# Lie Groups/Algebras, Notes 4: Lie Groups

Lie Groups:

Definition 1.  A Lie group is a manifold and group whose binary operation is a $C^\omega$ map between the manifolds $G\times G$ and $G.$

We will assume our Lie groups are Banach manifolds unless otherwise stated (locally homeomorphic to a Banach space $E$).  It turns out a Lie group is metrizable, and the metric space is complete.  Lie subgroups are subgroups which are also submanifolds and Lie groups.

Proposition 2.  Let $G$ be an $E$-Lie group (that is, a Lie group which is an $E$-manifold).  Then

1. If $E\in\{\mathbb{R},\mathbb{C}\},$ then $G$ is locally connected.
2. If $E\notin\{\mathbb{R},\mathbb{C}\},$ then $\dim(G)=\dim_F(E)=0.$
3. Suppose $E$ is locally compact, then $G$ is locally compact iff $\dim(G)<\infty.$
4. If $G$ is generated by a subspace whose topology admits a countable base, then the topology on $G$ admits a countable base.

Now suppose $X$ is a $C^\omega$-manifold with an analytic associative binary operation with identity.  Then the subset $G\subseteq X$ of invertible elements of $X$ is open in $X$ and is a Lie group.

Definition 3.  A Lie group morphism between Lie groups $G$ and $H$ is a $C^\omega$ group homomorphism $f:G\to H.$  $Aut(G)$ will denote the group of Lie group automorphisms of $G.$  A Lie group representation on a Banach space $E$ is a Lie group morphism $\rho:G\to Aut(E).$  (Note if $E$ is a Banach space over the field $F$, then $Aut(E)$ is a Lie group which is a manifold over $F$)

Definition 4.  Let $G$ be a topological group and $X$ be an analytic manifold.  We say that $X$ is a homogeneous space for $G$ if $X$ is a $G$-space (that is, continuous action) and $(g_1g_2)x=g_1(g_2x).$

This induces an equivalence relation on $X$ into orbits under the action of $G$ and a corresponding quotient space $X/G,$ called the orbit space of $X$ by $G.$

Proposition 5.  Let $X$ be Lie group and $G$ be a Lie subgroup.  Then:

1. The orbit space $X/G$ has a unique analytic manifold structure such that the canonical map $\pi:X\to X/G$ is a submersion.
2. If $G\unlhd X,$ then $X/G$ is a Lie group and $\pi$ is a Lie group morphism.

In case 1 above, we call the quotient $X/G$ the Lie homogeneous space of $X$ by $G,$ and in case 2 we call the quotient the Lie quotient group.

Let $G$ and $H$ be Lie groups.  Then there is a canonical correspondence $\tau$ between $T(G)\times T(H)$ and $T(G\times H).$  In particular, suppose we have two tangent vectors $u\in T_x(G),$ $v\in T_y(G),$ and $f\in C^\omega(G).$  Let us then define

$(uv)(f)=\tau(u,v)(f,0)+\tau(u,v)(0,f).$

It follows that $uv\in T_{xy}(G)$ for

$\begin{array}{lcl} (uv)(fg)&=&\tau(u,v)(fg,0)+\tau(u,v)(0,fg)\\&=&\tau(u,v)(f,0)\cdot g(xy)+f(xy)\tau(u,v)(g,0)+\tau(u,v)(0,f)\cdot g(xy)+f(xy)\tau(u,v)(0,g)\\&=&g(xy)\left(\tau(u,v)(f,0)+\tau(u,v)(0,f)\right)+f(xy)\left(\tau(u,v)(g,0)+\tau(u,v)(0,g)\right)\\&=&(uv)(f)g(xy)+f(xy)(uv)(g).\end{array}$

Thus we have a product on the tangent bundle of $G$ (note that $G$ being a group makes $xy$ an arbitrary element of $G$).  Hence $T(G)$ is a Lie group.

