# Lie groups I: Definition and examples

## 1 Introduction

This is part I of a series of articles about Lie groups. We define Lie groups and homomorphisms and give the most important examples. In following pages of this series we define the fundamental invariants like the Dynkin diagram, and report about the classification of compact Lie groups. We also report about exceptional Lie groups.

## 2 Definition and examples

Definition 2.1. A Lie group is a finite dimensional smooth manifold $G$${{Stub}} == Introduction == ; This is part I of a series of articles about Lie groups. We define Lie groups and homomorphisms and give the most important examples. In following pages of this series we define the fundamental invariants like the Dynkin diagram, and report about the classification of compact Lie groups. We also report about exceptional Lie groups. == Definition and examples == ; {{beginthm|Definition}} A Lie group is a finite dimensional smooth manifold G together with a group structure on G, such that the multiplication G \times G \to G and the attaching of an inverse g \mapsto g^{-1}: G \to G are smooth maps. A morphism between two Lie groups G and H is a map f:G \to H, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map f such that f and f^{-1} are morphisms. A Lie subgroup is a subgroup H in G such that H is also a smooth submanifold of G.{{endthm}} It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group. The simplest examples of Lie groups are countable groups, which with the discrete topology are a together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps.

A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$.

It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group.

The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$:

$\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$
are Lie groups.

They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups.

All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is
$\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$
and
$\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$

By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group:

$\displaystyle SO(n); SU(n).$

The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying:

$\displaystyle \dim SU(n) = n^2-1.$
The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$
$\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$

With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension

$\displaystyle \dim Sp(n) = n(2n+1).$
The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$.

As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???].

Some of the low dimensional Lie groups above occur in two or more ways, for example
$\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$

A special role in the world of Lie groups play the tori, the Lie groups

$\displaystyle T^n := (S^1)^n.$
As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus.

Here are a few fundamental examples of non-compact Lie groups:

1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$.

2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$.

The following theorems give a rough picture of all Lie groups:

Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???].

Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$.

Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???].

Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup.

## Further discussion

Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$.

Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones:

Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$.

$-dimensional Lie group. In particular all finite groups are $G$ together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps. A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$. It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group. The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$: $\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$ are Lie groups. They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups. All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is $\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ and $\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$ By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group: $\displaystyle SO(n); SU(n).$ The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying: $\displaystyle \dim SU(n) = n^2-1.$ The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$ $\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$ With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension $\displaystyle \dim Sp(n) = n(2n+1).$ The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$. As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???]. Some of the low dimensional Lie groups above occur in two or more ways, for example $\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$ A special role in the world of Lie groups play the tori, the Lie groups $\displaystyle T^n := (S^1)^n.$ As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus. Here are a few fundamental examples of non-compact Lie groups: 1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$. 2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$. The following theorems give a rough picture of all Lie groups: Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???]. Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$. Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???]. Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup. ## Further discussion Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$. Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones: Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$. ## 3 References$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$, $\mathbb C$ or the quternions $\mathbb H$: $$GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$$ are Lie groups. They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$, n^2$and n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups. All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$, the '''orthogonal group''', the hermitian matrices $U(n)$, the '''unitary group''', and the symplectic matrices $Sp(n)$, the '''symplectic group'''. The way to see this is to consider the map $A \mapsto A^t$ from $GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$ is a regular value. Thus the preimage of $E$, which is $O(n)$ is a smooth submanifold and so $O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$ from $GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$ and shows again that $E$ is a regular value. The dimension of $O(n)$ is $$\dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$$ and $$\dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$$ By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant G together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps.

A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$.

It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group.

The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$:

$\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$
are Lie groups.

They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups.

All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is
$\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$
and
$\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$

By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group:

$\displaystyle SO(n); SU(n).$

The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying:

$\displaystyle \dim SU(n) = n^2-1.$
The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$
$\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$

With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension

$\displaystyle \dim Sp(n) = n(2n+1).$
The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$.

As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???].

Some of the low dimensional Lie groups above occur in two or more ways, for example
$\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$

A special role in the world of Lie groups play the tori, the Lie groups

$\displaystyle T^n := (S^1)^n.$
As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus.

