Lie groups and algebras: adapted from lecture notes.

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

Last update: 15 July 2012

Lie algebras and the exponential map

A Lie group is a group that is also a manifold, i.e. a topological group that is locally isomorphic to ${ℝ}^n$

If $G$ is connected then $G$ is generated by the elements of $U$.

The exponential map is a smooth homomorphism $$\begin{array}{ccc}𝔤& \to & G\\ 0& \mapsto & 1\end{array}$$ which is a homeomorphism on a neighborhood of $0$. The Lie algebra $𝔤$ contains the structure of $G$ in a neighbourhood of the identity.

A one parameter subgroup of $G$ is a smooth group homomorphism $\gamma :ℝ\to G.$

Examples:

1. Define $$\begin{array}{rcl}\gamma :ℝ& \to & {\mathrm{GL}}_n\left(ℝ\right)\\ t& \mapsto & 1+t{E}_{ij}=x_{ij}\left(t\right),fori\ne j.\end{array}$$ Note that $$x_{ij}\left(t\right)x_{ij}\left(s\right)=x_{ij}(s+t)$$ since $$\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& s\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& t+s\\ 0& 1\end{array}\right).$$
2. Define $$\begin{array}{rrcl}\gamma :& ℝ& \to & {\mathrm{GL}}_n\left(ℝ\right)\\ & t& \mapsto & \left(\begin{array}{ccccccc}1& & & & & & \\ & \ddots & & & & & \\ & & 1& & & & \\ & & & e^t& & & \\ & & & & 1& & \\ & & & & & \ddots & \\ & & & & & & 1\end{array}\right)=h_i\left(e^t\right).\end{array}$$ Note that $$h_i\left(e^t\right)h_i\left(e^s\right)=h_i\left(e^{t+s}\right).$$

Let $G$ be a Lie group. The ring of functions on $G$ is $$C^{\infty }\left(G\right)= {f:G\to ℝ|fis\; smooth\; atgfor\; allg\in G}$$ where $$fis\; smooth\; atgif{ \frac{d^{k}f}{d{x}^k}| }_{x=g}exists\; for\; allk\in {\mathbb{Z}}_{>0}.$$

Let $g\in G$.

• A tangent vector to $G$ at $g$ is a linear map $\eta :C^{\infty }\left(G\right)\to ℝ$ such that $$\eta \left(f_1{f}_2\right)=f_1\left(g\right)\eta \left(f_2\right)+\eta \left(f_1\right)f_2\left(g\right),$$ for all $f_1,f_2\in C^{\infty }\left(G\right).$
• A vector field is a linear map $\partial :C^{\infty }\left(G\right)\to C^{\infty }\left(G\right)$ such that $$\partial \left(f_1{f}_2\right)=f_1\partial \left(f_2\right)+\partial \left(f_1\right)f_2,$$ 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 $$L_g\xi =\xi ,for\; allg\in G,$$ where $L_g:C^{\infty }\left(G\right)\to C^{\infty }\left(G\right)$ is given by $$\left(L_{g}f\right)\left(x\right)=f\left(g^{-1}x\right),forf\in C^{\infty }\left(G\right),g,x\in G.$$

The Lie algebra of $G$ is the vector space $𝔤$ of left invariant vector fields on $G$ with bracket $$[{\partial }_1,{\partial }_2] ={\partial }_1{\partial }_2-{\partial }_2{\partial }_1.$$

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 $$\frac{df\left(\gamma \right(t\left)\right)}{dt}=\underset{h\to 0}{\lim }\frac{f\left(\gamma \right(t+h\left)\right)-f\left(\gamma \right(t\left)\right)}{h}$$ and let ${\gamma }_1$ be the tangent vector at $1$ given by $${\gamma }_1\left(f\right)={\left. \frac{df\left(\gamma \right(t\left)\right)}{dt}\right|}_{t=0}.$$ Identify the vector spaces

