\chapter{Lie algebrae}


Finding continuous transformation groups on some space $X$ is usually tantamount to solving a Lie point symmetry group. 

In practice, one often starts with very limited information regarding the Lie group. Fortunately there is an extremely useful intermediate structure that proves vital in deducing data of the Lie group; the \emph{Lie algebra}.

As will be discussed, Lie algebrae and Lie groups have a very close correspondence with eachother; this can be summarized by the following:

\begin{itemize}
\item Every Lie group gives rise to a Lie algebra
\item Every Lie algebra gives rise to a Lie group
\item Isomorphic Lie groups correspond to isomorphic Lie algebrae
\item The Exponential map is the major tool used for obtaining data of the Lie group from the Lie algebra
\end{itemize}

This chapter will develop the motivation, definition, properties, and basic applications of Lie algebrae.



Irregardless of what we know about a specific Lie group $G$, left (or right) multiplication maps are a natural class of autodiffeomorphisms on a Lie group. Informally speaking, this represents a class of smooth symmetries for the Lie group itself; these maps are ways that the Lie group's elements can be smoothy permuted! 

In an attempt to gain insight to the Lie group, we take a look at the local behaviour of these maps on the identity element.

In what direction does each multiplication map $L_g$ "nudge" its image as it works on elements from $1_G$ along $\mathbf{v}_{1_G}$? This is essentially the evaluation of $(dL_{g})_{1_G}(\mathbf{v}_{1_G})$; it is useful to encode all these "nudges" by different $L_g$ associated with a single direction from the identity into one single vector field as such.

\[ \mathbf{V}(g) = (dL_{g})_{1_G}(\mathbf{v}_{1_G})\]

\begin{definition}[Infinitesimal generator]
Let $\mathbf{v}_{1_G}$ be a tangent vector in $T_{1_{G}} G$, then an \emph{infinitesimal generator of $G$} is a vector field $\mathbf{V}$ of the following form.
\[ \mathbf{V}(g)=(dL_{g})_{1_G}(\mathbf{v}_{1_G}) \]
Denote the set of all IGs of $G$ as $\mathfrak{g}$
\end{definition}

Essentially, each IG captures data related to how all the multiplication maps behave as they drift away from the identity in a certain direction. 
The reason we define infinitesimal generators as such is because it relies only on the tangent space at the identity, and because left (or right) multiplication maps being \emph{bijective} autodiffeomorphisms are most likely to capture the most group elements.

As it turns out, knowledge of the IGs alone is extremely powerful at recovering the structure of a Lie group; in special cases one can even determine the whole Lie group back! 



\begin{proposition}
Let $G$ be a Lie group, then $\mathfrak{g}$ is a linear space
\end{proposition}


We have seen that the IGs form a linear space, but what's more is that they are also closed under the Lie bracket, meaning they form not only an $F$-linear space, but more generally an \emph{$F$-algebra}!

\begin{proposition}
Let $G$ be a Lie group and $\mathbf{V}, \mathbf{U} \in \mathfrak{g}$, then $[\mathbf{V},\mathbf{U}] \in \mathfrak{g}$
\end{proposition}

Since the Lie bracket is an example of a bilinear operation, and it is closed on the linear spaces of infinitesimal generators, the linear spaces of infinitesimal generators can be upgraded to a structure called an \emph{algebra over a field}.


\begin{definition}[Lie algebra of a Lie group]
Let $G$ be a Lie group, then the \emph{Lie algebra of $G$} is the linear space $\mathfrak{g}$ of all \emph{infinitesimal generators}
\end{definition}


At the heart of an IG is some tangent vector at the identity that generates it, and one can show that these tangent vectors do more than just generate IGs, but characterize their structure completely!

Every tangent vector at $1_{G}$ produces its own linear space, and by the linearity of the differential, the addition of these tangent vectors is compatible with the addition of IGs and the zero vector of $T_{1_G}$ is associated with the zero vector of the Lie algebra.

Moreover, there is a bijective correspondence between IGs and these tangent vectors;
\[ (dL_{g})_{1_G}(\mathbf{u}_{1_G}) = (dL_{g})_{1_G}(\mathbf{v}_{1_G})\]
Since $L_{1_G}$ is the identity map on $G$, letting $g=1_{G}$ shows the following
\[ \mathbf{u}_{1_G} =\mathbf{v}_{1_G}\]
Therefore each tangent vector at the identity corresponds uniquely to some IG.