Here are a few fundamental examples of non-compact Lie groups:

1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$.

2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$.

The following theorems give a rough picture of all Lie groups:

Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???].

Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$.

Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???].

Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup.

## Further discussion

Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$.

Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones:

Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$.

$, the '''special orthogonal group''' or '''special unitary group''': $$SO(n); SU(n).$$ The first is just the component of$E$in$O(n)$, whereas the second is the preimage of the regular value G together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps. A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$. It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group. The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$: $\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$ are Lie groups. They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups. All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is $\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ and $\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$ By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group: $\displaystyle SO(n); SU(n).$ The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying: $\displaystyle \dim SU(n) = n^2-1.$ The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$ $\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$ With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension $\displaystyle \dim Sp(n) = n(2n+1).$ The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$. As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???]. Some of the low dimensional Lie groups above occur in two or more ways, for example $\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$ A special role in the world of Lie groups play the tori, the Lie groups $\displaystyle T^n := (S^1)^n.$ As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus. Here are a few fundamental examples of non-compact Lie groups: 1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$. 2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$. The following theorems give a rough picture of all Lie groups: Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???]. Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$. Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???]. Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup. ## Further discussion Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$. Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones: Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$. ## 3 References$ in $S^1$ and so has codimension G together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps.

A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$.

It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group.

The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$:

$\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$
are Lie groups.

They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups.

All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is
$\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$
and
$\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$

By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group:

$\displaystyle SO(n); SU(n).$

The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying:

$\displaystyle \dim SU(n) = n^2-1.$
The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$
$\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$

With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension

$\displaystyle \dim Sp(n) = n(2n+1).$
The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$.

As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???].

Some of the low dimensional Lie groups above occur in two or more ways, for example
$\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$

A special role in the world of Lie groups play the tori, the Lie groups

$\displaystyle T^n := (S^1)^n.$
As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus.

Here are a few fundamental examples of non-compact Lie groups:

1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$.

2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$.

The following theorems give a rough picture of all Lie groups:

Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???].

Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$.

Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???].

Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup.

## Further discussion

Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$.

Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones:

Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$.

$, implying: $$\dim SU(n) = n^2-1.$$ The symplectic group$Sp(n)$is not the group of isometries of a non-degenerate skew-symmetric$\mathbb H$-bilinear form, but the group of the hermitian form on$\mathbb H^n$$$:= \bar x_1y_1 + \dots + \bar x_ny_n.$$ With this definition one proceeds as in the examples above and show that$Sp(n)$is a Lie group of dimension $$\dim Sp(n) = n(2n+1).$$ The determinant of an element of$Sp(n)$is G together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps. A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$. It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group. The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$: $\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$ are Lie groups. They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups. All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is $\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ and $\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$ By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group: $\displaystyle SO(n); SU(n).$ The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying: $\displaystyle \dim SU(n) = n^2-1.$ The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$ $\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$ With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension $\displaystyle \dim Sp(n) = n(2n+1).$ The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$. As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???]. Some of the low dimensional Lie groups above occur in two or more ways, for example $\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$ A special role in the world of Lie groups play the tori, the Lie groups $\displaystyle T^n := (S^1)^n.$ As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus. Here are a few fundamental examples of non-compact Lie groups: 1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$. 2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$. The following theorems give a rough picture of all Lie groups: Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???]. Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$. Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???]. Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup. ## Further discussion Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$. Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones: Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$. ## 3 References$, thus we don't consider the groups $SSp(n)$. As mentioned above $SO(n)$ is the component of $E$ in $O(n)$. For $n>2$ it's fundamental group is $\mathbb Z/2$ \cite{???} generated by the inclusion $S^1 \cong SO(2) \to SO(n)$. If $G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$ for $n>2$ is a Lie group denoted by $Spin(n)$, the '''Spinor group'''. There is a more explicit construction of $Spin(n)$ in terms of Clifford algebras \cite{???}. Some of the low dimensional Lie groups above occur in two or more ways, for example $$SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$$ A special role in the world of Lie groups play the '''tori''', the Lie groups $$T^n := (S^1)^n.$$ As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus. Here are a few fundamental examples of non-compact Lie groups: 1.) The ''Lorentz group'' $O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$ given by $:= x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $. 2.) The Heisenberg Group$H $consisting of upper \times 3$ matrices with diagonal entries G together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps.