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

by the vector space isomorphisms $$\begin{array}{ccc}\{one\; parameter\; subgroups\}& \leftrightarrow & \{tangent\; vectors\; at1\}\\ \gamma & \mapsto & {\gamma }_1\end{array}$$ and $$\begin{array}{ccc}\{left\; invariant\; vector\; fields\}& \leftrightarrow & \{tangent\; vectors\; at1\}\\ \xi & \mapsto & {\xi }_1\end{array}$$ where $${\xi }_1\left(f\right)=\left(\xi f\right)\left(1\right).$$

The exponential map is $$\begin{array}{rcl}𝔤& \to & G\\ tX& \mapsto & e^{tX}where{e}^{tX}=\gamma \left(t\right),\end{array}$$ where $\gamma$ is the one-parameter subgroup corresponding to $X$.

Examples:

1. The Lie algebra ${\mathrm{𝔤𝔩}}_n$ is $${\mathrm{𝔤𝔩}}_n=\{x\in M_n(ℂ\left)\right\}$$ with bracket $$[x_1,x_2] =x_1{x}_2-x_2{x}_1.$$ Our favourite basis of ${\mathrm{𝔤𝔩}}_n$ is $${E_{ij}|1\le i,j\le n} .$$ The exponential map is $$\begin{array}{ccc}{\mathrm{𝔤𝔩}}_n& \to & {\mathrm{GL}}_n\\ tX& \mapsto & e^{tX},\end{array}$$ where $$e^A=1+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots$$ for a matrix $A$. In fact $$e^{t{E}_{ij}}=1+t{E}_{ij}=\left(\begin{array}{cccc}1& & & \\ & \ddots & t& \\ & & \ddots & \\ & & & 1\end{array}\right)fori\ne j,$$ where t is the $\left(i,j\right)$ matrix entry, and $$e^{t{E}_{ii}}=\left(\begin{array}{ccccccc}1& & & & & & \\ & \ddots & & & & & \\ & & 1& & & & \\ & & & e^t& & & \\ & & & & 1& & \\ & & & & & \ddots & \\ & & & & & & 1\end{array}\right)=h_i\left(e^t\right).$$
2. If $n=1$ the exponential map $$\begin{array}{ccc}ℂ& \to & {ℂ}^{\times }\\ tx& \mapsto & e^{tx}\end{array}$$ is a homeomorphism from a neighbourhood of $0$ to a neighbourhood of $1.$ In fact, if $e\left(t\right)=a_0+a_{1}t+a_2{t}^2+\cdots$ and $$e(s+t)=e\left(s\right)e\left(t\right),$$ then $$\begin{array}{rcl}e(s+t)& =& a_0+a_1(s+t)+a_2(s+t{)}^2+a_3(s+t{)}^3+\cdots \\ & =& \begin{array}{c}{a}_0+\\ a_{1}s+a_{1}t+\\ a_2{s}^2+2a_{2}st+a_2{t}^2\\ a_3{s}^3+3a_3{s}^{2}t+3a_{3}s{t}^2+a_3{t}^3+\\ ⋮\end{array}\end{array}$$ and $$\begin{array}{rcl}e\left(s\right)e\left(t\right)& =& (a_0+a_{1}s+a_2{s}^2+a_3{s}^3+\cdots )(a_0+a_{1}t+a_2{t}^2+a_3{t}^3+\cdots )\\ & =& \begin{array}{c}{a}_0^2+\\ a_0{a}_{1}s+a_0{a}_{1}t+\\ a_0{a}_2{s}^2+a_1^{2}st+a_0{a}_2{t}^2\\ a_0{a}_3{s}^3+a_2{a}_1{s}^{2}t+a_1{a}_{2}s{t}^2+a_0{a}_3{t}^3+\\ ⋮\end{array}\end{array}$$ Hence $e(s+t)=e\left(s\right)e\left(t\right)$ only if $$a_0{a}_1=a_1,2a_2=a_1^2,3a_3=a_1{a}_2,4a_3=a_1{a}_3,...$$ so that $$a_0=1,a_2=\frac{a_1^2}{2},a_3=\frac{a_1^3}{3!},a_4=\frac{a_1^4}{4!},...$$ and $$e\left(t\right)=1+a_{1}t+\frac{a_1^2}{2}{t}^2+\frac{a_1^3}{3!}{t}^3+\cdots =e^{a_{1}t}.$$ So $$\begin{array}{ccc}ℂ& \to & {ℂ}^{\times }\\ z& \mapsto & e^z\end{array}$$ is the "unique" smooth homomorphism $ℂ\to {ℂ}^{\times }.$

