Lie Groups Revisited
Disclaimer: Nothing in this note is original.
Let’s start with an example. Let $M(n,\mathbb{R})$ be the set of all $n\times n$ real matrices, it has $n\times n$ elements thus is topologically isomorphic to a $n\times n$ dimensional real space $\mathbb{R}^{n\times n}$. The real general linear group $\text{GL}(n,\mathbb{R})$ is the group of all invertible (with non-zero determinant) $n\times n$ real matrices. We can regard such matrices as a point $x$ in an abstract “space of matrices”, with the coordinates given by all its elements $x_ {ij}$. Obviously $\text{GL}(n,\mathbb{R})$ is a sub-space in $\mathbb{R}^{n\times n}$. The question is, what exactly is this subspace? Is it a codimension one open set or something else? To answer this question, notice that the determinant is a map from $M(n,\mathbb{R})$ to $\mathbb{R}$ which is continuous. The image of $GL(n,\mathbb{R})$ under the map is $\mathbb{R} - 0$ and is open in regular topology, thus the preimage of determinant function is also open, namely $GL(n,\mathbb{R})$ is a open subspace of $M(n,\mathbb{R})$. Product and inverse are also continuous functions defined on $GL(n,\mathbb{R})$, thus when we consider $GL(n,\mathbb{R})$ as a manifold and consider the matrix points, product and inverse as smooth functions on this manifold. This brings as to the definition of a Lie group:
Definition. A Lie group$G$ is a smooth manifold endowed with a product and an inverse. Product is a continuous map
making $G$ into a group. We demand that this map, as well as the “inversion map” be differentiable. Particularly Lie group is a topological group, meaning that it is a topological space with group structures such that the multiplication and inversion maps are continuous.
For example, $\mathbb{R}$ is a Lie group under addition. This group is commutative, or abelian. The positive real numbers $\mathbb{R}^{+}$ also form a abelian group under multiplication. The real and complex general linear matrices $GL(n,\mathbb{R})$ and $GL(n,\mathbb{C})$ are also Lie groups under the usual matrix multiplication. The special linear group, $SL(n,\mathbb{R})$ is the subset of $GL(n,\mathbb{R})$ that have determinant one. For any matrix group, the adjective special means that the determinant is equal to one. Other examples include $\mathbb{T}^{n}$ the $n$-torus group, which is the abelian group of diagonal matrices of the form
The $\mathbb{T}^{n}$ group is topologically homeomorphic to $\mathbb{S}^{1}\times\dots \times \mathbb{S}^{1}$, $n$ copies of them. Since the circles are connected, it follows that $\mathbb{T}^{n}$ is also connected.
As a manifold, a Lie group is very special for the following reason. A Lie group always has two families of diffeomorphisms, the left and right translations. For $g\in G$, these translations are defined by
\[L_ {g}: G \to G, \quad h \mapsto gh, \text{multiply from the left}\]and similarly
\[R_ {g}: G \to G, \quad h \mapsto hg, \text{multiply from the right.}\]Theorem. $U(N)$ is connected.
To see it, just notice that any unitary matrix can be diagonalized to a $\mathbb{T}^{n}$ under some similar transformation, we casually write it as $\mathbb{T}^{n} = g U(N) g^{-1}$ for some $g\in U(N)$. Then $U(N)=g^{-1}\mathbb{T}^{n}g$ is connected to $\mathbb{1}$ smoothly by varying $\mathbb{T}^{n}$ smoothly from $\mathbb{1}$.
The subgroup $\mathbb{T}^{n}$ of $U (n)$ given by (1) is called a maximal torus of $U(n)$. Any conjugate $h \mathbb{T}^{n} h^{-1}$ of this maximal torus is also called a maximal torus. By the same type of reasoning we may deal with the rotation group.
Theorem. $O(n)$ consists of two connected components, and $SO(n)$ is the part connected with the identity.
