Lie groups

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

and

Department of Mathematics
University of Wisconsin, Madison
Madison, WI 53706 USA
ram@math.wisc.edu

Last updates: 1 April 2010

Lie groups

The Lie group $S^1=ℝ/\mathbb{Z}=U_1\left(ℂ\right).$ A torus is a Lie group $G$ isomorphic to $S^1\times S^1\times \dots S^1$ ($k$ factors), for some $k\in {\mathbb{Z}}_{>0}.$

A connected Lie group is semisimple if $R\left(G\right)=\left\{1\right\}.$

Let $G$ be a Lie group and let $x\in G.$ A tangent vector at $x$ is a linear map ${\xi }_x{C}^{\infty }:\left(G\right)\to ℝ$ such that $${\xi }_x\left(f_1{f}_2\right)={\xi }_x\left(f_1\right)f_2\left(x\right)+f_1\left(x\right){\xi }_x\xi f_2,\text{for}{f}_1,f_2\in C^{\infty }\left(G\right).$$ A left invariant vector field on $G$ is a vector field $\xi :C^{\infty }\left(G\right)\to C^{\infty }\left(G\right)$ such that $$L_g\xi =\xi L_g,\text{for all }g\in G.$$ A one parameter subgroup of $G$ is a smooth group homomorphism $\gamma :ℝ\to G.$ If $\gamma$ is a one parameter subgroup of $G$ define $$\frac{d}{dx}f\left(\gamma \left(t\right)\right)=\underset{h\to 0}{\lim }\frac{f\left(\gamma \left(t+h\right)\right)-f\left(\gamma \left(t\right)\right)}{h}.$$ The following proposition says that we can identify three vector spaces

1. {left invariant vector fields on $G$},
2. {one parameter subgroups of $G$},
3. {tangent vectors at $1\in G$}.

The maps $$\begin{array}{rcl}\text{{left invariant vector fields}}& \to & \text{{tangent vectors at 1}}\\ \xi & \mapsto & {\xi }_1\end{array}$$ and $$\begin{array}{rcl}\text{{one parameter subgroups}}& \to & \text{{tangent vectors at 1}}\\ \gamma & \mapsto & {\gamma }_1\end{array}$$ where $${\xi }_{1}f=\left(\xi f\right)\left(1\right),\text{and}{\gamma }_1=\left(\frac{d}{dt}f\left(\gamma \left(t\right)\right)\right){|}_{t=0},$$ are vector space isomorphisms.

The Lie algebra $𝔤=\mathrm{Lie}\left(G\right)$ of the Lie group $G$ is the tangent space to $G$ at the identity with the bracket $\left[,\right]:𝔤\times 𝔤\to 𝔤$ given by $$\left[\xi ,x{i}_2\right]={\xi }_1{\xi }_2-{\xi }_2{\xi }_1,\text{for}{\xi }_1,{\xi }_2\in 𝔤.$$ Let $\phi :G\to H$ be a Lie group homomorphism and let $𝔤=\mathrm{Lie}\left(G\right)$ and $𝔥=\mathrm{Lie}\left(H\right).$ Then $$\begin{array}{rcl}{C}^{\infty }\left(H\right)& \stackrel{{\phi }^{*}}{\to }& C^{\infty }\left(G\right)\\ f& \mapsto & f\circ \phi \end{array}$$ and the differential of $\phi$ is the Lie group homomorphism $𝔤\stackrel{d\phi }{\to }𝔥$ given by $$\begin{array}{rcl}d\phi \left({\xi }_1\right)={\xi }_1\circ {\phi }^{*}& \text{if}& {\xi }_1\text{ is a tangent vector at the identity,}\\ d\phi \left(\xi \right)=\xi \circ {\phi }^{*}& \text{if}& \xi \text{ is a left invariant vector field,}\\ d\phi \left(\gamma \right)=\phi \circ \gamma & \text{if}& \gamma \text{ is a one parameter subgroup.}\end{array}$$

(Note: It should be checked that

1. the map $d\phi$ is well defined,
2. the three definitions of $d\phi$ are the same,
3. and that $d\phi$ is a Lie algebra homomorphism.