Maximal compact subgroups and maximal tori

Maximal compact subgroup examples:

Note that $$\begin{array}{rcl}{\mathrm{GL}}_1\left(ℂ\right)& =& {ℂ}^{\times }has\; maximal\; compact\; subgroup\\ U_1& =& S^1= {z\in {ℂ}^{\times }|z\stackrel{\_}{z}=1} \end{array}$$ and $$\begin{array}{rcl}{\mathrm{SL}}_1\left(ℂ\right)& =& \{\pm 1\}has\; maximal\; compact\; subgroup\\ {\mathrm{SU}}_1& =& \left\{1\right\}.\end{array}$$

Maximal tori:
${\mathrm{GL}}_n\left(ℂ\right)$ has maximal torus $${\left(\begin{array}{ccc}{x}_1& & 0\\ & \ddots & \\ 0& & x_n\end{array}\right)|x_1,...,x_n\in {ℂ}^{\times }} .$$ $U_n\left(ℂ\right)$ has maximal torus $${\left(\begin{array}{ccc}{z}_1& & 0\\ & \ddots & \\ 0& & z_n\end{array}\right)|z_1,...,z_n\in U\left(1\right)} .$$ ${\mathrm{SL}}_n\left(ℂ\right)$ has maximal torus $${\left(\begin{array}{ccc}{x}_1& & 0\\ & \ddots & \\ 0& & x_n\end{array}\right)|x_1,...,x_n\in {ℂ}^{\times },x_n=(x_1\cdots x_{n-1}{)}^{-1}} .$$ Small $n$: $$\begin{array}{rcl}U\left(1\right)& =& S^1= {z\in {ℂ}^{\times }|z\stackrel{\_}{z}=1} \\ \mathrm{SU}\left(1\right)& =& \left\{1\right\}\\ \mathrm{SU}\left(2\right)& =& {\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|g{\stackrel{\_}{g}}^t=1} \\ \mathrm{SL}\left(2\right)& =& {\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|ad-bc=1} \\ B& =& {\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)|a\in {ℂ}^{\times },b\in ℂ} \end{array}$$ $${\mathrm{SL}}_2/B=B\bigsqcup {x_{\alpha }\left(b\right)s_{\alpha }B|b\in ℂ} where{s}_{\alpha }=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)and{x}_{\alpha }\left(b\right)=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right)$$ $$\mathrm{SU}\left(2\right)= {\left(\begin{array}{cc}a& b\\ -\stackrel{\_}{b}& \stackrel{\_}{a}\end{array}\right)|a,b\in ℂ,|a{|}^2+|b{|}^2=1}$$ NOT SURE WHAT IS BEING SAID HERE ABOUT THE QUATERNIONS

It looks to me like what should be being said is the following:

Take the quaternions, and take the faithful [2]-representation given in the notes. Then SU(2) is isomorphic to the set of unit quaternions.

Next take the set of quaternions with scalar part 0. In our faithful representation, these are the 2x2 skew-Hermitian matrices with trace 0, which is the Lie algebra su_2(C).

So unit quaternions"="SU(2) and quaternions with no scalar part"="su_2.

Involution

$$\begin{array}{rrcl}\sigma :& {\mathrm{SL}}_2\left(ℂ\right)& \to & {\mathrm{SL}}_2\left(ℂ\right)\\ & x_{\alpha }\left(t\right)& \mapsto & x_{-\alpha }(-\stackrel{\_}{t})\\ & h_{\alpha }\left(t\right)& \mapsto & h_{\alpha }\left({\stackrel{\_}{t}}^{-1}\right)\\ & g& \mapsto & {\left(\stackrel{\_}{g}t\right)}^{-1}\\ & & & \\ & {\mathrm{SL}}_2(ℂ{)}^{\sigma }& =& {\mathrm{SU}}_2\left(ℂ\right)\end{array}$$ Let $K=G^{\sigma }.$ Then $G=BK$ and $K$ is maximal compact.

