These notes introduce Lie groups through matrices. The first goal is concrete: understand when familiar families of invertible matrices form smooth groups, and learn how curves through the identity turn nonlinear matrix equations into linear infinitesimal equations.

A second goal is structural. Once the infinitesimal object has been built, the notes ask how much local group multiplication, global topology, and representation theory it controls. The advanced tools needed for those questions are introduced only when they become necessary in the later chapters.

The course therefore moves from examples and calculations toward general structure. Matrix groups come first, Lie algebras and local formulas come next, and the final chapters explain how topology, representations, and integration enter the same story.

# Introduction

Lie groups are spaces where algebra and geometry meet: they are groups whose elements can be varied continuously and differentiably. This course begins with matrix examples, because matrix multiplication, determinants, eigenvalues, and invariant forms give concrete models in which the main ideas can be computed before the abstract language is introduced. The guiding problem is to understand how much of a group near the identity is controlled by its tangent space there, and how local infinitesimal data determines global structure.

The foundations developed here begin with closed subgroups of general linear groups, then pass to Lie algebras, the exponential map, local multiplication formulas, and covering groups. Differential geometry is used as the language of tangent spaces and smooth maps, but the first part of the course keeps the geometry tied to matrices. The main examples are rotations, unitary matrices, determinant-one matrices, symplectic matrices, and nilpotent upper-triangular groups such as the Heisenberg group.

## The Central Questions

The first question is why a group should carry a smooth structure at all. Finite groups have no interesting local geometry, while groups such as rotations, invertible matrices, and unitary matrices vary in continuous families. A Lie group is designed to capture precisely this compatibility between group operations and smooth geometry.

[definition: Lie Group] A Lie group is a smooth manifold $G$ equipped with a group structure such that the multiplication map $m:G\times G\to G$, $m(g,h)=gh$, and the inversion map $i:G\to G$, $i(g)=g^{-1}$, are smooth. [/definition]

This definition is the abstract endpoint of the course's first movement. The next problem is how to recognise Lie groups inside familiar linear algebra without first constructing abstract atlases, so we initially work with subgroups of matrix groups.

## Matrix Groups as First Examples

We write $M(n,\mathbb C)$ for the [vector space](/page/Vector%20Space) of complex $n\times n$ matrices and $GL(n,\mathbb C)$ for the group of invertible complex $n\times n$ matrices. Real matrix groups are viewed inside this complex ambient group when convenient.

Matrix Lie groups provide the concrete class of Lie groups studied in the first chapters: they are matrix subgroups whose topology is controlled well enough to support smooth structure. The closedness requirement is the key extra condition, because a subgroup of invertible matrices can otherwise be algebraically valid while behaving topologically like a dense curve rather than like a manifold of the expected dimension.

[definition: Matrix Lie Group] A matrix Lie group is a subgroup $G\le GL(n,\mathbb C)$ that is closed as a subset of $GL(n,\mathbb C)$ in the topology inherited from $M(n,\mathbb C)$. [/definition]

The closedness hypothesis is not cosmetic: without it, a subgroup can wind densely through a torus and fail to have the dimension suggested by its parametrisation. The next example records the basic pathology that the closedness condition excludes.

Here $S^1=\{z\in\mathbb C:|z|=1\}$ is the unit circle, $U(1)$ is the same group viewed as unitary $1\times1$ matrices, and $T^2=S^1\times S^1$ is the two-torus.

[example: Irrational Winding In A Torus] [claim]The subgroup $H=\{(e^{it},e^{i\alpha t}):t\in\mathbb R\}$ is dense in $T^2$ and is not closed.[/claim] [proof]Define $\phi:\mathbb R\to T^2$ by $\phi(t)=(e^{it},e^{i\alpha t})$. For $s,t\in\mathbb R$, \begin{align*} \phi(s+t)=(e^{i(s+t)},e^{i\alpha(s+t)})=(e^{is}e^{it},e^{i\alpha s}e^{i\alpha t})=\phi(s)\phi(t). \end{align*} Thus $H=\phi(\mathbb R)$ is a subgroup of $T^2$. To prove density, fix a target point $(e^{ia},e^{ib})\in T^2$. Points of $H$ with first coordinate exactly $e^{ia}$ are obtained by taking $t=a+2\pi n$ with $n\in\mathbb Z$, because \begin{align*} e^{i(a+2\pi n)}=e^{ia}e^{2\pi i n}=e^{ia}. \end{align*} For such $t$, the second coordinate is \begin{align*} e^{i\alpha(a+2\pi n)}=e^{i\alpha a}e^{2\pi i\alpha n}. \end{align*} So it remains to know that the set $\{e^{2\pi i\alpha n}:n\in\mathbb Z\}$ is dense in $S^1$. We verify this density by the pigeonhole principle. Since $\alpha$ is irrational, the fractional parts of $0,\alpha,2\alpha,\ldots,N\alpha$ are distinct. Divide $[0,1)$ into $N$ intervals of length $1/N$. Two of these $N+1$ fractional parts lie in the same interval, so for some integer $q$ with $1\le q\le N$ there is an integer $p$ such that \begin{align*} 0<|q\alpha-p|<1/N. \end{align*} Write $\varepsilon=q\alpha-p$. Then $e^{2\pi i q\alpha}=e^{2\pi i\varepsilon}$, and the step size $|\varepsilon|$ can be made arbitrarily small by choosing $N$ large. The multiples $0,\varepsilon,2\varepsilon,\ldots,m\varepsilon$ modulo $1$ then pass within $|\varepsilon|$ of every point of the circle before wrapping around, so the values $e^{2\pi i kq\alpha}$ are arbitrarily close to any prescribed point of $S^1$. Hence $\{e^{2\pi i\alpha n}:n\in\mathbb Z\}$ is dense in $S^1$. Therefore, for every $(e^{ia},e^{ib})\in T^2$ and every neighbourhood of it, one can choose $n$ so that \begin{align*} (e^{i(a+2\pi n)},e^{i\alpha(a+2\pi n)})=(e^{ia},e^{i\alpha a}e^{2\pi i\alpha n}) \end{align*} lies in that neighbourhood. Thus $H$ is dense in $T^2$. Since $H\ne T^2$, for example no $t$ satisfies both $e^{it}=1$ and $e^{i\alpha t}=e^{i\pi}$ because the first equation gives $t=2\pi m$ and then the second gives $e^{i\alpha t}=e^{2\pi i\alpha m}\ne -1$ when $\alpha$ is irrational, the dense subgroup $H$ is not closed.[/proof] This is the basic obstruction excluded by the closedness condition in the definition of a matrix Lie group: a subgroup can come from a smooth one-parameter parametrisation and still have the wrong [subspace topology](/page/Subspace%20Topology) to be an embedded Lie subgroup. [/example]

## Matrices Before Manifolds

The next problem is how to build enough examples before the full machinery of manifolds is available. Matrix groups are ideal because they sit inside $M(n,\mathbb F)$, where $\mathbb F$ is either $\mathbb R$ or $\mathbb C$, and smoothness can be checked by ordinary calculus on coordinate entries.

[quotetheorem:8769]