As our last example, consider the set of linear transformation of points in 1-dimension affine space. The points in 1-dimensional affine space are denoted by $(t,1)^{T}$ where $t$ is the coordinate and $1$ denotes that it is a point. A vector in affine space would be denoted by $(v,0)^{T}$, thus the difference of two points yields a vector. The linear transformations, $G=\mathbb{A}^{1}$, the affine group of lines. They consist of real $2\times 2$ matrices of form
\[\begin{pmatrix} x & y \\ 0 & 1 \end{pmatrix}.\]with $x>0$.
A matrix group is a subgroup of $Gl(n)$ that is also a submanifold of $Gl(n)$.
Invariant Vector Fields and Forms
Lie groups are special as manifolds in the sense that, given a tangent vector $X_ {e}$ at the unity $e$, we may use the group elements to translate $X_ {e}$ to each point of $G$, by left or right group action. The left-translated tangent vector $X_ {g}$ is given by
\[X_ {g} := L_ {g\ast}\, X_ {e}\]and the right-translated vector (denoted by the same notation)
\[X_ {g}:= R_ {g\ast }\, X_ {e}.\]In this way we have two non-vanishing vector fields all over $G$.
In fact, given a set of basis $X_ {i}$ of the tangent space $T_ {e}G$ at $e$, we can translate them to any point $g\in G$ thus yielding a basis $L_ {g\ast}\, X_ {i}$ of $T_ {g}G$. The importance of this claim has to do with the orientation of $G$ as a manifold. Recall that the orientation is defined by the order of basis, if we can translate the basis $X_ {i}$ to any point of the group then the group is orientable! We conclude that every Lie group is a orientable manifold.
Consider for instance, a closed orientable surface $M^{2}$ of genus $g$. We shall see that of these surfaces only the torus (genus $1$) can support even a single nonvanishing tangent vector field. To be specific, $\mathbb{T}^{2}$ supports two sets of vector fields $\partial_ {\theta}$ and $\partial_ {\phi}$ where $\theta$ and $\phi$ parametrize the circles in $\mathbb{T}\cong \mathbb{S}^{1}\times\mathbb{S}^{1}$. The Torus can be regarded as the abelian Lie group $\mathbb{S}^{1}\times\mathbb{S}^{1}$ with multiplication
\[(\theta,\phi)\times (\theta',\phi'):=(\theta+\theta',\phi+\phi').\]Topologically, the only compact Lie group of dimension 2 is the torus.
A vector field $X$ on $G$ is said to be left-invariant or right-invariant if it is invariant under left or right translation,
and similarly for right invariant.
We can define similar concept for differential forms. A differential form field $\omega$ on $G$ is said to be left invariant if
Note the direction of the pull-back.
Consider the example of Lie group $G:=Gl(n)$ of invertible matrices. Let $t\to h(t)$ be a curve in $G$, namely a G-valued curve parametrized by one parameter $t$. Let $h(0)=h$ and $h’(0)=X_ {h}$, where $h’$ is the $t$-derivative of $h$. This curve is simply a matrix whose entries $h_ {mn}$ are functions of $t$, $h(t)$ describes a curve in $n^{2}$ dimensional space. Then $X_ {h}$ is simple the matrix whose entries are $t$-derivatives of $h$ at $t=0$, $h’_ {mn}(t=0)$. Of course just because $h(0)$ is in $G$ does not mean $h’(0)$ is also in $G$. Then for the constant matrix $g$, the curve $gh(t)$ given by matrix multiplication is the left translated curve of $h(t)$ by $L_ {g}$. The tangent vector of $gh(t)$ at $t=0$ is given by $gX_ {h}$,
\[L_ {g\ast }X_ {h} = gX_ {h}\]which is again the simple matrix multiplication.