$K/T=G/B$

$$\begin{array}{rcl}\mathrm{SL}\left(2\right)& =& {\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)|ad-bc=1} \\ B& =& {\left(\begin{array}{cc}a& b\\ 0& a^{-1}\end{array}\right)|a\in {ℂ}^{\times },b\in ℂ} \\ {\mathrm{SL}}_2/B& =& B\bigsqcup B{n}_{\alpha }Bwhere\\ B\bigsqcup B{n}_{\alpha }B& =& {x_{\alpha }\left(b\right)n_{\alpha }B|b\in ℂ} with\end{array}$$ $$n_{\alpha }=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)and{x}_{\alpha }\left(b\right)=\left(\begin{array}{cc}1& b\\ 0& 1\end{array}\right).$$ Consider the involution $$\begin{array}{rrcl}\sigma :& {\mathrm{SL}}_2\left(ℂ\right)& \to & {\mathrm{SL}}_2\left(ℂ\right)\\ & g& \mapsto & {\left(\stackrel{\_}{g}t\right)}^{-1}\\ & x_{\alpha }\left(t\right)& \mapsto & x_{-\alpha }(-\stackrel{\_}{t})\\ & h_{\alpha }\left(t\right)& \mapsto & h_{\alpha }\left({\stackrel{\_}{t}}^{-1}\right).\end{array}$$ Then $$\begin{array}{rcl}K& =& {\mathrm{SL}}_2{\left(ℂ\right)}^{\sigma }= {g\in {\mathrm{SL}}_2|\sigma \left(g\right)=g} \\ & =& {\mathrm{SU}}_2= {g\in {\mathrm{SL}}_2|g{\stackrel{\_}{g}}^t=1} \\ & =& {\left(\begin{array}{cc}a& b\\ -\stackrel{\_}{b}& \stackrel{\_}{a}\end{array}\right)|a,b\in ℂ,{\left|a\right|}^2+{\left|b\right|}^2=1} .\end{array}$$ Then $$\begin{array}{rcl}T& =& {\mathrm{SU}}_2\cap B= {\left(\begin{array}{cc}a& b\\ -\stackrel{\_}{b}& \stackrel{\_}{a}\end{array}\right)|a,b\in ℂ,{\left|a\right|}^2+{\left|b\right|}^2=1,-\stackrel{\_}{b}=0} \\ & =& {\left(\begin{array}{cc}a& 0\\ 0& \stackrel{\_}{a}\end{array}\right)|a\in ℂ,{\left|a\right|}^2=1} \\ & =& {\left(\begin{array}{cc}{e}^{i\theta }& 0\\ 0& e^{-i\theta }\end{array}\right)|0\le \theta <2\pi } \simeq S^1.\end{array}$$ An element of ${\mathrm{SU}}_2$ is $$\left(\begin{array}{cc}r{e}^{i\theta }& s{e}^{i\phi }\\ -s{e}^{-i\phi }& r{e}^{-i\theta }\end{array}\right)with\begin{array}{rr}0\le \theta <2\pi ,& r,s\in {ℝ}_{>0}\\ 0\le \phi <2\pi ,& r^2+s^2=1.\end{array}$$ Then $$\left(\begin{array}{cc}r{e}^{i\theta }& s{e}^{i\phi }\\ -s{e}^{-i\phi }& r{e}^{-i\theta }\end{array}\right)\left(\begin{array}{cc}{e}^{i\phi }& 0\\ 0& e^{-i\phi }\end{array}\right)=\left(\begin{array}{cc}r{e}^{i(\theta +\phi )}& s\\ -s& r{e}^{-i(\theta +\phi )}\end{array}\right)$$ so that the cosets in ${\mathrm{SU}}_2/T$ have representatives $${\left(\begin{array}{cc}a& s\\ -s& \stackrel{\_}{a}\end{array}\right)|s\in {ℝ}_{>0},a\in ℂ,{\left|a\right|}^2+s^2=1} .$$ So we should have $${\mathrm{SU}}_2\cap x_{\alpha }\left(b\right)n_{\alpha }B=\left(\begin{array}{cc}a& s\\ -s& \stackrel{\_}{a}\end{array}\right)Tfor\; a\; uniquea\in ℂ,s\in {ℝ}_{>0}.$$ Since $$x_{\alpha }\left(b\right)n_{\alpha }B= {\left(\begin{array}{cc}-b& 1\\ -1& 0\end{array}\right)\left(\begin{array}{cc}c& d\\ 0& c^{-1}\end{array}\right)|c\in {ℂ}^{\times },d\in ℂ}$$ and $$x_{\alpha }\left(b\right)n_{\alpha }B\cap {\mathrm{SU}}_2= {\left(\begin{array}{cc}-bc& \stackrel{\_}{c}\\ -c& -\stackrel{\_}{bc}\end{array}\right)|c\in {ℂ}^{\times },-d=-\stackrel{\_}{bc},\stackrel{\_}{c}=c^{-1}-bd}$$ and since $$\left(\begin{array}{cc}-bc& \stackrel{\_}{c}\\ -c& -\stackrel{\_}{bc}\end{array}\right)\left(\begin{array}{cc}\frac{\stackrel{\_}{c}}{\left|c\right|}& 0\\ 0& \frac{c}{\left|c\right|}\end{array}\right)=\left(\begin{array}{cc}-b\left|c\right|& \left|c\right|\\ -\left|c\right|& -\stackrel{\_}{b}\left|c\right|\end{array}\right)$$ it follows that $$x_{\alpha }\left(b\right)n_{\alpha }B\cap {\mathrm{SU}}_2=\left(\begin{array}{cc}-bs& s\\ -s& -\stackrel{\_}{b}s\end{array}\right)Twiths=\sqrt{\frac{1}{1+{\left|b\right|}^2}}.$$

