Representation Theory

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

Last updated: 2 October 2014

Lecture 10

Lie algebras

A Lie group is a group that is also a manifold, i.e. a topological group that is locally isomorphic to ${ℝ}^n$ $$\begin{array}{c} image/svg+xml 1 U G U is an open neighbourhood of 1 in G 0 V R n V is an open neighbourhood of 0 in Rn G ℝ n U V 0 1 φ : U → ∼ V U is an open neighborhood of 1 in G V is an open neighborhood of 0 in ℝ n \end{array}$$ If $G$ is connected then $G$ is generated by the elements of $U\text{.}$

The exponential map is a smooth homomorphism $$\begin{array}{ccc}𝔤& ⟶& G\\ 0& ⟼& 1\end{array}$$ 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 $\gamma :ℝ\to G\text{.}$ $$\begin{array}{c} image/svg+xml 1 γ ( t ) 0 t ↦ γ \end{array}$$

(1)	$$\begin{array}{cccc}\gamma :& ℝ& ⟶& {GL}_n\left(ℝ\right)\\ & t& ⟼& 1+t{E}_{ij}& =& x_{ij}\left(t\right)\end{array}$$ for $i\ne j\text{.}$ Note that $$x_{ij}\left(t\right)x_{ij}\left(s\right)=x_{ij}(s+t)$$ since $$\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& s\\ 0& 1\end{array}\right)=\left(\begin{array}{cc}1& t+s\\ 0& 1\end{array}\right)\text{.}$$
(2)	$$\begin{array}{cccc}\gamma :& ℝ& ⟶& {GL}_n\left(ℝ\right)\\ & t& ⟼& \left(\begin{array}{c}1\\ & \ddots \\ & & 1\\ & & & e^t\\ & & & & 1\\ & & & & & \ddots \\ & & & & & & 1\end{array}\right)& =& h_i\left(e^t\right)\text{.}\end{array}$$ Note that $$h_i\left(e^t\right)h_i\left(e^s\right)=h_i\left(e^{t+s}\right)\text{.}$$

Let $G$ be a Lie group. The ring of functions on $G$ is $$C^{\infty }\left(G\right)=\{f:G\to ℝ\hspace{0.17em}|\hspace{0.17em}f\hspace{0.17em}\text{is smooth at}\hspace{0.17em}g\hspace{0.17em}\text{for all}\hspace{0.17em}g\in G\}$$ where $$f\hspace{0.17em}\text{is}\hspace{0.17em}\text{smooth at}\hspace{0.17em}g\hspace{0.17em}\text{if}\hspace{0.17em}\frac{d^{k}f}{d{x}^k}{|}_{x=g}\hspace{0.17em}\text{exists for all}\hspace{0.17em}k\in {\mathbb{Z}}_{>0}\text{.}$$

Let $g\in G\text{.}$ 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 $f_1,f_2\in C^{\infty }\left(G\right)\text{.}$ 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 $f_1,f_2\in C^{\infty }\left(G\right)\text{.}$ A left invariant vector field on $G$ is a vector field $\partial :C^{\infty }\left(G\right)\to C^{\infty }\left(G\right)$ such that $$L_g\xi =\xi L_g,$$ for all $g\in G,$ where $$L_g:C^{\infty }\left(G\right)⟶C^{\infty }\left(G\right)$$ is given by $$\left(L_{g}f\right)\left(x\right)=f\left(g^{-1}x\right),$$ for $f\in C^{\infty }\left(G\right),$ $g,x\in G\text{.}$