The Lie Algebra of a Lie Group:

Recall a vector field on a manifold $M$ is section of the vector fiber bundle $T(M)$ with $M$ as its base.  Hence it is a map $X:M\to T(M)$ satisfying $\pi\circ X=1$ where $\pi$ is the projection of the vector fiber bundle to its base.  Now suppose $X$ and $Y$ are two vector fields on $M,$ then we can define the vector fields

$(X+Y)(p)(f)=(X(p)+Y(p))(f)$

$(XY)(p)(f)=X(p)[Y\circ f]-Y(p)[X\circ f]$

where $(Y\circ f)(x)=Y(x)(f).$  This turns $V(M),$ the set of vector fields of $M,$ into a Lie algebra.

Now let $G$ be a Lie group and $\lambda_g:G\to G$ be left multiplication by $g\in G.$  These are certainly diffeomorphisms of $G$

Definition 6.  A vector field $X$ is left-invariant if

$(\lambda_g)_*(X)=X$

for all $g\in G$ where $(\lambda_g)_*(X)(x)=X(gx).$

Proposition 7.  The set $L(G)$ of left-invariant vector fields of $G$ is a Lie subalgebra of $V(G).$

Proof.  Let $X$ and $Y$ be left-invariant and $g\in G$.  Then

$\begin{array}{lcl}(\lambda_g)_*(X+Y)(x)(f)&=&(\lambda_g)_*X(x)(f)+(\lambda_g)_*Y(x)(f)\\&=&X(gx)(f)+Y(gx)(f)\\&=&(X+Y)(gx)(f)\end{array}$

and

$\displaystyle \begin{array}{lcl}(\lambda_g)_*(XY)(x)(f)&=&(\lambda_g)_*\left(X(x)[Y\circ f]-Y(x)[X\circ f]\right)\\&=&X(gx)[Y\circ f]-Y(gx)[X\circ f]\\&=&(XY)(gx)(f)\end{array}.$

The Lie subalgebra $L(G)$ of $V(G)$ is called the Lie algebra of $G.$  Note an obvious consequence is that if $v\in T_1(G),$ $X\in L(G),$ and $X(1)=v.$  Then

$X(g)=(\lambda_g)_*(X)(1).$

Hence it follows that left-invariant vector fields of $G$ are completely determined by where they send the identity element.  In this sense, we obtain the equivalence of the vector spaces $L(G)$ and $T_1(G).$

Lie Group of a Lie Algebra:

The standard convention is to discuss an “inverse” map from the Lie algebra of a Lie group back to the Lie group.  Let $X\in L(G).$  Then this left-invariant vector field corresponds to a tangent vector $v\in T_1(G).$  Let $\gamma_X:\mathbb{R}\to G$ be a path defined by

$\gamma_X(0)=1$

$X(\gamma_X(0))=X(1)=v$

$\gamma_X(t_1+t_2)=\gamma_X(t_1)\gamma_X(t_2).$

Note these uniquely define $\gamma_X,$ called a one-parameter subgroup of $G.$  We in turn define the map $\exp:L(G)\to G$ by

$\exp(X)=\gamma_X(1)$

or more generally

$\exp(tX)=\gamma_X(t).$

Hence $\exp(0)=1.$ It follows that $L(\exp(L(G)))=L(G).$

Suppose $G=GL_n(\mathbb{R}).$  Then $L(G)=M_n(\mathbb{R})$ and

$\displaystyle\exp(X)=\sum_{n=0}^\infty\frac{X^n}{n!}.$

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

[2]  Robert Milson, Thomas Foregger, Mike Fikes. “Lie group” (version 12). PlanetMath.org. Freely available at http://planetmath.org/?op=getobj;from=objects;id=1112

# Lie Groups/Algebras, Notes 3: Bialgebras and Free Lie Algebras

Recall the construction of an associative algebra $A$ over a field $K.$  We have a module homomorphism $\nabla:A\otimes A\to A$ such that

$\nabla\circ (1\otimes\nabla)=\nabla\circ(\nabla\otimes 1).$

If the algebra is unital, then we also a canonical map $\eta:K\to A$ with $\eta(1)$ defined as the identity of $A$ and corresponding assumption that

$\nabla\circ(1\otimes\eta)=\nabla\circ(\eta\otimes 1)=1$

on $A$ (with canonical identifications $A\otimes K=K\otimes A=A$).