A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$.

It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group.

The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$:

$\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$
are Lie groups.

They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups.

All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is
$\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$
and
$\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$

By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group:

$\displaystyle SO(n); SU(n).$

The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying:

$\displaystyle \dim SU(n) = n^2-1.$
The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$
$\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$

With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension

$\displaystyle \dim Sp(n) = n(2n+1).$
The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$.

As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???].

Some of the low dimensional Lie groups above occur in two or more ways, for example
$\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$

A special role in the world of Lie groups play the tori, the Lie groups

$\displaystyle T^n := (S^1)^n.$
As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus.

Here are a few fundamental examples of non-compact Lie groups:

1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$.

2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$.

The following theorems give a rough picture of all Lie groups:

Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???].

Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$.

Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???].

Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup.

## Further discussion

Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$.

Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones:

Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$.

## 3 References

$. It's dimension is$. The following theorems give a rough picture of all Lie groups: {{beginthm|Theorem}} Assume that the Lie group $G$ is semisimple. Then $G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$ for some $n$ \cite{???}. {{endthm}} Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$ be the [[Wikipedia:Adjoint_representation_of_a_Lie_group|Adjoint representation]] of $G$. If $G$ is semisimple, then the adjoint representation is faithful, thus $G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$. {{beginthm|Theorem}} A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace \cite{???}. {{endthm}} Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup. == Further discussion == ; Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying '''symmetries'''. Given a geometric object, say a closed smooth manifold $M$ with a Riemannian metric $g$, then one can consider the group of self isometries $Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$ mapping $(f,x)$ to $f(x)$ is a smooth map \cite{Kobayashi transformation groups ...}. Thus we have a smooth action of $Iso(M,g)$ on $M$. The size of this group is a measure for the symmetry of $(M,g)$. In turn if a compact Lie group $G$ acts smoothly on a closed smooth manifold $M$, then there is a Riemannian metric on $M$ such that $G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$ using a Haar measure). If the action is effective (meaning that if $g$ acts trivially, the $g =1$), then $G$ is a subgroup of $Iso(M,g)$. Motivated by these considerations one defines an invariant for closed smooth manifolds$M$, the '''degree of symmetry''' which is the largest dimension of a compact Lie group acting effectively on $M$ (or equivalently the largest dimension of $Iso(M,g)$ as $g$ varies over all Riemannian metrics of $M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones: {{beginthm|Theorem \cite{Frobenius-Birkhoff}}} The degree of symmetry of a clsoed manifold $M$ of dimension $m$ is $\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$, which acts on $S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$ then $M$ is diffeomorphic to $S^m$ or $\mathbb {RP}^m$. {{endthm}} == References == {{#RefList:}} [[Category:Manifolds]]G together with a group structure on $G$$G$, such that the multiplication $G \times G \to G$$G \times G \to G$ and the attaching of an inverse $g \mapsto g^{-1}: G \to G$$g \mapsto g^{-1}: G \to G$ are smooth maps.

A morphism between two Lie groups $G$$G$ and $H$$H$ is a map $f:G \to H$$f:G \to H$, which at the same time is smooth and a group homomorphism. An isomorphism is a bijective map $f$$f$ such that $f$$f$ and $f^{-1}$$f^{-1}$ are morphisms. A Lie subgroup is a subgroup $H$$H$ in $G$$G$ such that $H$$H$ is also a smooth submanifold of $G$$G$.

It is enough to require that the multiplication is smooth, the smoothness of the inverse map can be derived from this. Obviously the product of two Lie groups or a finite sequence of Lie groups is a Lie group.