Note that in any Lie group, if $X_ {i}$ is a basis for the left invariant vector fields and if $\sigma^{i}$ is the dual basis of 1-forms, then this dual basis is automatically left invariant. This can be shown using the relation $f^{\ast}\alpha(X)=\alpha(f_ {\ast}X)$. Also, note that if $\alpha$ and $\beta$ are invariant differential forms on $G$ then so are $d\alpha$ and $\alpha \wedge\beta$. A left invariant area form is also called a left Haar measure, which can be seen as a convenient way to use the left action $L_ {g\ast}$ to translate some well-defined infinitesimal areas about the entire group space to measure quantities defined on it.
One-Parameter Subgroups
Recall that a group homomorphism is a map between groups
\[f: G\to H\]that preserves the group multiplication. If it is further one-to-one (injective) and on-to (surjective) then it is an isomorphism. The exponential function is an example of group homomorphism, since $e^{ a }e^{ b }=e^{ a+b }$ for $a,b\in\mathbb{R}$ we have
\[\exp: (\mathbb{R},\times )\to (\mathbb{R}^{+},+).\]A 1-parameter subgroup of $G$ is by definition a differentiable homomorphism (a path)
of the additive group of the reals into the group $G$. Thus
\[g(s+t) = g(s)g(t) = g(t)g(s),\quad \text{(abelian)}.\]Next let’s look at the most important example, the one parameter subgroup of matrix group $G$. As matrices
\[g(t+s)=g(t)g(s),\quad \text{matrix multiplication.}\]To find $g$ that satisfy the condition, let’s take derivatives on both sides w.r.t. $s$ then put $s$ to zero, \(g'(t) = g(t)g'(0),\quad g'(0)=\text{const matrix.}\)
The solution is of exponential form,
\[g(t) =g(0) \exp \left\{ t g'(0) \right\} .\]Since $g(0)=e$ for any homomorphism, since homomorphism has to map identity to identity, we conclude that
\[g(t) = \exp \left\{ tg'(0) \right\} ,\quad g'(0)\in T_ {e}(G).\]This is also the most general one-parameter subgroup of matrix group.
The group multiplication property Eq. (2) tells us the how to proceed for arbitrary group. Regard $g’(0)$ as a tangent vector at $g(0)$ and regard the left multiplication by $g(t)$ as a map, Eq. (2) can be generalized to
\[g'(t) = L_ {g(t)\ast }\, g'(0), \quad L_ {g(t)\ast } \text{ is the induced map of }L_ {g(t)}.\]It says that, the tangent vector $X$ of the one-parameter subgroup (a curve, if you like) is left translated along the subgroup (curve). Thus, given a tangent vector $X_ {\mathbb{1}}$ at the identity $\mathbb{1}$ in $G$,
the 1-parameter subgroup of $G$ whose tangent at $\mathbb{1}$ is $X_ {\mathbb{1}}$ is the integral curve through $\mathbb{1}$ of the vector field $X$ on $G$ resulting from left translation of $X_ {\mathbb{1}}$ over all of $G$.
The vector $X_ {\mathbb{1}}$ is called the infinitesimal generator of the 1-parameter subgroup.
For any Lie group $G$ we shall denote the 1-parameter subgroup whose generator at $\mathbb{1}$ is $X$ by
\[g(t):= e^{ tX } \equiv \exp(tX).\]Example. The matrix
\[X= \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}\]can be considered a tangent vector at $\mathbb{1}$ of group $G=GL(2,\mathbb{R})$. The 1-parameter subgroup of $G$ is
\[\exp(tX) = \begin{bmatrix} \cos t & -\sin t \\ \sin t & \cos t \end{bmatrix}.\]Note that $X$ is a vector and $\exp X$ is a group element, different things fundamentally.
The Lie Algebra of a Lie Group
Let $G$ be a Lie group regarded as a manifold (with group structure), the tangent vector space at the identity $\mathbb{1}$ of $G$ plays an important role and is denoted by Gothic letter ${\frak g}$. Sometimes it is denoted by script letters, which we will not do here.
Definition. The Lie group ${\frak g}$ of group $G$ is
\[{\frak g}:= T_ {\mathbb{1}}\,G.\]The fancy notation might appears to be pretentious and uncomfortable at first, but once you get used to it, you’ll find them clear and precise. You’ll never mistake a point in the group space with a vector at the origin.