We can dually define a $K$coalgebra as a $K$-module $C$ together with a comultiplication map $\Delta:C\to C\otimes C$ such that

$(1\otimes\Delta)\circ\Delta=\Delta\circ(\Delta\otimes 1).$

We also call it unital if there is a map $\varepsilon:C\to K$ such that

$(1\otimes\varepsilon)\circ\Delta=(\varepsilon\otimes 1)\circ\Delta=1$

on $C$ (with the canonical identification $C\otimes K=K\otimes C=C$).

Definition 1.  A $K$bialgebra is a unital $K$-algebra and unital $K$-coalgebra such that

1. $\Delta\circ\nabla=(\nabla\otimes\nabla)\circ(1\otimes\tau\otimes 1)\circ(\Delta\otimes\Delta)$ ($\tau(x\otimes y)=y\otimes x$),
2. $\varepsilon\circ\nabla=\varepsilon\otimes\varepsilon,$
3. $\Delta\circ\eta=\eta\otimes\eta,$ and
4. $\varepsilon\circ\eta=1.$

Definition 2.  An element $x$ of a coalgebra is $u$primitive if $\Delta(x)=x\otimes u+u\otimes x.$  Elements are primitive if they are 1-primitive.

If $\sigma:\mathfrak{g}\to U(\mathfrak{g})$ is the embedding into the enveloping algebra, then the map $\Delta(\sigma(x))=\sigma(x)\otimes 1+1\otimes\sigma(x)$ gives $U(\mathfrak{g})$ a $K$-bialgebra structure.

Definition 3.  Let $X$ be a set and $A_K(X)$ be the free algebra over $X.$  We define the free Lie $K$algebra over $X$ as the quotient $A_K(X)/I$ where $I$ is the ideal generated by elements of the form $xy+yx$ and $x(yz)+z(xy)+y(zx).$  We will denote this Lie algebra by $L(X)$ and its product by $[x,y].$

The enveloping algebra $U(L(X))$ of the free Lie algebra $L(X)$ ends up being the free associative algebra over $X.$

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

# Lie Groups/Algebras, Notes 2: Representations and Types

Recall a representation of an algebra $A$ on an $R$-module $M$ is an algebra morphism $\rho:A\to End_R(M).$  When we have a Lie algebra $\mathfrak{g},$ a representation of it on an $R$-module $M$ is an algebra morphism

$\rho:\mathfrak{g}\to L(End_R(M)).$

Hence in particular we have $\rho(xy)(m)=\left(\rho(x)\rho(y)-\rho(y)\rho(x)\right)(m).$  We correspondingly call $M$ a $\mathfrak{g}$module.  A representation is faithful if it is injective.  It is simple/irreducible if its $\mathfrak{g}$-module is simple.

Let $\rho_1$ and $\rho_2$ be representations of Lie algebras $\mathfrak{g}_1$ and $\mathfrak{g}_2$ on $M_1$ and $M_2$ respectively.  Then $\rho_1\otimes\rho_2$ is a representation of $\mathfrak{g}_1\times\mathfrak{g}_2$ on $M_1\otimes M_2$ defined by

$\begin{array}{lcl}(\rho_1\otimes\rho_2)(x_1,x_2)(m_1\otimes m_2)&=&(\sigma_1(x_1)\otimes 1+1\otimes\sigma_2(x_2))(m_1\otimes m_2)\\&=&\sigma_1(x_1)(m_1)\otimes m_2+m_1\otimes\sigma_2(x_2)(m_2)\end{array}$

where $\sigma_1,\sigma_2$ are the canonical inclusions of $\mathfrak{g}_1$ in $U(\mathfrak{g}_1)$ and $\mathfrak{g}_2$ in $U(\mathfrak{g}_2).$  Hence $M_1\otimes M_2$ is both a $\mathfrak{g}_1\times\mathfrak{g}_2$-module and a $U(\mathfrak{g}_1)\otimes U(\mathfrak{g}_2)$-module.