The Lie algebra of $G$ is the vector space $𝔤$ of left invariant vector fields on $G$ with bracket $$[{\partial }_1,{\partial }_2]={\partial }_1{\partial }_2-{\partial }_2{\partial }_1\text{.}$$ A one-parameter subgroup of $G$ is a smooth group homomorphism $\gamma :ℝ\to G\text{.}$ If $\gamma$ is a one-parameter subgroup of $G$ define $$\frac{df\left(\gamma \left(t\right)\right)}{dt}=\underset{h\to 0}{\text{lim}}\frac{f\left(\gamma (t+h)\right)-f\left(\gamma \left(t\right)\right)}{h}$$ and let ${\gamma }_1$ be the tangent vector at $1$ given by $${\gamma }_1\left(f\right)=\frac{df\left(\gamma \left(t\right)\right)}{dt}{|}_{t=0}\text{.}$$ Identify the vector spaces $\left\{\text{left invariant vector fields on}\hspace{0.17em}G\right\},$
$\left\{\text{one parameter subgroups of}\hspace{0.17em}G\right\},$ and
$\left\{\text{tangent vectors at}\hspace{0.17em}1\right\},$ by the vector space isomorphisms $$\begin{array}{ccc}\left\{\text{one parameter subgroups}\right\}& ⟷& \left\{\text{tangent vectors at}\hspace{0.17em}1\right\}\\ \gamma & ⟼& {\gamma }_1\end{array}$$ and $$\begin{array}{ccc}\left\{\text{left invariant vector fields}\right\}& ⟷& \left\{\text{tangent vectors at}\hspace{0.17em}1\right\}\\ \xi & ⟼& {\xi }_1\end{array}$$ where $${\xi }_1\left(f\right)=\left(\xi f\right)\left(1\right)\text{.}$$

The exponential map is $$\begin{array}{ccc}𝔤& ⟶& G\\ tX& ⟼& e^{tX}\end{array}$$ where $e^{tX}=\gamma \left(t\right),$ where $\gamma$ is the $1\text{-parameter}$ subgroup corresponding to $X\text{.}$

(a)

The Lie algebra ${𝔤𝔩}_n$ is $${𝔤𝔩}_n=\{x\in M_n\left(ℂ\right)\}$$ with bracket $$[x_1,x_2]=x_1{x}_2-x_2{x}_1\text{.}$$ Our favourite basis of ${𝔤𝔩}_n$ is $$\left\{E_{ij}\hspace{0.17em}\right|\hspace{0.17em}1\le i,j\le n\}\text{.}$$ The exponential map is $$\begin{array}{ccc}{𝔤𝔩}_n& ⟶& {GL}_n\\ tX& ⟼& e^{tX},\end{array}$$ where $$e^A=1+A+\frac{A^2}{2!}+\frac{A^3}{3!}+\cdots$$ for a matrix $A\text{.}$ In fact $$e^{t{E}_{ij}}=1+t{E}_{ij}=\left(\begin{array}{c}1\\ & \ddots & t\\ & & & \\ & & & 1\end{array}\right)$$ for $i\ne j,$ where $t$ is the $(i,j)$ matrix entry, and $$e^{t{E}_{ii}}=\left(\begin{array}{c}1\\ & \ddots \\ & & 1\\ & & & e^t\\ & & & & 1\\ & & & & & \ddots \\ & & & & & & 1\end{array}\right)=h_i\left(e^t\right)\text{.}$$

(b)