Let $X_ {i}$ be a basis for ${\frak g}$, we will also use $X_ {i}$ to denote the left translation of this vector to all of $G$. Since any left invariant vector field is determined by its value at $\mathbb{1}$, the most general left invariant vector field is then of the form
\[X = \sum_ {i} v^{i}X_ {i},\quad v^{i} \text{ are constants.}\]Let $\sigma^{i}$ be the dual basis of left invariant 1-forms on $G$. They are determined by their values on vectors from the Lie algebra. The most general left invariant r-form on $G$ is of the form
\[\alpha^{r} = \sum_ {I_ {<}}\alpha_ {I} \, \sigma^{I}, \quad \sigma^{I} = \sigma^{I_ {1}}\wedge \sigma^{I_ {2}}\wedge \dots\]It is again determined by its value on a $r$-tuple of vectors from ${\frak g}$. The components $\alpha_ {I}$ are again constants.
It shouldn’t come as a surprise that,
Theorem. The Lie bracket $[X,Y]$ of two left invariant vector fields is again left invariant.
To prove it, notice that $X$ is left invariant iff $\left\langle X,\sigma \right\rangle=\text{const}$ for some left invariant 1-form $\sigma$. For such a $\sigma$, since $X,Y$ are left invariant we have $\left\langle X,\sigma \right\rangle=\text{const}$ and $\left\langle Y,\sigma \right\rangle=\text{const}$. So the question is, is $\left\langle [X,Y],\sigma \right\rangle=\text{const}$? Then since for general $\omega \in\Omega^{1}$
\(d\omega(X,Y)=-\omega([X,Y])+X(\omega(Y))-Y(\omega(X))\) we have
\[\left\langle [X,Y],\sigma \right\rangle =\sigma([X,Y]) =- d\sigma(X,Y)\]where, if $\sigma$ is left invariant then so is $d\sigma$ since $L_ {g}^{\ast}d\sigma \Large\mid_ {g} =dL_ {g^{\ast}}\sigma \Large\mid_ {g} =d \sigma \Large\mid_ {\mathbb{1}}$. Thus
\[\sigma([X,Y])=\text{const}.\]Q.E.D.
We may then write
\[[X_ {i},X_ {j}]=if_ {ij}^{k}X_ {k},\quad f_ {ij}^{k}=-f_ {ji}^{k}.\]$f_ {ij}^{k}$ is called the structure constants.
The structure constant also can be applied to the left invariant 1-forms $\sigma$.
Theorem. The Maurer-Cartan equations
hold, and $d^{2}\sigma^{i}=0$ yields the Jacobi identity \(f^{m}_ {na}f^{n}_ {bc}+f^{m}_ {nb}f^{n}_ {ca}+f^{m}_ {nc}f^{n}_ {ab}=0.\)
This Jacobi identity for left invariant I -forms is also a consequence of a general Jacobi identity for vector fields on any manifold $M^{n}$, which reads
\[[X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0.\]We now make the vector space $T_ {\mathbb{1}}G={\frak g}$ into a Lie algebra by defining a product
as follows.
Let $X,Y\in{\frak g}$, extend them to be vector fields $X’,Y’$ on all of $G$. Define the product of $X$ and $Y$ to be the Lie bracket
\[[X,Y]:= [X',Y']_ {\mathbb{1}}.\]Note that the Lie algebra product is not associative! Take $SO(3)$ for example, there exists three basis vectors $X,Y,Z$ satisfying $[X,Y]=Z$, $[Y,Z]=X$ and $[X,Z]=-Y$, then $[X,[X,Y]]=-Y\neq[[X,X],Y]=0$.
We shall consistently identify the Lie algebra ${\frak g}$ with the $N(= \text{dim} G)$ dimensional vector space of left invariant fields on $G$.