Definition 1$\mathfrak{g}$ is solvable if its derived series $D^n\mathfrak{g}=0$ for some $n.$  It is nilpotent if its lower central series $C_n\mathfrak{g}=0$ for some $n.$  The radical of $\mathfrak{g},$ $\sqrt{\mathfrak{g}},$ is the largest solvable ideal in $\mathfrak{g}.$  We say $\mathfrak{g}$ is semisimple if it has no nonzero abelian ideals.

Hence it follows that the center $Z(\mathfrak{g})$ of a nilpotent Lie algebra is nontrivial.  Also note that a nilpotent Lie algebra is solvable.  $\mathfrak{r}=\sqrt{\mathfrak{g}}$ is the smallest ideal such that $\mathfrak{g}/\mathfrak{r}$ has radical $0.$

Theorem 2 (Engel).  Let $V$ be a vector space and $\mathfrak{g}$ be a finite dimensional subalgebra of $\mathfrak{gl}(V).$.  If $V\neq 0,$ then there exists a $u\neq 0$ with $u\in\mathfrak{g}$ such that $xu=0$ for all $x\in\mathfrak{g}.$

It follows that a Lie algebra $\mathfrak{g}$ is nilpotent iff $\mbox{adj}\,x$ is nilpotent for all $x\in\mathfrak{g}.$

Note that $B(x,y)=\mbox{tr}(\mbox{adj}(x)\mbox{adj}(y))$ defines a symmetric bilinear form on $\mathfrak{g},$ called the Killing form of $\mathfrak{g}.$  The Killing form also satisfies $B(xy,z)=B(x,yz).$

Proposition 3.  $\mathfrak{g}$ is solvable iff $B(D\mathfrak{g},\mathfrak{g})=0.$

Proposition 4.  $\mathfrak{g}$ is semisimple iff $\mathfrak{g}=0$ iff $B$ is nondegenerate.

Theorem 5 (Weyl).  Every finite dimensional representation of a semisimple Lie algebra is completely reducible (i.e. the $\mathfrak{g}$-module is semisimple).

A Lie algebra is simple if its only ideals are $\mathfrak{g}$ and $0$ and noncommutative.

Proposition 6.  $\mathfrak{g}$ is semisimple iff it is a product of simple algebras.

A Lie algebra is reductive if its adjoint representation is semisimple.  That is, it is reductive if it can be written $\mathfrak{g}=\mathfrak{s}\oplus\mathfrak{a}$ where $\mathfrak{s}$ is semisimple and $\mathfrak{a}$ is abelian (since here we want $\mathfrak{g}$ as a module to be semisimple–so that as an algebra, it has an an abelian term).  This also turns out to be equivalent to $D\mathfrak{g}$ being semisimple.

Definition 7.  The nilpotent radical of $\mathfrak{g}$ is the intersection of the kernels of all finite dimensional simple representations of $\mathfrak{g}.$

For an element $x$ in the nilpotent radical, it follows that $\mbox{adj}(x)$ is nilpotent.  Now for any nilpotent endomorphism $u$ of an algebra, we have that

$\displaystyle e^u=\sum_{n=0}^\infty\frac{u^n}{n!}$

has a finite number of nonzero terms, and is hence well-defined.  If we can write $\mathfrak{g}=\sqrt{\mathfrak{g}}\ltimes_{e^{u\cdot}}\mathfrak{s}$ with $\mathfrak{s}$ semisimple and $u$ nilpotent, then we will call $\mathfrak{s}$ a Levi subalgebra of $\mathfrak{g}.$

Theorem 8 (Levi-Malcev).  Every Lie algebra has a Levi subalgebra (that is, it can be written $\mathfrak{g}=\sqrt{\mathfrak{g}}\ltimes\mathfrak{s}$ as above).  Moreover, every other Levi subalgebra $\mathfrak{s}'$ is the image of $\mathfrak{s}$ under $\exp(\mbox{adj}(x))$ for some $x$ in the nilpotent radical of $\mathfrak{g}.$

It turns out that a subalgebra of $\mathfrak{g}$ is a Levi subalgebra iff it is a maximal semisimple subalgebra.