The simplest examples of Lie groups are countable groups, which with the discrete topology are a $0$$0$-dimensional Lie group. In particular all finite groups are $0$$0$-dimensional Lie groups. The most basic Lie groups of positive dimension are matrix groups. The general linear groups over $R$$R$, $\mathbb C$$\mathbb C$ or the quternions $\mathbb H$$\mathbb H$:

$\displaystyle GL(n;\mathbb R) ; \,\, GL(n;\mathbb C);GL(n;\mathbb H),$
are Lie groups.

They are smooth manifolds as open subsets of the vector space of all corresponding matrices. Their dimension is $n^2$$n^2$, $2n^2$$2n^2$ and $4n^2$$4n^2$ resp. (note that Lie groups are real manifolds, this explains the formula for the dimension). $GL(n;\mathbb C)$$GL(n;\mathbb C)$ is also a complex manifold and one obtains a complex Lie group but we will here only consider real Lie groups.

All these groups are non-compact (for positive dimensions). They contain compact Lie subgroups given by the orthogonal matrices $O(n)$$O(n)$, the orthogonal group, the hermitian matrices $U(n)$$U(n)$, the unitary group, and the symplectic matrices $Sp(n)$$Sp(n)$, the symplectic group. The way to see this is to consider the map $A \mapsto A^t$$A \mapsto A^t$ from $GL(n;R)$$GL(n;R)$ to the vector space of symmetric matrices and to show that the unit matrix $E$$E$ is a regular value. Thus the preimage of $E$$E$, which is $O(n)$$O(n)$ is a smooth submanifold and so $O(n)$$O(n)$ is a Lie submanifold of $GL(n;\mathbb R)$$GL(n;\mathbb R)$. Similarly one considers the map $A \mapsto \bar A^t$$A \mapsto \bar A^t$ from $GL(n;\mathbb C)$$GL(n;\mathbb C)$ to the skew symmetric matrices over $\mathbb C$$\mathbb C$ and shows again that $E$$E$ is a regular value. The dimension of $O(n)$$O(n)$ is
$\displaystyle \dim O(n) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$
and
$\displaystyle \dim U(n) =2n^2-( n + \frac{2n(n-1)}{2}) = n^2.$

By definition these subgroups are closed subgroups. Since they are bounded subspaces of the vector space of all matrices they are compact. They contain the Lie subgroups of all matrices with determinant $1$$1$, the special orthogonal group or special unitary group:

$\displaystyle SO(n); SU(n).$

The first is just the component of $E$$E$ in $O(n)$$O(n)$, whereas the second is the preimage of the regular value $1$$1$ in $S^1$$S^1$ and so has codimension $1$$1$, implying:

$\displaystyle \dim SU(n) = n^2-1.$
The symplectic group $Sp(n)$$Sp(n)$ is not the group of isometries of a non-degenerate skew-symmetric $\mathbb H$$\mathbb H$-bilinear form, but the group of the hermitian form on $\mathbb H^n$$\mathbb H^n$
$\displaystyle := \bar x_1y_1 + \dots + \bar x_ny_n.$

With this definition one proceeds as in the examples above and show that $Sp(n)$$Sp(n)$ is a Lie group of dimension

$\displaystyle \dim Sp(n) = n(2n+1).$
The determinant of an element of $Sp(n)$$Sp(n)$ is $1$$1$, thus we don't consider the groups $SSp(n)$$SSp(n)$.

As mentioned above $SO(n)$$SO(n)$ is the component of $E$$E$ in $O(n)$$O(n)$. For $n>2$$n>2$ it's fundamental group is $\mathbb Z/2$$\mathbb Z/2$ [???] generated by the inclusion $S^1 \cong SO(2) \to SO(n)$$S^1 \cong SO(2) \to SO(n)$. If $G$$G$ is a Lie group then by construction of the universal covering this is a Lie group again. In particular the universal covering of $SO(n)$$SO(n)$ for $n>2$$n>2$ is a Lie group denoted by $Spin(n)$$Spin(n)$, the Spinor group. There is a more explicit construction of $Spin(n)$$Spin(n)$ in terms of Clifford algebras [???].