Classically the Lie algebra was known as the “infinitesimal group” of $G$, for classically a vector was thought of roughly as going from a point to an “infinitesimally nearby” point. Then the Lie algebra consisted of group elements infinitesimally near the identity! We shall not use this picture.
The Exponential Map
Theorem. For any matrix $A$,
\[\det \exp A = \exp \mathrm{Tr}\,A,\quad \text{or equivalently, } \det e^{ A }=e^{ \mathrm{Tr}\,A }.\]Theorem. The map
\[\exp: {\frak g}\to G,\quad A \mapsto e^{ A }\]is a diffeomorphism of some neighborhood of $0\in{\frak g}$ onto a neighborhood of $\mathbb{1}\in G$.
In a general Lie group, the 1-parameter subgroup $\text{exp}(t X)$ is the integral curve of a vector field on $G$, and thus it would seem that this need only be defined for $t$ small. In this case of a left invariant vector field on a group, it can be shown that the curve exists for all $t$, just as it does in the matrix case.
Examples of Lie Algebras
Consider our old friend, the matrix group $GL(n)$. Let $M(n\times n)$ be the vector space of all real $n\times n$ matrices, the dimension is clearly $n^{2}$. For $A\in M(n\times n)$, we have
\[\det e^{ A }=e^{ \mathrm{Tr}\,A }\geq 0.\]Therefore the matrix $e^{ A }$ is invertible, the exponential map maps any $n\times n$ matrix to an invertible matrix,
\[\exp: M(n\times n)\to GL(n).\]We usually use the same name of the group but all in lowercase Gothic font to denote the corresponding Lie algebra, for example the Lie algebra of $GL(n)$ group is denoted ${\frak gl}(n)$. Then
\[{\frak gl}(n)= M(n\times n).\]If $G$ is a matrix group, that is, a subgroup of $GL(n)$, including $SO(N)$, $SU(N)$, etc., then its Lie algebra ${\frak g}$ is the largest subspace of $M(n\times n)$ such that $\exp {\frak g}\to G$.
For special orthogonal matrices $SO(n)$, we simply state that its Lie algebra ${\frak so}(n)$ consists of anti-symmetric matrices.
The group of unitary matrices $U(n)$, on the other hand, has Lie algebra ${\frak u}(n)$ consists of anti-hermitian matrices, $A^{\dagger}=-A$. As a special case, the special unitary groups $SU(n)$ has Lie algebra ${\frak su}(n)$ made of anti-hermitian matrices with zero trace, since $\exp \mathrm{Tr}\,=\det$.
Do the 1-Parameter Subgroups Cover $G$?
The question is, can every $g\in G$ be generated by an $A\in {\frak g}$? In other words,
\[\text{is the map } \exp: {\frak g}\to G \text{ onto?}\]It can be shown that, $\exp$ is indeed onto if the group as a manifold is connected and compact. It is clear that a 1-parameter subgroup must lie in the connected piece of $G$ that contains the identity. The matrix group $SL(2,\mathbb{R})$ is not compact since the only constrain for an element
\[g=\begin{bmatrix} x & y \\ z & w \end{bmatrix}\]is that
\[\det g = xw-yz=1\]still $x$ can be arbitrary large even with this condition.
However, it can be shown that $SL(2,\mathbb{R})$ is connected. To see it, it is helpful to adopt a geometric viewpoint. $SL(2,\mathbb{R})$ can be pictured as a pair of column vectors $(x z)^{T}$ and $(y w)^{T}$ in $\mathbb{R}^{2}$ spanning a parallelogram of area $1$. Deform he lengths of both so that the first becomes a unit vector, keeping the area $1$. This deforms $SL(2,\mathbb{R})$ into itself. Next, “Gram-Schmidt” the second so that the columns are orthonormal. This can be done continuously. The resulting matrix is then in the subgroup $SO(2,\mathbb{R})$ That is, it represents a rotation of the plane. The $SO(2)$ group is nothing but the rotation in two-plane, parametrized by a single angle $\theta$, topologically a circle, which is connected.