Theorem 9 (Ado).  Let $\mathfrak{n}$ be the largest nilpotent ideal in $\mathfrak{g}.$  Then $\mathfrak{g}$ admits a finite dimensional faithful representation $\rho$ such that every element of $\rho(\mathfrak{n})$ is nilpotent.

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.

# Lie Groups/Algebras, Notes 1: Lie Algebras

Definition 1.  A Lie algebra $\mathfrak{g}$ is an $R$-algebra such that $xy+yx=0$ and $x(yz)+z(xy)+y(zx)=0.$

Hence $x^2=-x^2\Rightarrow 2x^2=0\Rightarrow x^2=0$ provided $R$ is not a ring of characteristic $2.$  Also, if $A$ is an $R$-algebra, then we can define a Lie algebra $L(A)$ whose product is defined by

$[x,y]=xy-yx.$

Hence in particular, $L(\mathfrak{g})$ simply doubles the product of $\mathfrak{g}.$  Recall that a derivation $d$ on an algebra satisfies

$d(ax+by)=ad(x)+bd(y)$

and

$d(xy)=d(x)y+xd(y).$

For $x\in\mathfrak{g},$ let us define the adjoint of $x$ as $\mbox{adj}_x\in\mbox{End}\mathfrak{g}$ defined by

$\mbox{adj}_xy=xy.$

Proposition 2.  For $x\in\mathfrak{g},$ $\mbox{adj}_x$ is a derivation on $\mathfrak{g}.$

Proof.  By tensorial properties of algebras we have

$\mbox{adj}_x(ay+bz)=x(ay+bz)=x(ay)+x(bz)=a(xy)+b(xz)=a\mbox{adj}_xy+b\mbox{adj}_xz.$

And

$\mbox{adj}_x(yz)=x(yz)=-z(xy)-y(zx)=(xy)z+y(xz)=(\mbox{adj}_xy)z+y(\mbox{adj}_xz).$

Proposition 3.  Let $\mathfrak{g}$ be a commutative Lie algebra over a ring $R$ with $\mbox{char}(R)\neq 2.$  Then $xy=0.$  If $xy=0,$ then $\mathfrak{g}$ is commutative, regardless of the ring.

Proof.  If $\mathfrak{g}$ is commutative, then $xy=yx=-xy\Rightarrow 2xy=0\Rightarrow xy=0.$  The second claim is trivial.

Let $I$ be a left ideal in $\mathfrak{g},$ $x\in I,$ and $y\in\mathfrak{g}.$  Then $yx=-xy\in I$ and hence $xy\in I.$  So $I$ is also a right ideal.  Hence we simply consider ideals of Lie algebras without reference to leftness or rightness.

Definition 4.  Let $I$ be a submodule of $\mathfrak{g}$ such that $dI\leq I$ for all derivations $d$ on $\mathfrak{g},$ then $I$ is called a characteristic ideal of $\mathfrak{g}.$

It’s easy to see that a characteristic ideal is actually an ideal in $\mathfrak{g}$ since we have

$\mbox{adj}_xI=xI\leq I$

for all $x\in\mathfrak{g}.$

Proposition 5.  Let $\mathfrak{a}$ be an ideal of $\mathfrak{g}$ and $\mathfrak{b}$ be a characteristic ideal of $\mathfrak{a},$ then $\mathfrak{b}$ is an ideal of $\mathfrak{g}.$

Proof.  If $d$ is a derivation on $\mathfrak{a},$ then $d\mathfrak{b}\leq\mathfrak{b}$ since $\mathfrak{b}$ is a characteristic ideal.  But multiplication by $x\in\mathfrak{g}$ is a derivation on $\mathfrak{a}.$  Hence $bx\in\mathfrak{b}$ for $b\in\mathfrak{b}.$