All these conditions come together to mean that one can treat the Lie algebra as the Tangent space at the identity!

\begin{proposition}
Let $G$ be a Lie group, then the tangent vectors at $1_{G}$ and Lie algebra $\mathfrak{g}$ are isomorphic as linear spaces.
\[\mathfrak{g} \cong T_{1_G} G\]
\end{proposition}

























\section{Lie algebrae}



As we have demonstrated, an algebra of infinitesimal generators can be generated from a given Lie groups, however the motivation for Lie algebrae is to construct a Lie group back. We will introduce a definition of a Lie algebra independent of a Lie group and eventually show how a generic Lie algebra has sufficient data to determine its own Lie group.


\begin{definition}[Lie algebra]
A \emph{Lie algebra} $\mathfrak{g}$ is an $F$-linear space with an alternating bilinear operation satisfying the Jacobi identity. 
	This operation is often called the \emph{Lie bracket of $\mathfrak{g}$}
\end{definition}

\begin{definition}[Lie subalgebra]
	linear subspace of Lie algebra closed under the Lie bracket of $\mathfrak{g}$
\end{definition}

Center of a Lie algebra

Lie bracket is commutative iff it is always zero.

Lie bracket is associative iff all Lie brackets are in the center.


\begin{definition}[Lie algebra homomorphism]
linear isomorphism such that
	\[ f([\mathbf{x},\mathbf{y}]) = [f(\mathbf{x}),f(\mathbf{y})] \]
\end{definition}


\begin{definition}[Adjoint Lie algebra map]
\[ \mathrm{ad}_{\mathbf{x}}(\mathbf{y}) = [\mathbf{x},\mathbf{y}] \]
\end{definition}

\begin{definition}[Adjoint homomorphism]
linear isomorphism from $\mathbf{g} \to \mathrm{Aut}_{\mathbf{Lin}}(\mathbf{g})$
\[ f(\mathbf{x}) = \mathrm{ad}_{\mathbf{x}} \]
\[ \mathrm{ad}_{\mathbf{x}}(\mathbf{y}) = [\mathbf{x},\mathbf{y}] \]
\end{definition}


Note that adjoint homomorphisms have the following representation.
\[ \mathrm{ad}_{[\mathbf{x},\mathbf{y}]}(\mathbf{z}) = [[\mathbf{x},\mathbf{y}],\mathbf{z}] \]
\[ \mathrm{ad}_{[\mathbf{x},\mathbf{y}]}(\mathbf{z}) = - [[\mathbf{y},\mathbf{z}],\mathbf{x}]  - [[\mathbf{z},\mathbf{x}],\mathbf{y}]\]
\[ \mathrm{ad}_{[\mathbf{x},\mathbf{y}]}(\mathbf{z}) = [\mathbf{x} , [\mathbf{y},\mathbf{z}]]  - [\mathbf{y},[\mathbf{x},\mathbf{z}]]\]
\[ \mathrm{ad}_{[\mathbf{x},\mathbf{y}]}(\mathbf{z}) = \mathrm{ad}_{\mathbf{x}} \circ \mathrm{ad}_{\mathbf{y}}(\mathbf{z}) - \mathrm{ad}_{\mathbf{y}} \circ \mathrm{ad}_{\mathbf{x}}(\mathbf{z}) \]

first isomorphism theorem

\subsection{Structure constants}

Being a vector space, all Lie algebrae have a (Hamel) basis.
By choosing some basis and applying linearity of the Lie bracket, the Lie bracket of any two elements can be reduced to a weighted sum of Lie brackets of basis vectors (just like the actual Lie bracket of differential topology).

We use the notion of \emph{structure constants} of a known basis to help calculate Lie brakcets; no matter how much Lie brackets of different spaces differ semantically, each of them can be reduced to a similar calculation method by using these structure constants. Furthermore, since the Lie bracket must obey skewsymmetry and the Jacobi identity, there are various conditions these structure constants follow.

%Since all Lie algebrae have a basis of vector fields, and any Lie brakcet of vector fields lies within the Lie algebra, then the Lie bracket of basis vector fields (which can be used to calculate any Lie bracket) is representable by the Hamel basis; we can find \emph{structure constants} for the Lie bracket to assist with computing Lie brackets and conveniently represent them using \emph{Commutator tables}.

%Structure constant and commutator tables examples here

Since the Lie bracket is bound by certain algebraic relations, these have a direct impact on the relationship between the structure constants.