In fact, Let $V^{r}$ be s submanifold of $M^{d}$, furthermore suppose $V$ is a deformation retract of $M$. A deformation retract means that there exists a 1-parameter, continuous map $r_ {t}: M\to M$ such that
- $r_ {0}$ is the identity map on $M$,
- $r_ {1}$ maps all of $M$ to $V$,
- $r_ {t}$ for all $t$ is the identity map on $V$.
Then, consider homology group of any coefficient group. The deformation retract will change continuous by definition, meaning during the deformation from $r_ {t}$ to $r_ {t+\Delta t}$ it sweeps over a area on $M$ whose boundary is $r_ {t+\Delta t}-r_ {t}$. Then the deformation process will not change the homology group, thus
Theorem. If $V \subset M$ is a deformation retract, then $V$ and $M$ have isomorphic homology groups, regardless of the coefficient group $G$,
\[H_ {p}(M;G)\cong H_ {p}(V; G).\]As an application of the above theorem, consider $SO(2)$ and $GL(2)$. Since $SO(2)$ is topologically a circle, we have
\[H_ {1}(SL(2,\mathbb{R})) \cong H_ {1}(\mathbb{S}^{1}) \cong \mathbb{Z}.\]Since we are talking about $2\times 2$ matrices, there exists an interesting result for $A\in 2\times 2$ matrices called the Cayley-Hamilton theorem, stating that
\(A^{2}-(\mathrm{Tr}\,A)A+\det A \,\mathbb{1}=0.\)
Subgroups and Subalgebras
Since the group $G$ is generated by its Lie algebra ${\frak g}$, at least for the connected component, there should be a way to find the subgroup of $G$ by looking at ${\frak g}$.
We will not distinguish between Lie algebra elements, namely the vectors at the identity of the group $X_ {\mathbb{1}}$, and the left-invariant vector field $X_ {g}$, since they are isomorphic to each other.
The exp map $\exp X$ gives the integral curve of $X$. Since $X$ is left invariant, the integral curve starting at $g$ is connected to the integral curve starting at $\mathbb{1}$, to be specific $g(t) = L_ {g}g(0)$. To be specif, notice with an infinitesimal parameter $dt$, the tangent vector $X$ will “move” the point $\mathbb{1}$ to $\mathbb{1}+dt X = \mathbb{1}(1+dt X)$, since $X$ is left-invariantly translated to $g$, the vector $X_ {g}$ will move the point $g$ to $g(1+dX)$. On the other hand, the $X$ vector field, like any vector field, generates a flow $\phi_ {t}(g)$ where $g$ is the starting point.
Convince yourself that the flow generated by the left invariant field $X$ is the 1-parameter group of right translations
\[\phi_ {t}(g) = g \exp(tX).\]Commutator of matrices
Recall that the Lie algebra ${\frak g}$ as a vector space, is simply the tangent space to $G$ at $\mathbb{1}$, but as an algebra it is identified with the left invariant vector fields on $G$. (Of course this is merely a convention; we could have used right invariant fields just as well.) If $X,Y\in{\frak g}$, then their Lie bracket is given by the Lie derivative since
\[[X,Y]=\mathcal{L}_ {X}Y.\]If $G$ is a matrix group, then the elements of ${\frak g}$ are matrices as well, not in $G$ but the derivative of $G$ at origin, and $[X,Y]$ is merely the commutator of two matrices.
Subgroup and subalgebra
The subset $H\subset G$ qualifies as a sub Lie group if H, if not embedded, is at least an immersed(locally embedded) submanifold of $G$. Furthermore, if the Lie algebra ${\frak h}$ is closed under Lie brackets, then ${\frak h}$ define a Lie algebra, called the subalgebra of ${\frak g}$.
In general, if $H$ is a subgroup of $G$, then the Lie algebra ${\frak h}$ of $H$ is a subalgebra of ${\frak g}$. The converse of this is also true and of immense importance.
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