For two ideals $\mathfrak{a},\mathfrak{b}\subseteq R,$ we have that $\mathfrak{a}+\mathfrak{b}$ and $\mathfrak{a}\cap\mathfrak{b}$ are ideals in $R$ and that $\mathfrak{a}\cup\mathfrak{b}\subseteq\mathfrak{a}+\mathfrak{b}$ and $\mathfrak{a}\mathfrak{b}\subseteq\mathfrak{a}\cap\mathfrak{b}.$  In the case of an $R$-algebra $A$, we will let $\mathfrak{a}\mathfrak{b}$ denote the submodule generated by elements $ab.$  This is in fact an ideal in a Lie algebra for if $d$ is a derivation, then

$d(ab)=d(a)b+ad(b)\in\mathfrak{a}\mathfrak{b}.$

Hence it is stable under adjoint action.

Definition 6.  The ideal $D^0\mathfrak{g}=\mathfrak{g}\mathfrak{g}$ is called the derived ideal of $\mathfrak{g}.$  Inductively we can define

$D^{n+1}\mathfrak{g}=D^n\mathfrak{g}D^n\mathfrak{g}.$

The sequence $\{D^n\mathfrak{g}\}$ is called the derived series of $\mathfrak{g}.$  Let us define $C_1\mathfrak{g}=\mathfrak{g}$ and

$C_{n+1}\mathfrak{g}=\mathfrak{g}C_n\mathfrak{g}.$

The sequence $\{C_n\mathfrak{g}\}$ is called the lower central series of $\mathfrak{g}.$

Definition 7.  Let $\mathfrak{g}$ be a Lie algebra, $T(\mathfrak{g})$ be its tensor algebra (with $\mathfrak{g}$ interpreted as a module), and $I$ be the ideal generated by elements of the form $x\otimes y-y\otimes x-xy.$  Then $L^{-1}(\mathfrak{g})=T/I$ is called the enveloping algebra of $\mathfrak{g}.$

Theorem 8.  The enveloping algebra $L^{-1}(\mathfrak{g})$ satisfies the universal property that if $f:\mathfrak{g}\to L(A)$ is an algebra homomorphism for some associative unital algebra $A$ and $\iota:\mathfrak{g}\to L(L^{-1}(\mathfrak{g}))$ is the canonical inclusion, then there exists a unique morphism $g:L^{-1}(\mathfrak{g})\to A$ and $\ell(g):L(L^{-1}(\mathfrak{g}))\to L(A)$ such that $f=\ell(g)\circ\iota.$

Proposition 9.  If $\mathfrak{g}_1$ and $\mathfrak{g}_2$ are Lie algebras, then $L^{-1}(\mathfrak{g}_1\times\mathfrak{g}_2)=L^{-1}(\mathfrak{g}_1)\otimes L^{-1}(\mathfrak{g}_2).$

Let us define $U_n(\mathfrak{g})=T^n(\mathfrak{g})/I$ with $I$ defined as above, $G_n=U_n/U_{n-1},$ and $G=\oplus G_n.$  It follows that $G$ is a filtered algebra, which we will call the associated filtered algebra to $L^{-1}(\mathfrak{g}).$

Theorem 10 (Poincare-Birkhoff-Witt)  Let $G$ be the associated filtered algebra to $L^{-1}(\mathfrak{g})$ and $S(\mathfrak{g})$ be the symmetric algebra of $\mathfrak{g}$ as a module.  Then if $\mathfrak{g}$ is free, $S(\mathfrak{g})=G.$

It follows that if $\mathfrak{g}$ is free, then the canonical homomorphism $f:\mathfrak{g}\to L^{=1}(\mathfrak{g})$ is injective.

[1]  Bourbaki, Nicholas.  Lie Groups and Lie Algebras, Chapters 1-3.  Springer-Verlag.  1971.