More examples

1. ${\mathrm{SL}}_2.$ Let $𝔤={\mathrm{𝔰𝔩}}_2=\mathrm{span} {X_{\alpha },H_{{\alpha }^{\vee }},X_{-\alpha }}$ with $$X_{\alpha }=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),X_{-\alpha }=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),H_{{\alpha }^{\vee }}=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ and $V$ the 2-dimensional representation of ${\mathrm{𝔰𝔩}}_2$ on column vectors. Then $$x_{\alpha }\left(f\right)=\left(\begin{array}{cc}1& f\\ 0& 1\end{array}\right),x_{-\alpha }=\left(\begin{array}{cc}1& 0\\ f& 1\end{array}\right),n_{\alpha }\left(g\right)=\left(\begin{array}{cc}0& g\\ -g^{-1}& 0\end{array}\right),and{h}_{{\alpha }^{\vee }}\left(g\right)=\left(\begin{array}{cc}g& 0\\ 0& g^{-1}\end{array}\right).$$ Then $G={\mathrm{SL}}_2\left(𝔽\right),$ $$h_{{\alpha }^{\vee }}\left(g\right)=1if\; and\; only\; if{g}^{ ⟨{\epsilon }_1,{\alpha }^{\vee }⟩ }=g^{ ⟨{\epsilon }_2,{\alpha }^{\vee }⟩ }=1if\; and\; only\; ifg=g^{-1}=1,$$ and $$Z\left(G\right)= {h_{{\alpha }^{\vee }}\left(g\right)|g^{ ⟨\alpha ,{\alpha }^{\vee }⟩ }=1} = {h_{{\alpha }^{\vee }}\left(1\right),h_{{\alpha }^{\vee }}\left(-,1\right)} .$$

Notes and References

These notes are adapted from the lecture notes of Arun Ram on representation theory, from 2008.

References

References?

page history