = x \ast y$, then the components of $z$ are smooth functions of $x$ and $y$:

$z_k = z_k(x_1, \dots, x_n; y_1, \dots, y_n)$

It is interesting and useful to look at the taylor expansion of $z_k$.

Theorem: Taylor Expansion of Z_k

$z_k = x_k + y_k + \sum_{ij} b_{ij,k} x_i y_j + \dots$ (terms of degree $\geq 3$) #-> Taylor expansion of $z_k$

similarly, if we consider the inversion map $x \to u = x^{-1}$ then the first terms for the Taylor expansion are:

$u_k = -x_k + \sum_{ij}b_{ij,k} x_i y_j + \dots$ (terms of degree $\geq 3$)

The constants $b_{ij,k}$ are specific to keep multiplication associative. If $b_{ij,k} = b_{ji,k}$ then the group will be commutative #-> because then swapping $x$ and $y$ doesn't change the sum

In general, the taylor expansion of an arbitrary smooth function $z_k = z_k(x_1, \dots, x_n; y_1, \dots, y_n)$ can be written as:

$z_k = a_0 + l_j x_j + m_j y_j + a_{ij} x_i x_j + b_{ij} x_i y_j + c_{ij} y_i y_j + \dots$

with summation over $i$ and $j$.

If we substitute $x = e$ ($= 1$), then we have $z = ey = y$, i.e.

$y_k = a_0 + m_j y_j + c_{ij} y_i y_j + \dots$

hence $a_0 = 0$, $m_j = 0$ for all $j$ except for $j = k$ and $m_k = 1$, and $c_{ij} = 0$.

Similarly, substituting $y = e$ gives $l_j = 0$ for all $j$ except $j = k$ and $l_k = 1$, and $a_{ij} = 0$.

So, $z_k = x_k + y_k + \sum_{ij} b_{ij} x_i y_j + ...$

Component of the identity

Remember that any smooth manifold $M$ can be presented as a disjoint union of connected manifolds called connected components of $M$:

$M = M_1 \bigcup M_2 \bigcup \dots, M_i \neq \{\}, M_i \bigcap M_j = \varnothing$

E.g. $M = S^1 \bigcup S^1$ is two disjoint circles; $M = S^1 \bigcup \mathbb{R}^1$ is a circle and an independent line

The second example is not a lie group because all connected components need to have the same topological structure #-> Because no operations between elements on a line and elements on a circle?

Theorem: Connected Components in Lie Groups

Let $G$ be a Lie group, $G = G_0 \bigcup G_1 \bigcup \dots$ be the decomposition into connected components, and $G_0 \subset G$ be the connected component that contains $e \in G$. In other words, G0 is a connected open-closed subset which contains e.

Notice that such a subset is unique and can be characterized as the set of those points which can be joined with e by a continuous path. #-> Unique because of the disjoint path-connected clause

.#-> In manifolds, connected = path connected

**Theorem**:

- $G_0$ is a normal subgroup in $G$;

- $G_0$ itself is a Lie group;

- any other connected component $G_i \subset G$ is diffeomorphic to $G_0$.

Proof:

Why subgroup: verify following conditions:

(i) $G_0$ is closed under multiplication

(ii) $G_0$ is closed under inversion

Proof of (i): If $a, b \in G_0$ and $G_0$ is path-connected, then there are continuous curves $a(t)$ and $b(t)$ which connect $e$ with $a$ and $b$ respectively, i.e. $a(0) = e$, $a(1) = a$, and $b(0) = e$, $b(1) = b$.

Then, $c(t) = a(t) b(t)$ is a continuous curve connecting $e$ with $ab$. Thus $ab$ belongs to the connected component that contains $e$, i.e. $G_0$. #-> Continuity ensures $c(t)$ is a path, and path-connectedness of $G_0$ ensures that $c(t)$ is on $G_0$

Proof of (ii) is similar; it suffices to consider the continuous curve $a(t)$ and $c(t) = a(t)^{-1}$.

Why $G_0$ is normal:

We just need to verify that for any $g \in G$, we have $gG_0 g^{-1} = G_0$.

Proof: Since the left multiplication by $g$ and right multiplication by $g^{-1}$ are both diffeomorphisms, $gG_0 g^{-1}$ is a certain connected component of $G$. #-> Why are they diffeomorphisms?

We know this because the left and right diffeomorphisms means that $gG_0 g^{-1}$ have the same topological structure as $G_0$. In addition, because $geg^{-1} = e$, we know that $e \in gG_0 g^{-1}$

Then, we can ensure that $gG_0 g^{-1} = G_0$ (because it is path-connected to $e$).