%theorems regarding skew symmetry and jacobi identity of sturcture constancs


%Given a Lie group, is there a "canonical" Lie algebra that can always be made? The answer is "yes" and this relationship is the backbone of most of Lie theory.

%Since Lie groups are smooth manifolds, each point has its own tangent space. If one considers the tangent space at the identity element

%On its own this correspondence seems slightly insipid, although there is a nice technique to make each tangent vector at $1_G$ actually \emph{extends uniquely to a special type vector field on all of $G$}.

%We construct these vector fields from the tangent vectors as such; consider the autodiffeomorphisms $f_g (h) = hg$ that exist on any Lie group (one could consider left-multiplication maps too), given some tangent vector of the Lie algebra $\mathbf{v}_{1_G}$, we can define $\mathbf{V}_{\mathbf{v}_{1_G}}(h) = d f_h (\mathbf{v}_{1_G}) $ as a vector field associated with $\mathbf{v}_{1_G}$. We call this the \emph{infinitesimal generator associated with $\mathbf{v}_{1_G}$}.

%Although it is interesting that one can naturally extend an indentity tangent vector to a whole vector field by this construction, what does the infinitesimal generator mean from a geometric, intuitive point of view? The left-multiplication are the natural autodiffeomorphisms on $G$ given the group structure; if one were to "nudge" all manifold point infinitesimally according to one of the left-multiplication maps, the image would be the same Lie group (i.e it is a continuous symmetry)). 

%An infinitesimal generator captures the "nudges" of one of these maps at all manifold points using a single vector field; it shows "how" you can transform a Lie group isomorphically.

%We made the claim that these tangent vectors ar the identity have a bijective correspondence to infinitesimal generators; let's prove this. Distinct tangent vector in $\mathfrak{g}$ generate distinct infinitesimal generators; this is easily seen by evaluating both infinitesimal generators at the identity element, giving us $\mathbf{V}_{\mathbf{v}_{1_G}}(1_G) = \mathbf{v}_{1_G}$ and $\mathbf{V}_{\mathbf{u}_{1_G}}(1_G) = \mathbf{u}_{1_G}$, which are distinct mappings at the same point on the manifold.

%Conversely, since infinitesimal generators evaluated at the identity return their corresponding vector, identical IGs return the same vector and hence identical IGs are formed from the same tangent vector of $\mathfrak{g}$, proving the bijection.

%For future, we consider vector fields that are invariant under these left-multiplication maps.

\section{Classical Lie algebrae}

An important class of Lie algebrae are that of the classical Lie groups. These examples are widespread, concrete examples of Lie algebrae, and working with them will help strengthen one's understanding of Lie algebrae in general.

We will first examine $\mathfrak{gl}(n)$; the Lie algebra of $\mathrm{GL}(n)$. Interpreting Lie algebrae from the perspective of a tangent space to the identity permits the best insight for its calculation; since the identity is $\mathbf{I}$, one may consider the Lie group as the tangent vectors from this element.

To calculate the permissible tangent vectors at the identity matrix, one considers arbitrary "smooth curves of invertible matrixes", say $\mathbf{A}(t)$, where $\mathbf{A}(0)= \mathbf{I}$. Since we take matrixes with the Euclidean topology on $\mathbb{R}^{n^2}$, one can directly differentiate this curve and evaluate at $0$ to obtain matrixes as the tangent vector at the identity matrix.

Futhermore, since we are working in $\mathrm{GL}(n)$, every matrix on the curve must be invertible; matrixes on the curve are bound by the expression $\mathrm{det}(\mathbf{A}(t))=c, c \neq 0$. Differentiating this expression gives a condition that elements of the Lie algebrae must obey, so the matrixes in the Lie algebrae are determined by $\mathrm{det}'(\mathbf{A}(t))=0$.



Results pertaining to Lie subalgebrae will allow the results of $\mathfrak{gl}(n)$ to extend to other classical Lie algebrae.

$[\mathbf{A},\mathbf{B}] = \mathbf{A}\mathbf{B} - \mathbf{B}\mathbf{A}$

The matrix commutator is the Lie bracket on any such Lie subalgebra.