Some of the low dimensional Lie groups above occur in two or more ways, for example
$\displaystyle SO(2) \cong U(1) = S^1, \,\, SU(2) \cong Spin(3) \cong Sp(1) \ = S^3.$

A special role in the world of Lie groups play the tori, the Lie groups

$\displaystyle T^n := (S^1)^n.$
As we will discuss in a later page, a connected abelian Lie group is isomorphic to a torus.

Here are a few fundamental examples of non-compact Lie groups:

1.) The Lorentz group $O(1,3)$$O(1,3)$ the group of the isometries of the Minkowski space, the isometries of the form on $\mathbb R^4$$\mathbb R^4$ given by $ := x_1y_1 - x_2y_2 - \dots - x_4y_4$$ := x_1y_1 - x_2y_2 - \dots - x_4y_4$. It's dimension is $6$$6$.

2.) The Heisenberg Group $H$$H$ consisting of upper $3 \times 3$$3 \times 3$ matrices with diagonal entries $1$$1$. It's dimension is $3$$3$.

The following theorems give a rough picture of all Lie groups:

Theorem 2.2. Assume that the Lie group $G$$G$ is semisimple. Then $G$$G$ is isomorphic to a Lie subgroup of $GL(n;\mathbb R)$$GL(n;\mathbb R)$ for some $n$$n$ [???].

Proof: Let $Ad:G\rightarrow GL\left(\underline{g}\right)$$Ad:G\rightarrow GL\left(\underline{g}\right)$ be the Adjoint representation of $G$$G$. If $G$$G$ is semisimple, then the adjoint representation is faithful, thus $G$$G$ is mapped isomorphically to a subgroup of $GL\left(\underline{g}\right)$$GL\left(\underline{g}\right)$.

Theorem 2.3. A subgroup of a Lie group is a Lie subgroup, if and only if it is closed as a topological subspace [???].

Thus it's easy to test, whether a subgroup of a Lie group is a Lie subgroup.

## Further discussion

Why are Lie groups interesting? There are many different answers to this question. Probably everybody will agree that Lie groups, in particular compact Lie groups, give the mathematical language for defining and studying symmetries. Given a geometric object, say a closed smooth manifold $M$$M$ with a Riemannian metric $g$$g$, then one can consider the group of self isometries $Iso(M,g)$$Iso(M,g)$. By a theorem of Myers and Steenrod \cite {???} this is a compact Lie group in such a way that the map $Iso(M,g) \times M \to M$$Iso(M,g) \times M \to M$ mapping $(f,x)$$(f,x)$ to $f(x)$$f(x)$ is a smooth map [Kobayashi transformation groups ...]. Thus we have a smooth action of $Iso(M,g)$$Iso(M,g)$ on $M$$M$. The size of this group is a measure for the symmetry of $(M,g)$$(M,g)$. In turn if a compact Lie group $G$$G$ acts smoothly on a closed smooth manifold $M$$M$, then there is a Riemannian metric on $M$$M$ such that $G$$G$ acts by isometries (choose an arbitrary Riemannian metric and average it over $G$$G$ using a Haar measure). If the action is effective (meaning that if $g$$g$ acts trivially, the $g =1$$g =1$), then $G$$G$ is a subgroup of $Iso(M,g)$$Iso(M,g)$.

Motivated by these considerations one defines an invariant for closed smooth manifolds$M$$M$, the degree of symmetry which is the largest dimension of a compact Lie group acting effectively on $M$$M$ (or equivalently the largest dimension of $Iso(M,g)$$Iso(M,g)$ as $g$$g$ varies over all Riemannian metrics of $M$$M$). The following result distinguishes the spheres and real projective spaces from all other manifolds as the most symmetric ones:

Theorem [Frobenius-Birkhoff] 4.1. The degree of symmetry of a clsoed manifold $M$$M$ of dimension $m$$m$ is $\le \frac{m(m+1)}{2}$$\le \frac{m(m+1)}{2}$ (the dimension of $O(m+1)$$O(m+1)$, which acts on $S^m$$S^m$), and if the degree of symmetry is $\frac{m(m+1)}{2}$$\frac{m(m+1)}{2}$ then $M$$M$ is diffeomorphic to $S^m$$S^m$ or $\mathbb {RP}^m$$\mathbb {RP}^m$.