Why any $G_i$ is diffeomorphic to $G_0$:

If we take $G_0$ as the subgroup of $G$, then you can see that $G_i$ is a left or right coset of $G_0$, so taking an arbitrary element $g \in G_i$, $G_i = gG_0$. #-> Why are they left/right cosets?

Theorem: Connected Component Generation

Let $U$ be any connected neighborhood of $e \in G$. Then, $G_0$ is generated by $U$ in the sense that any element $x \in G_0$ can be represented as $x = u_1 \ast u_2 \ast u_3 \ast\dots\ast u_k$ with $u_i \in U$.

Remark:

- If $U$ is a neighborhood of $e$, then $aU$ and $Ua$ are both neighborhoods of $a \in G$. These neighborhoods are naturally homeomorphic (and diffeomorphic) to $U$, but in general do not coincide.

- The set $U^{-1} = \{u^{-1} | u \in U\}$ is a neighborhood of the identity $e$.

**Proof**:

Consider subsets $U_1 = U, U^2 = U\ast U = \{u_1 \ast u_2 | u_1, u_2 \in U\}, \dots, U^k$ and take $L =$ infinite union of $U^k$ over $k$. We need to prove $L = G_0$.

Because $G_0$ is a connected component, it's a set that's open, closed, and connected.

1) Clearly, $L$ is an open set because it's a union of open subsets.

2) $L$ is closed, because $G\setminus L$ is open. Assume $a \in G\setminus L$ and consider its neighborhood $aU^{-1}$. Check that this neighborhood does not intersect with $L$.

By contradiction: if $aU^{-1} \bigcap L \neq \varnothing$ then there must exist $u$ and $u_1, \dots, u_k$ such that $au^{-1} = u_1 \ast \dots \ast u_k$ or equivalently $a = u_1 \ast \dots \ast u_k \ast u$.

That is, $a \in U^{k + 1} \subset L$. This is a contradiction, as we assumed $a \in G\setminus L$. #-> If every point in a set has an open neighborhood contained in the set, then the set is open?

3) $L$ is connected, because for any product $x = u_1 \ast u2 \ast \dots u_k \in L$, there exists a continuous path $\gamma(t)$ on $L$ which connects $x$ with the identity element $e$. This is obvious, because we can simply take $\gamma(t) = \gamma_1(t) \ast gamma_2(t) \ast \dots \ast \gamma_k(t)$ where $\gamma_i(t)$ is a continuous path that connects $e$ and $u_i$, i.e. $\gamma_i(0) = e$, $\gamma_i(1) = u_i$. Then obviously $\gamma(0) = e$ and $\gamma(1) = x$

Thus, $L$ is a connected open-closed subset which contains the identity $e \in G$, so by definition $L$ is the connected component $G_0$.

Definition: Algebraic Linear Group

An **algebraic linear group** $G$ is a subgroup in $GL(n, \mathbb{R})$ which is defined by a system of polynomial equations $p_m(a_{11}, a_{12}, \dots, a_{nn}) = 0$ where $a_{ij}$ are matrix components of $\mathbf{A} \in GL(n, \mathbb{R})$.

Examples:

- $SL(n, \mathbb{R})$ is given just by one equation, $\text{det } \mathbf{A} = 1$.

- $O(n)$ is given by one matrix equation $\mathbf{AA}^{T} = \text{id}$, which amounts to $\frac{n(n + 1)}{2}$ usual equations. In the case of $SO(n)$, one additional equation appears: $\text{det } \mathbf{A} = 1$.

etc.

Theorem: Subgroup of General Linear Group Is Linear Group

Any algebraic linear group $G \subseteq GL(n, \mathbb{R})$ is a Lie subgroup in $GL(n, \mathbb{R})$. In particular, $G$ is a Lie group.

Proof sketch:

The main problem is whether a subgroup $G \subseteq GL(n, R)$ has any singular points.

Take $V$ as a neighborhood of $e = \text{id} \in GL(n, \mathbb{R})$, and $U_e = G \bigcap V$ as the neighborhood of $e \in G$. Then, for any $x \in G$, $U_x = xU_e = x(G \bigcap V) = G\bigcap (xV)$ is the neighborhood of $x \in G$.

Then, because the left translation by $x$ is a diffeomorphism of $GL(n, \mathbb{R})$, $U_x$ and $U_e$ are absolutely "isomorphic" from both topological and diffeomorphic viewpoints.

Because of this, if locally $G$ has the structure of a smooth submanifold of $GL(n, \mathbb{R})$, then this condition should hold at any other point in $G$.