To calculate, we use the fact that $\mathfrak{gl}(n)$ is isomorphic to $T_{\mathbf{I}}\mathrm{GL}(n)$




















\section{Lie algebra ideals}


\begin{definition}[Lie algebra ideal]
	Lie subalgebra $I$ such that for all $\mathbf{i} \in I$, one has $ [\mathbf{x},\mathbf{i}] \in I$ for any $\mathbf{x}$ in $\mathfrak{g}$
\end{definition}

vector addition over ideals is an ideal
Lie bracket over ideals is an ideal
intersection of ideals is ideal

\subsection{Quotient lie algebrae}

Lie algebra of left cosets of a Lie algebra ideal

second isomorphism theorem
third isomorphism theorem

Let $J \leq I$ be an ideal of $\mathfrak{g}$, $I /J$ is an ideal of $\mathfrak{g} / J$ 

\section{Lie algebra constructions}

\begin{definition}[Derived Lie algebra]
\[ \mathfrak{g}'=[\mathfrak{g},\mathfrak{g}] \]
\end{definition}

Let $\mathbf{z} \in \mathfrak{g}'$, then $\mathrm{tr}(\mathrm{ad}_{\mathbf{z}})=0$



\section{Classifying Lie algebrae}

Linear spaces are classified purely by dimension, so the only way that Lie algebrae of the same dimension may not be isomorphic is by their Lie brackets differing in behaviour.

We analyze the behaviour of the Lie bracket by studying the center it creates, as well as the "derived" Lie algebra it produces.


For each dimension, there is one abelian Lie algebra.

There is only one Lie algebra of dimension 1.

There is only one nonabelian Lie algebra of dimension 2, and it has a basis such that $[\mathbf{e}_1,\mathbf{e}_2]=\mathbf{e}_1$.

\subsection{Lie algebrae of dimension 3}

There are is more diversity among Lie algebrae of dimension 3. 

Heisenberg algebra (derived has dimension 1 and is contained in center)
(derived has dimension 1 and is not contained in center)
(derived has dimension 2 and is not contained in center)


\begin{proposition}[Heisenberg algebra]
	Let $\mathfrak{g}$  be a nonabelian Lie algebra such that $\mathrm{dim}(\mathfrak{g})=3$ and $\mathfrak{g}' \leq Z(\mathfrak{g})$ has dimension 1, then we have a basis such that $[\mathbf{e}_1,\mathbf{e}_2]=\mathbf{e}_3$
\end{proposition}

Choose 2 vectors $\mathbf{x},\mathbf{y}$ such that $[\mathbf{x},\mathbf{y}]=\mathbf{z}$ where  $\mathbf{z}\neq \mathbf{0}$, which is possible since $\mathfrak{g}$ is nonabelian. Since $\mathfrak{g}' \leq Z(\mathfrak{g})$,  for any $\mathbf{v}$ one has $[\mathbf{v},\mathbf{z}]=\mathbf{0}$.

Let $c_1 \mathbf{x} + c_2 \mathbf{y} + c_3 \mathbf{z} =\mathbf{0}$, then $[ \mathbf{x} , c_1 \mathbf{x} + c_2 \mathbf{y} + c_3 \mathbf{z}] = c_2 \mathbf{z} = \mathbf{0}$, and since $\mathbf{z}\neq \mathbf{0}$, therefore $c_2=0$. One can use an identical argument to show that $c_1=0$, and since $\mathbf{z}\neq \mathbf{0}$, then $c_1=c_2=c_3=0$ so these vectors form a basis for $\mathfrak{g}$.































\section{Invariant vector fields}

There is an alternative way of arriving that the construction of an IG, This brings useful properties of IGs to light.

\begin{definition}[Left-invariant vector field]
Let $G$ be a Lie group.
$d f_h (\mathbf{V}(g))= \mathbf{V}(hg)$
\end{definition}

\begin{proposition}
Infinitesimal generators are precisely the left invariant vector field on Lie group.
\end{proposition}




\subsection{Vector flows on infinitesimal generators}


\begin{theorem}
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, the vector flow on an IG of $\mathfrak{g}$ is Lie isomorphic to $(\mathbb{R},+)$ or $\mathrm{SO}(2)$
\end{theorem}




















In the spirit of seeking symmetries, we seek for a natural class of autodiffeomorphisms on a Lie group to form an algebraic structure with. In group theory, the maps $f_h (g) = hg$ are general group automorphisms and autodiffeomorphisms 


\begin{definition}[Lie algebra of a Lie group]
Let $G$ be a Lie group, then its Lie algebra $\mathfrak{g}$ is the linear space of all right-invariant vector fields on $G$ (also called \emph{infinitesimal generators})
Let $G$ be a Lie group, then its Lie algebra $\mathfrak{g}$ is the linear space of all right-invariant vector fields on $G$ (also called \emph{infinitesimal generators})
\end{definition}