If $n=1$ the exponential map $$\begin{array}{ccc}ℂ& ⟶& {ℂ}^{\times }\\ tx& ⟼& e^{tx}\end{array}$$ is a homeomorphism from a neighborhood of $0$ to a neighborhood of $1\text{.}$ 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 $$\begin{array}{rcl}{\displaystyle e(s+t)}& =& {\displaystyle a_0+a_1(s+t)+a_2{(s+t)}^2+a_3{(s+t)}^3+\cdots }\\ & =& {\displaystyle \begin{array}{c}{a}_0+\\ a_{1}s+a_{1}t+\\ a_2{s}^2+2a_{2}st+a_2{t}^2\\ a_3{s}^3+3a_3{s}^{2}t+3a_{3}s{t}^2+a_3{t}^3+\\ ⋮\end{array}}\end{array}$$ and $$\begin{array}{rcl}{\displaystyle e\left(s\right)e\left(t\right)}& =& {\displaystyle (a_0+a_{1}s+a_2{s}^2+a_3{s}^3+\cdots )(a_0+a_{1}t+a_2{t}^2+a_3{t}^3+\cdots )}\\ & =& {\displaystyle \begin{array}{c}{a}_0^2+\\ a_0{a}_{1}s+a_0{a}_{1}t+\\ a_0{a}_2{s}^2+a_1^{2}st+a_0{a}_2{t}^2\\ a_0{a}_3{s}^3+a_2{a}_1{s}^{2}t+a_1{a}_{2}s{t}^2+a_0{a}_3{t}^3+\\ ⋮\end{array}}\end{array}$$ Hence $e(s+t)=e\left(s\right)e\left(t\right)$ only if $$a_0{a}_1=a_1,2a_2=a_1^2,3a_3=a_1{a}_2,4a_3=a_1{a}_3,\dots$$ so that $$a_0=1,a_2=\frac{a_1^2}{2},a_3=\frac{a_1^3}{3!},a_4=\frac{a_1^4}{4!},\dots$$ and $$e\left(t\right)=1+a_{1}t+\frac{a_1^2}{2}{t}^2+\frac{a_1^3{t}^3}{3!}+\cdots =e^{a_{1}t}\text{.}$$ So $$\begin{array}{ccc}ℂ& ⟶& {ℂ}^{\times }\\ z& ⟼& e^z\end{array}$$ is the "unique" smooth homomorphism $ℂ\to {ℂ}^{\times }\text{.}$