These checks are not immediate, but are nevertheless quite straightforward checks of the deifinitions.) The map $$\begin{array}{rcl}\text{the category of Lie groups}& \to & \text{the category of Lie algebras}\\ G& \mapsto & \mathrm{Lie}\left(G\right)\\ \phi & \mapsto & d\phi \end{array}$$ is a functor. This functor is not one to one; for example, the Lie groups $O_n\left(ℝ\right)$ and ${SO}_n\left(ℝ\right)$ have the same Lie algebra. On the other hand, the Lie algebra contains the structure of the Lie groups in a neighbourhood of the identity. The exponential map is $$\begin{array}{rcl}𝔤& \to & G\\ tX& \mapsto & e^{tX},\end{array}\text{where}{e}^{t}X=\gamma \left(t\right)$$ is the one parameter subgroup corresponding to $X\in 𝔤.$ This map is a homeomorphism from a neighbourhood of 0 in $𝔤$ to a neighbourgood of 1 in $G.$

(Lie's theorem) The functor $$\begin{array}{rcll}\text{Lie:}& \text{{connected simply connected Lie groups}}& \to & \text{{Lie algebras}}\\ & G& \mapsto & 𝔤=\mathrm{Lie}\left(G\right)=T_1\left(G\right)\end{array}$$ is an equivalence of categories.

If $𝔤$ is a Lie subalgebra of $𝔤{𝔩}_n$ then the matrices $$\left\{e^{tX}|t\in ℝ,X\in 𝔤{𝔩}_n\right\},\text{where}{e}^{t}X=\sum _{k\ge 0}\frac{t^k{X}^k}{k!},$$ form a group with Lie algebra $𝔤$.$$e^{tX}{e}^{tY}=e^{t\left(X+Y\right)}+\left(\frac{t^2}{2}\right)\left[X,Y\right]+\dots ,$$$$e^{tX}{e}^{tY}{e}^{-tX}=e^{t\left(Y\right)}+\left(\frac{t^2}{2}\right)\left[X,Y\right]+\dots ,$$$$e^{tX}{e}^{tY}{e}^{-tX}{e}^{-tY}=e^{\left(\frac{t^2}{2}\right)}\left[X,Y\right]+\dots ,$$

Let $G$ be a Lie group and let $𝔤=\mathrm{Lie}\left(G\right).$ Let $x\in G.$Then the differential of the Lie group homomorphism $$\begin{array}{rcll}{\mathrm{Int}}_x:& G& \to & G\\ & g& \mapsto & xg{x}^{-1}\end{array}$$ is a Lie algebra homomorphism $${\mathrm{Ad}}_x:𝔤\to 𝔤.$$ Since there is a map ${\mathrm{Ad}}_x$ for each $x\in G$, there is a map $$\begin{array}{rcll}\mathrm{Ad:}& G& \to & GL\left(𝔤\right)\\ & x& \mapsto & {\mathrm{Ad}}_x\end{array}\text{and}{\mathrm{Ad}}_x{\mathrm{Ad}}_y={\mathrm{Ad}}_{\mathrm{xy}},\text{for}x,y\in G,$$ since ${\mathrm{Int}}_x{\mathrm{Int}}_y={\mathrm{Int}}_{\mathrm{xy}}.$ The differential of Ad is $$\begin{array}{rcll}\mathrm{ad:}& 𝔤& \to & \mathrm{End}\left(𝔤\right)\\ & X& \mapsto & {\mathrm{ad}}_X\end{array},\text{where}\begin{array}{rcll}{\mathrm{ad}}_X:& 𝔤& \to & 𝔤\\ & Y& \mapsto & \left[X,Y\right]\end{array},$$ since Define a right action of $G$ on $C^{\infty }\left(G\right)$ by $$\left(R_{x}f\right)\left(g\right)=f\left(gx\right),\text{for}x\in G,f\in C^{\infty }\left(G\right),g\in G.$$ Then $${\mathrm{Ad}}_x\xi =R_x\xi R_{x^{-1}},\text{for all}x\in G,\xi \in 𝔤,$$ since, for $x\in G,{\mathrm{Int}}_x^{*}\left({\mathrm{Ad}}_x\xi \right)=\xi \circ {\mathrm{Int}}_x^{*}=\xi L_{x^{-1}}{R}_{x^{-1}}=L_{x^{-1}}\xi R_{x^{-1}}{L}_{x^{-1}}{R}_{x^{-1}}{R}_x\xi R_{x^{-1}}={\mathrm{Int}}_x^{*}\left(R_x\xi R_{x^{-1}}\right).$