Consider "right-invariant" vector fields on a lie group. note that they are entirely characterized by $\mathbf{F}(1_G)$, so any tangent vector of $T_{1_G} G$ corresponds to an infinitesimal generator; this means we can colloquially think of a Lie algebra as $T_{1_G} G$







\subsection{Exponential map}


As preluded to, the Lie algebra can be used to reconstruct (at least partially) the related Lie group. Interpreting the Lie algebra as the tangent space to the identity, 


%Given a Lie group $G$ and its Lie algebra $\mathfrak{g}$, there is a natural map one can always form from the $\mathfrak{g}$ to $G$. In 


%Say we have a Lie algebra $\mathfrak{g}$ representing all infinitesimal generators of $G$ (vector fields on $M$ whose vector flow at $G$)

%It would be useful if there were some general function from the Lie algebra to the Lie group that always exist. Note that any infinitesimal generator of $\mathfrak{g}$ gives a corresponding vector flow in $G$; if one takes 

Given that the elements of $\mathfrak{g}$ are smooth vector fields on $G$, one can recover group elements by considering vector flows from the identity; choosing an IG and its vector flow for $1$ unit of time is how we define the \emph{exponential map}.

\begin{definition}[Exponential map]
Let $G$ be a Lie group with Lie algebra of IGs $\mathfrak{g}$, the \emph{exponential map} is the map from the Lie algebra to its Lie group defined as such, where $\psi_{\mathbf{V}}$ is the vector flow of $\mathbf{V}$
$\exp : \mathfrak{g} \to G$
	\[\exp(\mathbf{V}) = \psi_{\mathbf{V}}(1,1_{G})\]
\end{definition}


Scaling an IG $\mathbf{V}$ by $t$ can be used to find other times for the vector flow on $\mathbf{V}$; indeed one has the following expression

\[\exp(t\mathbf{V}) = \psi_{\mathbf{V}}(t,1_{G})\]



This map serves as a direct tool to recover data of the Lie group using its corresponding Lie algebra.


What makes this map "exponential" is that for the Lie group $(\mathbb{R}_{+},\cdot)$ the exponential map is exactly the exponential function.


The Lie algebra of $(\mathbb{R}_{+},\cdot)$ is spanned only by the infinitesimal generators is $x\partial_x$, and so the vector flow for $c_1 x \partial_x$ is solved to be $\psi(\varepsilon , x)= e^{c_1 \varepsilon}x$, and so the exponential function is $\exp ( c_1 x \partial_x ) = e^{c_1}$.


Please consider the flow through the identity of a Lie group for some vector field, $\psi_{\mathbf{v}}(t,1_{G})$; this actually forms a 1-dimensional Lie subgroup (isomorphic to either $\mathbb{R}$ or $\mathrm{SO}(2)$).


\begin{proposition}
Given a right-invariant vector field $\mathbf{F}$ on Lie group $G$, its integral curve at the identity is a one-parameter Lie subgroup. Given vector flow $\psi$, elements of the form $\psi(t,1_G)$  for any $t$ form this Lie subgroup.
\end{proposition}

Essentially just taking the integral curve through the identi


Notice that any space of dimension 1 has trivial Lie bracket.




























\section{Lie subalgebrae}

Linear subspace of a Lie algebra closed under the Lie bracket.

\begin{theorem}[Lie subalgebra-Lie subgroup correspondence]
Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$, then any Lie subgroup $H \leq G$ is associated with some Lie subalgebra $\mathfrak{h} \leq \mathfrak{g}$. Conversely, any Lie subalgebra $\mathfrak{h} \leq \mathfrak{g}$ is of some Lie subgroup $H \leq G$.
\end{theorem}

\begin{theorem}[Ado's theorem]
	Let $\mathfrak{g}$ be a $n$-dimensional Lie algebra, $\mathfrak{g}$ is isomorphic to a Lie subalgebra of $\mathfrak{gl}(n)$
\end{theorem}
This allows a proof of a more general theorem

\begin{theorem}[Lie algebra-Lie group correspondence]
\end{theorem}


\subsection{Lie algebrae of local Lie groups}


There is a proposition that easily allows the calculation of a basis for such Lie algebrae.

\begin{proposition}
	\[ \mathbf{F}_{k}(x) = \sum \xi^{i}_{k}(x) \partial_{x^i}\]
\end{proposition}

Notice that the right-invariant maps can be used to represent the group operation. Using the properties of right-invariant vector fields gives the following.

Lie derivative