The Lie groups ${SL}_2$ and ${SU}_2$

$${SL}_2=\left\{\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}ad-bc=1\}$$ One parameter subgroups are $$x_{12}\left(t\right)=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right),x_{21}\left(t\right)=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right),h_{{\alpha }_1}\left(e^t\right)=\left(\begin{array}{cc}{e}^t& 0\\ 0& e^{-t}\end{array}\right)$$ and the Lie algebra $${𝔰𝔩}_2=\{x\in {𝔤𝔩}_2\hspace{0.17em}|\hspace{0.17em}\text{tr}\hspace{0.17em}x=0\}$$ has basis $\{x,y,h\}$ where $$x=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right),y=\left(\begin{array}{cc}0& 0\\ 1& 0\end{array}\right),h=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)\text{.}$$ The Lie algebra ${𝔰𝔩}_2$ is presented by generators $x,y,h$ and relations $$[x,y]=h,[h,x]=2x\text{and}[h,y]=-2y\text{.}$$ The group ${SL}_2$ is presented by generators $$x_{\alpha }\left(t\right)=\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right),x_{-\alpha }\left(t\right)=\left(\begin{array}{cc}1& 0\\ t& 1\end{array}\right),t\in 𝔽$$ with relations $$\begin{array}{rcl}{\displaystyle x_{\alpha }(s+t)}& =& {\displaystyle x_{\alpha }\left(s\right)x_{\alpha }\left(t\right),}\\ {\displaystyle h_{\alpha }\left(c_1{c}_2\right)}& =& {\displaystyle h_{\alpha }\left(c_1\right)h_{\alpha }\left(c_2\right),\text{and}}\\ {\displaystyle n_{\alpha }\left(t\right)x_{\alpha }\left(u\right)n_{\alpha }(-t)}& =& {\displaystyle x_{-\alpha }(-t^{-2}n)}\end{array}$$ where $$\begin{array}{rcl}{\displaystyle n_{\alpha }\left(t\right)}& =& {\displaystyle x_{\alpha }\left(t\right)x_{-\alpha }(-t^{-1})x_{\alpha }\left(t\right)\text{and}}\\ {\displaystyle h_{\alpha }\left(t\right)}& =& {\displaystyle n_{\alpha }\left(t\right)n_{\alpha }(-1)\text{.}}\end{array}$$ Note that $$\begin{array}{rcl}{\displaystyle n_{\alpha }\left(t\right)}& =& {\displaystyle \left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ -t^{-1}& 1\end{array}\right)\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cc}0& t\\ -t^{-1}& 1\end{array}\right)\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)}\\ & =& {\displaystyle \left(\begin{array}{cc}0& t\\ -t^{-1}& 0\end{array}\right)}\end{array}$$ and $$h_{\alpha }\left(t\right)=\left(\begin{array}{cc}0& t\\ -t^{-1}& 0\end{array}\right)\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)=\left(\begin{array}{cc}t& 0\\ 0& t^{-1}\end{array}\right)\text{.}$$ The maximal compact subgroup of ${SL}_2\left(ℂ\right)$ is $$\begin{array}{rcl}{\displaystyle {SU}_2}& =& {\displaystyle \{g=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)\hspace{0.17em}|\hspace{0.17em}g{\bar{g}}^t=1,\text{det}\left(g\right)=1\}}\\ & =& {\displaystyle \left\{\left(\begin{array}{cc}a& b\\ -\bar{b}& a\end{array}\right)\hspace{0.17em}\right|\hspace{0.17em}a,b\in ℂ,{\mid a\mid }^2+{\mid b\mid }^2=1\}}\end{array}$$ since $$g^{-1}=\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)\text{and}{\bar{g}}^t=\left(\begin{array}{cc}\bar{a}& \bar{c}\\ \bar{b}& \bar{d}\end{array}\right)\text{.}$$ The Lie algebra ${𝔰𝔲}_2$ is $$\begin{array}{rcl}{\displaystyle {𝔰𝔲}_2}& =& {\displaystyle \{x\in {𝔤𝔩}_2\left(ℂ\right)\hspace{0.17em}|\hspace{0.17em}\text{tr}\hspace{0.17em}x=0\hspace{0.17em}\text{and}\hspace{0.17em}x+{\bar{x}}^t=0\}}\\ & =& {\displaystyle ℝ\text{-span}\{i{\sigma }^x,i{\sigma }^y,i{\sigma }^z\},}\end{array}$$ where $${\sigma }^x=\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right),{\sigma }^y=\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right),{\sigma }^z=\left(\begin{array}{cc}1& 0\\ 0& -1\end{array}\right)$$ are the Pauli matrices and $$[{\sigma }^z,{\sigma }^y]=2i{\sigma }^z,[{\sigma }^y,{\sigma }^z]=2i{\sigma }^z,[{\sigma }^z,{\sigma }^x]=2i{\sigma }^y\text{.}$$ Then ${𝔰𝔩}_2\left(ℂ\right)$ is the complexification of ${𝔰𝔲}_2,$ $${𝔰𝔩}_2\left(ℂ\right)=ℂ{\otimes }_{ℝ}{𝔰𝔲}_2$$ and the change of basis is given by $$\begin{array}{c}{\sigma }^x=x+y,{\sigma }^y=-ix+iy,{\sigma }^z=h,\\ x=\frac{1}{2}({\sigma }^x+i{\sigma }^y),y=\frac{1}{2}({\sigma }^x-i{\sigma }^y)\text{.}\end{array}$$ Note that $${GL}_1\left(ℂ\right)={ℂ}^{\times }$$ has maximal compact subgroup $$U\left(1\right)=S^1=\{z\in {ℂ}^{\times }\hspace{0.17em}|\hspace{0.17em}z\bar{z}=1\}$$ and $${SL}_1\left(ℂ\right)=\{\pm 1\}$$ has maximal compact subgroup $$SU\left(1\right)=\left\{1\right\}\text{.}$$

Notes and References

These are a typed copy of Lecture 10 from a series of handwritten lecture notes for the class Representation Theory given on October 14, 2008.

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