Recall that the adjoint representation of $G$ is $$\begin{array}{rcll}\mathrm{Ad}:& G& \to & \mathrm{GL}\left(𝔤\right)\\ & x& \mapsto & {\mathrm{Ad}}_x\end{array}\text{where}\begin{array}{rcll}{\mathrm{Ad}}_x:& 𝔤& \to & fg\\ & \xi & \mapsto & R_x\xi R_{x^{-1}}\end{array}$$ is the differential of $$\begin{array}{rcll}\mathrm{Int}:& G& \to & G\\ & g& \mapsto & xg{x}^{-1}\end{array}.$$ The coadjoint representation of $G$ is the dual of the adjoint representation, ie the action of $G$ on $𝔤*=\mathrm{Hom}\left(𝔤,ℂ\right)$ given by $$\left(g\phi \right)\left(X\right)=\phi \left({\mathrm{Ad}}_{g^{-1}}X\right),\text{for}g\in G,\phi \in 𝔤*,X\in 𝔤.$$

The coadjoint orbitis the set produced by the action of $G$ on an element $\phi \in 𝔤*,$ ie $G\phi \subseteq 𝔤*$ is a coadjoint orbit. Let $G$ be a Lie group and let $𝔤$ be its Lie algebra. Then $G^0$ is nilpotent if and only if $\mathrm{Lie}\left(G\right)$ is nilpotent, and $G^0$ is solvable iff $\mathrm{Lie}\left(G\right)$ is solvable. A semisimple Lie group is a connected Lie group with semisimple Lie algebra.

The class of reductive Lie groups is the largest class of Lie groups which contains all the semisimple Lie groups and parabolic subgroups of them and for which the representation theory is still controllable. A real Lie group is reductive if there is a linear algebraic group $G$ over $ℝ$ whose identity component (in the Zariski topology) is reductive and a morphism $\nu :G\to \mathrm{GL}\left(ℝ\right)$ with finite kernel, whose image is an open subgroup of $\mathrm{GL}\left(ℝ\right).$ Fpr the definition of Harish-Chandra class see Knapp's article.

1. $U\left(n\right)=\left\{x\in M_n\left(ℂ\right)|x{\stackrel{-}{x}}^t=\mathrm{id}\right\}.$
2. $\mathrm{Sp}\left(2n,ℂ\right)=\left\{A\in M_n\left(ℂ\right)|A^{t}JA=J\right\}.$
3. ${\mathrm{Sp}}_{2n}=\mathrm{Sp}\left(2n,ℂ\right)\cap U\left(2n\right).$

The simple compact Lie groups are

1. (Type A) $S{U}_n\left(ℂ\right)$
2. (Type $B_n$) $S{O}_{2n+1}\left(ℝ\right),n\ge 1$
3. (Type $C_n$) S p 2n ℂ ∩ U n ,n≥1,
4. (Type $D_n$) $S{O}_{2n}\left(ℝ\right),n\ge 4,$
5. (Type E) ???????

If $G$ is a maximal Lie group such that $G/G^0$ is finite then

1. $G$ has a maximal compact subgroup,
2. Any two maximal compact subgroups are conjugate
3. $G$ is homeomorphic to $K\times {ℝ}^m$ under the map $$\begin{array}{rcl}K\times 𝔭& \to & G\\ \left(k,x\right)& \mapsto & k{e}^x\end{array}$$ where $K$ is a maximal compact subgeroup of $G$ and $𝔭=$??????.
4. If $G$ is a semisimple Lie group then $$K=\left\{g\in G|\Theta \left(g\right)=g\right\},$$ where $\Theta$ is the Cartan involution on $G,$ a maximal compact subgroup of $G.$ For matrix groups $$\begin{array}{rcll}\Theta :& G& \to & G\\ & g& \mapsto & \end{array}$$ g -1 t is the Cartan involution.

On the Lie algebra level $$\begin{array}{rcll}\theta :& 𝔤& \to & fg\\ & x& \mapsto & -{\stackrel{-}{x}}^t\end{array},$$$$𝔱=\left\{x\in 𝔤|\theta x=x\right\},$$$$𝔭=\left\{s\in 𝔤|\theta x=-x\right\},$$$$𝔤=𝔱\oplus 𝔭,$$$$𝔲=𝔱\oplus i𝔭,$$$${𝔤}_{ℂ}=𝔤\oplus i𝔤=𝔲\oplus i𝔲.$$

1. There is an equivalence of categories $$\begin{array}{rcl}\text{{compact connected Lie groups}}& \leftrightarrow & \text{{connected reductive algebraic groups over }ℂ\text{}}\\ U& \leftrightarrow & G\end{array}$$ where $U$ is the maximal compact subgroup of $G$ and $G$ is the algebraic group with coordinate ring $C{\left(U\right)}^{\mathrm{rep}}.$ The group $G$ is the complexification of $U.$
2. The functor $${\mathrm{Res}}_K^G:\text{{holomorphic representations of }G\text{}}\to \text{{representations of }K\text{}}$$ is an equivalence of categories.

Proof (a) The point of (a) is that for compact groups the continuous functions separate the points of $G$ and for algebraic groups the polynomial functions separate the points of $G$, and, for $ℂ$ and $ℝ$ the polynomial functions are dense in the continuous functions.

Examples: Under the equivalence of ???

1. semisimple algebraic groups correspond exactly to the Lie groups with fiinte center,

algebraic tori correspond exactly to geometric tori.

irreducible finite dimensional representations of $G$ correspond exactly to irreducible finite dimensional representations of $U.$ $$\begin{array}{rcl}{U}_n& \leftrightarrow & G{L}_n\left(ℂ\right)\\ S{U}_n& \leftrightarrow & S{L}_n\left(ℂ\right)\\ S{O}_{2n+1}\left(ℝ\right)& \leftrightarrow & S{O}_{2n+1}\left(ℂ\right)\\ S{p}_{2n}& \leftrightarrow & S{p}_{2n}\left(ℂ\right)\\ S{O}_{2n}\left(ℝ\right)& \leftrightarrow & S{O}_{2n}\left(ℂ\right)\end{array}$$ Other examples are $G{L}_n\left(ℂ\right),S{L}_n\left(ℂ\right),\mathrm{PG}{L}_n\left(ℂ\right),O_n\left(ℂ\right),S{O}_n\left(ℂ\right),{\mathrm{Pin}}_n,{\mathrm{Spin}}_n,S{p}_{2n}\left(ℂ\right),P{\mathrm{Sp}}_{2n}\left(ℂ\right),U_n\left(ℂ\right),S{U}_n\left(ℂ\right),U_n\left(ℂ\right)/Z\left(U_n\left(ℂ\right),\right),O_n\left(ℝ\right),{SO}_n\left(ℝ\right),\dots$

Equivalences: $$\begin{array}{rcl}\text{{compact Lie groups}}& \leftrightarrow & \text{{complex semisimple Lie groups}}\\ & \leftrightarrow & \text{{semisimple algebraic groups}}\\ & \to & \text{{complex semisimple Lie algebras}}\end{array}$$

A representation of $G$ is an action of $G$ on a vector space by linear transformations. The words representation and $G$-module are used interchangeably. A complex representation is a representation where $V$ is a vector space over $ℂ.$ In order to distinguish the group element $g$ from the linear transformation of $V$ given by the action of $g$ write $V\left(g\right)$ for the linear transformation. Then $$V:G\to GL\left(V\right)$$ and the statement that the representation is a group action means that $$V\left(xy,=,V,\left(x\right),V,\left(y\right),,,,\text{for all},,x,,,y,\in ,G,.\right)$$ Unless otherwise stated we shall assume that all representations of $G$ are Lie group homomorphisms. A hlomorphic representation is a representation in the category of complex Lie groups.

A representation is irreducible, or simple, if it has no subrepresentations except 0 and itself. In the case when $V$ is a topolgical vector space then a subrepresentation is required to be a closed subspace of $V.$ The trivial $G$-module is the representation $$\begin{array}{rcll}1:& G& \to & {ℂ}^{*}={\mathrm{GL}}_1\left(ℂ\right)\\ & g& \mapsto & 1\end{array}$$ If $V$ and $W$ are $G$-modules then the tensor product is the action of $G$ on $V\otimes W$ given by $$g\left(v\otimes w\right)=gv\otimes gw,\text{for}v\in V,w\in W,g\in G.$$ If $V$ is a $G$-module then the dual $G$-module to $V$ is the action of $G$ on $V*=\mathrm{Hom}\left(V,ℂ\right)$ (linear maps $\psi :V\to ℂ$) given by $$\left(g\psi \right)\left(v\right)=\psi \left(g^{-1}v\right),\text{for}g\in G,\psi \in V*,v\in V.$$ The maps $$\begin{array}{rcl}1\otimes V& \tilde{\to }& V\\ 1\otimes v& \mapsto & v\end{array}$$ and $$\begin{array}{rcl}V\otimes 1& \tilde{\to }& V\\ v\otimes 1& \mapsto & v\end{array}$$ are $G$-module isomorphisms for any $V.$ The maps $$\begin{array}{rcl}V*\otimes V& \to & 1\\ \phi \otimes v& \mapsto & \phi \left(v\right)\end{array}$$ and $$\begin{array}{rcl}1& \to & V\\ 1& \mapsto & \sum _i{b}_i\otimes {\beta }_i^{*}\end{array}$$ where $\left\{b_i\right\}$ is a basis of $V$ and $\left\{{\beta }_i^{*}\right\}$ is the dual basis in $V*$ and are $G$-module homomorphisms.

If $V:G\to \mathrm{GL}\left(V\right)$ is a homomorphism of Lie groups then the differential of $V$ is a map $$dV:𝔤\to \mathrm{End}\left(V\right)$$ which satisfies $$dV\left(\left[x,y\right]\right)=\left[dV\left(x\right)\right]dV\left(y\right)=dV\left(x\right)dV\left(y\right)-dV\left(y\right)dV\left(\mathrm{y=x}\right),$$ for $x,y\in 𝔤.$ A representation of a Lie algebra $𝔤$, or $𝔤$-module, is an action of $𝔤$ on a vector space $V$ by linear transformations, ie a linear map $\phi :𝔤\to \mathrm{End}\left(V\right)$ such that $$V\left(\left[x,y\right]\right)=\left[V\left(x\right),V\left(y\right)\right]=V\left(x\right)V\left(y\right)-V\left(y\right)V\left(x\right),\text{for all}x,y\in 𝔤,$$ where $V\left(x\right)$ is the linear transformation of $V$ determined by the action of $x\in 𝔤.$ The trivial representation of $𝔤$ is the map $$\begin{array}{rcll}1:& 𝔤& \to & ℂ\\ & x& \mapsto & 0\end{array}.$$ If $V$ is a $𝔤$-module, the dual $𝔤$-module is the $𝔤$-action on $V*=\mathrm{Hom}\left(V,ℂ\right)$given by $$\left(x\phi \right)\left(v\right)=\phi \left(-xv\right),\text{for}x\in 𝔤,\phi \in V*,v\in V.$$ If $V$ and $W$ are $𝔤$-modules then the tensor product of $V$ and $W$ is the $𝔤$-action on $V\otimes W$ given by $$x\left(v\otimes w\right)=xv\otimes w+v\otimes xw,x\in 𝔤,v\in V,w\in W.$$

The definitions of the trivial, dual and tensor product $𝔤$-modules are accounted for by the following formulas: $$\frac{d}{dt}1{|}_{t=0}=\frac{d}{dt}{e}^{t\times 0}{|}_{t=0,}$$$$\frac{d}{dt}{\left(e^{tX}\right)}^{-1}{|}_{t=0}=\frac{d}{dt}{e}^{-tX}-1{|}_{t=0}=-X,$$$$\frac{d}{dt}\left(e^{tX}\otimes e^{tX}\right){|}_{t=0}=\frac{d}{dt}\left(1+tX+\frac{t^2{X}^2}{2!}+\dots \right)\otimes \left(1+tX+\frac{t^2{X}^2}{2!}+\dots \right){|}_{t=0}=\frac{d}{dt}\left(1\otimes 1+t\left(X\otimes 1+1\otimes X\right)+\dots \right){|}_{t=0}=X\otimes 1+1\otimes X.$$

References [PLACEHOLDER]

[BG] A. Braverman and D. Gaitsgory, Crystals via the affine Grassmanian, Duke Math. J. 107 no. 3, (2001), 561-575; arXiv:math/9909077v2, MR1828302 (2002e:20083)

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